elementary differential equations 6th edition - edwards and penny_high

ELEMENTARYDIFFERENTIALEQUATIONS)))

ELEMENTARYDIFFERENTIALEQUATIONS)SixthEdition)

C.Henry EdwardsDavidE.Penney)TheUniversity ofGeorgia)

with the assistanceofDavidCalvis)Baldwin-WallaceCollege)

PEARSON----------Prentice

flaIl) UpperSaddleRiver,NJ 07458)))

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@2008,2004,2000,1996by Pearson Education, Inc.Pearson Education, Inc.Upper Saddle River, New Jersey 07458)

All rights reserved. No part of this book may bereproduced, in any form or by any means,without permission in writing from the publisher.)

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Preface vii)

1)

First-OrderDifferentialEquations 11 .1 Differential Equationsand MathematicalModels1 .2 Integralsas Generaland ParticularSolutions 1 01 .3 SlopeFieldsand SolutionCurves 191 .4 SeparableEquationsand Applications 321.5 LinearFirst-OrderEquations 461 .6 SubstitutionMethodsand ExactEquations 591 .7 PopulationModels 741.8 Acceleration-VelocityModels 85)

CHAPTER)

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LinearEquationsof HigherOrder 1002.1 Introduction:Second-OrderLinearEquations 1002.2 GeneralSolutionsof LinearEquations 1132.3 HomogeneousEquationswith ConstantCoefficients1242.4 MechanicalVibrations 1352.5 NonhomogeneousEquationsand UndeterminedCoefficients1482.6 ForcedOscillationsand Resonance1622.7 ElectricalCircuits 1732.8 Endpoint Problemsand Eigenvalues 180)

CHAPTER)

CHAPTER) PowerSeriesMethods) 194)

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3.13.23.33.43.53.6)

Introductionand Reviewof PowerSeries 194SeriesSolutionsNearOrdinary Points 207RegularSingularPoints 218Methodof Frobenius:The ExceptionalCases233Bessel'sEquation 248Applicationsof BesselFunctions 257)

v)))

.VI Contents)

4)LaplaceTransformMethods 2664.1 LaplaceTransforms and InverseTransforms 2664.2 Transformationof Initial ValueProblems 2774.3 Translationand Partial Fractions2894.4 Derivatives,Integrals,and Productsof Transforms 2974.5 Periodicand PiecewiseContinuousInput Functions 3044.6 Impulsesand DeltaFunctions 316)

CHAPTER)

5)LinearSystemsof DifferentialEquations5.1 First-OrderSystemsand Applications 3265.2 The Methodof Elimination 3385.3 Matricesand LinearSystems 3475.4 The EigenvalueMethodfor HomogeneousSystems 3665.5 Second-OrderSystemsand MechanicalApplications 3815.6 MultipleEigenvalueSolutions 3935.7 Matrix Exponentialsand LinearSystems 4075.8 NonhomogeneousLinearSystems 420)

326)CHAPTER)

CHAPTER) NumericalMethods) 430)

6)6.1 NumericalApproximation:Euler'sMethod 4306.2 A CloserLookat the Euler Method 4426.3 The Runge-KuttaMethod 4536.4 NumericalMethodsfor Systems 464)

7)NonlinearSystemsand Phenomena 4807.1 Equilibrium Solutionsand Stability 4807.2 Stabilityand the PhasePlane 4887.3 Linearand Almost LinearSystems 5007.4 EcologicalModels:Predatorsand Competitors5137.5 NonlinearMechanicalSystems 5267.6 Chaosin DynamicalSystems 542)

CHAPTER)

Referencesfor FurtherStudy 555Appendix:ExistenceandUniquenessofSolutions559AnswerstoSelectedProblems573Index 1-1)))

P'REFACE)

The evolution of the presenttext in successiveeditions is basedon experienceteaching the introductory differential equations coursewith an emphasis on

conceptual ideas and the use of applications and projectsto involve students in

active problem-solving experiences.At various points our approach reflects the

widespreaduseof technicalcomputingenvironments likeMaple,Mathematica,andMATLAB for the graphical, numerical, or symbolicsolution of differential equa-tions.Nevertheless,we continue to believe that the traditional elementaryanalyticalmethods of solution are important for students to learn and use.One reason is that

effective and reliableuseof computer methods often requirespreliminary analysisusing standard symbolictechniques;the construction of a realistic computationalmodeloften is basedon the study of a simpleranalyticalmodel.)

PrincinalFeaturesofThisRevision.....N\037\" .../N;....//lfll\"'.M...'NN'..N'N\"/h'//..\037N;I'In' ...I::/\"...N/\"/N//N//..#NI'NJIVI'N\"\"\"\"......./H'N Nd.lh./hNh./\037\"'\" /I'Nk>' nN>Vfl \037N'..fl_ nv..;.;\302\253/.........x'..y,\037....'/Y.'h;;I.N'N/.9\037//JP...I#//N/N..\"\"'*NhYn.N'N'.IN\"'N\037 N.lN.N'.\302\253.t'N'/N.lN'.I/NIN'.IN..Nh>vVN.I-v.,/Nv'''''-vN.l-v.,/N.N'NhV>.... N.lN.lN.I'XVN.N.N\"'....-><N .v\"/,/\"\"\",/\"\"/N'.IN.I/hnv,,..MN'I<NN.I\"NN-v\"\"-' N :Jo.'.dN\"\"'N\037\"'\037N'-'..\037N'-'..I<N\037.. I'MJV.I .....,...... \037 I.JVN.lMv><NY\"\"1' N..;\\V\".....\"\"\"'..,.......................... ....,... <'.............\037\"'.....v.v............... W -.<v..... ....., ..............'\"' ,..........,., ........ ...............\".....,..,_...\037 \"\" \037,\037 ,\302\253'i'.\037.''''' \037)

While the successfulfeatures of precedingeditions have beenretained, the exposi-tion has beensignificantly enhanced in everychapterand in most individual sectionsof the text. Both new graphicsand new text have beeninserted where neededfor

improved student understanding of key concepts.However, the solidclass-testedchapter and sectionstructure of the bookis unchanged, so classnotes and syllabiwill not require revisionfor useof this new edition.The followingexamplesof this

revision illustrate the way the localstructure of the text has beenaugmented and

polishedfor this edition.)

Chapter1:New Figures1.3.9and 1.3.10showing directionfields that illus-trate failure of existenceand uniquenessof solutions(page24);new Problems34 and 35 showing that small changesin initial conditionscan make big dif-ferencesin results,but bigchangesin initial conditionsmay sometimesmake

only smalldifferencesin results(page30);new Remarks 1 and 2 clarifying the

conceptof implicit solutions (page35);new Remark clarifying the meaningof homogeneityfor first-orderequations (page61);additionaldetails insertedin the derivationof the rocketpropulsion equation (page95),and new Prob-lem 5 insertedto investigate the liftoff pauseof a rocket on the launch padsometimesobservedbefore blastoff (page97).

Chapter2: New explanation of signs and directions of internal forces in

mass-springsystems(page101);new introduction of differential operatorsand clarificationof the algebra of polynomial operators (page127);new in-troduction and illustration of polar exponential forms of complexnumbers

(page132);fuller explanationof method of undeterminedcoefficientsin Ex-amples1 and 3 (page149-150);new Remarks 1 and 2 introducing \"shooting\"

terminology,and new Figures2.8.1and 2.8.2illustrating why someendpoint)..VII)))

...VIII Preface)

value problemshave infinitely many solutions,while others have no solutionsat all (page181);new Figures2.8.4and 2.8.5illustrating different types ofeigenfunctions(pages183-184).Chapter3:New Problem 35 on determination of radii of convergenceofpowerseriessolutions of differential equations (page218);new Example3just before the subsectionon logarithmiccasesin the method of Frobenius,toillustrate first the reduction-of-orderformula with a simplenon-seriesproblem(page239).Chapter4:New discussionclarifying functions of exponentialorderand ex-istenceof Laplacetransforms (page273);new Remark discussingthe me-chanicsof partial-fraction decomposition(page279);new much-expandeddiscussionof the proof of the Laplace-transformexistencetheorem and itsextensionto include the jump discontinuities that play an important rolein

many practical applications(page286-287).Chapter5:New Problems20-23for student exploration of three-railway-carssystemswith different initial velocityconditions(page392);new Remarkillustrating the relation betweenmatrix exponential methods and the gener-alizedeigenvalue methods discussedpreviously (page416);new expositioninsertedat endof sectionto explain the connection between matrix variationof parametershere and (scalar)variation of parametersfor second-orderequa-tions discussedpreviously in Chapter 3 (page427).

Chapter6:New discussionwith new Figures6.3.11and 6.3.12clarifyingthe differencebetweenrotating and non-rotatingcoordinatesystemsin moon-earth orbit problems(page473).Chapter7: New remarks on phase plane portraits, autonomous systems,and critical points (page488-490);new introduction of linearized systems(page502);new 3-dimensionalFigures6.5.18and 6.5.20illustrating Lorenzand Rosslertrajectories (page552-553).)

Throughout the text, almost 550computer-generatedfigures show studentsvivid picturesof directionfields,solutioncurves,and phaseplaneportraits that bring

symbolicsolutionsof differentialequations to life.About 15applicationmodulesfollow key sectionsthroughout the text. Their

purposeis to add concreteappliedemphasisand to engagestudents is more exten-sive investigationsthan affordedby typical exercisesand problems.

A solidnumerical emphasisprovided where appropriate (as in Chapter 6 onNumericalMethods)by the inclusionof genericnumericalalgorithms and a limitednumber of illustrative graphing calculator,BASIC,and MATLAB routines.)

Organizationand C?Iltent)

The traditional organizationof this text still accommodatesfresh new material andcombinationsof topics.For instance:). The final two sectionsof Chapter 1 (on populationsand elementary mechan-

ics)offer an early introduction to mathematicalmodelingwith significant ap-plications.. The final sectionof Chapter 2 offers unusually early exposureto endpoint)))

Preface) .IX)

problemsand eigenvalues, with interesting applications to whirling stringsand buckledbeams.. Chapter 3 combinesa completeand solidtreatment of infinite seriesmethodswith interestingapplicationsof Besselfunctions in its final section.. Chapter 4 combinesa completeand solid treatment of Laplacetransformmethods with brief coverage of delta functions and their applicationsin its

final section.. Chapter 5 provides an unusually flexible treatment of linear systems.Sec-tions 5.1and 5.2offer an early, intuitive introduction to first-ordersystemsand models.The chapter continues with a self-contained treatment of the

necessarylinear algebra, and then presentsthe eigenvalueapproachto linear

systems.It includesan unusual number of applications (ranging from brinetanks to railway cars)of all the various casesof the eigenvaluemethod. Thecoverageof exponentialmatrices in Section5.7is expandedfrom earlieredi-ti ons.. Chapter 6 on numerical methods beginsin Section6.1with the elementaryEuler method for singleequations and ends in Section6.4with the Runge-Kutta method for systemsand applicationsto orbits of cometsand satellites.. Chapter7 on nonlinearsystemsand phenomenaranges from phaseplaneanal-ysisto ecologicaland mechanicalsystemsto an innovative concludingsectionon chaosand bifurcation in dynamical systems.Section7.6presentsan ele-mentary introduction to such contemporarytopicsas period-doublingin bio-logicaland mechanicalsystems,the pitchfork diagram,and the Lorenzstrangeattractor (all illustrated with vivid computergraphics).)

This bookincludesadequatematerial for different introductory coursesvary-ing in length from a singleterm to two quarters. The longer version, ElementaryDifferential Equations with Boundary Value Problems(0-13-600613-2),containsadditionalchapters on Fourier seriesmethods and partial differential equations(in-cluding separationof variablesand boundary valueproblems).)

:,\\p\037li\037 \037\037__.______)To samplethe range of applications in this text, take a lookat the following ques-tion s:)

.What explainsthe commonlyobservedlag time between indoor and outdoordaily temperatureoscillations?(Section1.5).What makesthe differencebetweendoomsdayand extinctionin alligatorpop-ulations? (Section1.7).How do a unicycle and a two-axle car react differently to road bumps? (Sec-tions 2.6and 5.5).Why are flagpoleshollowinstead of solid?(Section3.6). If a mass on a spring is periodicallystruck with a hammer, how does the

behavior of the massdependon the frequency of the hammer blows?(Sec-tion 4.6). If a moving train hits the rear end of a train of railway carssitting at rest, how

can it happen that just a singlecar is \"popped\" off the front end of the secondtrain? (Section5.5))))

X Preface)

. Howcan you predictthe time of next perihelionpassageof a newly observedcomet?(Section6.4). What determineswhether two specieswill live harmoniously together, orwhether competition will result in the extinction of one of them and the sur-vival of the other? (Section7.4). Why and when doesnon-linearity leadto chaosin biologicaland mechanicalsystems?(Section7.6))

Applicationsand SolutionsManuals(\"h'/\037'\302\273'I\037\",Z\037'>W//\"//hW\"\"NV,:V/hV/.\302\253>'//h':.'..w\"'h\"..:-:?:>'h/\"'/hhlh\"''''Yh////hlh/'''''''h/\"<'1'('v/Y/lhl.-wI\"hW//n'I\302\273\"\037h/lY/h//I'MYl'h//n'...:q\037.\302\253\302\253(o\"/.o(.Q'\037\037'>mY/..mm'MX-V.\302\253\302\253(o\"\037..vh;\302\253,..wI\302\273\",c.y'-\302\273\302\273y\037\"M\"\"\037..w\037o\302\273\302\273\302\273(\037.....v...\037\037..:.w..w\037....WIY/.../_..w\037\037..w'(O\037W/..wh\037Y....../..:vhV/..:.w..w.\302\253r\302\273\302\273'\037)

The answer sectionhas beenexpandedconsiderablyto increaseits valueasa learn-ing aid. It now includesthe answersto most odd-numberedproblemsplusa goodmany even-numberedones.The 60S-pageInstructor'sSolutionsManual (0-13-600614-0)accompanying this bookprovides worked-out solutions for most of the

problemsin the book,and the 345-pageStudentSolutionsManual (0-13-600615-9)contains solutionsfor most of the odd-numberedproblems.

The approximately15applicationmodulesin the text containadditionalprob-lem and projectmaterial designedlargely to engagestudents in the explorationand

applicationof computationaltechnology.Theseinvestigationsare expandedconsid-erably in the 320-pageApplicationsManual (0-13-600679-5)that accompaniesthetext and supplementsit with about 30 additionalapplicationsmodules.Each sectionin this manual has parallel subsections\"Using Maple,\"\"Using Mathematica,\" and

\"Using MATLAB\" that detail the applicablemethods and techniques of each sys-tem, and will afford student usersan opportunity to comparethe merits and stylesof differentcomputationalsystems.)

t\\\037o\037\037m\037\037____)In preparing this revision we profitedgreatly from the advice and assistanceof the

following very capableand perceptivereviewers:

Raymond A. Claspadle, Irfan UI-Haq,University of Memphis University of Wisconsin-Platteville

SemionGutman, Carl Lutzer,University of Oklahoma RochesterInstitute of Technology

MiklosBona, SigalGittlieb,University of Florida University ofMassachusetts,Dartmouth

It is a pleasure to (onceagain) credit Dennis Kletzing and his extraordinaryTEXpertisefor the attractive presentation of both the text and the art in this book.Finally, but far from least,I am especiallyhappy to acknowledgea new contrib-utor to this effort, David Calvis,who assistedin every aspectof this revision andcontributed tangibly to the improvementof every chapter in the book.)

[email protected])))

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First-OrderDifferentialEquations)

11IIDiff'\037\037\037\037ti\0371.\0379\037.\037.tion_s_ \037.\037<!__\037.\037th.:\037_\"1.;}t.i<:al \037o<!els.)

The laws of the universe are written in the language of mathematics. Algebrais sufficient to solve many static problems,but the most interesting natural

phenomena involve change and are describedby equations that relate changing

quantities.Becausethe derivativedxjdt= f'(t)of the function f is the rate at which

the quantity x = f(t) is changing with respectto the independentvariable t, it

is natural that equations involving derivativesare frequently usedto describethe

changing universe. An equation relating an unknown function and one or more ofits derivativesiscalleda differential equation.)

Example1)

......

The differentialequation)dx-= x2 + t 2dt

involvesboth the unknown function x(t) and its first derivative x/(t)= dxjdt.Thedifferentialequation)

d2y dy-+3-+7y=Odx2 dx

involves the unknown function y of the independent variable x and the first two

derivativesy/ and y\" of y. .)

The study of differentialequationshas three principalgoals:

1.To discover the differential equation that describesa specifiedphysicalsituation.

2.To find-eitherexactly or approximately-theappropriate solution of that

equation.3.To interpret the solution that is found.)

1)))

2 Chapter1 First-OrderDifferential Equations)

Example2)

Example3)

Temperature A)

Temperature T)

FIGURE1.1.1.Newton's law ofcooling,Eq.(3),describesthe

coolingofa hot rock in water.)

Example4)

In algebra,we typically seekthe unknown numbers that satisfy an equationsuchasx3 +7x2-llx+41= O.By contrast, in solvinga differentialequation,weare challengedto find the unknown functions y = y(x) for which an identity suchasy'(x)= 2xy(x)-thatis, the differentialequation)

dy-= 2xydx-holdson some interval of real numbers. Ordinarily, we will want to find allsolutions of the differentialequation, if possible.)

If C is a constant and)x2

y(x) = Ce ,) (1))

then)

\037\037

= C (2xex2

)= (2x)(ce

x2)

= 2xy.

Thus every function y(x) of the form in Eq. (1)satisfies-andthus is a solutionof-thedifferentialequation)

dy-= 2xydxfor all x. In particular, Eq. (1)defines an infinite family of different solutions ofthis differential equation, one for eachchoiceof the arbitrary constant C.By themethod of separation of variables (Section1.4)it can be shown that every solutionof the differentialequation in (2)is of the form in Eq.(1). .)

(2))

DifferentialEquationsandMathematicalModelsThe followingthree examplesillustrate the processof translating scientificlaws and

principlesinto differential equations. In eachof theseexamplesthe independentvariable is time t, but we will seenumerousexamplesin which somequantity otherthan time is the independentvariable.)

Newton'slaw of coolingmay be stated in this way: The time rate of change (therate of change with respectto time t) of the temperature T(t) of a body is propor-tional to the differencebetweenT and the temperatureA of the surrounding medium

(Fig.1.1.1).That is,)dT-= -k(T-A),dt)

(3))

where k is a positive constant. Observethat if T > A, then dTjdt < 0,so the

temperature is a decreasingfunction of t and the body is cooling.But if T < A,then dTjdt >0,so that T is increasing.

Thus the physicallaw is translated into a differentialequation. If we are giventhe values of k and A, we should be ableto find an explicitformula for T(t), andthen-with the aid of this formula-wecan predictthe future temperature of the

bod\037 .)...... ...... ....

Torricelli'slaw impliesthat the time rate of change of the volume V of water in adraining tank (Fig.1.1.2)is proportional to the squareroot of the depth y of waterin the tank:)

dV-= -k\037,dt)

(4))))

FIGURE1.1.2.Torricelli'slawofdraining, Eq.(4),describesthe

draining ofa water tank.)

Example6)

1 .1 Differential Equationsand MathematicalModels 3)

where k is a constant. If the tank is a cylinderwith verticalsidesand cross-sectionalarea A, then V = Ay, sodV/dt = A \302\267 (dy/dt). In this caseEq.(4)takes the form

dy = -h..jY,dt)

(5))

where h = k/A is a constant.) .)

The time rate of change of a population pet) with constantbirth and death rates is,in many simplecases,proportional to the sizeof the population.That is,

dP-= kP (6)dt 'where k is the constant of proportionality. .

Letus discussExample5 further. Note first that each function of the form

pet) = Cekt(7))

is a solutionof the differentialequation

dP-=kPdt

in (6).We verify this assertionasfollows:

P'(t)= Ckekt = k (Cekt) = kP(t)

for all real numbers t. Becausesubstitution of eachfunction of the form given in

(7)into Eq.(6)producesan identity, all such functions are solutionsof Eq.(6).Thus, even if the value of the constant k is known, the differential equation

dP/dt = kPhas infinitely many differentsolutionsof the form pet) = Cekt , oneforeachchoiceof the \"arbitrary\" constant C.This is typical of differential equations.It is alsofortunate, becauseit may allow us to useadditional information to selectfrom among all thesesolutionsa particularone that fits the situation under study.)

Supposethat pet) = Cekt is the population of a colony of bacteria at time t, that

the population at time t = 0 (hours, h) was 1000,and that the populationdoubledafter 1 h. This additional information about pet) yieldsthe followingequations:

1000= P(O)= Ceo= C,2000= P(l)= Cek.

It follows that C = 1000and that ek = 2,so k = In 2 \037 0.693147.With this valueof k the differentialequation in (6)is

dPdt

= (In2)P \037 (0.693147)P.Substitution of k = In 2 and C = 1000in Eq.(7)yieldsthe particularsolution

pet) = 1000e(ln2)t= 1000(e1n2)t = 1000.2t (becausee1n2 = 2))

that satisfies the given conditions.We can use this particular solution to predictfuture populations of the bacteria colony. For instance, the predictednumber ofbacteria in the population after one and a half hours (when t = 1.5)is)

P(I.5)= 1000.23/2 \037 2828.) .)))


8642

Q... 0-2-4-6-8-2

C= -12)

C= 12 C= 6 C= 3)

t)

FIGURE1.1.3.Graphs ofP(t)= Cekt with k = In 2.)

The condition P(O) = 1000in Example6 is calledan initial conditionbe-causewe frequently write differential equations for which t = 0 is the \"startingtime.\" Figure 1.1.3showsseveral different graphs of the form P(t) = Cekt with

k = In 2.The graphs of all the infinitely many solutions of dPjdt = kP in fact fill

the entire two-dimensionalplane,and no two intersect.Moreover, the selectionofanyone point Po on the P-axisamounts to a determinationof P(O).Becauseex-actly one solutionpassesthrough eachsuch point, we seein this casethat an initial

condition P(O)= Podeterminesa unique solution agreeing with the given data.)

MathematicalModels)Our brief discussionof populationgrowth in Examples5 and 6 illustratesthe crucialprocessof mathematical modeling(Fig.1.1.4),which involvesthe following:

1.The formulation of a real-world problemin mathematical terms; that is, theconstructionof a mathematicalmodel.

2.The analysis or solution of the resulting mathematicalproblem.3.The interpretation of the mathematical resultsin the context of the original

real-worldsituation-forexample,answering the questionoriginally posed.)

FIGURE1.1.4.Theprocessofmathematical modeling.)

In the population example,the real-world problemis that of determining the

population at somefuture time. A mathematicalmodelconsistsof a list of vari-ables(P and t) that describethe given situation, togetherwith one or moreequationsrelating thesevariables (dPjdt = kP,P(O)= Po) that are known or are assumedtohold.The mathematicalanalysis consistsof solving theseequations (here,for P asa function of t). Finally, we apply thesemathematicalresultsto attempt to answerthe original real-worldquestion.

As an exampleof this process,think of first formulating the mathematicalmodelconsistingof the equations dPjdt = kP, P(O)= 1000,describingthe bac-teria population of Example6.Then our mathematical analysis there consistedofsolving for the solution function P(t) = 1000e(1n2)t= 1000.2t as our mathemat-ical result. For an interpretation in terms of our real-world situation-theactualbacteria population-wesubstituted t = 1.5to obtain the predictedpopulation ofP(1.5)\037 2828bacteria after 1.5hours. If, for instance, the bacteria population isgrowing under idealconditions of unlimited spaceand food supply, our predictionmay be quite accurate, in which casewe concludethat the mathematicalmodel isquite adequate for studying this particularpopulation.

On the other hand, it may turn out that no solution of the selecteddifferentialequation accurately fits the actual population we'restudying. For instance, for nochoiceof the constantsC and k doesthe solution P(t) = Cekt in Eq. (7)accurately)))

Example7)

1.1Differential Equationsand MathematicalModels 5)

describethe actual growth of the human population of the world over the past fewcenturies.We must concludethat the differential equation dPjdt = kP is inad-equate for modeling the world population-,-which in recent decadeshas \"leveledoff\" as comparedwith the steeplyclimbing graphs in the upper half (P > 0) ofFig.1.1.3.With sufficient insight, we might formulate a new mathematical modelincluding a perhapsmore complicateddifferentialequation,one that that takes into

account such factors asa limited food supply and the effectof increasedpopulationon birth and death rates.With the formulation of this new mathematical model,wemay attempt to traverseonceagain the diagram of Fig.1.1.4in a counterclockwisemanner. If we can solve the new differential equation, we get new solution func-tions to comparewith the real-world population. Indeed,a successfulpopulationanalysis may require refining the mathematicalmodel still further as it is repeatedlymeasuredagainst real-worldexperience.

But in Example6 we simply ignoredany complicatingfactors that might af-fect our bacteria population. This made the mathematical analysis quite simple,perhapsunrealisticallyso.A satisfactorymathematicalmodel is subjectto two con-tradictory requirements:It must be sufficiently detailed to representthe real-worldsituation with relativeaccuracy,yet it must be sufficiently simpleto make the math-ematical analysis practical. If the model is so detailed that it fully representsthe

physicalsituation, then the mathematicalanalysis may be too difficult to carry out.If the modelis too simple,the resultsmay be so inaccurateas to be useless.Thusthere is an inevitabletradeoffbetweenwhat isphysicallyrealisticand what ismath-

ematically possible.The construction of a model that adequatelybridgesthis gapbetweenrealism and feasibility is therefore the most crucial and delicate step in

the process.Ways must be found to simplify the model mathematically without

sacrificingessentialfeatures of the real-worldsituation.Mathematical modelsare discussedthroughout this book. The remainderof

this introductory sectionisdevotedto simpleexamplesand to standardterminologyusedin discussingdifferentialequations and their solutions.)

ExamplesandTerminology)If C is a constant and y(x) = 1j(C-x), then

dy 1 2dx (C-x)2

=Y)

if xi-C.Thus)1

y(x) = C-x) (8))

definesa solution of the differentialequation)

dy 2-=ydxon any interval of real numbers not containing the point x = C.Actually, Eq.(8)defines a one-parameterfamily of solutions of dyjdx = y2, one for each value ofthe arbitrary constant or \"parameter\" C.With C = 1 we get the particular solution)

(9))

1y(x) = I-x)

that satisfiesthe initial condition y(O) = 1.As indicated in Fig.1.1.5,this solutionis continuouson the interval (-00,1)but has a vertical asymptoteat x = 1. .)))

6 Chapter1 First-OrderDifferential Equations)NUl Nu,\"' M,,\",.., , \"\"\"''''''I'. \"\"', , UW u,.,.

Example8) Verify that the function y(x) = 2xl/2-x1/2 lnx satisfiesthe differentialequation)

4x2y\" + y = 0) (10))

for all x > o.)

Solution First we compute the derivatives)

y/(x) = _!x-1/21nx and y\"(x) = ix-3/2Inx- 4x-3/2.)

Then substitution into Eq.(10)yields)

4x2y\" + y = 4x2 (ix-3/2Inx- !x-3/2

) +2xl/2-x1/2lnx = 0)

if x is positive, so the differentialequation is satisfiedfor all x >o.) .)The fact that we can write a differential equation is not enough to guarantee

that it has a solution. For example,it is clearthat the differentialequation)

(y/)2 + y2 = -1) (11))

has no (real-valued)solution, becausethe sum of nonnegativenumbers cannot benegative. For a variation on this theme, note that the equation)

(y/)2 + y2 = 0) (12))

obviously has only the (real-valued)solution y(x) = O. In our previous examplesany differentialequation having at leastone solution indeedhad infinitely many.

The orderof a differentialequation is the orderof the highest derivative that

appearsin it. The differential equation of Example8 is of secondorder, those in

Examples2 through 7 are first-orderequations, and)

y(4) +x2y

(3) +x5y = sin x)

is a fourth-order equation. The most general form of an nth-orderdifferential

equation with independent variablex and unknown function or dependentvariabley=y(x)is)

F (/\"

Cn\302\273

) 0x,y,y,y,...,y=,) (13))

where F is a specificreal-valued function of n +2 variables.Our use of the word solution has beenuntil now somewhat informal. To be

precise,wesay that the continuousfunction u = u (x)isa solutionof the differentialequation in (13)on the interval I provided that the derivativesu',u\", ..., u Cn ) existon I and)

F (/\"

Cn\302\273

) 0X,U,U,u,...,u=)

for all x in I. For the sakeof brevity, we may say that u = u (x) satisfiesthedifferentialequation in (13)on I.

Remark:Recall from elementarycalculusthat a differentiablefunction onan open interval is necessarilycontinuous there. This is why only a continuousfunction can qualify as a (differentiable)solution of a differential equation on aninterval. .)))


\"\" \"\" \037 A'''NA'n\037N;,'Nn''''nn,.., ,____ '\"\" \"\"\"'''''''''''''...0''.... '\" .......\037....... \"\"'.., '-' '\" \"\"\"'\",\"\"\",, .........,.,....... \"\"'.......\" .... \"\"',..........\"'\" \"\"'''''....... \"\" _, ...........\" '\" \"'\" ... \"\" ... ..,.. _... \"A\" ... .\037.\037. \037 ... oJ . _\"\"._.\"..\037_ ...\"....... \037\037 _\037 \037'n'\037\"\"\"\"'\"

Figure 1.1.5showsthe two \"connected\"branchesof the graph y = 1/(1-x).TheContinued left-hand branch is the graph of a (continuous)solutionof the differential equation

y/ = y2 that is definedon the interval (-00,1).The right-hand branch is the graphof a different solution of the differential equation that is defined (and continuous)on the different interval (1,(0).Sothe singleformula y (x) = 1/(1 -x) actuallydefines two different solutions (with different domains of definition) of the samedifferentialequation y/ = y2. .)

Example7)

Example9)

5)

x=l)

;;:.-. 0)

-5-5) o

x)

FIGURE1.1.5.Thesolution ofy/ = y2 defined byy(x) = 1/(1-x).)

5)

;;:.-.0)

-5-3) o

x)

FIGURE1.1.6.Thethreesolutions YI (x)= 3cos3x,Y2(X) = 2 sin 3x, and

Y3(X) = -3cos3x + 2 sin 3x ofthe differential equationy\" + 9y = O.)

\037 .\"....... ..... ..,......... ..... .....

If A and B are constants and)

y(x) = A cos3x+ B sin 3x,) (14))

then two successivedifferentiationsyield)

y/ (x)= -3Asin 3x +3Bcos3x,y\" (x)= -9Acos3x-9Bsin 3x = -9y(x))

for all x. Consequently,Eq. (14)defines what it is natural to call a two-parameterfamily of solutionsof the second-orderdifferentialequation)

5) y\" +9y = 0) (15))

3)

on the whole real number line. Figure 1.1.6showsthe graphs of several suchsolutions. .

Although the differentialequations in (11)and (12)are exceptionsto the gen-eral rule, we will seethat an nth-order differential equation ordinarily has an n-parameter family of solutions-oneinvolving n different arbitrary constantsor pa-rameters.

In both Eqs.(11)and (12),the appearanceof y/ as an implicitly definedfunc-tion causescomplications.For this reason,we will ordinarily assumethat any dif-ferentialequationunder study can besolvedexplicitlyfor the highest derivative that

appears;that is, that the equation can be written in the so-callednormalform(n) -G (

/\" (n-I)) (16)y - x,y,y,y,...,y ,

where G is a real-valued function of n + 1 variables. In addition, we will alwaysseekonly real-valuedsolutionsunlesswe warn the readerotherwise.

All the differentialequations we have mentioned so far are ordinarydiffer-ential equations, meaning that the unknown function (dependentvariable)dependson only a single independent variable. If the dependentvariable is a function oftwo or more independentvariables,then partial derivativesare likely to be involved;if they are, the equation is calleda partialdifferentialequation. For example,the

temperature u = u (x,t) of a long thin uniform rod at the point x at time t satisfies(under appropriatesimpleconditions)the partial differentialequation

au a2u--k-at-

ax2 'where k is a constant (calledthe thermal diffusivity of the rod). In Chapters 1

through 7 we will be concernedonly with ordinary differential equationsand will

refer to them simply asdifferentialequations.In this chapterwe concentrateonfirst-orderdifferentialequationsof the form

dydx

= f(x,y).) (17))))


Example10)

We also will samplethe widerange of applications of such equations.A typicalmathematicalmodelof an appliedsituation will be an initial value problem,con-sistingof a differentialequation of the form in (17)together with an initial condi-tion y(xo) = Yo. Note that we call y(xo)= Yo an initial condition whether or notXo = O.To solvethe initial value problem

dydx

= f(x,y), y(xo)= Yo)\037) (18))

means to find a differentiable function y = y (x) that satisfies both conditions in

Eq.(18)on someinterval containingXo.)

.....

Given the solution y(x) = I/(C- x) of the differential equation dyldx = y2discussedin Example7,solve the initial value problem

dy 2dx

= y, y(l) = 2.)

Solution We needonly find a value of C so that the solution y(x) = I/(C-x) satisfiestheinitial condition y(l) = 2.Substitution of the valuesx = 1 and y = 2 in the givensolution yields)

5)I

I

I

I

I

I

Ix =3/2)

y = 2/(3-2x)

\\

(1,2))

:>-..0)

-5-5) o

x)

FIGURE1.1.7.Thesolutions ofy/ = y2 defined byy(x) = 2/(3- 2x).)

12 = y(l) =

C _ 1'

1,and henceC =\037.

With this value of C we obtain the desired)so 2C- 2solution)

1y(x) = 3--x

2)

23 -2x)

Figure 1.1.7showsthe two branchesof the graph y = 2/(3-2x).The left-handbranch is the graph on (-00,

\037)of the solution of the given initial value problem

y/ = y2, Y (1)= 2.The right-hand branch passesthrough the point (2,-2) and istherefore the graph on

(\037, (0) of the solution of the different initial value problemy/ = y2, y(2) = -2. .)

In Problems 1 through 12,verify by substitution that eachgiven function is a solution of the given differential equation.Throughout theseproblems, primes denotederivatives with re-spectto x.1.y/ = 3x2; Y = x3 + 72.y/ + 2y = 0; y = 3e-2x

3.y\" + 4y = 0; YI = COS2x, Y2 = sin 2x4. y\" = 9y; YI = e3x , Y2 = e-3x)

5)

The centralquestionof greatest immediateinterest to us is this:If we are givena differentialequation known to have a solution satisfying a given initial condition,how do we actuallyfind or compute that solution? And, oncefound, what canwe do with it? We will seethat a relatively few simpletechniques-separationof variables (Section1.4),solution of linear equations (Section1.5),elementarysubstitution methods (Section1.6)-areenough to enableus to solve a variety offirst-orderequations having impressiveapplications.)

-\037\302\243\037\037\037 \037\037\037 ---------,------------------------------------------------------------------------)5. y/ = Y + 2e-x ; y = eX -e-X6.y\" + 4y' + 4y = 0; YI = e-2x , Y2 = xe-2x

7. y\"-2y' + 2y = 0; YI = eX cosx,Y2 = eX sin x

8.y\" +y = 3cos2x,YI = cosx-cos2x,Y2 = sinx-cos2x19.y/ + 2xy

2 = 0; y =1+X2

110.x2y\" + xy/

-Y = In x;YI = X - In x,Y2 = -- In x

x)))


11 2'\" 1 In x\302\267 x y + 5xy + 4y = 0; YI = 2'Y2 =

\037X x12.x2

y\"-xy'+ 2y = 0; YI = X cos(lnx),Y2 = x sin(lnx))

In Problems 13through 16,substitute y = erx into the givendifferential equation to determine all values of the constant rfor which y = erx is a solution of the equation.)

13.3y' = 2y15.y\" + y' -2y = 0)

14.4y\" = y

16.3y\" + 3y'-4y = 0)

In Problems 17through 26,first verify that y (x) satisfiesthe

given differential equation. Then determine a value ofthe con-stant Csothat y(x) satisfiesthe given initial condition. Useacomputer or graphing calculator(if desired)to sketch severaltypical solutions of the given differential equation, and high-light the onethat satisfiesthe given initial condition.)

17.y' + y = 0; y(x) = Ce-x, y(O)= 218.y' = 2y;y(x) = Ce2x, y(O)= 319.y' = y + 1;y(x) = Cex -1,y(O)= 520.y' = x -y; y(x) = Ce-X +x -1,y(O)= 1021.y' + 3x2

y = 0; y(x) = Ce-x3 , y(O)= 722.eYy' = 1;y(x) = In(x + C),y(O)= 0

dy23.x-+ 3y = 2x5; y(x) =\037x5

+ Cx-3, y(2) = 1dx

24. xy'-3y = x3; y(x) = x3(C+ In x),y(l) = 1725.y' = 3x2

(y2 + 1);y(x) = tan(x3 + C),y(O)= 1

26.y' + y tan x = cosx;y(x) = (x + C)cosx,y(n) = 0)

In Problems27 through 31,afunction y = g(x)is describedby somegeometricproperty of its graph. Write a differentialequation of the form dyjdx = f (x,y) having the function gas its solution (orasoneof its solutions).

27.Theslopeof the graph of g at the point (x,y) is the sumofx and y.

28.Theline tangent to the graph ofg at the point (x,y) inter-sectsthe x-axisat the point (xj2,0).

29.Every straight line normal to the graph ofg passesthroughthe point (0,1).Can you guesswhat the graph of such afunction g might look like?

30.The graph of g is normal to every curve of the formy = x2 + k (k is a constant) where they meet.

31.Theline tangent to the graph ofg at (x,y) passesthroughthe point (-y,x).)

In Problems 32 through 36, write-in the manner ofEqs.(3)through (6) of this section-adifferential equation that is amathematical modelof the situation described.32.Thetime rate of change of a population P is proportional

to the square root of P.33. The time rate of change of the velocity v of a coasting

motorboat is proportional 'to the squareof v.34. Theaccelerationdvjdt of a Lamborghini is proportional

to the differencebetween250kmJh and the velocity of thecar.)

35.In a city having a fixed population of P persons,the time

rate ofchangeof the number N ofthosepersonswho have

heard a certain rumor is proportional to the number ofthose who have not yet heard the rumor.

36.In a city with a fixed population of P persons,the time rateofchangeof the number N of thosepersonsinfected with

a certain contagious diseaseis proportional to the productof the number who have the diseaseand the number who

donot.)

In Problems 37 through 42, determine by inspection at leastone solution of the given differential equation. That is, useyour knowledge of derivatives to make an intelligent guess.Then test your hypothesis.

37.y\" = 0 38.y' = y

39.xy'+ y = 3x2 40. (y')2 + y2 = 141.y' + y = eX 42. y\" + y = 043. (a) If k is a constant, show that a general(one-parameter)

solution of the differential equation)

dx = kx 2dt

is given by x(t)= Ij(C-kt), where C is an arbitraryconstant.

(b) Determineby inspection a solution of the initial value

problem x'= kx 2, X (0)= O.44. (a) Continuing Problem43,assumethat k is positive, and

then sketch graphs of solutions ofx'= kx 2 with sev-eral typical positive values ofx(O).

(b) How would these solutions differ if the constant k

were negative?45. Supposea population P of rodents satisfiesthe differen-

tial equation dPjdt = kp2. Initially, there are P(O)= 2rodents, and their number is increasing at the rate ofdPjdt = 1 rodent per month when there are P = 10ro-dents. How long will it take for this population to growto a hundred rodents? Toa thousand? What's happeninghere?

46. Supposethe velocity v of a motorboat coasting in water

satisfiesthe differential equation dvjdt = kv 2.Theinitial

speedof the motorboat is v(O) = 10meters per second(m1s), and v is decreasingat the rate of 1m/S2 when v = 5m1s.How long doesit take for the velocity of the boat to

decreaseto 1 m1s?To /omls? When doesthe boat cometo a stop?

47. In Example 7 we saw that y(x) = Ij(C-x) defines aone-parameterfamily of solutions of the differential equa-tion dyjdx = y2. (a) Determine a value of C so that

y(10)= 10.(b)Is there a value of C such that y(O) = O?Can you nevertheless find by inspection a solution ofdyjdx = y2 such that y(O) = O?(c)Figure 1.1.8shows

typical graphs of solutions of the form y(x) = Ij(C-x).Doesit appearthat thesesolution curves fill the entire xy-plane? Can you concludethat, given any point (a,b) in

the plane, the differential equation dyjdx = y2 has ex-actly onesolution y(x) satisfying the condition y(a) = b?)))

C=-2C=-1c=oC=1C=2C=33)

\037 0C=-4

-1)

2)

1) C=4)

10080604020

\037 0-20-40-60-80-100-5-4-3-2-10 1 2 3 4 5

x)

-2)

FIGURE1.1.8.Graphsof solutions of the

equation dy/dx = y2.)

FIGURE1.1.9.Thegraph y = Cx4forvarious values of C.)

48. (a) Show that y(x) = Cx4 defines a one-parameterfam-

ily of differentiable solutions of the differential equationxy' = 4y (Fig. 1.1.9).(b)Show that

y(x) ={-x4 if x < 0,

x4 if x > 0)

definesa differentiable solution ofxy' = 4y for all x,but

is not of the form y(x) = Cx4. (c) Given any two realnumbers a and b, explain why-in contrast to the situa-

tion in part (c)ofProblem47-thereexist infinitely manydifferentiable solutions of xy' = 4y that all satisfy the

condition y(a) = b.)

1IIJ._.!!l!\037g\037\037!S_ \037S._\037\037\037\037\302\243\037\037...\037J?:\037}>..\037\037!!\037\037\037.\037\037__\037\037\037_!\037\037_\037_.______________________)

The first-order equation dy/dx = I(x,y) takesan especiallysimpleform if the

right-hand-sidefunction Idoesnot actually involve the dependentvariabley, so)

dy-= I(x).dxIn this specialcaseweneedonly integrateboth sidesof Eq. (1)to obtain)

\037) (1))

\037) y(x) = f f(x)dx+ c.) (2))

This is a generalsolutionof Eq.(1),meaning that it involvesan arbitrary constantC,and for every choiceof C it is a solution of the differential equation in (1).IfG(x)is a particular anti derivativeof I-thatis, if G'(x)= I(x)-then)

y(x) = G(x)+ C.) (3))

The graphs of any two such solutions Yl (x) G(x)+ Cl and Y2(X) =G(x)+C2on the sameinterval I are \"parallel\" in the senseillustrated by Figs.1.2.1and 1.2.2.There we seethat the constant C is geometrically the vertical distancebetweenthe two curves y(x) = G(x)and y(x) = G(x)+C.

To satisfy an initial condition y(xo)= Yo, we needonly substitute x = Xo and

y = Yo into Eq. (3)to obtain Yo = G(xo)+C,so that C = Yo-G(xo).With this

choiceof C,we obtain the particularsolutionofEq.(1)satisfying the initial value

problem)

\037)

dydx

= I(x), y(xo)= Yo.)))

Example1)

1.2Integralsas Generaland ParticularSolutions 11)

4321)

6)

C=-1C=-2)

4)

2)

\037 0) \037 0)

-1-2-3-4-4-3-2-1 0 1 2 3 4

x)

-2)

-4) C=-4-6-6 -4 -2 0 2 4 6

x)

FIGURE1.2.1.Graphs ofy =

\037

x2 + C for various values of C.)FIGURE1.2.2.Graphs ofy = sin x + C for various values of C.)

We will seethat this is the typicalpattern for solutionsof first-orderdifferential

equations.Ordinarily, we will first find a generalsolution involving an arbitraryconstant C.We can then attempt to obtain, by appropriatechoiceof C,aparticularsolution satisfying a given initial condition y (xo) = Yo.)

Remark:As the term isusedin the previousparagraph,a generalsolutionof a first-orderdifferentialequation is simply a one-parameterfamily of solutions.A natural question is whether a given general solution contains every particularsolution of the differential equation. When this is known to be true, we call it

the general solution of the differential equation. For example,becauseany two

antiderivativesof the samefunction f (x)can differ only by a constant, it followsthat every solutionof Eq.(1)is of the form in (2).Thus Eq.(2)servesto definethegeneral solutionof (1). .)

Solvethe initial valueproblem)

dy-= 2x+3, y(l) = 2.dx)

Solution Integrationof both sidesof the differentialequationas in Eq.(2)immediately yieldsthe general solution)4

2

0

-2\037

-4-6-8-10-6 -4 -2 0 2 4)

x)

FIGURE1.2.3.Solution curvesfor the differential equation in

Example 1.)

y(x}= f (2x+3}dx= x2 +3x+C.)

Figure 1.2.3showsthe graph y = x2 + 3x + C for various values of C. Theparticular solution we seekcorrespondsto the curve that passesthrough the point

(1,2),thereby satisfying the initial condition)

y(l) = (1)2+3 .(1)+C = 2.)

It follows that C = -2,so the desiredparticular solution is)

y(x)=x2 +3x-2.) .)))


Second-orderequations.The observation that the special first-order equationdyjdx = f(x) isreadily solvable(providedthat an anti derivativeof f can befound)extendsto second-orderdifferentialequationsof the specialform)

d2y

dx2 = g(x),) (4))

in which the function g on the right-hand sideinvolvesneither the dependentvari-abley nor its derivativedyjdx. We simply integrateonceto obtain)

\037\037

= f y\"(x)dx= f g(x)dx= G(x)+ C\\,)

where G is an antiderivative of g and C1 is an arbitrary constant. Then anotherintegration yields)

y(x) = f y'(x)dx= f [G(x)+Cddx= f G(x)dx+C\\x + C2,)

where C2 is a secondarbitrary constant. In effect, the second-orderdifferentialequation in (4) is one that can be solved by solving successivelythe first-orderequations)

dvdx

= g(x) and)dy-= vex).dx)

VelocityandAccelerationDirectintegration is sufficient to allow us to solve a number of important problemsconcerning the motion of a particle (or masspoint) in terms of the forcesactingon it. The motion of a particle along a straight line (the x-axis)is describedby itspositionfunction)

x = f(t)) (5))

giving its x-coordinateat time t.The velocity of the particle is defined to be)

\037) vet) = f' (t);) that is,)dxv= -.dt)

(6))

ItsaccelerationaCt) is aCt) = v'(t)= x\"(t);in Leibniznotation,)

\037)

dV d2xa - --

dt-

dt2 .) (7))

Equation (6)is sometimesappliedeither in the indefinite integral form x(t) =J vet) dt or in the definite integral form)

x(t) = x(to) +it

v(s)ds,to)

which you should recognizeas a statementof the fundamental theoremof calculus(preciselybecausedxjdt= v).)))

. . . .Example2)


Newton'ssecondlaw of motion says that if a force F(t) acts on the particleand is directedalong its line of motion, then)

ma(t)= F(t);) that is,) (8))F=ma,)

where m is the massof the particle. If the force F is known, then the equationx\"(t) = F(t)lmcan be integrated twice to find the position function x(t) in termsof two constants of integration. Thesetwo arbitrary constantsare frequently deter-mined by the initial positionXa = x(O)and the initial velocity Va = v(O) of the

particle.

Constantacceleration.For instance, supposethat the force F,and thereforethe

accelerationa = F1m, are constant. Then we beginwith the equation)

dv-= a (a is a constant)dt)

(9))

and integrateboth sidesto obtain)

v(t) = f adt=at+C!.)

We know that v = Va when t = 0,and substitution of this information into the

precedingequationyieldsthe fact that C1= Va. So)

dxvet) =-= at + Va.

dt)(10))

A secondintegration gives)

x(t) = f v(t) dt = f (at + vo) dt = 4at2 + vot +C2,)

and the substitution t = 0,x = Xa gives C2 = Xa. Therefore,)

X(t) = 4at2 + vat +Xa.) (11))

Thus, with Eq.(10)wecan find the velocity,and with Eq.(11)the position,ofthe particle at any time t in terms of its constant accelerationa, its initial velocityVa, and its initial position Xa.)

A lunar lander is falling freely toward the surface of the moon at a speedof 450meters per second(m/s). Its retrorockets,when fired, provide a constant decel-eration of 2.5meters per secondper second(m/s2) (the gravitational accelerationproducedby the moon is assumedto be included in the given deceleration).At what

height above the lunar surface should the retrorocketsbe activatedto ensurea \"soft

touchdown\" (v = 0 at impact)?)))

Solution We denoteby x(t) the height of the lunar lander above the surface, as indicatedin Fig.1.2.4.We let t = 0 denotethe time at which the retrorocketsshould befired. Then Vo = -450(mjs,negativebecausethe height x(t) is decreasing),anda = +2.5,becausean upward thrust increasesthe velocity v (although it decreasesthe speedIvl). Then Eqs.(10)and (11)become)

Lunar surface)

FIGURE1.2.4.Thelunar landerofExample 2.)

vet) = 2.5t-450) (12))

and)

x(t) = 1.25t2-450t+xo,) (13))

where Xo is the height of the lander above the lunar surface at the time t = 0 when

the retrorocketsshouldbe activated.FromEq.(12)weseethat v = 0 (soft touchdown)occurswhen t = 450j2.5=

180s (that is, 3 minutes); then substitution of t = 180,x = 0 into Eq.(13)yields)

Xo = 0 - (1.25)(180)2+450(180)= 40,500)

meters-thatis, Xo = 40.5km \03725\037

miles.Thus the retrorocketsshould be acti-vated when the lunar lander is40.5kilometersabove the surfaceof the moon, and it

will touch down softly on the lunar surfaceafter 3 minutes of deceleratingdescent..)

PhysicalUnits)

Numerical work requiresunits for the measurementof physicalquantities such asdistanceand time. We sometimesuse ad hoc units-suchas distancein milesorkilometersand time in hours-inspecialsituations (suchas in a probleminvolvingan auto trip). However, the foot-pound-second(fps) and meter-kilogram-second(mks)unit systemsare usedmore generally in scientificand engineeringproblems.In fact, fps units are commonly used only in the United States(and a few othercountries),while mks units constitute the standard international systemof scientificuni ts.)

.\037:,:;::\037':';:':;)

>'.;:;il'>;\037'.? U'lY\"iiil\037 s';';;\"i\037i,T;1';!

..

':....'.;...\"..,1,......'......

..

,.

........'.'............,...............:.'

..

....................;.m',.

...

.

.'....

...

.

.

.;

.

.

.

.

.....

.'...

.

.

.

.

....1;,;

..

.

.'..

......

.... ..'''''...........

..

..........................u

..

.......

.

...

.

......... ....n....

.

........'...........40

...

'..

.

.....8..

.

..

..............,

.--. -.-;- :\037,<

'])/>1(1 \037

\037>r:; ...:,.<:(\037,>t/y>c '.

\"\" ','a\", ',':L:

?\037:r:\037\037\037:::\\.- '<>::::;',\302\253\\:<\037-';: :\037/i:t'\037p;:,:::?r- ::.:_::{;;)l ';)

\037:':\\/:T/e/(>:,::; \"\037:_::;

\037,:..:\037)

>.:.-'::\037:\037{>}h:/:.:::;\037)

}>;r<r\037:\037fHSJ;::}{<;-;un:i;\\;',-';'-'- :'\"

;\":\"':,; -.\037' '-;.-.,)

Force pound (lb) newton (N)Mass slug kilogram (kg)Distance foot (ft) meter (m)Time second(s) second(8)g 32ft/s 2 9.8mls 2)

The last line of this table gives values for the gravitational accelerationg atthe surface of the earth. Although theseapproximate values will suffice for mostexamplesand problems,more precisevalues are 9.7805mjs2 and 32.088ftj s2 (atsealevel at the equator).

Both systemsare compatiblewith Newton'ssecondlaw F = ma.Thus 1 N is(by definition) the force requiredto impart an accelerationof 1 mjs2 to a massof 1

kg. Similarly, 1 slug is (by definition) the massthat experiencesan accelerationof1 ftj s2 under a force of lIb.(We will usemks units in all problemsrequiring massunits and thus will rarely needslugsto measure mass.))))


Inchesand centimeters (aswell as milesand kilometers)alsoare commonlyusedin describingdistances.For conversionsbetweenfps and mks units it helpstorememberthat)

1 in.= 2.54cm (exactly) and lIb \037 4.448N.)

For instance,)

1 ft = 12in. x 2.54\037m = 30.48cm,In.)

and it follows that)

1 mi = 5280ft x 30.48cm = 160934.4cm \037 1.609km.ft)

Thus a postedU.S.speedlimit of 50mi/hmeans that-ininternational terms-thelegalspeedlimit is about 50x 1.609\037 80.45lan/h.)

VerticalMotionwith GravitationalAcceleration)The weight W of a body is the force exertedon the body by gravity. Substitution

of a = g and F = W in Newton'ssecondlaw F = ma gives)

W=mg) (14))

for the weight W of the massm at the surfaceof the earth (whereg \037 32ft/s2\037 9.8

m/s2). For instance,a massof m = 20kg has a weight of W = (20kg)(9.8mjs2)=196N. Similarly, a massm weighing 100poundshas mks weight)

W = (100Ib)(4.448N/lb) = 444.8N,)

so its massis)W 444.8N

m = -= \037 45.4kg.g 9.8m/s2

To discussvertical motion it is natural to choosethe y-axisas the coordinatesystemfor position,frequently with y = 0 correspondingto \"ground level.\" If wechoosethe upward directionas the positivedirection, then the effectof gravity on averticallymoving bodyis to decreaseits height and alsoto decreaseits velocityv =dy/dt. Consequently,if we ignore air resistance,then the accelerationa = dv/dt ofthe body is given by)

\037)

dv---gdt- .) (15))

This acceleration equation provides a starting point in many problemsinvolvingverticalmotion. Successiveintegrations(as in Eqs.(10)and (11))yield the velocityand height formulas)

vet) = -gt+ va) (16))

and)

yet) =_\037gt2 + vat + Yo.) (17))

Here,Yo denotesthe initial (t = 0)height of the body and va its initial velocity.)))


Example3)

y-axis)

(-a,0)) (a,0)x-axis)

VR)

Vs)

FIGURE1.2.5.A swimmer'sproblem (Example 4).)

Example4)

(a) Supposethat a ball is thrown straight upward from the ground (Yo = 0)with

initial velocity Vo = 96(ftjs, so weuseg = 32ftj s2 in fps units). Then it reachesits maximum height when its velocity (Eq.(16))is zero,)

vet) = -32t+96= 0,)

and thus when t = 3 s.Hencethe maximum height that the ball attains is)

y(3) = -4.32.32 +96.3+0 = 144(ft)

(with the aid of Eq.(17)).(b) If an arrow is shot straight upward from the ground with initial velocityVo = 49(mjs,sowe useg = 9.8mjs2 in mks units), then it returns to the ground when)

yet) = -4.(9.8)t2 +49t= (4.9)t(-t+ 10)= 0,)

and thus after 10s in the air.) .)

A Swimmer'sProblemFigure 1.2.5showsa northward-flowing river of width w = 2a.The linesx = :f:arepresentthe banksof the river and the y-axisits center.Supposethat the velocityvR at which the water flows increasesasoneapproachesthe center of the river, andindeedis given in terms of distancex from the center by)

VR = Vo (1-

=:) .) (18))

You can use Eq. (18)to verify that the water does flow the fastest at the center,where VR = vo, and that VR = 0 at eachriverbank.

Supposethat a swimmerstarts at the point (-a,0)on the westbank and swimsdueeast(relativeto the water)with constantspeedVs. As indicated in Fig.1.2.5,his

velocity vector (relative to the riverbed) has horizontal component Vs and verticalcomponent VR. Hencethe swimmer'sdirection anglea is given by)

VRtan a = -.

Vs)

Becausetan a = dyjdx, substitution using (18)gives the differentialequation)

dy Vo

(X2

)dx=

Vs1-

a 2)(19))

for the swimmer'strajectory y = y(x) ashe crossesthe river.)

\037.....,...........,..\"\"\"\"'\" N \"\"'\" \037\037\037\"\"\"\"\"'__,_\"\"\"\"\"'\037_\"'....'_\037\037........ \"\"'''''''V\037\037\037\037'''''''''''''''\037\037N ....,\037\"\"'_\"'\" v.....,..\037\037'_....'N ''''\037.....,..''''''''''''.... N'\037\"\",,,,,,,,,,,, \"\",\037\037r-v \037\037\037\"\",,,,,,.... .....,.._......\037\037\"\"'.....,...... _....._,_,.\",\037 ....,_...\037\037\037... ............

Supposethat the river is 1 mile wideand that its midstreamvelocity is Vo = 9 mijh.If the swimmer'svelocity is Vs = 3 mijh,then Eq.(19)takesthe form

dy = 3(1_ 4x2).dx)

Integrationyields)

y(x) = f (3 - 12x2)dx = 3x-4x3 + C)))

for the swimmer'strajectory.The initial condition y (-4)= 0 yieldsC = 1,so)

Then)

y(x) = 3x -4x3 + 1.)

y (4) = 3 (4)-4 (4)3+ 1 ==2,so the swimmerdrifts 2 milesdownstreamwhile he swims1 mile acrossthe river.

.)

11IIIProblems)In Problems 1through 10,find a function y = f (x)satisfy-ing the given differential equation and the prescribedinitial

condition.

dy1.-=2x+l;y(0)=3dxdy2.-= (x -2)2;y(2) = 1dx

dy3.-= -IX;y(4) = 0dxdy 14. -= -;y(l)=5dx x2

dy 15.-= ; y(2) = -1dx -Vx + 2

6. dy = X -VX2 + 9; y(-4)= 0dxdy 107.- ; y(O)= 0dx x2 + 1

dy 19.-= ; y(O)= 0dx -V I -x 2)

dy8.-= cos2x;y(O)= 1dx

dy10.-=xe-x; y(O)= 1dx)

In Problems11through 18,find the positionfunction x(t) ofamoving particlewith the given accelerationa(t), initial posi-tion Xo = x (0),and initial velocity Vo = v (0).11.a(t) = 50,Vo = 10,Xo = 2012.a(t) = -20,Vo = -15,Xo = 513.a(t) = 3t, Vo = 5,Xo = 014.a(t) = 2t + 1,Vo = -7,xo= 415.a(t) = 4(t + 3)2,Vo = -1,Xo = 1

116.a(t) = , Vo = -1,xo== 1-vt + 4

117.a(t) ==3 ' Vo = 0,Xo = 0

(t + 1)18.a(t) = 50sin 5t, Vo = -10,Xo == 8)

In Problems19through 22,a particlestarts at the origin andtravels along the x-axiswith the velocity function v(t) whosegraph is shown in Figs. 1.2.6through 1.2.9.Sketch the graphofthe resulting positionfunction x(t)for 0 < t < 10.)

19. 10

8

6\037

4

2 . . . . .. . .

00 2 4)

(5,5))

8)6) 10)t)

20.)

FIGURE1.2.6.Graph of the

velocity function v(t) ofProblem 19.10)

8)

6)\037)

(5,5).)

4)

2)

2) 4 6t)

8) 10)


velocity function v(t) ofProblem 20.21. 10)

8)

6)(5,5):)

\037)

4)

2)

2) 8)4) 6) 10)t)


velocity function v(t) ofProblem 21.)))

1822. 10

8

6;:::..

4)

Chapter1 First-OrderDifferential Equations)

. (3,5) (7,5))

2) 4) 6) 8) 10)t)


velocity function vet) ofProblem22.23.What is the maximum height attained by the arrow ofpart

(b) ofExample 3?24. A ball is dropped from the top of a building 400 ft high.

How long does it take to reachthe ground? With what

speeddoesthe ball strike the ground?25.Thebrakes of a carareapplied when it is moving at 100

km/h and provide a constant decelerationof 10meterspersecondpersecond(m/s2).How far doesthe cartravel be-fore coming to a stop?

26.A projectileis fired straight upward with an initial veloc-ity of 100ml s from the top of a building 20 m high andfalls to the ground at the baseof the building. Find (a) itsmaximum height abovethe ground; (b) when it passesthe

top of the building; (c) its total time in the air.27.A ball is thrown straight downward from the top of a tall

building. Theinitial speedof the ball is 10mls.It strikesthe ground with a speedof 60mls.How tall is the build-

ing?28.A baseballis thrown straight downward with an initial

speedof 40 ft/s from the top of the Washington Monu-ment (555ft high). How long doesit take to reachthe

ground, and with what speeddoesthe baseballstrike the

ground?29.A dieselcar gradually speedsup so that for the first 10s

its accelerationis given by)

dv-= (0.12)t2 + (0.6)t (ft/s2).

dt)

If the car starts from rest (xo = 0, Vo = 0), find the dis-tance it has traveled at the end of the first 10s and its

velocity at that time.30.A car traveling at 60 mi/h (88ft/s) skids 176ft after its

brakes are suddenly applied.Under the assumption thatthe braking system provides constant deceleration,whatis that deceleration?For how long doesthe skid continue?

31.Theskid marks made by an automobile indicated that itsbrakes were fully applied for a distanceof 75 m beforeitcameto a stop.Thecarin question is known to have a con-stant decelerationof20m/s2 under theseconditions. Howfast-in km/h-wasthe car traveling when the brakeswerefirst applied?)

32.Supposethat a car skids 15m if it is moving at 50km/hwhen the brakes are applied. Assuming that the car hasthe sameconstant deceleration,how far will it skid if it ismoving at 100km/h when the brakesareapplied?

33.On the planet Gzyx, a ball dropped from a height of 20 ft

hits the ground in 2 s.If a ball is dropped from the top ofa 200-ft-tallbuilding on Gzyx, how long will it take to hit

the ground? With what speedwill it hit?

34. A personcan throw a ball straight upward from the sur-

faceof the earth to a maximum height of 144ft. How

high could this personthrow the ball on the planet GzyxofProblem29?

35.A stone is droppedfrom rest at an initial height h abovethe surfaceof the earth. Show that the speedwith which it

strikes the ground is v = ,J2gh .36.Supposea woman has enough \"spring\" in her legsto jump

(on earth) from the ground to a height of 2.25feet. Ifshejumps straight upward with the same initial velocityon the moon-wherethe surfacegravitational accelerationis (approximately) 5.3 ft/s 2-howhigh abovethe surfacewill sherise?

37.At noon a car starts from rest at point A and proceedsat

constant accelerationalong a straight road toward pointB. If the car reachesB at 12:50P.M.with a velocity of60mi/h, what is the distance from A to B?

38.At noon a carstarts from rest at point A and proceedswith

constant accelerationalong a straight road toward point C,35miles away. If the constantly acceleratedcararrives at

C with a velocity of 60 mi/h, at what time doesit arrive

at C?39.If a = 0.5mi and Vo = 9 mi/h as in Example 4, what

must the swimmer's speedVs be in order that he drifts

only 1 mile downstream ashe crossesthe river?

40. Supposethat a = 0.5mi, Vo = 9 mi/h, and vs = 3 mi/has in Example 4, but that the velocity of the river is given

by the fourth-degree function)

VR = Vo (1-

\037: ))

rather than the quadratic function in Eq.(18).Now find

how far downstream the swimmer drifts ashe crossesthe

flver.

41.A bomb is dropped from a helicopterhovering at an alti-

tude of 800feetabove the ground. From the ground di-rectly beneath the helicopter,a projectileis fired straight

upward toward the bomb, exactly 2secondsafter the bombis released.With what initial velocity should the projectilebe fired, in order to hit the bomb at an altitude of exactly400feet?

42. A spacecraftis in freefall toward the surfaceof the moonat a speedof 1000mph (mi/h). Its retrorockets, when

fired, provide a constant decelerationof 20,000mi/h 2.At

what height abovethe lunar surface should the astronauts

fire the retrorockets to insure a soft touchdown? (As in

Example 2, ignore the moon'sgravitational field.))))

1.3SlopeFieldsandSolutionCurves 19)

43. Arthur Clarke'sThe Wind from the Sun (1963)describesDiana, a spacecraftpropelledby the solarwind. Its alu-minized sail provides it with a constant accelerationof0.001g= 0.0098m1s2. Supposethis spacecraftstartsfrom rest at time t = 0 and simultaneously fires a pro-jectile(straight ahead in the samedirection) that travels atone-tenth of the speedc = 3 x 108 m1sof light. How longwill it take the spacecraftto catch up with the projectile,)

and how far will it have traveled by then?

44. A driver involved in an accidentclaimshe was going only25mph. When policetested his car, they found that when

its brakes wereapplied at 25 mph, the car skidded only45 feet beforecoming to a stop. But the driver's skidmarks at the accidentscenemeasured 210feet. Assum-

ing the same(constant) deceleration,determine the speedhe was actually traveling just prior to the accident.)

_ Sl\037peFieldsand SolutionCurves)

y)

FIGURE1.3.1.A solution curvefor the differential equationy' = x -

y together with tangentlines having. slopeml = Xl - YI at the

point (Xl, YI);. slopem2 = X2-

Y2 at the

point (X2, Y2); and. slopem3 = X3 -Y3 at the

point (X3, Y3).)

Example1)

Considera differentialequationof the form)

\037)

dydx

= I(x,y)) (1))

x)

where the right-hand function I(x,y) involvesboth the independentvariablex andthe dependentvariable y. We might think of integrating both sidesin (1)with re-spectto x, and hencewrite y(x) = J I(x,y(x\302\273

dx+C.However,this approachdoesnot leadto a solutionof the differentialequation,becausethe indicatedintegralinvolvesthe unknown function y(x) itself, and thereforecannotbeevaluatedexplic-itly. Actually, there existsno straightforward procedureby which a generaldifferen-tial equationcan be solvedexplicitly.Indeed,the solutionsof sucha simple-lookingdifferential equation as y' = x2 + y2 cannot be expressedin terms of the ordinary

elementary functions studiedin calculustextbooks.Nevertheless,the graphicalandnumerical methods of this and later sectionscan be usedto constructapproximatesolutionsof differentialequations that sufficefor many practicalpurposes.)

SlopeFieldsandGraphicalSolutionsThere is a simplegeometric way to think about solutions of a given differential

equation y' = I(x,y). At each point (x,y) of the xy-plane, the value of I(x,y)determinesa slopem = I(x,y). A solutionof the differentialequationis simply adifferentiablefunction whosegraph y = y(x) has this \"correctslope\"at eachpoint

(x,y(x\302\273 through which it passes-thatis, y'(x) = I(x,y(x\302\273. Thus a solutioncurve of the differential equation y' = I(x,y)-thegraph of a solution of the

equation-issimply a curve in the xy-planewhosetangent line at each point (x,y)has slopem = I(x,y). For instance, Fig.1.3.1showsa solution curve of the

differential equation y' = x - y together with its tangent linesat three typicalpoints.

This geometric viewpoint suggestsa graphicalmethod for constructing ap-proximate solutions of the differential equation y' = I(x,y). Through each of arepresentativecollectionof points (x,y) in the plane we draw a short line segmenthaving the properslopem = I(x,y). All theseline segmentsconstitutea slopefield (ora directionfield) for the equation y' = I(x,y).)

\302\245..... ............. ..... ......... ......... .............. ......... .........-....u.............. ...... ..... .............. ........ \302\245-.......... .-. ...... ............ .........\037..... ...........

Figures1.3.2(a)-(d) show slopefieldsand solutioncurvesfor the differential equa-tion)

dy = kydx)

(2))

with the valuesk = 2, 0.5,-1,and -3of the parameterk in Eq.(2).Note that eachslopefield yieldsimportant qualitative information about the set of all solutions)))

4321

\037 0-1-2)

o 1 234x)

4321

\037O

-1-2-3

\\ \\ \\ \\ \\ \\ \\ \\-4-4-3 -2 -1 0 1 2 3 4x)

FIGURE1.3.2(a)Slopefieldand solution curves for y' = 2y.)

FIGURE1.3.2(b)Slopefieldand solution curves for

y' = (O.5)y.)

4 43 32 21 1

\037O \037O

-1 -1-2 -2)

o 1 234x)

o 1 234x)

FIGURE1.3.2(c)Slopefieldand solution curves for y' = -

y.)

FIGURE1.3.2(d) Slopefieldand solution curvesfor y' = -3y.)

of the differential equation. For instance, Figs.1.3.2(a)and (b) suggestthat eachsolution y(x) approaches:l::ooas x \037 +00if k > 0,whereasFigs.1.3.2(c)and(d) suggestthat y(x) \037 0 as x \037 +00if k < O. Moreover, although the signof k determinesthe direction of increaseor decreaseof y(x),its absolute value Ikl

appearsto determine the rate of change of y(x). All this is apparent from slopefields like those in Fig.1.3.2,even without knowing that the general solution ofEq. (2)is given explicitlyby y(x) = Cekx. .

A slopefield suggestsvisually the general shapesof solution curves of thedifferential equation. Through eachpoint a solution curve should proceedin sucha direction that its tangent line is nearly parallel to the nearby line segmentsof the

slopefield. Starting at any initial point (a,b),we can attempt to sketchfreehand an

approximate solution curve that threads its way through the slopefield, followingthe visible line segmentsascloselyaspossible.)

Example2) Construct a slopefield for the differentialequation y/ = X -Y and use it to sketch

an approximatesolutioncurve that passesthrough the point (-4,4).)

Solution Solution Fig.1.3.3showsa table of slopesfor the given equation. The numericalslopem = x -

y appearsat the intersection of the horizontalx-rowand the ver-tical y-column of the table. If you inspectthe pattern of upper-left to lower-rightdiagonalsin this table, you can seethat it was easily and quickly constructed.(Of)))

43

21

\037 0-1-2-3-4)

-4 -3 -2 -10 1 2 3 4)

x)

FIGURE1.3.6.Slopefield and

typical solution curves fory' = x -

y.)

1.3SlopeFieldsand SolutionCurves 21)

x\\y -4 -3 -2 -1 0 1 2 3 4-4 0 -1 -2 -3 -4 -5 -6 -7 -8-3 1 0 -1 -2 -3 -4 -5 -6 -7-2 2 1 0 -1 -2 -3 -4 -5 -6-1 3 2 1 0 -1 -2 -3 -4 -5

0 4 3 2 1 0 -1 -2 -3 -41 5 4 3 2 1 0 -1 -2 -32 6 5 4 3 2 1 0 -1 -23 7 6 5 4 3 2 1 0 -14 8 7 6 5 4 3 2 1 0)

FIGURE1.3.3.Values of the slopey' = x - y for -4< x, y < 4.)

5) 54321

;::.... 0-1-2-3-4-5

-5)

;::.... 0)

\\ \\ '\" -\\ '\" -\"'-)

\\ \\ '\" -\\ '\" - /'\" - / /- / / /)

\\

\\ '\" - /'\" - / /- / / /)

-5-5) o

x)

5)o 5x)

FIGURE1.3.4.Slopefield fory' = x -

y correspondingto thetable ofslopesin Fig. 1.3.3.)

FIGURE1.3.5.Thesolutioncurve through (-4,4).)

course,a more complicatedfunction f(x,y) on the right-hand sideof the differen-tial equation would necessitatemore complicatedcalculations.)Figure 1.3.4showsthe correspondingslopefield, and Fig.1.3.5showsan approximatesolution curvesketchedthrough the point (-4,4) so as to follow as this slopefield as closelyaspossible.At eachpoint it appearsto proceedin the directionindicatedby the nearbyline segmentsof the slopefield. .)

Although a spreadsheetprogram (for instance) readily constructs a table ofslopesas in Fig.1.3.3,it can be quite tedious to plot by hand a sufficient numberof slopesegmentsas in Fig.1.3.4.However,most computer algebra systemsin-cludecommandsfor quickand ready constructionof slopefieldswith as many linesegmentsas desired;such commands are illustrated in the applicationmaterial forthis section.The more line segmentsare constructed, the more accuratelysolutioncurves can be visualized and sketched.Figure 1.3.6showsa \"finer\" slopefield forthe differential equation y/ = X -

Y of Example2,together with typical solutioncurves treading through this slopefield.

If you lookcloselyat Fig.1.3.6,you may spota solutioncurve that appearsto be a straight line!Indeed,you can verify that the linear function y = x - 1 isa solution of the equation y/ = X - y, and it appearslikely that the other solutioncurves approach this straight line as an asymptote as x \037 +00.This inferenceillustrates the fact that a slopefieldcan suggesttangible information about solutionsthat is not at all evidentfrom the differentialequation itself. Can you, by tracing the)))


Example3)

400 \"\"\"\"\"\"\", , , , , , , , , , , , ,300 ','\037 \037 \037 \037 \037 \037 \037 \037 \037 \037 \037

.....\"\"'=\037 ..................................................)\037 200)

\037\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"\"

100 / / / \037 \037 \037 \037 \037 \037 \037 \037 \037 \037 \037/ / / / / / / / / / / / // / / / / / / / / / / / / /)

5 10 15 20 25)t)


typical solution curves forv' = 32-O.16v.)

appropriate solutioncurve in this figure, infer that y(3) \037 2 for the solutiony(x) ofthe initial value problemy' = x- y, y(-4)= 4?)

ApplicationsofSlopefieldsThe next two examplesillustrate the useof slopefields to glean useful informationin physicalsituations that are modeledby differentialequations.Example3 isbasedon the fact that a baseballmoving through the air at a moderate speedv (lessthan

about 300ftls) encounters air resistancethat is approximatelyproportional to v. Ifthe baseballis thrown straight downward from the top of a tall building or from ahovering helicopter,then it experiencesboth the downwardaccelerationof gravityand an upward acceleration of air resistance.If the y-axisis directeddownward,then the ball'svelocity v = dy/dt and its gravitational accelerationg = 32 ftls 2 areboth positive, while its accelerationdue to air resistanceis negative.Henceits total

accelerationis of the form)

dv-= g-kvdt

.) (3))

A typical value of the air resistanceproportionalityconstant might be k = 0.16.)

Supposeyou throw a baseballstraight downward from a helicopterhovering at analtitude of 3000feet. You wonder whether someonestanding on the ground belowcouldconceivablycatch it. In orderto estimate the speedwith which the ball will

land, you can use your laptop'scomputer algebra system to construct a slopefieldfor the differentialequation)

dv-= 32 -0.16v.dt)

(4))

The result is shown in Fig.1.3.7,together with a number of solution curvescorrespondingto different values of the initial velocity v(O)with which you mightthrow the baseballdownward.Note that all thesesolutioncurvesappear to approachthe horizontal line v = 200as an asymptote. This impliesthat-howeveryouthrow it-thebaseballshould approach the limiting velocity v = 200ftls insteadof accelerating indefinitely (as it would in the absenceof any air resistance).Thehandy fact that 60milh = 88 ftls yields)

ft 60milh miv = 200-x \037 136.36-.

s 88 ftls h)

Perhapsa catcher accustomedto 100milh fastballs would have somechance offielding this speedingball. .

Comment If the ball'sinitial velocity is v(O) = 200,then Eq. (4)givesv' (0) = 32 - (0.16)(200)= 0,so the ball experiencesno initial acceleration.Its

velocity therefore remains unchanged,and hencev(t) = 200is a constant \"equilib-rium solution\" of the differentialequation. If the initial velocity is greater than 200,then the initial accelerationgiven by Eq. (4)is negative,so the ball slowsdown as it

falls.But if the initial velocity is lessthan 200,then the initial accelerationgiven by

(4) is positive, so the ball speedsup as it falls. It therefore seemsquite reasonablethat, becauseof air resistance,the baseballwill approach a limiting velocity of 200ftls-whateverinitial velocity it starts with. You might like to verify that-intheabsenceof air resistance-thisball would hit the ground at over300mi/h. .)))

Example4)

300250200

Q.. 15010050o

o) 25) 50t)

75)


typical solution curves forP'= O.06P-O.0004p2.)

Example5)


In Section1.7we will discussin detail the logisticdifferential equation)

dP-= kP(M-P)dt)

(5))

that often is used to model a population P(t) that inhabits an environment with

carrying capacity M. This means that M is the maximum populationthat this envi-ronment can sustain on a long-termbasis(in terms of the maximum availablefood,for instance).)

If we take k = 0.0004and M = 150,then the logisticequation in (5)takes the form)

dP 2-= 0.0004P(150- P) = 0.06P-0.0004P.dt)

(6))

100)

The positive term 0.06Pon the right in (6)correspondsto natural growth at a 6%annual rate (with time t measuredin years).The negative term -0.0004p2

repre-sentsthe inhibition of growth due to limited resourcesin the environment.

Figure 1.3.8showsa slopefieldfor Eq. (6),togetherwith a number of solutioncurves correspondingto possibledifferent values of the initial population P(0).Note that all thesesolution curves appear to approach the horizontal line P == 150asan asymptote. This impliesthat-whatever the initial population-thepopulationpet) approachesthe limiting population P = 150as t \037 00. .

Comment If the initial population is P(O)= 150,then Eq. (6)gives)

P'(O)= 0.0004(150)(150- 150)= 0,)

sothe populationexperiencesno initial (instantaneous)change.It thereforeremains

unchanged, and henceP(t) = 150is a constant \"equilibrium solution\" of the dif-ferentialequation. If the initial populationis greater than 150,then the initial rate ofchange given by (6)is negative,so the population immediatelybeginsto decrease.But if the initial population is lessthan 150,then the initial rate of changegiven by

(6)is positive, so the population immediatelybeginsto increase.It thereforeseemsquite reasonableto concludethat the population will approach a limiting value ofISO-whateverthe (positive)initial population. .)ExistenceandUniquenessofSolutionsBeforeone spendsmuch time attempting to solve a given differential equation, it

is wiseto know that solutions actually exist. We may alsowant to know whether

there is only one solution of the equation satisfying a given initial condition-thatis, whether its solutionsare unique.)

(a) [Failureof existence]The initial value problem)

, 1y=-x)

y(O)= 0) (7))

has no solution, becauseno solution y (x) = J(l/x)dx = In Ix I + C of the differ-ential equation is defined at x = O. We seethis graphically in Fig.1.3.9,which

showsa direction field and sometypical solution curves for the equationy' = l/x.It is apparent that the indicated direction field \"forces\"all solution curvesnear the

y-axisto plunge downwardso that none can passthrough the point (0,0).)))

24 Chapter1 First-OrderDifferential Equations

2/ / / / / / // / / / / / // / / / / / // / / / / / /

>-0 / / / /Yl(X) =X2

;::.... / / / / / /: // / / / / / // //////)

o)

//\037//////Y2(X)

= 0)

ox)

1)

(0,0)o)

x)

FIGURE1.3.9.Direction fieldand typical solution curves forthe equation y' = l/x.)

FIGURE1.3.10.Directionfield and twodifferent solution curvesfor the initial value

problem y' = 2-JY,y(O)= O.)

(b) [Failureof uniqueness]On the other hand, you can readily verify that the initial

value problem)y/ ==2\037, y(O) ==0) (8))

has the two different solutions Yl (x) ==x2 and Y2(X) = 0 (seeProblem27).Figure1.3.10showsa direction field and thesetwo different solutioncurves for the initial

value problemin (8).We seethat the curve Yl (x) ==x2 threads its way through theindicateddirection field,whereasthe differentialequation y/ ==2,JYspecifiesslopey/ = 0 along the x-axisY2(X) ==o. .

Example5 illustrates the fact that, before we can speakof \"the\" solution ofan initial value problem,we need to know that it has one and only one solution.Questionsof existenceand uniquenessof solutions also bear on the processofmathematicalmodeling.Supposethat we are studying a physical systemwhosebe-havior is completely determinedby certain initial conditions,but that our proposedmathematical modelinvolves a differential equation not having a unique solutionsatisfying thoseconditions.This raisesan immediate question as to whether themathematicalmodeladequatelyrepresentsthe physicalsystem.

The theorem statedbelowimpliesthat the initial valueproblem y/ ==I(x,y),yea) = b has one and only one solutiondefinednear the point x = a on the x-axis,provided that both the function I and its partial derivative aljay are continuousnear the point (a,b) in the xy-plane.Methodsof proving existenceand uniquenesstheorems are discussedin the Appendix.)

Y)

b)

THEOREM1 ExistenceandUniquenessof SolutionsSupposethat both the function I(x,y) and its partial derivativeDyl(x,y) arecontinuous on somerectangle R in the xy-planethat contains the point (a,b)in its interior. Then, for someopeninterval I containing the point a, the initial

value problem)

x) \037)

dydx

= I(x,y), yea) = b) (9))FIGURE1.3.11.TherectangleR and x-interval I ofTheorem1,and the solution curve y = y (x)through the point (a,b).)

has one and only one solution that is definedon the interval I. (As illustrated in

Fig.1.3.11,the solution interval I may not be as \"wide\" as the originalrectangleR of continuity; seeRemark 3 below.))))

6)

4)y = 1/(1-x))

;::.... 2) --,IRI

I

I)

o)

-2-4) -2) o

x)

2)

FIGURE1.3.12.Thesolutioncurve through the initial point(0,1)leavesthe rectangle Rbeforeit reachesthe right sideofR.)

Example6)


Remark 1:In the caseof the differentialequationdy/dx = -y of Exam-pIe 1 and Fig.1.3.2(c),both the function I(x,y) = -y and the partial derivativeaIjay = -1are continuous everywhere,so Theorem 1 impliesthe existenceof aunique solution for any initial data (a,b). Although the theoremensuresexistenceonly on someopeninterval containingx = a, each solutiony(x) = Ce-x actuallyis definedfor all x.

Remark 2: In the caseof the differential equation dyjdx = -2,JYofExample5(b)and Eq. (8),the function I(x,y) = -2,JYis continuouswherevery > 0,but the partial derivativealjay = Ij,JYis discontinuouswhen y = 0, andhenceat the point (0,0).This is why it is possiblefor there to existtwo different

solutions Y1 (x) = x2 and Y2 (x) = 0,eachof which satisfies the initial conditiony(O)= o.)

Remark 3: In Example7 of Section1.1we examinedthe especiallysim-pledifferentialequationdyjdx = y2. Herewe have I(x,y) = y2 and aljay = 2y.Both of thesefunctions are continuous everywhere in the xy-plane,and in partic-ular on the rectangle -2< x < 2,0 < y < 2.Becausethe point (0,1)lies in

the interior of this rectangle, Theorem 1 guaranteesa unique solution-necessarilya continuousfunction-ofthe initial value problem)

\037)

dy 2dx

= y, y(O)= 1) (10))

on someopenx-interval containinga = O.Indeedthis is the solution)

1y(x) =

1 -x)

4)

that we discussedin Example7. But y(x) = Ij(1-x) is discontinuousat x = 1,soour unique continuoussolutiondoesnot existon the entire interval -2<x < 2.Thus the solution interval I of Theorem 1 may not be as wide as the rectangleR

where I and aljay are continuous. Geometrically,the reasonis that the solutioncurve provided by the theorem may leave the rectangle-whereinsolutionsof thedifferentialequation are guaranteedto exist-beforeit reachesthe one or both endsof the interval (seeFig.1.3.12). .)

The following exampleshowsthat, if the function I(x,y) and/or its partialderivative aljay fail to satisfy the continuity hypothesis of Theorem 1,then the

initial value problemin (9)may have either no solution ormany-eveninfinitely

many-solutions.).....,..... \037\"A\"'\" ........h \"...... \037 .. \037 .... ..... , .......A....... ........ ,...... ..,,, ......

Considerthe first-orderdifferentialequation)

dyxdx= 2y.) (11))

Applying Theorem 1 with I(x,y) = 2yjx and aljay 2jx,we conclude that

Eq. (11)must have a unique solution near any point in the xy-planewherex i= O.Indeed,we seeimmediatelyby substitution in (11)that)

y(x) = Cx2)

(12))))

(0,b) (0,0))

2)

4)

;::.,. 0)

-2)

-4-2) -1) o

x)

FIGURE1.3.13.Thereareinfinitely many solution curvesthrough the point (0,0),but nosolution curves through the point(0,b) if b 1=o.)

4)

2)

;::.,. 0)

-2)

-4-2) o

x)

-1) 1)

FIGURE1.3.14.Thereareinfinitely many solution curvesthrough the point (1,-1).)

satisfiesEq. (11)for any value of the constant C and for all values of the variablex. In particular, the initial value problem)

dyxdx= 2y, y(O)= 0) (13))

2)

has infinitely many differentsolutions,whosesolutioncurves are the parabolasy =Cx2 illustrated in Fig.1.3.13.(In caseC = 0 the \"parabola\" is actually the x-axisy = 0.)

Observethat all theseparabolaspass through the origin (0,0), but none ofthem passesthrough any other point on the y-axis. It follows that the initial valueproblemin (13)has infinitely many solutions,but the initial value problem)

X\037\037

=2Y, y(O)=b) (14))

has no solution if b i= O.Finally, note that through any point off the y-axisthere passesonly one of the

parabolasy = Cx2.Hence,if a i= 0,then the initial value problem)

dyxdx= 2y, yea) = b) (15))

has a unique solutionon any interval that contains the point x = a but not the originx = 0 In summary, the initial valueproblemin (15)has). a unique solution near (a,b) if a i= 0;. no solution if a = 0 but b i= 0;. infinitely many solutions if a = b = O.) .)

Still more can be said about the initial value problemin (15).Consideratypical initial point off the y-axis-forinstance the point (-1,1)indicated in Fig.1.3.14.Then for any value of the constant C the function definedby)

2)

Ix2

y(x) = 2Cx)(16))

if x <0,if x > 0)

is continuous and satisfiesthe initial value problem)

dyx-= 2y, y(-I)= 1.dx)

(17))

For a particular value of C,the solution curve defined by (16)consistsof the lefthalf of the parabola y = x2 and the right half of the parabola y = Cx2. Thus the

unique solution curve near (-1,1)branchesat the origin into the infinitely manysolution curves illustrated in Fig.1.3.14.

We therefore seethat Theorem 1 (if its hypothesesare satisfied) guaranteesuniquenessof the solution near the initial point (a,b),but a solutioncurve through

(a,b) may eventually branch elsewhereso that uniquenessis lost. Thus a solution

may existon a larger interval than one on which the solutionisunique. For instance,the solution y(x) = x2 of the initial value problemin (17)existson the whole x-axis,but this solution is unique only on the negativex-axis-00< x < O.)))


lIB)ProbleD1s)In Problems1through 10,we have provided the slopefieldofthe indicated differential equation, together with oneor moresolution curves.Sketch likely solution curves through the ad-ditional points marked in eachslopefield.

dy1.dx)

dy4. -= x -Y

dx)

3)\\ \\ \\ \\

\\ \\ \\ \\

\\. ..\\\\ \\ \\ \\

\\ \\ \\ \\ \\

\\ \\ \\ \\ \\ \\

\\ \\ \\ \\ \\ ,\\ \\ \\ \\ ,,\\ \\ \\ ,, ,)

\\\\\\\\\",--\\\\\\\",--..-\\e\\

,,..--..-/\\\",--..-/\",--..-/)

2)

\\ \\ \\ \\ \\ \\ \\ \\ \\

\\ \\ \\ \\ \\ \\ \\ \\ \\

\\e\\,,.'\\ ...\\\\\"\"\"'\\\"\"\"'\

\\\\ \\\\ \\\\\\\\\\ \\\\ \\\\\\\\e\\ \\. \\..\\

\\ \\ \\ \\ \\ \\ \\

\\ \\ \\ \\ \\ \\ \\ \\ \\

\\ \\ \\ \\ \\ \\ \\ \\ \\

\\ \\ \\ \\ \\ \\ \\ \\,, \\ \\ \\ \\ \\ ,,\"\"\"\

\037 0)

1)

I I

I I

le\\\\ \\

\\ \\

\\ \\

\\\037

\\ \\

\\ \\)

2)

-y- sIn x)

3)

-1)

/ / I I I'Y / / I / I I/ / I I / I I/ I I / / I I11111111111111I I \037 I I leI111111/111////)

1) -2)

\037 0) //P/////// / / / / / / /-1 /11111/// I I I I I I I /I I I / / / I I II I / / / / / I I-2 / I \037 I I Ie/ / I1111111111/11/1111-3-3 -2 -1 0

X)

-2 -1) oX)

1) 2) 3)

//////////////////I / \037 / / /e/ / II I I / / / / I I/ I I I I I I I I)

FIGURE1.3.18.)

1 2) 3)

FIGURE1.3.15.) 5.\037\037

= y-X + 1)

dy2.-=x+ydx)

3)

......--\",\\\\\\--\",\\\\\\\\-,,, \\\\e\\ \\ \\,,, \\ \\ \\ \\ \\ \\,, \\ \\ \\ \\ \\ \\ \\, \\ \\ \\ \\ \\ \\ \\ \\

\\ \\ \\e\\ \\ \\\037 \\ \\

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\)

1)

11111//1111//el I I .I1/1111I I I I I I/ I I / I I/ / I I I I/ / / I I I/ / / / I I)

/ / /

/ / // Ie // / /I / /I I II I II I II I I)

/ /

/ /2/// // / // /

1/ 11 ,e, ,

/ I I/ I I

\037 0)

3)

2)

-1)\037 0)

\\ \\

\\ \\-1 \\ e\\\\ \\

\\ \\

\\ \\-2 \\ \\

I \\

\\ \\-3-3 -2)

\\ \\ \\ , \037 ,\\ \\ \\ \\ , \037\\.\\\\.'\\ \\ \\ \\ \\ \\

\\ \\ \\ \\ \\ \\

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\)

-2)

-3-3 -2 -1) oX)

1 2) 3)

-1) oX)

1) 2) 3)FIGURE1.3.19.)

FIGURE1.3.16.)

dy3.-= y- sIn x

dx)

dy6.-= x -y + 1

dx)

3)

1)

3)

\037 0) \037 0)

2)2)

111111111111lieIIleiI I I II I I I

I)

I I I I / / I I II I I / / / / I II / /e/ / /fI / I/////////////////\037

/)

1)

1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\e\\ \\ \\.\\ \\.\\

\\ \\ \\ \\ \\ \\ \\ \\ ,\\ \\ \\ \\ \\ \\ \\ ,,\\ \\ \\ \\ \\ \\ ,,,\\e...\\\\\\\",-\\\\\\\\\",--\\ \\ \\ \",--......)

-1) ,................-..-..-.................,) -1)/IIII

I /\037/

/ // /)

III//

/ /

leI/ /

/ /)

\",\",','\\\"\"\"'\\-2 \\\\,e\",\037\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\

\\ \\ \\ \\ \\ \\ \\ \\ \\-3-3 -2 -1)

-2)

ox)

1) 2) 3) -2) -1) ox)

1) 2) 3)

FIGURE1.3.17.) FIGURE1.3.20.)))


dy. .7.-= SIn x + SIn ydx)

O dy 2 .1 .-= -x + SIn ydx)

3)

1 /------...-/)

3)

1)

---........................................--..-.-...--.......----,)2 ,.-------...-/)/...,....-----//) 2)/...-...----,.-/)

\037 0) \037 0)\"\"\"'-.;:\037 \\,, \\e\\ \\ \\ \\ ,, \\-1 '\\ \\ \\ \\ \\ \\ \\ , ........--.--.....---....... -1 ,\\ \\ \\ \\ \\ \\ \\ \\ , ,.......-------........ ,\\ \\ \037\\ \\ \\ \\ \\ \\ ,.....----........, \\

-2 \\ \\ \\ \\ \\ \\ \\ \\ , ,.......-------............... -2 , \\

\\ \\ \\ \\ \\ \\ \\ , ........---------......... \\\037, \\ ,-\\ \\ \\ \\ ,, '-----.-- \\ \\\"'\\\\\\'\" -------///,..--\\ \\-3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3-3 -3

x x)

FIGURE1.3.21.) FIGURE1.3.24.)

In Problems11through 20,determine whether Theorem 1doesordoesnot guarantee existenceofa solution ofthe given initial

value problem. If existenceis guaranteed, determine whether

8. dy Theorem 1doesor doesnot guarantee uniqueness of that so--= x2 -y lution.dx

3 11.dy y(1)= -1-= 2x2y2;/1//\"\\\\ dx/ / / / -, \\ \\ \\

2 / / / / -,, \\ \\

/ / / /,.-.....,\\ \\

dy/ / 1//-\", 12. y(1)= 11 / / I / / -.....\" -= x In y;/., \037 / ,/...,e, dx/ / /1//--'I / / I / / /--

\037 0 dy/ 13.-=\037; y(O)= 1/-1 I dx/

/

dy-2 / 14. y(O)= 0/

dx= \037;

/

/-3 -2 -1 0 1 2 3-3 dyx 15.-= y'x - y; y(2) = 2dx

FIGURE1.3.22.16.dy

y(2) = 1-= y'x - y;dx

17. dyy(O)= 1y- = x -1;dx

9. dy 2dy-=X -y-2dx 18.y-=x-l; y(l) = 0dx

3

19.dyy(O)= 02 -= In(1+ y2);dx

1 dy20._ = x2 _ y2; y(O)= 1\037 0 dx

I / / / /-,, \\ /////-\", /

In Problems 21and 22, first use the method of Example 2-1 / /1/--\" /

/ / ////-\" /

to construct a slopefieldfor the given differential equation./ / /1//--' /

-2 / / / I / / /-- / /

Then sketch the solution curve correspondingto the given ini-/ It / / ,/ / ..-- ,e,/ / / / 1 / / / / / /

tial condition. Finally, use this solution curve to estimate the/ / / / / / / / / / /-3-3 -2 -1 0 1 2 3 desiredvalue of the solution y (x).x

FIGURE1.3.23. 21.y' = x + y, y(O)= 0; y(-4)=?22.y' = y

-x, y(4)= 0; y(-4)=?)))

Problems 23 and 24 are like Problems 21and 22, but nowusea computer algebrasystem to plot and print out a slopefieldfor the given differential equation. Ifyou wish (and know

how), you cancheckyour manually sketchedsolution curve by

plotting it with the computer.

23.y/ = x2 + y2 -1, y (0)= 0; y (2)= ?124. y/=x+\"2y2 , y(-2)=0; y(2)=?

25.You bailout of the helicopterof Example 3 and pull the

ripcord of your parachute. Now k = 1.6in Eq. (3), soyour downward velocity satisfiesthe initial value problem)

dv

dt= 32- 1.6v, v(O) = O.)

26.)

In order to investigate your chancesof survival, constructa slopefield for this differential equation and sketch the

appropriate solution curve. What will your limiting veloc-ity be?Will a strategically locatedhaystack do any good?How long will it take you to reach95%of your limiting

velocity?Supposethe deerpopulation P(t)in a small forest satisfiesthe logistic equation)

dPdt

= 0.0225P-0.0003p2.)

Construct a slopefield and appropriate solution curve toanswer the following questions: If there are 25 deer attime t = 0 and t is measured in months, how long will

it take the number of deer to double? What will be the

limiting deerpopulation?)

Thenext sevenproblems illustrate the fact that, if the hypothe-sesof Theorem 1are not satisfied, then the initial value prob-lem y/ = f(x,y), yea) = b may have either no solutions,finitely many solutions, or infinitely many solutions.

27.(a) Verify that if c is a constant, then the function defined

piecewiseby)

y(x) ={\037x

_ C)2)

for x < c,for x > c)

satisfies the differential equation y/ = 2-JY for all x (in-cluding the point x = c).Construct a figure illustrating thefact that the initial value problem y/ = 2-JY,y (0)= 0 hasinfinitely many different solutions. (b) For what values ofb doesthe initial value problem y/ = 2-JY,y(O)= b have(i) no solution, (ii) a unique solution that is defined for all

x?28.Verify that if k is a constant, then the function y (x) = kx

satisfies the differential equation xy/ = y for all x. Con-struct a slopefield and severalof thesestraight line so-lution curves.Then determine (in terms of a and b) how

many different solutions the initial value problem xy/ = y,yea)= b has-one,none, or infinitely many.)


29.Verify that if c is a constant, then the function defined

piecewiseby)

y(x) ={o

3(x - c))

for x < c,for x > c)

satisfiesthe differential equation y/ = 3y2/3 for all x.Canyou alsouse the \"left half\" of the cubic y = (x -c)3in

piecing together a solution curve of the differential equa-tion? (SeeFig. 1.3.25.)Sketcha variety of such solution

curves. Is there a point (a, b) of the xy-plane such that

the initial value problem y/ = 3y2/3, yea) = b has eitherno solution or a unique solution that is defined for all x?Reconcileyour answer with Theorem 1.)

y)

y =(x-c)3)

c)x)

FIGURE1.3.25.A suggestion for Problem 29.)

30.Verify that if c is a constant, then the function defined

piecewiseby)

1

+1y(x) = cos(x-c)

-1)

.f <1 X _ c,if c < x < c + T{,if x > c + T{)

satisfiesthe differential equation y/ = -JI - y2 for all x.(Perhapsa preliminary sketch with c = 0 will behelpful.)Sketcha variety of such solution curves. Then determine

(in terms ofa and b) how many diffe rent solutions the ini-

tial value problem y/ = -JI - y2, yea)= b has.31.Carry out an investigation similar to that in Problem 30,

exceptwith the differential equation y/ = +JI - y2.Doesit suffice simply to replacecos(x-c) with sin(x -c)in piecing together a solution that is defined for all x?

32.Verify that if c > 0, then the function defined piecewiseby)

{0 if x2 < C,

y(x) = -(x2 -c)2 if x2 > c

satisfies the differential equation y/ = 4x-JY for all x.Sketcha variety of such solution curves for different val-ues ofc.Then determine (in terms ofa and b) how manydifferent solutions the initial value problem y/ = 4xJY,yea)= b has.)))

33.If c i= 0, verify that the function defined by y (x) =xj(cx- 1)(with graph illustrated in Fig. 1.3.26)satisfiesthe differential equation x2

y/ + y2 = 0 if x i= 1jc.Sketcha variety of such solution curves for different values ofc.Also, note the constant-valued function y(x) = 0 thatdoesnot result from any choiceof the constant c.Finally,determine (in terms ofa and b) how many different solu-tions the initial value problem x2

y/ + y2 = 0, yea) = bhas.)

34. (a) Usethe direction field of Problem 5 to estimate the

values at x = 1 of the two solutions of the differ-

ential equation y/ = Y- x + 1 with initial values

y(-l)= -1.2and y(-l)= -0.8.)(b) Usea computer algebra system to estimate the val-

ues at x = 3 of the two solutions of this differen-tial equation with initial values y(-3)= -3.01and

y(-3)= -2.99.)\" \\ \\ \\

\" \" \\ \\

..........\"\"\\--..........\";::..,.

---..........\"-....................\"\\

..........\"\"\\ \\

\" \" \\ \\ \\

\" \\ \\ \\ \\

x)

I

1\\\\ \\ \"

1\\ \\\"\"1\\ \"\"..........

\\

I

\037 --.....-==. :::--=--)

Thelessonof this problem is that small changesin initial

conditions can make big differencesin results.)

35.(a) Usethe direction field of Problem 6 to estimate thevalues at x = 2 of the two solutions of the differ-ential equation y/ = X -

y + 1 with initial values

y(-3)= -0.2and y(-3)= +0.2.)

\\ \" \" ..........

\\ \\ \" \"

\\ \\ \\ \

(b) Usea computer algebra system to estimate the val-ues at x = 3 of the two solutions of this differen-tial equation with initial values y(-3)= -0.5and

y(-3)= +0.5.The lessonof this problem is that big changes in initial

conditions may make only small differencesin results.)

\"..........--)

FIGURE1.3.26.Slopefield for x2y/ + y2 = 0

and graph ofa solution y(x) = xj(cx- 1).)

1.3Application)\037hN_\037\037'\037PN\037\037\037,\037 :-Q\037N\037\037\037\037t\037N\037__\037\037!\037p\037m,\037\037elds

and SolutioI?-Curv\037,s)

Widely availablecomputer algebrasystemsand technical computing environmentsincludefacilities to automate the constructionof slopefieldsand solutioncurves,asdo somegraphing calculators (seeFig.1.3.27).)

FIGURE1.3.27.Slopefield and solution curves for the differential

equation)dy .-= sln(x - y)dx

with initial points (0,b),b = -3,-1,-2,0,2,4and window-5< x,y < 5 on a TI-89graphing calculator.)

The applications manual accompanying this textbook includesdiscussionofMaple\342\204\242, Mathematica\342\204\242, and MATLAB\342\204\242 resourcesfor the investigation of dif-ferential equations.For instance, the Maplecommand

wit.h(DEt.ools):DEplot.(diff(y(x),x)=sin(x-y(x\302\273,y(x), x=-S..S,y=-S..S);)))

o 1 2 345x)

FIGURE1.3.28.Computer-generatedslopefield and solutioncurvesfor the differential equationy' = sin (x - y).)


and the Mathematica command

\302\253 Graphics\\Plot.Field.mPlot.Vect.orField[{l,Sin[x-y]},{x,-5,5},{y, -5,5}]

produceslope fields similar to the one shown in Fig.1.3.28.Figure 1.3.28it-self was generated with the MATLAB program dfield[John Polking and DavidArnold, Ordinary DifferentialEquations UsingMATLAB, 2nd edition,UpperSad-dle River, NJ: Prentice Hall, 1999]that is freely available for educationaluse(math.rice.edul\037dfield).When a differential equation is entered in the dfieldsetup menu (Fig.1.3.29),you can (with mousebutton clicks)plot both a slopefieldand the solution curve (orcurves) through any desiredpoint (or points).Another

freely available and user-friendly MATLAB-basedODEpackagewith impressivegraphicalcapabilitiesis lode(www.math.uiuc.eduliode).)

FIGURE1.3.29.MATLAB dfieldsetup to construct slopefield and solution curvesfor y' = sin(x - y).)

Usea graphing calculatoror computer system in the followinginvestigations.You might warm up by generating the slopefields and somesolution curves forProblems1 through lOinthis section.)

INVESTIGATIONA: Plot a slopefieldand typical solutioncurvesfor the differen-tial equationdy/dx = sin(x-y), but with a larger window than that of Fig.1.3.28.With -10< x < 10,-10<

Y< 10,for instance, a number of apparent straight

line solutioncurves shouldbe visible.

(a) Substitute y = ax+b in the differentialequation to determinewhat the coeffi-cientsa and b must be in orderto get a solution.

(b) A computer algebra systemgives the general solution)

(X-2-C

)y(x)=x-2tan-1 .x-c)Plot this solutionwith selectedvaluesof the constantC to comparethe resultingsolutioncurves with thoseindicated in Fig.1.3.28.Can you seethat no value ofC yieldsthe linear solution y = x-n/2 correspondingto the initial conditiony(n/2) = O? Are there any values of C for which the correspondingsolutioncurves liecloseto this straight line solutioncurve?)))


INVESTIGATIONB: For your own personalinvestigation,let n be the smallestdigit in your student ID number that is greater than 1,and considerthe differential

equation)dy 1-= -cos(x-ny).dx n

(a) First investigate(asin part (a) of InvestigationA) the possibilityof straight linesolutions.

(b) Then generate a slopefield for this differentialequation, with the viewing win-dow chosenso that you can picture someof thesestraight lines,plusa sufficientnumber of nonlinear solution curves that you can formulate a conjectureaboutwhat happensto y(x) asx --* +00.State your inferenceas plainly as you can.Given the initial value y(O) = Yo, try to predict(perhapsin terms of Yo) howy(x) behaves asx --* +00.

(c) A computer algebrasystem gives the general solution)

y(x) =\037 [x

+2 tan- 1

(x \037 c)J.)

Can you make a connection between this symbolicsolution and your graphi-cally generatedsolution curves (straight linesor otherwise)?)

....SeparableEquationsandApplications-.JIIIIiI,\"_ _ ___\" , __ ____ \037 _\037_ \037 ___ _\037\"___ ___'_'Nm'_N______\037 _N_ \",\" ______O\"\"N'N_N 0 _ ____ __o\037 _ ,__ _\037,__ \"0' \"Nm_____w_,,_,_ \037m,_ _mN____ ,_ _m ___\"' NN N _ _ _ __ ___ ____ _\" _,.w __ _ __\"N'\" ___ ___ __N_ N_O\037\"y\"_'_.'.N\037..'NY' N)

The first-orderdifferentialequation)

dy-= H(x,y)dx)

(1))

iscalledseparableprovided that H(x,y) can bewritten as the productof a function

of x and a function of y:)

\037)

dy g(x)-= g(x)h(y)= ,dx fey))

where h (y) = 1/f (y). In this casethe variablesx and y can beseparated-isolatedon oppositesidesof an equation-bywriting informally the equation)

fey) dy = g(x)dx,)

which we understand to beconcisenotation for the differentialequation)

dyf(y)-= g(x).dx)

(2))

It is easy to solve this specialtype of differential equation simply by integratingboth sideswith respectto x:)

f f(y(x\302\273\037\037

dx= f g(x)dx+ c;)))

Example1)

1.4SeparableEquationsand Applications 33)

equivalently,)

f fey) dy = f g(x)dx+c.) (3))

All that is requiredis that the antiderivatives)

F(y) = f f(y)dy and G(x)= f g(x)dx)

can be found. To seethat Eqs.(2)and (3)are equivalent,note the following conse-quenceof the chain rule:)

, , dyDx[F(y(x\302\273J = F (y(x\302\273y (x)= f(y)-= g(x)= Dx[G(x)J,

dx)

which in turn is equivalentto)

F(y(x\302\273= G(x)+C,) (4))

becausetwo functions have the samederivativeon an interval if and only if theydiffer by a constant on that interval.)

Solvethe initial value problem)

dydx

= -6xy, y(O)= 7.)

Solution Informally, we divide both sidesof the differentialequationby y and multiply eachsideby dx to get)

8I

6 I

I

4 I

I2 I

;;:.... 0 I /\\ \"-

-2 \\

\\-4I

-6 I

I

-8 I

.-2 1 2x (0,-4))

FIGURE1.4.1.Slopefield andsolution curves for y/ = -6xyin

Example 1.)

dy-= -6xdx.y)

Hence)

f d;= f (-6x)dx;

In Iyl = -3x2+C.)

We seefrom the initial condition y(O) = 7 that y(x) is positivenear x = 0, sowemay deletethe absolutevalue symbols:)

lny = -3x2+ C,)

and hence)y(x) = e-3x2+c = e-3x2eC = Ae-3x2

,

where A = eC.The condition y (0)= 7 yieldsA = 7,so the desiredsolution is)

y(x) = 7e-3x2.)

This is the upperemphasizedsolutioncurve shown in Fig.1.4.1.) .)))


\037 ,,\"\" ,,,,nn\"'\"'' In\" \"\"\" nn\"n,,,',

Example2)

Remark:Suppose,instead,that the initial condition in Example1had beeny(O) = -4.Then it would follow that y(x) is negative near x = O. We shouldtherefore replaceIyl with -y in the integrated equation In Iyl = -3x2+ C toobtain

In(-y) = -3x2+C.The initial condition then yieldsC = In4,soIn(-y) = -3x2+In4,and hence)

y(x) = _4e-3x2.)

This is the lower emphasizedsolutioncurve in Fig.1.4.1.) .)

Solvethe differentialequation)

dy 4 -2xdx

-3y2-5.) (5))

Solution When we separatethe variablesand integrateboth sides,we get)

6) . .....- -------: ----,-- :--- -- --). . ,\"\",.\".-\"'-\"'---------------)

4 .,.'., ..(1 3)

.....',.//,.,.......-----, .-----........\"\"\", . .)

;;:.... 0)

-2)

-4). . .-----------------)

-6-6 -4 -2 0 2 4 6 8

x)

FIGURE1.4.2.Slopefield andsolution curves for

y/ = (4-2x)/(3y 2 -5) in

Example 2.)

f (3l-5)dy = f (4-2x)dx;

y3 _ 5y = 4x -x2 +C.) (6))

This equation is not readily solvedfor y as an explicitfunction of x.) .)As Example2 illustrates, it mayor may not be possibleor practical to solve

Eq. (4) explicitly for y in terms of x. If not, then we call (4) an implicit solutionof the differential equation in (2). Thus Eq. (6)gives an implicit solution of the

differentialequation in (5).Although it is not convenientto solveEq.(6)explicitlyin terms of x,weseethat each solution curve y = y(x) lieson a contour (or level)curve where the function)

H(x,y) = x2 -4x + y3-5y)

is constant. Figure 1.4.2showsseveralof thesecontour curves.)

------'--'--\037--------\037-'---._---------,-\037----,------,---,---\037-------- _. ,,------_..,--,\037--,..,----,-----------------.---_ To solve the initial value problem)

dy 4 -2x-- , y(l) = 3,dx 3y2- 5)(7))

we substitute x = 1 and y = 3 in Eq.(6)and get C = 9.Thus the desiredparticularsolution y(x) is defined implicitly by the equation)

y3_ 5y = 4x -x2 +9.) (8))

The correspondingsolution curve y = y(x) lieson the uppercontour curve in

Fig.1.4.2-theone passingthrough (1,3). Becausethe graph of a differentiablesolution cannot have a vertical tangent line anywhere, it appearsfrom the figurethat this particular solution is definedon the interval (-1,5)but not on the interval

(-3,7). .)))

2015105--0

\037

-5-10-15-20-6 -4 -2 0 2 4 6

y)

FIGURE1.4.3.Graph offey) = y3

-5y - 9.)

3

2)

1)

\037 0)

-1)

-2)

o 1 2 3x)

FIGURE1.4.4.Slopefield andsolution curves for y/ = -xjy.)


Remark 1:When a specificvalue of x is substituted in Eq.(8),we canattempt to solve numericallyfor y. For instance,x = 4 yieldsthe equation)

fey) = y3 -5y -9 = o.)

Figure 1.4.3showsthe graph of f. With a graphing calculatorwe can solvefor the

singlereal root y \037 2.8552.This yieldsthe value y(4) \037 2.8552of the particularsolution in Example3.

Remark 2: If the initial condition in (7)is replacedwith the conditiony(l) = 0,then the resulting particular solution of the differential equation in (5)lieson the lower \"half\" of the oval contour curve in Fig.1.4.2.It appears that this

particular solution through (1,0) is defined on the interval (0,4) but not on the

interval (-1,5).On the other hand, with the initial conditiony(l) = -2we get the

lower contour curve in Fig.1.4.2.This particular solution is definedfor all x.Thusthe initial condition can determine whether a particular solution is definedon the

whole real line or only on someboundedinterval. With a computeralgebrasystemone can readily calculate a table of valuesof the y-solutionsof Eq. (8)for x-valuesat desiredincrements from x = -1to x = 5 (for instance).Sucha table of valuesserveseffectivelyas a \"numerical solution\" of the initial value problem in (7). .)

Implicit,General,andSingularSolutionsThe equation K (x,y) = 0 is commonlycalledan implicitsolutionof a differential

equation if it is satisfied (on some interval) by some solution y = y (x) of the

differential equation. But note that a particular solution y = y (x)of K(x,y) = 0mayor may not satisfy a given initial condition. For example,differentiation ofx2 + y2 = 4 yields)

dyx +y- = 0,dxsox2 + y2 = 4 is an implicit solutionof the differentialequationx +yy' = O.But

only the first of the two explicitsolutions)

y(x) = +J4 -x2 and y(x) = -J4 -x 2)

satisfiesthe initial condition y (0)= 2 (Fig.1.4.4).Remark 1:You should not assumethat everypossiblealgebraicsolution

y = y (x) of an implicit solution satisfies the samedifferential equation. For in-stance,if we multiply the implicit solutionx2 + y2 -4 = 0 by the factor (y -2x),then we get the new implicit solution)

(y -2x)(x2 + y2 -4) = 0)

that yields (or \"contain s\") not only the previously noted explicitsolutions y

+-J4 -x2 and y = -,y'4 -x2 of the differentialequationx+yy' = 0,but also the

additional function y = 2xthat doesnot satisfy this differentialequation.

Remark 2: Similarly, solutions of a given differential equation can beeither gained or lost when it is multiplied or divided by an algebraic factor. Forinstance,considerthe differentialequation)

dy(y -2x)y-= -x(y-2x)

dx)(9))))


75)

50);>..,)

25)

o)

-15-10-5 0 5 10 15x)

FIGURE1.4.5.Thegeneralsolution curves y = (x -C)2andthe singular solution curve y = 0of the differential equation(y/)2 = 4y.)

Example4)

Solution)

4)

2)

;>.., 0)

-2)

-4)

-2 -1 0)

x)

FIGURE1.4.6.Generaland

singular solution curves fory/ == 6x(y - 1)2/3.)

having the obvious solution y = 2x. But if we divide both sidesby the commonfactor (y -2x),then we get the previouslydiscusseddifferentialequation)

dyy dx

= -x,) dyor x +y- = 0,dx)

(10))

of which y = 2xis not a solution. Thus we \"lose\" the solution y = 2xof Eq. (9)upon its division by the factor (y -2x);alternatively, we \"gain\" this new solutionwhen we multiply Eq. (10)by (y -2x).Suchelementary algebraicoperations to

simplify a given differential equation before attempting to solve it are common in

practice,but the possibilityof lossor gain of such \"extraneoussolutions\" shouldbekeptin mind. .

A solution of a differentialequation that contains an \"arbitrary constant\" (likethe constant C in the solution of Examples1 and 2) is commonlycalleda generalsolutionof the differentialequation; any particularchoiceof a specificvalue for Cyieldsa singleparticular solution of the equation.

The argument precedingExample1 actually sufficesto show that everypartic-ular solution of the differentialequation f (y) y' = g (x) in (2)satisfiesthe equationF(y(x))= G(x)+C in (4).Consequently,it is appropriateto call (4)not merelyageneral solution of (2),but the general solutionof (2).

In Section1.5we shall seethat everyparticular solutionof a linearfirst-orderdifferential equation is contained in its general solution. By contrast, it is com-mon for a nonlinear first-order differential equation to have both a general solu-tion involving an arbitrary constant C and one or several particular solutions that

cannot be obtained by selectinga value for C. Theseexceptionalsolutions arefrequently calledsingularsolutions.In Problem 30 we ask you to show that the

general solution of the differentialequation (y')2= 4y yieldsthe family of parabo-las y = (x-C)2 illustrated in Fig. 1.4.5,and to observethat the constant-valuedfunction y(x) = 0 is a singular solution that cannot be obtained from the generalsolution by any choiceof the arbitrary constant C.)

................. ...... .Find all solutionsof the differentialequation)

dy-= 6x(y- 1)2/3.dx)

Separationof variables gives)! 12/3 dY =!2XdX;

3(y- 1)

(y - 1)1/3= x2 +C;

y(x) = 1 + (x2 +C)3.)

2)

Positive values of the arbitrary constant C give the solution curves in Fig.1.4.6that lieabove the line y = 1,whereasnegative values yield those that dip belowit. The value C = 0 gives the solution y(x) = 1 +x6, but no value of C givesthe singular solution y(x) = 1 that was lost when the variables wereseparated.Note that the two different solutions y(x) = 1 and y(x) = 1 + (x2 - 1)3both

satisfy the initial condition y(l) = 1.Indeed,the whole singular solution curvey = 1 consistsof points where the solution is not unique and where the function

f (x,y) = 6x(y - 1)2/3is not differentiable. .)))


Natural GrowthandDecayThe differentialequation)

\037)

dx = kx (k a constant)dt)

(11))

servesasa mathematicalmodelfor a remarkably widerangeof natural phenomena-any involving a quantity whosetime rate of change isproportionalto its current size.Hereare someexamples.)

POPULATIONGROWTH: Supposethat pet) is the number of individuals in a

population (of humans, or insects,or bacteria)having constant birth and death rates{3 and 8 (in births or deaths per individual per unit of time). Then, during a short

time interval \037t, approximately{3P(t)\037t births and 8P(t) \037t deathsoccur,so the

change in pet) is given approximatelyby)

\037P \037 ({3-8)P(t) \037t,)

and therefore)dPdt)

\037Plim -= k P,

\037t--+O \037t)

(12))

where k = {3- 8.)

COMPOUNDINTEREST: LetA(t) be the number of dollarsin a savingsaccountat time t (in years),and supposethat the interest is compoundedcontinuously at

an annual interest rate r. (Note that 10%annual interest means that r = 0.10.)Continuouscompoundingmeans that during a short time interval \037t, the amount ofinterest addedto the account is approximately \037A = r A(t) \037t, so that)

dA \037A-= lim -= r A.dt \037t--+O \037t)

(13))

RADIOACTIVEDECAY: Considera sampleof material that containsN(t) atomsof a certainradioactiveisotopeat time t.It has beenobservedthat a constantfractionof those radioactiveatoms will spontaneouslydecay(into atomsof another elementor into another isotopeof the sameelement)during each unit of time. Consequently,the samplebehavesexactlylikea populationwith a constantdeathrate and no births.

To write a modelfor N(t), we use Eq. (12)with N in placeof P,with k > 0 in

placeof 8, and with {3 = O.We thus get the differentialequation)

dN-= -kN.dt)

(14))

The value of k dependson the particular radioactiveisotope.The key to the method of radiocarbondating is that a constant proportion

of the carbon atoms in any living creature is made up of the radioactiveisotope14Cof carbon.This proportion remains constant becausethe fractionof 14Cin the

atmosphere remains almost constant, and living matter is continuously taking upcarbon from the air or isconsumingother living matter containingthe sameconstantratio of 14Catoms to ordinary 12Catoms. This sameratio permeatesall life,becauseorganicprocessesseemto make no distinctionbetweenthe two isotopes.)))

The ratio of 14Cto normal carbonremainsconstant in the atmospherebecause,although 14Cis radioactive and slowly decays,the amount is continuously replen-ishedthrough the conversionof 14N(ordinary nitrogen) to 14Cby cosmicrays bom-barding the upperatmosphere.Over the long history of the planet, this decay and

replenishmentprocesshas comeinto nearly steady state.Of course,when a living organism dies,it ceasesits metabolism of carbon

and the process'ofradioactivedecaybeginsto depleteits 14Ccontent. There is no

replenishmentof this 14C,and consequentlythe ratio of 14Cto normal carbonbeginsto drop. By measuring this ratio, the amount of time elapsedsincethe death of the

organism can be estimated.For such purposesit is necessaryto measure the decayconstantk.For 14C,it is known that k \037 0.0001216if t is measured in years.

(Mattersare not assimpleaswe havemade them appear.In applying the tech-nique of radiocarbondating, extremecaremust be taken to avoidcontaminating the

samplewith organic matter or even with ordinary fresh air. In addition, the cosmicray levels apparently have not beenconstant, so the ratio of 14Cin the atmospherehas varied over the pastcenturies.By using independent methods of dating sam-

ples,researchersin this area have compiledtablesof correction factors to enhancethe accuracy of this process.))

DRUGELIMINATION: In many casesthe amount A(t) of a certain drug in the

bloodstream,measuredby the excessover the natural level of the drug, will declineat a rate proportional to the current excessamount. That is,)

dA-== -AA,dt)

(15))

where A > O.The parameter A iscalledthe elimination constantof the drug.)

The NaturalGrowthEquationThe prototype differential equation dxjdt == kx with x(t) > 0 and k a constant

(either negative or positive) is readily solved by separating the variables and inte-

grating:)

f \037

dx= f k dt;

In x ==k t + C.)

Then we solve for x:)

e1nx ==ekt +C; x ==x(t) ==eCekt ==Aekt

.)

BecauseC is a constant, so is A == eC. It is alsoclearthat A ==x(0)== Xo, so the

particular solution of Eq.(11)with the initial conditionx(O)==Xo is simply)

\037) x(t) ==xoekt.) (16))

Becauseof the presenceof the natural exponentialfunction in its solution, the

differentialequation)

\037)

dx-==kxdt)

(17))

is often calledthe exponentialor naturalgrowth equation.Figure 1.4.7showsatypical graph of x(t) in the casek >0;the casek < 0 is illustrated in Fig.1.4.8.)))


x) x)

x =Xo ekt

(k> 0)) Xo)

Xo)

t)

FIGURE1.4.7.Natural growth.) FIGURE1.4.8.Natural decay.)

Example5) According to data listedat www.censlls.gov,the world'stotal populationreached6 billion personsin mid-1999,and was then increasing at the rate of about 212thousand personseachday. Assuming that natural population growth at this ratecontinues, we want to answer thesequestions:(a) What is the annual growth rate k?(b) What will be the world populationat the middle of the 21st century?(c) How long will it take the world population to increasetenfold-therebyreach-ing the 60billion that somedemographersbelieve to be the maximum for which the

planet can provide adequate food supplies?)

Solution (a) We measure the world population pet) in billionsand measure time in years.We take t ==0 to correspondto (mid) 1999,so Po ==6.The fact that P is increasingby 212,000,or 0.000212billion, personsperday at time t ==0 means that)

p'(0)== (0.000212)(365.25)\037 0.07743)

billion per year. From the natural growth equation P' == k P with t == 0 we now

obtain)

k == P'(O)\037

0.07743\037 0.0129.

P(O) 6

Thus the worldpopulationwas growing at the rate of about 1.29%annually in 1999.This value of k gives the world population function)

P(t) ==6eO.0129t.)

(b) With t ==51we obtain the prediction)

P(51)==6e(0.0129)(51)\037 11.58(billion))

for the world population in mid-2050(sothe population will almost have doubledin the just over a half-centurysince1999).(c) The world populationshould reach 60 billion when)

60==6eO.0129t;)In 10

that is, when t == \037 178;0.0129)

and thus in the year 2177.) .)))


Note: Actually, the rate of growth of the world population is expectedtoslow somewhatduring the next half-century,and the bestcurrent predictionfor the

2050population is \"only\" 9.1billion. A simplemathematical model cannot beexpectedto mirror preciselythe complexityof the real world.

The decayconstant of a radioactive isotopeis often specifiedin terms of an-other empirical constant, the half-lifeof the isotope,becausethis parameter is moreconvenient. The half-life 1:of a radioactiveisotopeis the time required for half ofit to decay.To find the relationship betweenk and 1:,we set t = 1:and N = 4No

in the equation N (t) = Noekt , so that 4No = NoekT.When we solve for 1:,we find

that)In21:=-.

k)(18))

For example,the half-life of 14Cis 1:\037 (In 2)/(0.0001216),approximately5700years.)

Example6) A specimenof charcoal found at Stonehengeturns out to contain 63% as much 14Cas a sampleof present-daycharcoal of equal mass.What is the ageof the sample?)

Solution We take t = 0 as the time of the death of the tree from which the Stonehengecharcoal was made and No as the number of 14Catoms that the Stonehengesamplecontained then. We are given that N = (0.63)Nonow, so we solve the equation(0.63)No= Noe-kt with the value k = 0.0001216.Thus we find that)

In(0.63)t = - \037 3800(years).0.0001216)

Thus the sampleisabout3800yearsold.If it has any connectionwith the buildersofStonehenge,our computationssuggestthat this observatory,monument, or temple-whichever it may be-datesfrom 1800B.C.or earlier. .)

CoolingandHeatingAccording to Newton'slaw of cooling(Eq. (3)of Section1.1),the time rate ofchange of the temperature T(t) of a body immersedin a medium of constant tem-perature A is proportional to the differenceA -T.That is,)

dT-= k(A -T),dt)

(19))

where k is a positiveconstant. This is an instanceof the linear first-orderdifferential

equation with constant coefficients:)

\037)

dx-= ax+b.dt)

(20))

It includesthe exponential equation as a specialcase(b = 0) and is alsoeasy tosolve by separationof variables.)

Example7) A 4-lb roast, initially at 50\302\260F, is placedin a 375\302\260F oven at 5:00P.M. After 75minutes it is found that the temperature T(t) of the roast is 125\302\260F. When will the

roast be 150\302\260F (medium rare)?)))

Solution We take time t in minutes, with t = 0 correspondingto 5:00P.M.We alsoassume(somewhat unrealistically) that at any instant the temperature T(t) of the roast isuniform throughout. We have T(t) < A = 375,T(O) = 50,and T(75)= 125.Hence)

dT-= k(375-T);dt

f1

dT = f k dt.375-T

'

-In(375- T) = kt + C;)

375- T = Be-kt.)

Now T(O) = 50impliesthat B = 325,so T(t) = 375 -325e-kt . We alsoknow

that T = 125when t = 75.Substitution of thesevalues in the precedingequationyields)

k = -7

1

5 In ( \037\037\037 ) \037 0.0035.Hencewe finally solve the equation)

150= 375- 325e(-o.0035)t)

for t = -[In(225j325)]j(0.0035)\037 105(min), the total cookingtime required.Becausethe roast was placedin the oven at 5:00P.M.,it shouldberemovedat about6:45P.M. .)Torricelli'sLaw)

Supposethat a water tank has a hole with area a at its bottom, from which water isleaking.Denoteby yet) the depth of water in the tank at time t, and by Vet) the

volumeof water in the tank then. It isplausible-andtrue, under idealconditions-that the velocity of water exiting through the hole is)

v = J2gy,) (21))

which is the velocitya dropof water would acquire in falling freely from the surfaceof the water to the hole (seeProblem35 of Section1.2).Onecan derivethis formula

beginning with the assumptionthat the sum of the kineticand potentialenergyof the

systemremainsconstant. Underreal conditions,taking into accountthe constrictionof a waterjetfrom an orifice,v = c,J2gy, wherec is an empiricalconstantbetweeno and 1 (usually about 0.6for a small continuous stream of water). For simplicitywe take c = 1 in the followingdiscussion.

As a consequenceofEq.(21),we have)

\037)

dV-= -av = -aJ2gy;dt)

(22a))

equivalently,)

\037)

dV-= -k,JY where k = aJ2i.dt)

(22b))

This is a statementof Torricelli'slaw for a draining tank. Let A(y) denotethe hori-zontal cross-sectionalarea of the tank at height y. Then, appliedto a thin horizontal)))


Example8)

sliceof water at height y with area A (y ) and thicknessdy , the integral calculusmethod of crosssectionsgives)

v (y) = 1Y

A (y ) dy \302\267)

The fundamental theorem of calculustherefore impliesthat dVjdyhencethat)

A(y) and)

dV dV dy dydt

=dy

.dt

= A(y) dt\302\267) (23))

From Eqs.(22)and (23)we finally obtain)

dy \037A(y)-= -ay 2gy = -k\037,dt)

(24))

an alternativeform of Torricelli'slaw.) .)\037 .. ....

A hemisphericalbowl has top radius 4 ft and at time t = 0 is full of water. At that

moment a circular hole with diameter 1 in. isopenedin the bottom of the tank. Howlong will it take for all the water to drain from the tank?)

Solution From the right triangle in Fig.1.4.9,we seethat)

Positive y-values)

t

I)

FIGURE1.4.9.Draining ahemispherical tank.)

A(y) = Jrr 2 = Jr [16- (4-y)2] = Jr(8y _ y2).)

With g = 32 ftj s2,Eq.(24)becomes)

2 dy(

1

)2 /Jr(8y - Y )-= -Jr 24 V 2.32y;dt

f (8y I/2 -l/2)dy = -f A dt;

16y

3/2 _ \037

y5/2--1..t + C3 5-

72 .)

Now y(O) = 4, so)

C - 16.43/2 _ \037 .45/2 _ 448-3 5

- 15.)

The tank is empty when y = 0,thus when)

t = 72.41\0378

\037 2150(s);)

that is, about 35 min 50s. So it takes slightly lessthan 36 min for the tank todrain. .)))

__ Problems)

Find generalsolutions (implicit if necessary,explicit if conve-nient) of the differential equations in Problems1through 18.Primesdenotederivatives with respectto x.)

dy1.-+ 2xy = 0dxdy .3.dx

= y SIn x

5.2..jXdy = Jl_

y2dxdy7.-= (64xy)lj3dx

9.(1-x2)dy = 2ydx

11.y' = xy3

dy13.y3-= (y4 + 1)cosxdx

15.dy (x - 1)y5dx x2(2y3 - y)

17.y' = 1+x+ y +xy (Suggestion:Factor the right-handside.)

18.x2y' = 1-x2 + y2_ x2

y2)

dy2.-+ 2xy2 = 0dx

dy4. (1+ x)dx= 4y

dy6.dx

=3vfXY

dy8.-= 2x secydx

10.(1+ X)2dy = (1+ y)2dx

12.yy' = X(y2 + 1)

14.dy = 1+ ..jXdx 1+ ,JY)

16.(x2 + l)(tan y)y' = x)

Find explicit particular solutions of the initial value problemsin Problems19through 28.

dy19.dx

= yeX , y(O)= 2e

dy20.dx

= 3x2(y2 + 1), y(O)= 1

21.2ydy = x

, y(5) = 2dx -VX2 - 16

dy22.dx

= 4x3y-y, y(l) = -3

dy23.-+ 1 = 2y, y(l) = 1dxdy

(I

)I24. (tan x)dx

= y, Y\"2]\"[ =\"2]\"[

dy25.x--y = 2x2y, y(l) = 1

dxdy26.dx

= 2xy2 + 3x2y2 , Y(1)= -1

27.\037:

= 6e2x-Y , y(O)= 0

dy28. 2..jX-= cos2 y, y(4) = ]\"[/4dx29. (a) Find a general solution of the differential equation

dy/dx = y2. (b) Find a singular solution that is not in-cluded in the general solution. (c)Inspecta sketch of typi-calsolution curvesto determine the points (a,b) for whichthe initial value problem y' = y2, y(a) = b has a uniquesolution.)


30.Solvethe differential equation (dy/dx)2 = 4y to verify the

general solution curves and singular solution curve that

are illustrated in Fig. 1.4.5.Then determine the points(a, b) in the plane for which the initial value problem(y')2 = 4y, yea) = b has (a) no solution, (b) infinitely

many solutions that are defined for all x, (c) on someneighborhood of the point x = a, only finitely many solu-ti 0ns.

31.Discussthe differencebetween the differential equations

(dy/dx)2 = 4y and dy/dx = 2,JY. Do they have the

same solution curves?Why or why not? Determine the

points (a, b) in the plane for which the initial value prob-lem y' = 2,JY, yea)= b has (a)no solution, (b) a uniquesolution, (c) infinitely many solutions.

32.Find a general solution and any singular solution s of the

differential equation dymyslashdx = y .Jy2-1.Deter-

mine the points (a,b) in the plane for which the initial

value problem y' = y .Jy2-1,yea)= b has (a)no solu-

tion, (b) a unique solution, (c) infinitely many solutions.

33.(Population growth) A certain city had a population of25000in 1960and a population of30000in 1970.Assumethat its population will continue to grow exponentially at aconstant rate. What population can its city planners expectin the year 2000?

34. (Population growth) In a certain culture of bacteria, the

number of bacteriaincreasedsixfold in 10h. How longdid it take for the population to double?

35.(Radiocarbon dating) Carbon extracted from an ancient

skull contained only one-sixth as much 14Cas carbon ex-tracted from present-day bone.How old is the skull?

36.(Radiocarbondating) Carbon taken from a purported relicof the time of Christ contained 4.6x 1010 atoms of 14Cper gram. Carbon extracted from a present-day specimenof the samesubstancecontained 5.0x 1010atoms of 14Cper gram. Compute the approximate ageofthe relic.What

is your opinion as to its authenticity?

37. (Continuously compounded interest) Upon the birth oftheir first child, a coupledeposited$5000in an account

that pays 8% interest compounded continuously. The in-

terest payments are allowed to accumulate. How much

will the account contain on the child'seighteenth birth-

day?38.(Continuously compounded interest) Supposethat you

discoverin your attic an overdue library book on which

your grandfather oweda fine of30cents 100yearsago.Ifan overdue fine grows exponentially at a 5%annual rate

compounded continuously, how much would you have to

pay if you returned the book today?

39.(Drug elimination) Supposethat sodium pentobarbital isused to anesthetize a dog. Thedog is anesthetized when

its bloodstream contains at least45 milligrams (mg) ofsodium pentobarbitol per kilogram of the dog'sbody)))


weight. Supposealso that sodium pentobarbitol is elim-inated exponentially from the dog'sbloodstream, with ahalf-life of 5 h. What single doseshould be administeredin order to anesthetize a 50-kgdog for 1h?

40.Thehalf-life of radioactive cobaltis 5.27years.Supposethat a nuclear accidenthas left the level of cobaltradia-tion in a certain region at 100times the level acceptablefor human habitation. How long will it be until the regionis again habitable? (Ignore the probablepresenceof otherradioactive isotopes.)

41.Suppose that a mineral body formed in an ancient

cataclysm-perhapsthe formation of the earth itself-originally contained the uranium isotope238U (which hasa half-life of4.51x 109 years)but no lead,the end productof the radioactive decayof 238U. If today the ratio of 238Uatoms to leadatoms in the mineral body is 0.9,when didthe cataclysm occur?

42. A certain moon rock was found to contain equal numbersof potassium and argon atoms. Assume that all the argonis the result of radioactive decayofpotassium (its half-lifeis about 1.28x 109 years)and that oneofevery nine potas-sium atom disintegrations yields an argon atom. What isthe ageof the rock, measured from the time it contained

only potassium?43. A pitcher of buttermilk initially at 25\302\260C is to be cooled

by setting it on the front porch, where the temperature is0\302\260 C. Supposethat the temperature of the buttermilk hasdroppedto 15\302\260C after 20min. When will it be at 5 \302\260C?

44. When sugar is dissolvedin water, the amount A that re-mains undissolved after t minutes satisfies'the differential

equation dAjdt = -kA (k > 0).If 25%of the sugar dis-solvesafter 1 min, how long does it take for half of the

sugar to dissolve?45.The intensity 1of light at a depth of x meters below

the surface of a lake satisfies the differential equationd1jdx = (-1.4)I.(a) At what depth is the intensity halfthe intensity 10at the surface (where x = O)?(b) Whatis the intensity at a depth of 10m (as a fraction of lo)?(c) At what depth will the intensity be 1% of that at thesurface?

46. Thebarometric pressurep (in inches of mercury) at analtitude x miles above sea level satisfies the initial value

problem dpjdx= (-0.2)p, p(O)= 29.92.(a) Calculatethe barometric pressureat 10,000ft and again at 30,000ft. (b) Without prior conditioning, few peoplecan sur-vive when the pressuredrops to lessthan 15 in. of mer-cury. How high is that?

47. A certain pieceof dubious information about phenylethy-lamine in the drinking water began to spreadoneday in acity with a population of 100,000.Within a week, 10,000peoplehad heard this rumor. Assume that the rate of in-creaseof the number who have heard the rumor is propor-tional to the number who have not yet heard it. How longwill it beuntil half the population of the city has heard therumor?)

48. According to one cosmologicaltheory, there were equalamounts of the two uranium isotopes235U and 238U at the

creationof the universe in the \"big bang.\" At presentthereare 137.7atoms of 238U for eachatom of 235U. Using the

half-lives 4.51x 109 years for 238U and 7.10x 108 yearsfor 235U, calculatethe ageof the universe.

49. A cakeis removed from an oven at 210\302\260 F and left to coolat room temperature, which is 70\302\260 F. After 30 min the

temperature of the cakeis 140\302\260F. When will it be 100\302\260F?

50.The amount A (t) of atmospheric pollutants in a certain

mountain valley grows naturally and is tripling every 7.5years.(a) If the initial amount is 10pu (pollutant units), write

a formula for A (t) giving the amount (in pu) presentafter t years.

(b) What will be the amount (in pu) of pollutants presentin the valley atmosphere after 5 years?

(c) If it will be dangerous to stay in the valley when the

amount of pollutants reaches100pu, how long will

this take?51.An accidentat a nuclear powerplant has left the surround-

ing areapolluted with radioactive material that decaysnat-

urally. The initial amount of radioactive material presentis 15su (safeunits), and 5 months later it is still 10suo

(a) Write a formula giving the amount A(t) ofradioactivematerial (in su) remaining after t months.

(b) What amount ofradioactive material will remain after

8 months?

(c) How long-total number of months or fraction

thereof-will it be until A = 1 su, so it is safefor

peopleto return to the area?52.Therearenow about 3300different human \"language fam-

ilies\" in the whole world. Assume that all these are de-rived from a single original language, and that a language

family developsinto 1.5language families every 6 thou-

sand years. About how long agowas the single originalhuman language spoken?

53.Thousands ofyearsagoancestorsof the Native Americanscrossedthe Bering Strait from Asia and entered the west-ern hemisphere. Sincethen, they have fanned out acrossNorth and South America. The single language that the

original Native Americans spokehas sincesplit into manyIndian \"language families.\" Assume (as in Problem 52)that the number of theselanguage families has beenmul-

tiplied by 1.5every 6000years.Therearenow 150Native

American language families in the western hemisphere.About when did the ancestorsof today's Native Ameri-cansarrive?

54. A tank is shaped like a vertical cylinder; it initially con-tains water to a depth of9 ft, and a bottom plug is removedat time t = 0 (hours). After 1 h the depth of the water has

dropped to 4 ft. How long doesit take for all the water to

drain from the tank?

55.Supposethat the tank of Problem 48 has a radius of 3 ft

and that its bottom hole is circular with radius 1 in. How)))

long will it take the water (initially 9 ft deep)to drain com-pletely?)

56.At time t == 0 the bottom plug (at the vertex) ofa full con-icalwater tank 16ft high is removed. After 1 h the waterin the tank is 9 ft deep.When will the tank be empty?)

57.Supposethat a cylindrical tank initially containing Vo gal-lons ofwater drains (through a bottom hole) in T minutes.UseTorricelli'slaw to show that the volume of water in

the tank after t < T minutes is V == Vo [1- (t/T)]2.)

58.)A water tank has the shapeobtained by revolving the curvey == X4/3 around the y-axis. A plug at the bottom is re-moved at 12noon, when the depth of water in the tank is12ft. At 1 P.M. the depth of the water is 6 ft. When will

the tank be empty?)

59. A water tank has the shapeobtained by revolving the

parabola x2 == by around the y-axis.Thewater depth is4 ft at 12noon, when a circular plug in the bottom of thetank is removed.At 1 P.M. the depth of the water is 1 ft.

(a) Find the depth yet) of water remaining after t hours.

(b) When will the tank be empty? (c) If the initial radiusof the top surfaceof the water is 2 ft, what is the radius ofthe circularholein the bottom?)

60. A cylindrical tank with length 5 ft and radius 3 ft is sit-uated with its axis horizontal. If a circular bottom holewith a radius of 1 in. is openedand the tank is initiallyhalf full of xylene, how long will it take for the liquid todrain completely?)

61.A spherical tank of radius 4 ft is full of gasolinewhen acircular bottom hole with radius 1 in. is opened.How longwill berequired for all the gasolineto drain from the tank?)

62. Supposethat an initially full hemispherical water tank ofradius 1 m has its flat sideas its bottom. It has a bottomholeofradius 1cm.If this bottom hole is openedat 1P.M.,when will the tank be empty?)

63. Considerthe initially full hemispherical water tank ofEx-ample 8,exceptthat the radius r of its circularbottom holeis now unknown. At 1 P.M. the bottom hole is openedandat 1:30P.M.the depth of water in the tank is 2 ft. (a) UseTorricelli'slaw in the form dV/dt == -(0.6)nr2,J2gy(taking constriction into account) to determine when thetank will be empty. (b) What is the radius of the bottomhole?)

64. (Theclepsydra,or water clock)A 12-hwater clockis tobe designedwith the dimensions shown in Fig. 1.4.10,shaped like the surface obtained by revolving the curvey == f (x) around the y-axis.What should be this curve,and what should be the radius of the circularbottom hole,in orderthat the water level will fall at the constant rate of4 inchesperhour (in./h)?)


y)

4ft)

orx =g(y))

t Water flow

I)

x)

FIGURE1.4.10.Theclepsydra.)

65.Just beforemidday the body of an apparent homicide vic-tim is found in a room that is kept at a constant tempera-ture of 70\302\260F. At 12noon the temperature of the body is80\302\260F and at 1P.M.it is 75\302\260F. Assume that the temperatureof the body at the time ofdeath was 98.6\302\260F and that it has

cooledin accordwith Newton's law. What was the time

ofdeath?66.Early onemorning it began to snow at a constant rate. At

7 A.M. a snowplow set off to cleara road. By 8 A.M. it

had traveled 2 miles, but it took two more hours (untillOA.M.) for the snowplow to go an additional 2 miles.(a) Let t == 0 when it began to snow and let x denotethe

distance traveled by the snowplow at time t. Assumingthat the snowplow clearssnow from the road at a constant

rate (in cubicfeetper hour, say), show that)

dx 1k-=-dt t)

where k is a constant. (b) What time did it start snowing?(Answer: 6 A.M.)

67.A snowplow setsoff at 7 A.M. as in Problem 66.Supposenow that by 8 A. M. it had traveled 4 miles and that by9 A.M. it had moved an additional 3 miles. What time did

it start snowing? This is a more difficult snowplow prob-lem becausenow a transcendental equation must besolvednumerically to find the value ofk. (Answer: 4:27A.M.)

68.Figure 1.4.11shows a bead sliding down a frictionlesswire from point P to point Q.Thebrachistochrone prob-lem asks what shape the wire should be in order to min-

imize the bead'stime of descentfrom P to Q. In June

of 1696,John Bernoulli proposedthis problem as a pub-lic challenge,with a 6-month deadline (later extended to

Easter 1697at GeorgeLeibniz'srequest). IsaacNewton,then retired from academiclife and serving as Warden

of the Mint in London, receivedBernoulli's challengeon

January 29, 1697.The very next day he communicatedhis own solution-the curve ofminimal descenttime is an)))


arc of an inverted cycloid-tothe Royal Societyof Lon-don. For a modem derivation of this result, supposethebeadstarts from rest at the origin P and let y = y (x) bethe equation of the desiredcurve in a coordinatesystemwith the y-axispointing downward. Then a mechanicalanalogue ofSnell'slaw in opticsimplies that) 69.)

of the cycloidthat is generatedby a point on the rim

of a circular wheel of radius a as it rolls along the x-axis. [SeeExample 5 in Section9.4of Edwards and

Penney, Calculus:Early Transcendentals,7th edition

(UpperSaddleRiver, NJ: PrenticeHall, 2008).]Supposea uniform flexible cable is suspendedbetweentwo points (:l::L,H) at equal heights locatedsymmetri-cally on either sideof the x-axis(Fig. 1.4.12).Principlesofphysicscan beusedto show that the shapey = y (x) ofthe hanging cablesatisfiesthe differential equation)

sIn a-= constant,v)

(i))

where a denotesthe angle of deflection (from the verti-

cal)of the tangent line to the curve-socota = y' (x)(why?)-andv = ...j2gyis the bead'svelocity when it hasdescendeda distance y vertically (from KE = !mv

2

mgy = -PE).)

d2ya-=

dx 2) (dy

)

2

1+dx ')

p) where the constant a = Tjp is the ratio of the cable'stension T at its lowest point x = 0 (where y' (0) = 0) and its (constant) linear density p. If we substitute

v = dymyslashdx, dvjdx = d2yjdx

2 in this second-orderdifferential equation, we get the first-order equation)

(a) First derive from Eq.(i) the differential equation

dY =J2a

Y

dx y)(ii))

adv = J1+v2.dx

Solve this differential equation for y' (x) = v(x)sinh(xja).Then integrate to get the shapefunction

y(x)=acoshC)+Cof the hanging cable.This curve is calleda catenary, fromthe Latin word for chain.)

FIGURE1.4.11.A beadsliding down awire-thebrachistochroneproblem.)

Y)

where a is an appropriate positive constant.

(b) Substitute y = 2a sin 2 t, dy = 4a sin t cost dt in (ii)to derive the solution)

(-L,H)) (L,H))

x = a(2t - sin 2t), y = a(l-cos2t)) (iii)) Yo)

for which t = Y = 0 when x = O.Finally, the sub-stitution of e = 2a in (iii) yields the standard para-metric equations x = ace - sin e),y = a(l-cose))

x)

FIGURE1.4.12.Thecatenary.)

lID)LinearFirst-OrderEquations........... .....)

In Section1.4we saw how to solve a separabledifferentialequation by integrating

after multiplying both sidesby an appropriate factor. For instance, to solve the

equation)dy-= 2xy (y >0),dx

we multiply both sidesby the factor l/y to get

1 dy-.-=2x;that is, Dx(lny)= Dx(x2).

y dxBecauseeachsideof the equation in (2)is recognizableasa derivative (with respectto the independent variable x), all that remains are two simpleintegrations,which)

(1))

(2))))

1.5LinearFirst-OrderEquations 47)

yield In y = x2 +C.For this reason,the function p(y) = l/y is calledan integrat-ing factorfor the original equation in (1).An integratingfactor for a differential

equation is a function p(x,y) such that the multiplication of each sideof the differ-ential equation by p(x, y) yieldsan equation in which each sideis recognizableasa derivative.

With the aid of the appropriateintegrating factor,there isa standard techniquefor solving the linearfirst-orderequation)

\037)

dy-+ P(x)y = Q(x)dx)

(3))

on an interval on which the coefficientfunctions P(x)and Q(x)are continuous.We

multiply eachsidein Eq.(3)by the integrating factor)

\037) p(x) = efP(x)dx.) (4))

The result is)

ef P(x)dx dy + P(x)ef P(x)dxy = Q(x)ef P(x)dx.dx)

(5))

Because)

Dx[IP(x)dx] = P(x),)

the left-handsideis the derivativeof the producty(x) .ef P(x)dx, soEq.(5)isequiv-alent to)

Dx [y(x).efP(X)dX]

= Q(x)ef P(x)dx.

Integration of both sidesof this equation gives)

y(x)ef P(x)dx = 1(Q(x)ef

P(X)dX)dx+C.)

Finally, solvingfor y, we obtain the generalsolutionof the linearfirst-orderequationin (3):)

y(x) = e-f P(x)dx [I(Q(x)ef

P(X)dX)dx+c].) (6))

This formula should not be memorized.In a specificproblem it generallyis

simplerto use the method by which we developedthe formula. That is, in orderto solve an equation that can be written in the form in Eq. (3)with the coefficientfunctions P(x)and Q(x)displayedexplicitly,you should attempt to carry out the

following steps.)

METHOD:SOLUTIONOFFIRST-ORDEREQUATIONS

1.Beginby calculating the integrating factor p(x)= efP(x)dx.

2.Then multiply both sidesof the differentialequationby p(x).3.Next, recognizethe left-hand sideof the resulting equation as the derivative

of a product:)Dx [p(x)y(x)]= p(x)Q(x).)))


4. Finally, integrate this equation,)

p(x)y(x)= f p(x)Q(x)dx+C,)

then solve for y to obtain the general solutionof the originaldifferentialequa-tion.)

Remark 1: Given an initial condition y(xo) = Yo, you can (as usual)substitute x = Xo and y = Yo into the general solution and solve for the valueof Cyielding the particular solution that satisfies this initial condition.

Remark 2: You neednot supply explicitlya constant of integration whenyou find the integrating factor p(x).For if we replace

f P(x)dx with f P(x)dx+ K)

in Eq.(4),the result is)

p(x) = eK+!P(x)dx = eKe!P(x)dx.

But the constant factor eK doesnot affect materially the result of multiplying bothsidesof the differential equation in (3)by p(x),so we might as well take K = O.You may therefore choosefor f P(x)dx any convenient anti derivative of P(x),without bothering to add a constant of integration. .)

Example1)

.-. .....-. \"\" '\" \037 .....

Solvethe initial value problem

dy 11 -x/3dx -y=ge , y(O)=-l.)

Solution Herewe have P(x)= -1and Q(x)= 18

1 e-x/3, so the integrating factor is

p(x) = e!(-I)dx= e-x .Multiplicationof both sidesof the given equation by e-x yields)

-xdy -x 11 -4x/3e dx-e y = ge ,) (7))

which we recognizeas)d_ (e-Xy)= 1

81 e-4x/3.

dxHenceintegration with respectto x gives

e-Xy= f1

81 e-4x/3dx= _

\037\037

e-4x/3+C,

and multiplication by eX gives the general solution

y(x) = Cex -j\037

e-x/3.) (8))

Substitution of x = 0 and y = -1now gives C =3

1

2' so the desiredparticularsolution is)

y(x) =3

1

2eX -j\037

e-x/3 =3

1

2 (eX-33e-x/3

) .) .)))

o)

;:.... -1)

-2-3

-4)

y =_33exp(-x/3)32)

-1 0 2 3 4 5x)

FIGURE1.5.1.Slopefield andsolution curves fory' = y +

18

1e-x/3.)

Example2)


Remark:Figure 1.5.1showsa slopefield and typical solution curves forEq. (7),including the one passingthrough the point (0,-1).Note that somesolu-tions grow rapidly in the positivedirectionasx increases,while othersgrow rapidlyin the negative direction.The behavior of a given solution curve is determinedbyits initial condition y(O) = Yo. The two types of behaviorare separatedby the par-ticular solution y(x) = -

\037\037

e-x/3 for which C = 0 in Eq.(8),so Yo = -\037\037

for thesolution curve that is dashedin Fig.1.5.1.If Yo > -

\037\037

, then C > 0 in Eq.(8),sothe term eX eventually dominates the behavior of y(x),and hencey(x) \037 +00asx \037 +00.But if Yo < -

\037\037

, then C < 0,so both terms in y(x) are negative andthereforey(x) \037 -00asx \037 +00.Thus the initial conditionYo = -

\037\037

iscriticalin the sensethat solutions that start above -

\037\037

on the y-axisgrow in the positivedirection,while solutions that start lower than -

\037\037

grow in the negativedirectionasx \037 +00.The interpretation of a mathematicalmodel oftenhingeson finding sucha critical condition that separatesone kind of behaviorof a solution from a different

kind of behavior. .)

Find a general solution of)

dy(x2 +1)-+ 3xy = 6x.dx)

(9))

Solution After division of both sidesof the equation by x2 + 1,we recognizethe result)

76543

;::.... 21

o-1-2-3

-5-4-3-2-10 1 2 3 4 5x)

FIGURE1.5.2.Slopefield andsolution curves for the differential

equation in Eq.(9).)

dy 3x 6x-+ y=dx x2 + 1 x2 + 1)

as a first-orderlinear equation with P(x)= 3x/(x2 + 1)and Q(x)= 6x/(x2 + 1).Multiplicationby)

p(x) = exp(Ix23:1 dX)= exp(\037ln(x2 + 1))= (x2 + 1)3/2)

yields)

(x2 + 1)3/2dy +3x(x2 + 1)1/2y= 6x(x2 + 1)1/2,dx)

and thus)

Dx [(x2 + 1)3/2y ] = 6x(x2 + 1)1/2.)

Integration then yields)

(x2 + 1)3/2y= 16x(x2 + 1)1/2dx= 2(x2 + 1)3/2+ c.)

Multiplicationof both sidesby (x2 + 1)-3/2 gives the general solution)

y(x) = 2 + C(x2 + 1)-3/2.) (10).)

Remark:Figure 1.5.2showsa slopefield and typical solution curves forEq. (9). Note that, as x \037 +00,all other solution curves approach the constantsolution curve y(x) = 2 that correspondsto C = 0 in Eq. (10).This constant)))


solution can bedescribedas an equilibrium solution of the differentialequation,be-causey(O) = 2 impliesthat y(x) = 2 for all x (and thus the value of the solutionremains forever where it starts).Moregenerally, the word \"equilibrium\" connotes\"unchanging,\" so by an equilibrium solution of a differential equation is meant aconstant solution y(x) = c,for which it follows that y'(x) = O. Note that substi-tution of y' = 0 in the differential equation (9)yields3xy = 6x,so it follows that

y = 2 if x i= O.Hencewe seethat y (x)= 2 is the only equilibrium solutionof this

differentialequation, as seemsvisually obvious in Fig.1.5.2. .)A CloserLookat theMethod)The precedingderivationof the solution in Eq.(6)of the linear first-orderequationy' + Py = Q bearscloserexamination. Supposethat the coefficientfunctions P(x)and Q(x) are continuous on the (possiblyunbounded) open interval I. Then theantiderivatives

f P(x)dx and f (Q(x)ef

P(X)dX)dx

existon I.Our derivationof Eq. (6)showsthat if y = y (x) is a solutionof Eq.(3)on I, then y(x) is given by the formula in Eq. (6)for somechoiceof the constantC.Conversely,you may verify by direct substitution (Problem31)that the function

y(x) given in Eq. (6)satisfiesEq. (3).Finally, given a point Xo of I and any num-ber Yo, there is-aspreviously noted-aunique value of C such that y(xo) = Yo.

Consequently,we have proved the followingexistence-uniquenesstheorem.)

THEOREM1 TheLinearFirst-OrderEquationIf the functions P(x)and Q(x)are continuouson the openinterval I containingthe point Xo, then the initial value problem)

\037)

dydx + P(x)y = Q(x), y(xo)= Yo) (11))

has a unique solution y(x) on I, given by the formula in Eq. (6)with an appro-priate value of c.)

Remark 1:Theorem 1 givesa solutionon the entire interval I for a lineardifferential equation, in contrast with Theorem 1 of Section1.3,which guaranteesonly a solution on a possiblysmaller interval.

Remark2: Theorem 1 tellsus that every solutionofEq.(3)is included in

the general solution given in Eq. (6).Thus a linearfirst-orderdifferentialequationhas no singular solutions.

Remark3: The appropriatevalueof the constant C in Eq.(6)-asneededto solve the initial value problemin Eq. (1I)-canbe selected\"automatically\" bywri ting)

p(x) = exp(1:P(t)dt) ,

y(x) = 1

[Yo + 1x

p(t)Q(t)dt].

P(x)XQ)

(12))))

Example3)


The indicated limits Xo and x effect a choiceof indefinite integrals in Eq. (6) that

guarantees in advance that p(xo)= 1 and that y(xo)= Yo (asyou can verify directlyby substituting x = Xo in Eqs.(12)). .)


2dy .1)x -+ xY = sInx, y ( = Yo.

dx)(13))

Solution Divisionby x2 gives the linear first-orderequation)

3

2)

;::.... 0

-1-2-3 (1,-3)

0 5 10 15 20x)

FIGURE 1.5.3.Typicalsolutioncurves defined by Eq.(15).)

dy 1 sinx-+ -y =-dx X x2)

with P(x)= Ijxand Q(x)= (sinx)jx2.With Xo = 1 the integrating factor in (12)IS)

p(x) = exp(IX \037 dt)= exp(lnx)= x,

so the desiredparticular solution is given by)

1

[[X sin t

]y(x) =x Yo + 11t dt \302\267) (14))

In accordwith Theorem 1,this solution is definedon the wholepositivex-axis..)

Comment:In general,an integral such as the one in Eq.(14)would (forgiven x)needto beapproximatednumerically-usingSimpson'srule, for instance-to find the value y (x)of the solution at x. In this case,however,we have the sineintegral function)

lx sin tSi(x)= -dt,

o t

which appearswith sufficient frequency in applications that its values have beentabulated. A good set of tablesof specialfunctions is Abramowitz and Stegun,Handbookof Mathematical Functions (New York: Dover, 1965).Then the particu-lar solution in Eq.(14)reducesto)

1

[ lx sin t 11 sin t

]1 . .y(x)=- Yo+ -dt- -dt =-[Yo+SI(x)-SI(I)].x 0 tot x)

(15))

The sineintegral function is available in most scientificcomputing systemsand canbe used to plot typical solution curves defined by Eq. (15).Figure 1.5.3showsaselectionof solution curves with initial values y (1)= Yo ranging from Yo = -3to

Yo = 3.It appearsthat on each solution curve, y(x) \037 0 asx \037 +00,and this isin fact true becausethe sineintegral function is bounded. .

In the sequelwe will seethat it is the exception-ratherthan the rule-whenasolutionof a differentialequationcan beexpressedin terms of elementaryfunctions.We will study various devicesfor obtaining goodapproximationsto the valuesofthe nonelementaryfunctions we encounter. In Chapter 6 we will discussnumerical

integration of differentialequations in somedetail.)))


Input: Yi LIs, Ci g/L)

Amount x(t)Volume V(t)Concentration co(t)=

\037)

Output:

fYo Lis,Co g/L)

FIGURE1.5.4.Thesingle-tankmixture problem.)

MixtureProblems)As a first applicationof linear first-orderequations,we considera tank containingasolution-amixture of solute and solvent-suchas salt dissolvedin water.There isboth inflow and outflow, and we want to compute the amount x(t) of solute in thetank at time t, given the amount x(0) = Xa at time t = O. Supposethat solutionwith a concentration of Ci grams of solute per liter of solution flows into the tank

at the constant rate of ri liters per second,and that the solution in the tank-keptthoroughly mixedby stirring-flowsout at the constant rate of r0 liters persecond.

To set up a differential equation for x(t), we estimate the change \037x in xduring the brief time interval [t, t + \037t]. The amount of solute that flows into thetank during \037t secondsis riCi \037t grams.To checkthis, note how the cancellationof dimensionschecksour computations:)

(liters

) (grams

)ri Ci . (\037t seconds)second lIter)

yieldsa quantity measuredin grams.The amount of solute that flows out of the tank during the sametime interval

dependson the concentrationco(t)of solute in the solutionat time t.But asnoted in

Fig.1.5.4,co(t)= x(t)/V(t),where Vet) denotesthe volume (not constantunlessri = r0) of solution in the tank at time t.Then)

\037x = {gramsinput}- {gramsoutput} \037 rici \037t

- roco \037t.

We now divide by \037t:)

\037x-\037 rici - roco.\037t

Finally, we take the limit as \037t \037 0;if all the functions involved are continuousand x(t) is differentiable,then the error in this approximationalsoapproacheszero,and we obtain the differentialequation

dx-= rici - roco,dt)

\037) (16))

in which ri,Ci,and r0 are constants, but Codenotesthe variableconcentration)

x(t)co(t)=-

Vet))(17))

of solute in the tank at time t. Thus the amount x(t) of solute in the tank satisfiesthe differentialequation)

dx ro-= rici --x.dt V

If Va = V(O), then Vet) = Va + (ri - ro)t, so Eq. (18)is a linear first-orderdifferentialequation for the amount x(t) of solute in the tank at time t.

Important: Equation (18)neednot be committed to memory. It is the pro-cesswe used to obtain that equation-examinationof the behavior of the systemover a short time interval [t, t + \037t]-that you should strive to understand,becauseit is a very useful tool for obtaining all sortsof differentialequations.

Remark:It was convenientfor us to useg/Lmass/volumeunits in derivingEq. (18).But any other consistentsystem of units can beusedto measure amountsof solute and volumes of solution. In the following examplewe measure both in

cubickilometers. .)

(18))))


Example4) Assumethat LakeEriehas a volume of 480 kIn 3 and that its rate of inflow (fromLakeHuron) and outflow (to LakeOntario)are both 350kIn 3 peryear. Supposethat

at the time t ==0 (years),the pollutant concentrationof LakeErie-causedby pastindustrial pollution that has now beenorderedto cease-isfive times that of LakeHuron.If the outflow henceforthisperfectlymixed lakewater, how long will it taketo reducethe pollution concentration in LakeErieto twice that of LakeHuron?)

Solution Herewe have)

v ==480 (km3),

ri ==r0 ==r ==350(kIn3/yr),

Ci ==C(the pollutant concentrationof LakeHuron),and

xo ==x(O)==5cV,)

and the question is this: When is x(t) == 2cV?With this notation, Eq.(18)is the

separableequation)dx r-==rc- -xdt V

') (19))

which we rewrite in the linear first-orderform)

dx-+ px ==qdt)

(20))

with constant coefficientsp == r/V, q == rc, and integrating factor p == ePt . You

can either solve this equationdirectly or apply the formula in (12).The latter gives)

x(t) = e-pt

[xo+ itqePt

dt]= e-pt

[xo+ ;(ePt -1)]

= e-rt / V

[SCV+

\037\037

(ert/v -

1)];

x(t) ==cV +4cV e-rt/V.) (21))

To find when x(t) ==2cV,we thereforeneedonly solve the equation)

cV +4cVe-rt/V ==2cV)V 480

for t ==-;In 4 ==350In 4 \037 1.901(years).) .)

Example5) A 120-gallon(gal)tank initially contains 90 lb of salt dissolvedin 90 gal of water.Brinecontaining2 lb/gal of salt flows into the tank at the rate of 4 gal/min, and the

well-stirredmixture flows out of the tank at the rate of 3 gal/min.How much saltdoesthe tank contain when it is full?)

Solution The interesting feature of this exampleis that, due to the differing rates of inflow

and outflow, the volume of brine in the tank increasessteadily with V (t) ==90 + t

gallons.The change \037x in the amount x of salt in the tank from time t to timet + \037t (minutes) is given by)

\037X\037(4)(2)\037t-3 (x

) \037t,90+t)))


_ Problems)

soour differentialequation is)

dx 3-+ x = 8.dt 90 + t)

An integrating factor is)

p(x) = exp(I3

dt) = e31n(90+t) = (90+ t)3,90 + t)

which gives)

Dt [(90+ t)3x]= 8(90+ t)3;

(90+ t)3x = 2(90+ t)4 + c.)

Substitution of x(0) = 90gives C = -(90)4,so the amount of salt in the tank attime t is)

904x(t) = 2(90+ t) - (90+ t)3

.The tank is full after 30min, and when t = 30,we have)

904x(30)= 2(90+30)-_3\037 202(lb)120)

of salt in the tank.) .)

21.xy' = 3y + x4 cosx,y(2n) = 022.y' = 2xy + 3x2 exp(x2),y(O)= 523.xy' + (2x- 3)y = 4x4

24. (x2 + 4)y'+ 3xy =x,y(O) = 1dy25.(x2+1)-+3x3y=6xexp(-\037x2),y(0)=1dx)

Find generalsolutions of the differential equations in Prob-lems 1 through 25. If an initial condition is given, find the

corresponding particular solution. Throughout, primes denotederivatives with respectto x.)

1.y' + y = 2, y (0)= 02.y' - 2y = 3e2x , y(O)= 03.y' + 3y = 2xe-3x

24. y' - 2xy = eX

5.xy' + 2y = 3x, y (1)= 56.xy' + 5y = 7x2, y(2) = 57. 2xy'+ y = 10,JX8.3xy'+ y = 12x9.xy' - y =x,y(l) = 710.2xy'- 3y = 9x3

11.xy' + y = 3xy, y(l) = 012.xy' + 3y = 2x5, y(2) = 113.y' + y = eX, y(O)= 114.xy' - 3y = x3, y(l) = 1015.y' + 2xy =x,y(O)= -216.y' = (1-y)cosx,y(n) = 217.(1 +x)y'+ y = cosx,y(O)= 118.xy' = 2y + x3cosx19.y' + y cotx = cosx20.y' = 1+ x + y + xy, y(O) = 0)

Solvethe differential equations in Problems26 through 28 by

regarding y as the independent variable rather than x.)

dy27.(x + yeY)-= 1dx)

26.(1-4xy2)dy = y3dx

28.(1+ 2xy)dy = 1+ y2dx

29.Expressthe general solution ofdy/dx = 1+ 2xy in termsof the errorfunction)

2 (X 2erf(x)= .j7r10

e-t dt.)

30.Expressthe solution of the initial value problem)

dy2x-= y + 2xcosx, y(l) = 0dx)

as an integral as in Example 3 of this section.)))

Problems 31and 32 illustrate-for the specialcaseoffirst-order linear equations-techniquesthat will be importantwhen we study higher-orderlinear equations in Chapter 3.)

31.(a) Show that)

yc(x) = Ce-J P(x)dx

is a general solution ofdy/dx + P(x)y= O. (b) Showthat)

Yp(x) = e-f P(x)dx [I(Q(x)ef

P(X)dX)dx]

is a particular solution of dy/dx + P(x)y = Q(x).(c)Supposethat yc(x) is any general solution ofdy/dx +P (x)y = 0 and that y p (x) is any particular solution ofdy/dx + P(x)y= Q(x).Show that y(x) = yc(x)+ yp(x)is a general solution ofdy/dx + P(x)y= Q(x).

32.(a) Find constants A and B such that y p (x) = A sin x +Bcosxis a solution ofdy/dx + y = 2sinx. (b) Usetheresult ofpart (a) and the method ofProblem31to find the

general solution of dy/dx + y = 2sinx. (c) Solvetheinitial value problem dy/dx + y = 2sinx,y(O)= 1.

33.A tank contains 1000liters (L)of a solution consisting of100kg of salt dissolvedin water. Pure water is pumpedinto the tank at the rate of 5 L/s, and the mixture-keptuniform by stirring-is pumped out at the samerate.How

long will it beuntil only 10kg of salt remains in the tank?

34. Considera reservoir with a volume of 8 billion cubicfeet(ft3) and an initial pollutant concentration of0.25%.Thereis a daily inflow of 500million ft 3 of water with a pollu-tant concentration of0.05%and an equal daily outflow ofthe well-mixed water in the reservoir.How long will it

take to reducethe pollutant concentration in the reservoirto 0.10%?

35. Rework Example 4 for the caseof LakeOntario, which

empties into the St.LawrenceRiver and receivesinflow

from LakeErie (via the Niagara River). Theonly differ-encesare that this lake has a volume of 1640Ian 3 and aninflow-outflow rate of410Ian 3/year.

36. A tank initially contains 60 gal of pure water. Brinecontaining 1 lb of salt per gallon enters the tank at2 gal/min, and the (perfectly mixed) solution leavesthetank at 3 gal/min; thus the tank is empty after exactly 1h.(a) Find the amount of salt in the tank after t minutes.

(b) What is the maximum amount ofsalt ever in the tank?

37. A 400-galtank initially contains 100gal of brine contain-

ing 50 lb of salt. Brine containing 1 lb of salt pergallonenters the tank at the rate of 5 galls,and the well-mixedbrine in the tank flows out at the rate of 3 gal/so Howmuch salt will the tank contain when it is full ofbrine?

38. Considerthe cascadeof two tanks shown in Fig. 1.5.5,with VI = 100(gal) and V2 = 200(gal) the volumes ofbrine in the two tanks. Each tank also initially contains50 lb of salt. The three flow rates indicated in the fig-ure areeach5 gal/min, with pure water flowing into tank1.(a) Find the amount x(t)of salt in tank 1 at time t.(b) Supposethat y(t) is the amount of salt in tank 2 at)


time t. Show first that

dy 5x 5y---dt 100 200'

and then solvefor y(t), using the function x(t) found in

part (a). (c) Finally, find the maximum amount of saltever in tank 2.)

Tank 1

Volume VI

Amount x)

Tank 2Volume V2Amount y)

f)

FIGURE1.5.5.A cascadeof two tanks.

39.Supposethat in the cascadeshown in Fig. 1.5.5,tank 1initially contains 100gal of pure ethanol and tank 2 ini-

tially contains 100gal of pure water. Pure water flowsinto tank 1 at 10gal/min, and the other two flow ratesarealso10gal/min. (a) Find the amounts x(t) and y(t)of ethanol in the two tanks at time t > O. (b) Find themaximum amount ofethanol ever in tank 2.

40. A multiple cascadeis shown in Fig. 1.5.6.)

\037)

FIGURE1.5.6.A multiple cascade.At time t = 0, tank 0 contains 1 gal of ethanol and 1 galofwater; all the remaining tanks contain 2gal ofpure wa-ter each. Pure water is pumped into tank 0 at 1 gal/min,)))


and the varying mixture in eachtank is pumped into theone below it at the samerate. Assume, as usual, that themixtures arekept perfectly uniform by stirring. LetXn (t)denote the amount of ethanol in tank n at time t.(a) Show that xo(t) = e-t/2.(b) Show by induction on nthat)

t ne-t/2xn(t) = for n > O.

n! 2n

(c) Show that the maximum value of Xn (t) for n > 0 isMn = xn (2n) = nne-njnL (d) Concludefrom Stirling'sapproximation n! \037 nne-n -J2nn that Mn \037 (2nn)-1/2.41.A 30-year-oldwoman acceptsan engineering positionwith a starting salary of $30,000per year. Her salaryS(t) increasesexponentially, with S(t) = 30et/20 thou-sand dollars after t years.Meanwhile, 12%of her salaryis depositedcontinuously in a retirement account, whichaccumulates interest at a continuous annual rate of 6%.(a) Estimate \037A in terms of \037t to derive the differential

equation satisfied by the amount A (t) in her retirementaccount after t years. (b) Compute A (40), the amountavailable for her retirement at age70.

42. Supposethat a falling hailstone with density 8 = 1 startsfrom rest with negligible radius r = O.Thereafter its ra-dius is r = kt (k is a constant) as it grows by accretionduring its fall. UseNewton's secondlaw-accordingtowhich the net forceF acting on a possibly variable massm equals the time rate of change dpjdt of its momentump = mv-tosetup and solvethe initial value problem

ddt (mv) = mg, v(O) = 0,

where m is the variable mass of the hailstone, v = dyjdtis its velocity, and the positive y-axispoints downward.Then show that dvjdt = gj4.Thus the hailstone falls asthough it were under one-fourth the influence of gravity.

43. Figure 1.5.7shows a slope field and typical solutioncurves for the equation y' = x -

y.)

108642

:>-. 0-2-4-6-8

-10-5) o

x)

5)

FIGURE1.5.7.Slopefield and solutioncurves for y' = x - y.

(a) Show that every solution curve approachesthe

straight line y = x - 1 as x \037 +00. (b) For eachof the five values Yl = 3.998,3.999,4.000,4.001,and

4.002,determine the initial value Yo (accurateto four dec-imal places)such that y(5) = Yl for the solution satisfyingthe initial condition y(-5)= Yo.)

44. Figure 1.5.8shows a slope field and typical solutioncurves for the equation y' = x + y. (a) Show that everysolution curve approachesthe straight line y = -x- 1asx \037 -00.(b) For eachof the five values YI = -10,-5,0,5,and 10,determine the initial value Yo (accuratetofive decimal places)such that y(5) = YI for the solution

satisfying the initial condition y(-5)= Yo.)

108642

:>-. 0-2-4-6-8

-10-5) o

x)

5)

FIGURE1.5.8.Slopefield and solutioncurves for y' = x + y.)

Problems 45 and 46 deal with a shallow reservoirthat hasa one squarekilometer water surfaceand an averagewater

depth of2 meters. Initially it is filled with fresh water, but attime t = 0 water contaminated with a liquid pollutant begins

flowing into the reservoirat the rate of 200 thousand cubicmeterspermonth. Thewell-mixed water in the reservoirflowsout at the samerate. Your first task is to find the amount x (t) ofpollutant (in millions of liters) in the reservoirafter t months.)

45.)The incoming water has a pollutant concentration ofc(t) = 10liters per cubic meter (L/m3). Verify that

the graph of x(t) resemblesthe steadily rising curve in

Fig. 1.5.9,which approachesasymptotically the graph ofthe equilibrium solution x(t) = 20 that correspondsto the

reservoir'slong-term pollutant content. How long doesit

take the pollutant concentration in the reservoir to reach5 L/m3

?)

25 x)

20) x=20)

15)

\037\037Problem 46

Problem 45)

10)

5)

10 20 30 40 50 60

FIGURE1.5.9.Graphs of solutions in

Problems45 and 46.)

46. The incoming water has pollutant concentration c(t)1O(1+ cost) L/m 3 that varies between0 and 20,with an

averageconcentration of 10L/m3 and a period of oscilla-tion of slightly over6* months. Doesit seempredictablethat the lake'spolutant content should ultimately oscillateperiodically about an average level of 20 million liters?)))


Verify that the graph of x (t) does,indeed,resemblethe

oscillatory curve shown in Fig. 1.5.9.How long does it)

take the pollutant concentration in the reservoir to reach5 L/m3

?)

1.5Application)Indoo\037!\037\037l:\037\037rat\037re _ _9scill\037otions)

For an interesting appliedproblemthat involves the solution of a linear differen-tial equation, considerindoor temperature oscillationsthat are driven by outdoortemperatureoscillationsof the form)

A(t) = ao +al coswt+b i sinwt.) (1))

If w = 7T/12,then theseoscillationshave a periodof 24 hours (so that the cycleofoutdoor temperaturesrepeatsitself daily) and Eq.(1)providesa realisticmodelforthe temperatureoutsidea houseon a day when no change in the overallday-to-dayweather pattern is occurring.For instance, for a typical July day in Athens, GAwith a minimum temperature of 70\302\260F when t = 4 (4 A.M.) and a maximum of90\302\260F when t = 16(4 P.M.),we would take)

A(t) = 80 - 10cosw(t-4) = 80-5 coswt-5,J3sinwt. (2))

We derived Eq. (2)by using the identity cos(a- (3) = cosa cosf3 + sin a sin,B to

get ao = 80,al = -5,and b i = -5,J3in Eq.(1).If we write Newton'slaw of cooling(Eq. (3)of Section1.1)for the corre-

spondingindoor temperature u(t) at time t, but with the outside temperatureA(t)given by Eq. (1) instead of a constant ambient temperature A, we get the linearfirst-orderdifferentialequation)

du

dt= -k(u-A(t));)

that is,)

dub

.-+ ku = k(ao +al coswt + I slnwt)dt)

(3))

with coefficient functions P(t) = k and Q(t) = kA(t). Typical values of the

proportionalityconstant k range from 0.2to 0.5(although k might be greater than

0.5for a poorly insulated building with openwindows,or lessthan 0.2for a well-insulatedbuilding with tightly sealedwindows).)

SCENARIO: Supposethat our air conditioner fails at time to = 0 one midnight,and we cannot afford to have it repaireduntil payday at the end of the month. Wetherefore want to investigatethe resulting indoor temperaturesthat we must endurefor the next severaldays.

Beginyour investigation by solving Eq. (3)with the initial conditionu(O) ==

Uo (the indoor temperature at the time of the failure of the air conditioner). You

may want to use the integral formulas in 49 and 50of the endpapers,or possiblyacomputer algebra system.You should get the solution)

u(t) = ao +coe-kt +CI coswt + dl sin wt,) (4))))


10095 I

I

Uo=95

1

90 I

I

85 I,-..I

OI)

\037 80 I

'-\"\037 75

I

I I

I I

70 I I

Uo=65:

I

65 I

It = 12 It = 36600 10 20 30 40

t (h))

FIGURE1.5.10.Solution curvesgiven by Eq.(5)with Uo = 65,68,71,...,92,95.)

100959085)

,,-.....

OI)

\037 80'-\"

\037 75

70

6560

0)

I

,\\1

Indoor

temperatureI

It = 3610 20 30 40

t (h)

FIGURE1.5.11.Comparisonofindoor and outdoor temperatureoscillations.)

where)

k2al - kwh lCo= Uo - aO-

k2 +W2)

CI =) k2al - kwh l

k2 +W2)d _ kwat +k2bt

I -k2 +W2)

with w = n/12.With ao = 80,al = -5,hI = -5,J3(as in Eq. (2)),w = n/12,and k = 0.2

(for instance),this solutionreduces(approximately)to)

nt ntu(t) = 80 +e-t/5 (uo -82.3351)+ (2.3351)cos-- (5.6036)sin-. (5)12 12)

Observefirst that the \"damped\" exponential term in Eq. (5)approacheszeroas t --* +00,leaving the long-term \"steady periodic\"solution)

nt ntusp(t) = 80 + (2.3351)cos-- (5.6036)sin-.12 12)

(6))

Consequently, the long-term indoor temperatures oscillateevery 24 hours aroundthe sameaverage temperature 80\302\260F as the averageoutdoor temperature.

Figure 1.5.10showsa number of solution curves correspondingto possibleinitial temperatures Uo ranging from 65\302\260F to 95\302\260F. Observethat-whatever theinitial temperature-theindoor temperature \"settlesdown\" within about 18hoursto a periodicdaily oscillation.But the amplitude of temperature variation is lessindoorsthan outdoors.Indeed,using the trigonometric identity mentioned earlier,Eq.(6)can be rewritten (verify this!)as)

u(t) = 80 - (6.0707)cos(7\037

-1.9656)

n= 80 - (6.0707)cos-(t-7.5082).12)

(7))

Doyou seethat this impliesthat the indoor temperaturevariesbetween a minimumof about 74\302\260 F and a maximum of about 86\302\260 F?

Finally, comparisonof Eqs.(2)and (7)indicates that the indoor temperaturelagsbehind the outdoor temperatureby about7.5082-4 \037 3.5hours, as illustratedin Fig.1.5.11.Thus the temperature insidethe housecontinues to riseuntil about7:30P.M.each evening, so the hottest part of the day insideis early evening ratherthan late afternoon (as outside).

For a personalproblemto investigate, carry out a similar analysis using av-erageJuly daily maximum/minimum figures for your own localeand a value of k

appropriate to your own home.You might alsoconsidera winter day instead ofa summer day. (What is the winter-summerdifference for the indoor temperatureproblem?)You may wish to explorethe use of available technologyboth to solvethe differentialequationand to graph its solutionfor the indoor temperaturein com-parisonwith the outdoor temperature.)))

1.6SubstitutionMethodsand ExactEquations 59)

11IIISubstitutionMethodsand ExactEq\037ations)

The first-orderdifferential equations we have solved in the previoussectionshaveall beeneither separableor linear. But many applicationsinvolve differential equa-tions that are neither separablenor linear. In this sectionwe illustrate (mainly with

examples)substitution methods that sometimescan be used to transform a givendifferentialequation into one that we already know how to solve.

For instance, the differentialequation)

dy-= f(x,y),dx)

(1))

with dependentvariable y and independentvariablex, may contain a conspicuouscombination)

v=a(x,y)) (2))

of x and y that suggestsitself asa new independentvariablev. Thus the differential

equation)dy = (x+ y +3)2dx

practicallydemandsthe substitution v = x+ y +3 of the form in Eq.(2).If the substitution relation in Eq.(2)can be solved for)

y = f3 (x,v),) (3))

then application of the chain rule-regardingv as an (unknown) function of x-yields)

dy af3 dx af3 dv dv

dx=

axdx +av dx

= f3x + f3v dx ') (4))

where the partial derivativesaf3jax = f3x(x, v) and af3jav = f3v(x, v) are known

functions of x and v. If we substitute the right-hand sidein (4)for dyjdx in Eq.(1)and then solve for dvjdx, the result is a new differentialequationof the form)

dv-= g(x,v)dx)

(5))

with new dependentvariable v. If this new equation is either separableor linear,then we can apply the methodsof precedingsectionsto solve it.

If v = v(x) is a solution of Eq.(5),then y = f3 (x,v (x))will be a solution ofthe original Eq. (1).The trick; is to selecta substitution such that the transformed

Eq. (5)is one we can solve. Even when possible,this is not always easy;it mayrequire a fair amount of ingenuity or trial and error.)

Example1) Solvethe differentialequation)

dy 2-= (x+ y +3) .dx)))

Solution As indicated earlier, let'stry the substitution)

108642

\037 0-2-4-6-8-10

-5) ox)

5)

FIGURE1.6.1.Slopefield andsolution curves fory' = (x + y + 3)2.)

v = x + y + 3;) that is, y = v -x- 3.)

Then)dy dv-= --1,dx dx)

so the transformedequation is)

dv-= 1 + v2.dx)

This is a separableequation, and we have no difficulty in obtaining its solution)

fdvx =

2 = tan- 1 v +C.l+v)

Sov = tan(x -C).Becausev = x + y +3,the general solution of the originalequation dy/dx = (x+ y + 3)2 isx + y + 3 = tan(x -C);that is,)

y(x) = tan(x -C)-x- 3.) .)Remark:Figure 1.6.1showsa slopefield and typical solution curves for

the differentialequationof Example1.We seethat, although the function f (x,y) =(x+y +3)2iscontinuouslydifferentiablefor all x and y, each solutioniscontinuousonly on a boundedinterval. In particular,becausethe tangent function iscontinuouson the open interval (-T[/2,T[/2), the particular solution with arbitrary constantvalue C iscontinuouson the interval where -T[/2 <x-C <T[/2;that is,C-T[/2 <x < C + T[/2.This situation is fairly typical of nonlinear differentialequations, in

contrast with linear differentialequations,whosesolutionsare continuouswhereverthe coefficientfunctions in the equation are continuous. .

Example1 illustrates the fact that any differentialequationof the form)

dy-= F(ax+by+c)dx)

(6))

can be transformed into a separableequation by useof the substitution v = ax+by + c (seeProblem 55). The paragraphs that follow deal with other classesoffirst-order equations for which there are standard substitutions that are known tosucceed.)

HomogeneousEquationsA homogeneousfirst-order differential equation is one that can be written in theform)

\037)

dy = F (Y

) .dx x)(7))

If we make the substitutions)

yv = -, y = vx,

x)

dy dv-=v+x-,dx dx)(8))))


then Eq.(7)is transformed into the separableequation

dvx-= F(v)- v.dx)

Thus everyhomogeneousfirst-orderdifferentialequationcan bereducedto an inte-gration problemby means of the substitutions in (8).

Remark:A dictionary definition of \"homogeneous\"is \"of a similar kind

or nature.\" Considera differentialequationof the form)

dyAxmyn_ = BxPyq+Cxrys

dx)(*))

whosepolynomial coefficientfunctions are \"homogeneous\"in the sensethat eachof their terms has the sametotal degree,m + n = p + q = r + s = K. If wedivide each sideof (*)by xK, then the result-becausexm

yn Ixm +n = (yIx)n,andsoforth-isthe equation)

(y

)n dy _

(y

)q

(Y

)S

A - --B - +C-x dx x x)

which evidently can be written (by another division) in the form of Eq. (7).Moregenerally,a differentialequationof the form P(x,y) y/ = Q(x,y) with polynomialcoefficientsP and Q ishomogeneousif the terms in thesepolynomialsall have the

sametotal degreeK.The differential equation in the followingexample is of this

form with K = 2.)

Example2) Solvethe differentialequation)

dy2xy-= 4x2 +3y 2.dx)

Solution This equation is neither separablenor linear, but we recognizeit asa homogeneousequation by writing it in the form)

dydx)

4x2 +3y2 =2(X

) + 3(

Y

) .2xy y 2 x)

The substitutions in (8)then take the form)

y = vx,)dy dv-=v+x-,dx dx)

y 1v = -, andx v)

x)

y)

Theseyield)dv 2 3v+x-=-+-v,dx v 2)

and hence)

dv 2 V v2 +4X - + - .

dx-

v 2-

2v')

f2v

f1

dv = -dx;v2 +4 x)

In(v2 +4) = In Ixl + In C.)))

6

4 V2 +4 ==Clxl;

2 y2-+4==Clxl;\037 0 X 2

-2y2 +4x2 ==kx3.

-4)

o 2 4 6x)

FIGURE1.6.2.Slopefield andsolution curves for2xyy' = 4x2 + 3y 2.)

Example3)

We apply the exponentialfunction to both sidesof the last equation to obtain)

Note that the left-hand sideof this equation is necessarilynonnegative. It followsthat k > 0 in the caseof solutions that are defined for x > 0,while k < 0 forsolutionswherex <O.Indeed,the family of solutioncurves illustrated in Fig.1.6.2exhibitssymmetry about both coordinate axes.Actuall y, there are positive-valuedand negative-valuedsolutions of the forms y(x) == -::1::. .vkx3 -4x2 that are definedfor x >4/k if the constant k is positive,and for x <4/k if k is negative. .)


dyxdx= y +Jx2 -y2, y(xo)= 0,)

where Xo >O.)

Solution We divide both sidesby x and find that)

5040302010

\037 0-10-20-30-40-50

o) 10 20 30 40 50x)

FIGURE1.6.3.Soluti on curves

for xy' = y + Jx2 -y2.)

dy = Y +)1-(Y

)2,dx x x)

so we make the substitutions in (8);we get)

v +x\037:

= v +J1- v2 ;

f1

dv ==f\037 dx;

.vI - v2 x)

sin-1 v == In x +C.)

We neednot write In Ix Ibecausex > 0near x == Xo > O. Now note that v(xo)

y(xo)/xo==0,so C ==sin-1 0-lnxo==-lnxo.Hence)

v == y ==sin (In x- In xo) ==sin (In \037

) ,x Xo)

and therefore)

y(x) = xsin(In:J

is the desiredparticular solution. Figure 1.6.3showssometypical solution curves.Becauseof the radical in the differentialequation, thesesolutioncurvesare confinedto the indicated triangular region x > Iyl. You can checkthat the boundary linesy ==x and y ==-x(for x >0)are singular solutioncurves that consistof points oftangency with the solutioncurves found earlier. .)))

Example4)


BernoulliEquationsA first-orderdifferentialequationof the form)

\037)

dy-+ P(x)y = Q(x)yndx)

(9))

is calleda Bernoulliequation.If either n = 0or n 1,then Eg.(9)is linear.Otherwise,as we askyou to show in Problem56,the substitution)

\037) v = yl-n) (10))

transforms Eq.(9)into the linear equation)

dv-+ (1-n)P(x)v= (1-n)Q(x).dx)

Rather than memorizing the form of this transformedequation, it is more efficientto make the substitution in Eq.(10)explicitly,as in the followingexamples.)

............... ........ ..... ....

If we rewrite the homogeneousequation 2xyy' = 4x2 + 3y2 of Example2 in the

form)dy 3 2x---y=-,dx 2x y

we seethat it is also a Bernoulli equation with P(x)n = -1,and 1- n = 2.Hencewe substitute)

-3/(2x),Q(x)) 2x,)

v = y2,)dy dy dv 1 -1/2dv-=--=-v -dx dv dx 2 dx)

y = vI /2,) and)

This gives)1 dv 3_v- 1/2_ _ _v 1/2 = 2xv-I/2.2 dx 2x

Then multiplication by 2v 1/2 producesthe linear equation)

dv 3-- -v = 4xdx x)

with integrating factor p = e!(-3/x)dx= x-3.Sowe obtain)

-3 4Dx(x v) = 2\";x)

4x-3v = --+C;x)

4x-3y2 = --+C;

x)

y2 = -4x2+Cx3.) .)))


Example5)

Example6)

The equation)dyx-+6y ==3Xy

4/3dx

is neither separablenor linear nor homogeneous,but it is a Bernoulli equation with

n ==\037,

1 - n ==-1.The substitutions)

v ==y-I/3,)dy dy dv _4dv-==--==-3v -dx dv dx dx)

-3Y == v ,) and)

transform it into)dv-3xv-4-+6v-3 ==3xv-4.dx

Division by -3xv-4 yieldsthe linear equation)

dv 2---v==-1dx x)

with integrating factor p ==eI(-2/x)dx ==x-2.This gives)

-2 1Dx(x v) ==-2\";x)

1x-2v ==-+C;x)

v ==x+Cx2;)

and finally,)

1y(x) =

(x+Cx2)3') .)

The equation)

dy2xe2y-==3x4 +e2y

dx)(11))

is neither separable,nor linear,nor homogeneous,nor is it a Bernoulliequation. Butwe observethat y appearsonly in the combinationse2y and Dx (e2y ) ==2e2y y,. Thisprompts the substitution)

v ==e2y,)

dv2y dy-==2e -dx dx)

that transformsEq.(11)into the linear equationxv/ex)==3x4 + vex); that is,)

dv 1 3-- -v ==3x .dx x)

After multiplying by the integrating factor p == l/x,we find that)

1

f2 3

xv== 3x dx==x +c,) so e2Y ==v==x4 +cx,)

and hence)

y(x) == 41nIx4 +cxl.) .)))

y)

y =I(x))//'/'Va/'/'/')

(a,0) x)

FIGURE1.6.4.Theairplaneheadedfor the origin.)

\037

:,y)

Yx2+y2 /\037/'/' Iy/'/' I

/'/' I/' I

/'/' 8 I

x)

FIGURE1.6.5.Thecomponentsof the velocity vectorof the

airplane.)


FlightTrajectoriesSupposethat an airplanedepartsfrom the point (a,0)locateddue eastof its intendeddestination-anairport locatedat the origin (0,0).The plane travels with constant

speedVo relative to the wind, which is blowing due north with constant speedw.As indicated in Fig.1.6.4,we assumethat the plane'spilot maintains its headingdirectly toward the origin.

Figure 1.6.5helpsus derive the plane'svelocity components relative to the

ground. They are)

dx VOX--= -vocose= -dt Jx2 + y2

'

dy. voY-= -Vo Sin e + w = - +w.

dt Jx2 + y2)

Hencethe trajectory y = f (x)of the plane satisfiesthe differential equation)

dy dy/dt 1( J )

--= =- voY- w x2 + y2 .

dx dx/dt VOX)

(12))

If we set)

wk =-,

Vo)

(13))

the ratio of the windspeedto the plane'sairspeed,then Eq.(12)takes the homoge-neousform)

\037\037

=\037 _k[1+(\037 )2]'/2.) (14))

The substitution y = XV, y' = V +xv' then leadsroutinely to)

fdv --

fk

dx,Jl + v2

- x.) (15))

By trigonometric substitution, or by consulting a table for the integral on the left,we find that)

In (v +J1 + v2)

= -kInx +C,

and the initial condition v(a) = y(a)/a = 0 yields)

(16))

C = kIn a.) (17))

As we askyou to show in Problem68,the result of substituting (17)in Eq.(16)andthen solving for V is)

v =\037 [Grk -

G )k

l) (18))

Becausey = XV, we finally obtain)

_ a

[(X

)l-k

(X

)l+k

]y(x) -- - --2 a a)

(19))

for the equationof the plane'strajectory.)))

Example7)

Note that only in the casek < 1 (that is, w < vo) doesthe curve in Eq. (19)pass through the origin, so that the plane reachesits destination. If w == Vo (sothat k == 1),then Eq. (19)takesthe form y(x) ==

\037a(l-x2ja2), so the plane's

trajectory approachesthe point (0,aj2) rather than (0,0).The situation is evenworseif w > Vo (sok > I)-inthis caseit follows from Eq.(19)that y \037 +00asx \037 O.The three casesare illustrated in Fig.1.6.6.)

If a ==200mi, Vo ==500mijh,and w == 100mijh,then k == wjvo ==\037,

so the planewill succeedin reaching the airport at (0,0).With thesevalues,Eq.(19)yields)

[(X

)4/5

(X

)6/5

]y(x) == 100 200-

200 .) (20))

Now supposethat we want to find the maximum amount by which the plane isblown off courseduring its trip. That is, what is the maximum value of y(x) foro<x <200?)

y)

Solution Differentiationof the function in Eq.(20)yields)

(a,0)x)(0,0)

FIGURE1.6.6.Thethree casesW < Va (planevelocity exceedswind velocity), W = Va (equalvelocities),and w > Va (wind isgreater).)

dy == \037

[4

(\037)-1/5_ 6

(\037)

1/5

]dx 2 5 200 5 200 ')

and we readily solve the equation y/ (x)==0to obtain (xj200)2/5==\037.

Hence)

[(2

)2

(2

)3

]400

Ymax == 1003

-3

==27

\037 14.81.)

Thus the plane isblown almost 15mi north at one point during its westwardjoumey.(The graph of the function in Eq. (20)is the one usedto construct Fig.1.6.4.Thevertical scalethere is exaggeratedby a factor of 4.) .)ExactDifferentialEquationsWe have seenthat a general solution y(x) of a first-order differential equation isoften defined implicitly by an equation of the form)

F(x,y(x))== C,) (21))

where C is a constant. On the other hand, given the identity in (21),we can recoverthe original differentialequation by differentiating each sidewith respectto x. Pro-vided that Eq. (21)implicitly defines y as a differentiablefunction of x, this givesthe original differentialequation in the form)

aF aF dy-+--==0;ax ay dx)

that is,)dyM(x,y) +N(x,y)- ==0,dx)

(22))

where M(x,y) == Fx(x,y) and N(x,y) == Fy(x,y).)))

Example8)


It is sometimesconvenientto rewriteEq.(22)in the more symmetric form)

\037) (23))M(x,y) dx+N(x,y) dy = 0,)

calledits differential form. The general first-order differential equation y' ==

f (x,y) can be written in this form with M = f (x,y) and N = -1.The pre-cedingdiscussionshowsthat, if there existsa function F(x,y) such that)

aF-= M andax)

aF-=Nay

,)

then the equation)

\037) F(x,y) = C)

implicitly defines a general solution of Eq. (23).In this case,Eg.(23)is calledan

exactdifferential equation-thedifferential)

dF= Fxdx+ Fydy)

of F(x,y) is exactly Mdx+ N dy.Natural questionsare these:How can we determine whether the differential

equation in (23)is exact?And if it is exact,how can we find the function F suchthat Fx = M and Fy

= N? To answer the first question, let us recall that if the

mixedsecond-orderpartial derivatives Fxy and Fyx are continuouson an open setin the xy-plane,then they are equal: Fxy

= Fyx. If Eq. (23)is exact and M and Nhave continuouspartial derivatives,it then follows that)

aM aN-=Fxy

=Fyx

=-.ay ax)

Thus the equation)

\037)

aM aN

ay ax)(24))

is a necessarycondition that the differentialequation Mdx+ N dy = 0 be exact.That is, if My =1= Nx, then the differentialequation in question is not exact, soweneednot attempt to find a function F(x,y) such that Fx = M and Fy

= N-thereis no such function.)

The differentialequation)

y3 dx+3xy2

dy = 0) (25))

is exactbecausewe can immediately seethat the function F(x,y) = xy3 has the

property that Fx = y3 and Fy= 3xy 2. Thus a general solutionof Eq.(25)is)

xy3 = c;)

if you prefer, y(x) = kx-I /3.) .)))

But supposethat we divide each term of the differentialequation in Example8 by y2 to obtain)

y dx+3x dy = o.) (26))

This equation is not exactbecause,with M = y and N = 3x,we have)

aM aN-= 1 =1= 3 =-.ay ax)

Hencethe necessarycondition in Eq.(24)is not satisfied.We are confronted with a curious situation here.The differentialequations in

(25)and (26)are essentiallyequivalent, and they have exactly the samesolutions,yet one is exactand the other is not. In brief, whether a given differentialequationisexactor not is related to the preciseform Mdx+N dy = 0 in which it is written.

Theorem 1 tellsus that (subjectto differentiability conditionsusually satisfiedin practice)the necessarycondition in (24)is alsoa sufficient condition for exact-ness.In other words,if My = Nx , then the differentialequation Mdx+ N dy = 0is exact.)

THEOREM1 Criterionfor Exactness)

Supposethat the functions M(x,y) and N (x,y) are continuous and have con-tinuous first-order partial derivatives in the open rectangle R: a < x < b,c < y <d.Then the differentialequation)

M(x,y)dx +N(x,y)dy = 0) (23))

is exactin R if and only if)

aM aNay ax)

(24))

at eachpoint of R. That is,there existsa function F(x,y) defined on R with

aFjax = M and aFjay = N if and only if Eq.(24)holdson R.)

Proof:We have seenalready that it is necessaryfor Eq. (24) to hold if

Eq. (23)is to beexact.To prove the converse,we must show that ifEq.(24)holds,then we can construct a function F(x,y) such that aFjax= M and aFjay = N.Note first that, for any function g(y), the function)

F(x,y) = f M(x,y)dx +g(y)

satisfiesthe condition aFjax= M. (In Eq.(27),the notation f M(x,y) dx denotesan antiderivative of M(x,y) with respectto x.)We plan to chooseg(y) so that)

(27))

N = aF = (\037

f M(x,y) dX ) +g'(y)ay ay)

as well;that is, so that)

, a

fg (y) = N -- M(x,y)dx.ay)

(28))))


Toseethat there is such a function of y, it sufficesto show that the right-hand sidein

Eq. (28)is a function of y alone.We can then find g(y) by integrating with respectto y. Becausethe right-hand sidein Eq.(28)is definedon a rectangle,and henceonan interval as a function of x, it sufficesto show that its derivative with respectto xis identicallyzero.But)

\037

(N - \037 JM(x,Y)dX ) ==aN - \037\037 JM(x,y)dxax ay ax axay

aN a a

J==---- M(x,y)dxax ayaxaN aM==---==0ax ay)

by hypothesis.So we can, indeed,find the desiredfunction g(y) by integrating

Eq.(28).We substitute this result in Eq.(27)to obtain)

F(x,y)= JM(x,y)dx+J (N(X,y)-aay JM(X,y)dX)dY (29))

as the desiredfunction with Fx ==M and Fy==N. \037)

Insteadof memorizingEq.(29),it is usually better to solve an exactequationMdx+N dy ==0 by carrying out the processindicatedby Eqs.(27)and (28).Firstwe integrate M(x,y) with respectto x and write)

F(x,y)= JM(x,y)dx +g(y),)

thinking of the function g (y) as an \"arbitrary constant of integration\" as far as the

variable x is concerned.Then we determine g (y) by imposing the condition that

aFjay ==N (x,y). This yieldsa general solution in the implicit form F(x,y) == C.)

Example9) Solve the differentialequation)

(6xy- y3) dx+ (4y +3x2-3xy2)dy ==O.) (30))

Solution Let M(x,y) == 6xy - y3 and N(x,y) == 4y + 3x2 - 3xy 2. The given equation isexactbecause)

aM 2 aN-==6x-3y ==-.ay ax

Integrating aFjax ==M(x,y) with respectto x,we get)

F(x,y) = J(6xy-i)dx= 3x2y -xl+g(y).)

Then we differentiatewith respectto y and setaFjay ==N(x,y). This yields)

aF = 3x2 _ 3xi+g'(y)= 4y +3x2-3xi,ay)))


54321

\037 0-1-2-3-4-5

-5-4-3-2-10 1 2 3 4 5x)

FIGURE1.6.7.Slopefield andsolution curves for the exactequation in Example 9.)

and it follows that g'(y) ==4y. Henceg(y) ==2y2 +C1 , and thus)

F(x,y) ==3x2y-xy3 +2y

2 +C1 .)

Therefore, a general solution of the differentialequation is definedimplicitly by theequation)

3x2y-xy3 +2y

2 == C) (31))

(wehave absorbedthe constant C1 into the constant C).) .)

Remark:Figure 1.6.7showsa rather complicatedstructure of solutioncurves for the differential equation of Example9.The solution satisfying a giveninitial condition y(xo)== Yo is defined implicitly by Eq.(31),with C determinedby

substituting x ==Xo and y == Yo in the equation. For instance, the particularsolutionsatisfying y(O) == 1 is defined implicitly by the equation 3x2

y- xy3 + 2y

2 == 2.The other two specialpoints in the figure-at(0,0)and near (0.75,2.12)-areoneswhere both coefficientfunctions in Eq. (30)vanish, so the theorem of Section1.3doesnot guarantee a unique solution. .)ReducibleSecond-OrderEquationsA second-orderdifferentialequation involvesthe secondderivativeof the unknown

function y(x),and thus has the general form)

F(x,y, y', y\") ==O.) (32))

If either the dependentvariable y or the independent variablex is missingfrom asecond-orderequation, then it is easilyreducedby a simplesubstitution to a first-

orderequation that may be solvableby the methods of this chapter.

Dependentvariable y missing. If y is missing,then Eq.(32)takesthe form)

\037) F(x,y', y\") ==O.) (33))

Then the substitution)

\037)I dy

p == y ==dx') (34))

dpdx)

\"y)

resultsin the first-orderdifferentialequation)

F(x,p,p') ==O.)

If we can solve this equation for a general solution p(x,C1 ) involving an arbitraryconstant C1,then we needonly write)

y(x) = Jy/(x)dx= Jp(x,Cddx+ C2)

to get a solutionof Eq.(33)that involvestwo arbitrary constantsCI and C2(asis tobeexpectedin the caseof a second-orderdifferential equation).)))

252015105

;>, 0)

Example10)


Solvethe equation xy\" +2y' ==6x in which the dependentvariabley is missing.)

Solution The substitution defined in (34)gives the first-orderequation)

-5-10-15-20-25-5-4-3-2-1012345

x)

FIGURE1.6.8.Solution curvesC]of the form y == x2 + -forx

C1= 0, \0373, \03710, \03720, \03735, \03760,

:f:100.)

dpxdx +2p==6x;)dp 2

that is,-+-p ==6.dx x)

Observing that the equationon the right here is linear,we multiply by its integratingfactor p ==exp(f(2Ix)dx) ==e21nx ==x2 and get)

Dx(x2p) ==6x2,

x2P ==2x3 + C1 ,

dy CIP ==-==2x+-.dx x2)

A final integration with respectto x yieldsthe general solution)

2 CIy(x) ==x +-+C2

x)

of the second-orderequation xy// + 2y' == 6x. Solution curves with CI == 0 but

C2 i= 0 are simply vertical translatesof the parabola y ==x2 (for which CI == C2 ==

0).Figure 1.6.8showsthis parabola and sometypical solutioncurves with C2 == 0but C1 i= O.Solution curves with C1 and C2both nonzero are verticaltranslatesofthose(other than the parabola) shown in Fig.1.6.8. .)

Independentvariable xmissing. If x is missing,then Eq.(32)takes the form)

\037) F(y, y', y\") ==O.) (35))

Then the substitution)

\037),

p==y)dydx')

\" dp dp dy dpY- - -p- -- -- - -

dx dy dx dy)(36))

resultsin the first-orderdifferentialequation)

F(y,

p,p\037\037 ) = 0)

for p asa function of y. If we can solve this equationfor a generalsolution p(y, C1)involving an arbitrary constant CI,then (assuming that y' i= 0) we needonly write)

Jdx

J1

J1

J dyx(y)== -dy== dy== -dy== +C2.

dy dyldx p p(y, CI ))

If the final integral P ==f(11p)dy can beevaluated,the result isan implicit solutionx(y) == P(y, C1)+ C2 of our second-orderdifferentialequation.)))

Example11),.....\037. .....

Solvethe equation yy\" = (y')2in which the independentvariablex is missing.)

Solution We assumetemporarily that y and y' are both nonnegative,and then point out at theend that this restriction is unnecessary.The substitution defined in (36)gives thefirst-orderequation)

54321

;>, a-1-2-3-4-5

-5 -4 -3 -2 -1a 1 2 3 4 5x)

FIGURE1.6.9.Thesolutioncurves y = AeBx with B = 0 andA = 0, :f:1arethe horizontal linesy = 0, :f:1.Theexponential curveswith B > 0 and A = :f:1are in

color,those with B < 0 andA = :f:1areblack.)

IIIIIJ Problems)

dp 2yp- = P .dy)

Then separationof variablesgives)

f d:= f d;,In p = In y + C (becausey >0 and p = y' >0),

p = C1y)

where C1 = eC.Hence)

dx 1 1

dy P C1 y'

C1x= f d;= loy + C1 .)

The resulting general solution of the second-orderequation yy\" = (y')2is)

y(x) = exp(C1x-C2) = AeBx,)

where A = e-C2 and B = C1.Despiteour temporary assumptions,which implythat the constants A and B are both positive, we readily verify that y(x) = AeBx

satisfiesyy\" = (y')2 for all real values of A and B. With B = 0 and differentvalues of A, we get all horizontal linesin the plane as solution curves.The upperhalf of Fig.1.6.9showsthe solution curves obtained with A = 1 (for instance)anddifferent positive values of B. With A = -1thesesolution curves are reflectedin the x-axis,and with negative values of B they are reflected in the y-axis. In

particular, we seethat we get solutions of yy\" = (y')2,allowing both positiveand

negativepossibilitiesfor both y and y'. .)

Find generalsolutions of the differential equations in Prob-lems 1through 30.Primes denotederivatives with respectto xthroughout.)

2.2xyy' = x2 + 2y2

4. (x - y)y' = x + y

6.(x + 2y) y' = Y

8.x2y' = xy + x2eY/x

10.xyy' = x2 + 3y2)

1.(x + y)y' = x -Y

3.xy' = y + 2,JXY5.x(x+ y)y' = y(x - y)7. xy2y' = x3 + y3

9.x2y' = xy + y2

11.(x2 -y2)y' = 2xy12.xyy' = y2 + xJ4x2 + y2

13.xy' = y + Jx2 + y2

14.yy' + x = Jx2 + y2)

15.x(x+ y)y' + y(3x + y) = 016.y' = ,Jx + y + 1 17.y' = (4x + y)218.(x + y)y' = 1 19.x2y' + 2xy = 5y

3

20.y2y' + 2xy3 = 6x 21.y' = y + y3

22.x2y' + 2xy = 5y4 23. xy' + 6y = 3xy 4/3

24. 2xy'+ y3e-2x = 2xy

25.y2(xy' + y)(l + X4)1/2= x26.3y2y' + y3 = e-x27.3xy2y' = 3x4 + y3

28.xeYy' = 2(eY + x3e2x )29.(2xsin y cosy) y' = 4x2 + sin 2

y

30.(x + eY)y' =xe-Y - 1)))

In Problems 31through 42, verify that the given differentialequation is exact;then solveit.)

31.(2x + 3y) dx + (3x+ 2y) dy = 032. (4x - y) dx + (6y -x)dy = 033. (3x2 + 2y2)dx + (4xy + 6y2)dy = 034. (2xy

2 + 3x2) dx + (2x2y+ 4y3)dy = 0

35. (x3+\037 )dx+(y2+1nx)dy=O36. (1+ yexy ) dx + (2y + xexy ) dy = 0

37.(cosx+lny)dx+(\037 +eY)dY=O38. (x + tan-1

y) dx +x + y

dy = 01+ y2

39.(3x2y3 + y4) dx + (3x3

y2 + y4 + 4xy3)dy = 040. (eX sin y + tan y) dx + (eX cosy + x sec2 y) dy = 0

41.(2X _ 3y2

)dx+ (

2Y _ X2+\037

) dY=Oy x4 x3 y2 \037

2X5/2 - 3y5/3 3y5/3 - 2X 5/242.

2X 5/2y2/3dx+

3X 3/2y5/3dy = 0

Find a generalsolution ofeachreduciblesecond-orderdiffer-ential equation in Problems43-54. Assume x, y and/or y/

positive where helpful (asin Example 11).43. xy\" = y/ 44. yy\" + (y/)2 = 045. y\" + 4y = 0 46. Xy\" + y/ = 4x47. y\" = (y/)2 48.x2

y\" + 3xy'= 249. yy\" + (y/)2 = yy/ 50.y\" = (x + y/)251.y\" = 2Y(y/)3 52. y3 y\" = 153. y\" = 2yy' 54. yy\" = 3(y')255. Show that the substitution v = ax + by + c transforms

the differential equation dyldx = F(ax+ by + c) into aseparableequation.

56. Supposethat n 1= 0 and n 1= 1.Show that the sub-stitution v = y

I-n transforms the Bernoulli equationdy Idx + P (x)y = Q(x)yn into the linear equation

dv

dx+ (1-n)P(x)v(x)= (1-n)Q(x).

57. Show that the substitution v = In y transforms the differ-ential equation dyldx + P(x)y = Q(x)(yIn y) into thelinear equation dvldx + P(x)= Q(x)v(x).

58.Usethe ideain Problem57 to solvethe equation

dy 2X

dx-4x y + 2y In y = o.

59.Solvethe differential equation

dy x -y- 1

dx x + y + 3

by finding hand k so that the substitutions x = u + h,

y = v + k transform it into the homogeneousequation

dv u - v

du u + v)


60.Usethe method in Problem 59 to solvethe differential

equation)dy 2y -x + 7dx 4x- 3y - 18

61.Make an appropriate substitution to find a solution of the

equation dyldx = sin(x -y). Doesthis general solutioncontain the linear solution y (x) = x - ]f12that is readilyverified by substitution in the differential equation?

62.Show that the solution curvesof the differential equation)

dy y(2x3 _ y3)dx x(2y

3 -x3))

areof the form x3 + y3 = 3Cxy.63.Theequation dyldx = A(X)y2 + B(x)y+ C(x)is called

a Riccatiequation.Supposethat oneparticular solution

Yl (x) of this equation is known. Show that the substitution)

1y = Yl + -

v)

transforms the Riccati equation into the linear equation)

dv

dx+ (B+ 2AYl)V = -A.)

Usethe method ofProblem63 to solve the equations in Prob-lems64and 65,given that Yl (x) = x is a solution ofeach.

dy64.-+ y2 = 1+ x2dx

dy65.-+ 2xy = 1+ x2 + y2dx

66.An equation of the form)

y = xy/ + g(y/)) (37))

is called a Clairaut equation. Show that the one-parameter family of straight linesdescribedby)

y(x) = Cx+ g(C)) (38))

is a general solution ofEq.(37).67.Considerthe Clairaut equation)

y = xy/-

i(y')2)

for which g(y/) = -i(y/)2 in Eq. (37).Show that the line)

y = Cx- iC2)

is tangent to the parabola y = x2 at the point (\037C, iC2).Explain why this implies that y = x2 is a singular solu-tion of the given Clairaut equation. This singular solution

and the one-parameterfamily of straight line solutions areillustrated in Fig. 1.6.10.)))


71.A river 100ft wide is flowing north at w feetper second.A dog starts at (100,0)and swims at Va = 4 ft/s, always

heading toward a treeat (0,0) on the west bank directlyacrossfrom the dog'sstarting point. (a) If w = 2 ft/s,show that the dog reachesthe tree. (b) If w = 4 ft/s,show that the dog reachesinstead the point on the westbank 50 ft north of the tree. (c) If w = 6 ft/ s, show that

the dog never reachesthe west bank.72. In the calculusofplane curves, onelearns that the curva-

ture K of the curve y = y (x) at the point (x,y) is given

by)

FIGURE1.6.10.Solutions of the Clairaut

equation ofProblem67.The\"typical\" straightline with equation y = Cx -

\037

C2 is tangent to

the parabolaat the point (4c,\037

C2).68.DeriveEq.(18)in this sectionfrom Eqs.(16)and (17).69.In the situation of Example 7, supposethat a = 100mi,

Va = 400mi/h, and w = 40 mi/h. Now how far north-ward doesthe wind blow the airplane?

70.As in the text discussion,supposethat an airplane main-tains a heading toward an airport at the origin. If Va = 500mi/h and w = 50mi/h (with the wind blowing due north),and the plane begins at the point (200,150),show that its

trajectory is describedby

y + Jx2 + y2 = 2(200x9)1/10.)

ly\"(x)1K=[1+ Y/(X)2]3/2'

and that the curvature of a circleof radius r is K = 1/r .[SeeExample 3 in Section11.6of Edwards and Penney,Calculus:Early Transcendentals,7th edition (UpperSad-dle River, NJ: PrenticeHall, 2008).]Conversely, substi-

tute p = y/ to derivea generalsolution ofthe second-orderdifferential equation)

ry\" = [1+ (y/)2]3/2)

(with r constant) in the form)

(x -a)2+ (y -b)2 = r2.)

Thus a circleofradius r (ora part thereof) is the only planecurve with constant curvature l/r.)

lIB P()pulationModels)

In Section1.4we introduced the exponential differential equation dPjdt = kP,with solution P(t) = Poekt , as a mathematical model for natural population

growth that occursas a result of constant birth and death rates. Herewe presenta more general population modelthat accommodatesbirth and death rates that arenot necessarilyconstant. As before,however,our population function P(t) will bea continuous approximationto the actual population,which of coursechangesonly

by integral increments-thatis, by one birth or death at a time.Supposethat the population changesonly by the occurrenceof births and

deaths-thereis no immigration or emigration from outsidethe country or envi-ronment under consideration.It is customary to track the growth or declineof apopulation in terms of its birth rate and death rate functions definedas follows:). fJ (t) is the number of births perunit of populationperunit of time at time t;. 8(t) is the number of deaths perunit of populationperunit of time at time t.)

Then the numbers of births and deaths that occurduring the time interval

[t, t + \037t] is given (approximately)by)

births: fJ (t) .P(t) . \037 t ,) deaths: 8(t) .P(t) . \037 t .)

Hencethe change \037P in the populationduring the time interval [t, t + \037t] oflength \037t is

\037P = {births}- {deaths}\037 fJ(t) .P(t) . \037t-8(t).P(t) .

\037t,)))

Example1)

1.7PopulationModels 75)

so)\037P-\037 [,8(t)-8(t)]P(t).

\037t

The error in this approximationshould approachzero as \037t -+ 0,so-takingthe limit-weget the differentialequation)

\037)

dPdt

= (,8-8)P,) (1))

in which we write ,8 = ,8(t),8 = 8(t),and P = P(t) for brevity. Equation (1)isthe generalpopulationequation.If ,8 and 8 are constant, Eq. (1)reducesto the

natural growth equation with k = ,8-8.But it alsoincludesthe possibilitythat f3

and 8 are variable functions of t. The birth and death rates neednot be known in

advance; they may well dependon the unknown function P(t).)

Supposethat an alligator populationnumbers 100initially, and that its death rate is8 = 0 (sonone of the alligators is dying).If the birth rate is ,8 = (0.0005)P-andthus increasesas the populationdoes-thenEq.(1)gives the initial value problem)

dP-= (0.0005)P2, P(O)= 100

dt)

(with t in years).Then upon separating the variableswe get)

f ;2dP = f (0.0005)dt;

1-P

= (0.0005)t+C.

Substitution of t = 0,P = 100gives C = -1/100,and then we readilysolvefor)

2000P(t) = .20- t)

For instance, P(10)= 2000/10= 200,so after 10years the alligatorpopu-lation has doubled.But we seethat P -+ +00as t -+ 20,so a real \"population

explosion\"occursin 20years.Indeed,the directionfieldand solution curvesshown

in Fig.1.7.1indicate that a population explosionalways occurs,whatever the sizeof the (positive) initial population P(0) = Po. In particular, it appearsthat the

populationalwaysbecomesunbounded in afinite periodof time. .)

500)

400)

300Q..

200(0,100)

100)

o 10 20 30 40 50t)

FIGURE1.7.1.Slopefield and solution curves for the equationdP/dt = (0.0005)p2 in Example 1.)))


Example2)

BoundedPopulationsandtheLogisticEquationIn situationsasdiverseas the human populationof a nation and a fruit fly populationin a closedcontainer, it is often observedthat the birth rate decreasesas the popu-lation itself increases.The reasonsmay range from increasedscientificor cultural

sophistication to a limited food supply. Suppose,for example,that the birth ratefJ is a lineardecreasingfunction of the population sizeP,so that fJ == fJo

-fJ 1 P,

where fJo and fJI are positive constants.If the death rate 8 == 80 remains constant,then Eq.(1)takes the form)

dPdt

== (fJo - fJI P -80)P;)

that is,\037)

dP 2-==aP-bPdt ') (2))

where a == fJo -80 and b == fJI.If the coefficientsa and b are both positive, then Eq. (2)is calledthe logistic

equation.For the purposeof relating the behavior of the population P(t) to thevalues of the parameters in the equation, it is useful to rewrite the logisticequationin the form)

\037)

dPdt

==kP(M- P),) (3))

where k ==bandM ==a/b are constants.)

\"....\" \"

In Example4 of Section1.3we exploredgraphically a population that is modeledby the logisticequation)

dP-==0.0004P(150-P) ==0.06P-0.0004p2.dt)

(4))

To solve this differentialequation symbolically,we separatethe variablesand inte-grate.We get)

J 5\037

= J0.0004dt,P(1 -P)

1\0370 J ( \037

+ 1501_ P) dP = J0.0004dt [partialfractions],

In IPI-In1150-PI==0.06t+C,P

==:f:eceO.06t == BeO.06t [where B ==:f:ec].150- P)

If we substitute t ==0 and P == Po i= 150into this last equation, we find that

B == Po/(150- Po).Hence)

P150- P)

PoeO.06t150- Po)

Finally, this equation is easyto solve for the population)

150PoP(t) ==

Po+ (150- Po)e-0.06t)(5))))

p)

300

240

180150120

6020)

t100)25) 50) 75)

FIGURE1.7.2.Typicalsolutioncurves for the logistic equationP' == O.06P-O.0004p2.)

p)

M)

M/2)P= M/2)

FIGURE1.7.3.Typicalsolutioncurves for the logistic equationP' == kP(M- P).Each solutioncurve that starts below the lineP == M/2 has an inflection pointon this line. (SeeProblem 34.))

Example3)


at time t in terms of the initial population Po == P(0).Figure 1.7.2showsa numberof solutioncurvescorrespondingto different valuesof the initial population rangingfrom Po ==20to Po ==300.Note that all thesesolution curves appear to approachthe horizontal line P == 150as an asymptote. Indeed,you should be able to seedirectly from Eq.(5)that limt\037oo P(t) == 150,whatever the initial value Po > O..)

LimitingPopulationsandCarryingCapacityThe finite limiting population noted in Example2 is characteristicof logisticpop-ulations. In Problem 32 we askyou to use the method of solution of Example2 toshow that the solutionof the logisticinitial value problem)

dPdt

==kP(M-P), P(O)== Po) (6))

IS)

MPP(t) == 0 .Po+ (M-Po)e-kMt) (7))

Actual animal populations are positive valued. If Po == M, then (7)reducesto the unchanging (constant-valued)\"equilibrium population\" P(t) = M. Other-wise, the behavior of a logisticpopulation dependson whether 0 < Po < M orPo > M. If 0 < Po < M, then we seefrom (6)and (7)that P'> 0 and)

MPoP(t) ==Po+ (M-Po)e-kMt)

MPo

Po+ {pos.number})

MPo<-== M.Po)

However,if Po > M, then we seefrom (6)and (7)that P'< 0 and)

MPoP(t) ==

Po+ (M-Po)e-kMt)

MPo MPo>-==M.Po+ {neg.number} Po)

In either case,the \"positive number\" or \"negative number\" in the denominator hasabsolutevalue lessthan Po and-becauseof the exponentialfactor-approaches0as t \037 +00.It follows that)

MPolim P(t) == ==M.

t\037+oo Po+ 0)(8))

Thus a population that satisfies the logisticequation doesnot grow without

bound like a naturally growing population modeledby the exponentialequationP' == k P. Instead,it approachesthe finite limiting populationM as t --+ +00.As illustrated by the typical logisticsolution curves in Fig.1.7.3,the populationP(t) steadily increasesand approachesM from belowif 0 < Po < M, but steadilydecreasesand approachesM from above if Po > M. SometimesM is called the

carryingcapacityof the environment, consideringit to bethe maximum populationthat the environment can supporton a long-termbasis.)

Supposethat in 1885the population of a certain country was 50 million and wasgrowing at the rate of 750,000peopleper year at that time. Supposealso that in

1940its population was 100million and was then growing at the rate of 1 million

peryear. Assume that this populationsatisfiesthe logisticequation.Determineboth

the limiting population M and the predictedpopulation for the year 2000.)))


Solution We substitute the two given pairsof data in Eq.(3)and find that)

Example4)

Example5)

0.75==50k(M-50), 1.00== 100k(M- 100).)

We solve simultaneouslyfor M ==200and k ==0.0001.Thus the limiting popula-tion of the country in question is 200million. With thesevalues of M and k, andwith t == 0 correspondingto the year 1940(in which Po == 100),we find that-accordingto Eq.(7)-thepopulation in the year 2000will be)

100.200P(60)== 100+ (200- 100)e-(O.OOOl)(200)(60)

,)

about 153.7million people.) .)

HistoricalNote)

The logisticequation was introduced (around 1840)by the Belgianmathematicianand demographerP. F. Verhulst as a possiblemodel for human populationgrowth.In the next two exampleswe comparenatural growth and logisticmodel fits to the

19th-centuryU.S.population censusdata, then compareprojectionsfor the 20thcentury.)

The U.S.population in 1800was 5.308million and in 1900was 76.212million. Ifwe take Po ==5.308(with t ==0 in 1800)in the natural growth model P(t) == Poert

and substitute t == 100,P ==76.212,we find that)

76.212==5.308eIOOr,)

1 76.212so r ==-In \037 0.026643.100 5.308)

Thus our natural growth modelfor the U.S.populationduring the 19thcentury)IS)

P(t) == (5.308)e(O.026643)t) (9))

(with t in years and P in millions).BecauseeO.026643\037 1.02700,the averagepopu-lation growth between1800and 1900was about 2.7%peryear. .)

The U.S.population in 1850was 23.192million. If we take Po == 5.308and sub-stitute the data pairst ==50,P ==23.192(for 1850)and t == 100,P ==76.212(for1900)in the logisticmodelformula in Eq.(7),we get the two equations)

(5.308)M ==23.1925.308+ (M-5.308)e-50kM '

(5.308)M ==762125.308+ (M-5.308)e-IOOkM.)

(10))

in the two unknowns k and M. Nonlinear systemslike this ordinarily are solvednumerically using an appropriate computer system. But with the right algebraictrick (Problem36 in this section)the equations in (10)can be solved manually fork ==0.000167716,M == 188.121.Substitution of thesevalues in Eq.(7)yieldsthe

logisticmodel)998.546

P(t) == .5.308+ (182.813)e-(O.031551)t)( 11))))

Error)

40%)

20%) Exponential)

-40%)

FIGURE1.7.5.Percentageerrors in the exponential and

logistic population models for1800-1950.)


The table in Fig.1.7.4comparesthe actual 1800-1990U.S.censuspopula-tion figures with those predictedby the exponential growth model in (9)and the

logisticmodel in (11).Both agreewell with the 19th-centuryfigures. But the ex-ponential model diverges appreciably from the censusdata in the early decadesof the 20th century, whereas the logisticmodel remains accurate until 1940.Bythe end of the 20th century the exponential model vastly overestimatesthe actualU.S.population-predictingover a billion in the year 2000-whereasthe logisticmodelsomewhatunderestimatesit.)

.:.......';.......:...);\037c\037\037Q\037:......:..........>......:..]\037\037DQJ.1\037D.tial

\302\245\037at;;\\:.\\:i.;\037\037\037\037:illftR\037>,:;)

:,;;;;.cii:Mfjeleli)

ExponentialError)

Logistic LogisticModel Error

5.308 0.0007.202 0.0389.735 -0.097

13.095 -0.23417.501 -0.43723.192 0.00030.405 1.03839.326 -0.76850.034 0.15562.435 0.54576.213 -0.00190.834 1.394105.612 0.410119.834 3.369132.886 -0.721144.354 6.972154.052 25.271161.990 41.312168.316 58.226173.252 76.458177.038 104.384)

180018101820183018401850186018701880189019001910192019301940195019601970198019902000)

5.3087.2409.63812.86117.06423.19231.44338.55850.18962.98076.21292.228106.022123.203132.165151.326179.323203.302226.542248.710281.422)

5.3086.9299.04411.80515.40920.11326.25334.26844.73058.38776.21299.479

129.849169.492221.237288.780376.943492.023642.236838.308

1094.240)

0.0000.3110.5941.0561.6553.0795.1904.2905.4594.5930.000

-7.251-23.827-46.289-89.072

-137.454-197.620-288.721-415.694-589.598-812.818)

FIGURE1.7.4.Comparisonof exponential growth and logisticmodelswith U.S.censuspopulations (in millions).)

The two modelsare comparedin Fig.1.7.5,where plots of their respectiveerrors-asa percentageof the actual population-areshown for the 1800-1950period.We seethat the logisticmodel tracks the actual populationreasonablywell

throughout this ISO-yearperiod. However, the exponential error is considerablylarger during the 19thcentury and literally goesoff the chart during the first half ofthe 20th century.

In orderto measure the extent to which a given model fits actualdata, it iscus-tomary to define the averageerror(in the model)as the squareroot of the averageof the squaresof the individual errors(the latter appearing in the fourth and sixth

columns of the table in Fig.1.7.4).Usingonly the 1800-1900data, this definition

gives 3.162for the averageerror in the exponentialmodel,while the averageerrorin the logisticmodel is only 0.452.Consequently,even in 1900we might well have

anticipated that the logisticmodel would predictthe U.S.populationgrowth duringthe 20th century more accurately than the exponentialmodel. .)))


The moral of Examples4 and 5 is simply that one shouldnot expecttoo muchof modelsthat are basedon severely limited information (suchas just a pair of datapoints). Much of the scienceof statisticsis devoted to the analysis of large \"data

sets\" to formulate useful (and perhapsreliable)mathematicalmodels.)

MoreApplicationsof theLogisticEquationWe next describesomesituations that illustrate the varied circumstancesin which

the logisticequation is a satisfactorymathematicalmodel.)

1.Limited environment situation. A certain environment can support a popula-tion of at most M individuals. It is then reasonableto expectthe growth ratefJ- 8 (the combinedbirth and death rates)to be proportional to M -P,be-

causewe may think of M - P as the potential for further expansion.ThenfJ- 8 ==k(M -P),so that

dPdt

== (fJ -8)P ==kP(M- P).)

The classicexampleof a limited environment situation is a fruit fly populationin a closedcontainer.

2.Competitionsituation. If the birth rate fJ is constant but the death rate 8 isproportional to P,so that 8 ==aP,then)

dP-== (fJ -aP)P==kP(M-P).dt)

This might be a reasonableworking hypothesis in a study of a cannibalisticpopulation, in which all deaths result from chance encounters between indi-viduals. Of course,competitionbetween individuals is not usually sodeadly,nor its effects so immediate and decisive.

3.Joint proportionsituation. Let pet) denote the number of individuals in aconstant-sizesusceptiblepopulation M who are infected with a certain con-tagious and incurable disease.The diseaseis spreadby chance encounters.Then P'(t) should beproportional to the product of the number P of individ-uals having the diseaseand the number M - P of those not having it, andtherefore dPjdt == kP(M- P).Again we discoverthat the mathematicalmodelis the logisticequation. The mathematicaldescriptionof the spreadofa rumor in a populationof M individuals is identical.)

Example6)..... .... ..........,.. ..............

Supposethat at time t ==0,10thousand peoplein a city with population M == 100thousand peoplehave heard a certain rumor. After 1 week the number pet) ofthose who have heard it has increasedto P(l)==20thousand. Assuming that pet)satisfiesa logisticequation, when will 80% of the city'spopulation have heard therumor?)

Solution Substituting Po == 10and M == 100(thousand) in Eq.(7),we get

1000pet) == .10+90e-lOOkt

Then substitution of t == 1,P ==20gives the equation)

(12))

20==1000

10+90e-100k)))


that is readily solvedfor

e-100k ==\037,

so k == 160In\037

\037 0.008109.With pet) ==80,Eq.(12)takes the form)

100080 == 10+90e-100kt

'which we solve for e-100kt==

31

6. It follows that 80% of the populationhas heardthe rumor when)

In36t==

lOOk

thus after about 4 weeksand 3 days.)

In369 \037 4.42,

In -4)

.)

DoomsdayversusExtinctionConsidera population pet) of unsophisticatedanimals in which femalesrely solelyon chanceencounters to meet males for reproductivepurposes.It is reasonableto

expectsuch encounters to occurat a rate that is proportional to the productof the

number P/2 of males and the number P/2 of females, henceat a rate proportionalto p2. We therefore assumethat births occurat the rate kP2 (perunit time, with

k constant). The birth rate (births/time/population)is then given by f3 == kP. Ifthe death rate 8 is constant, then the general population equation in (1)yields the

differentialequation)

\037)

dP 2-==kP -8P ==kP(P-M)dt)

(13))

(where M ==8/k >0)as a mathematicalmodel of the population.Note that the right-hand sidein Eq.(13)is the negativeof the right-hand side

in the logisticequation in (3).We will seethat the constant M is now a thresholdpopulation,with the way the populationbehaves in the future dependingcriticallyon whether the initial population Po is lessthan or greater than M.)

Example7) Consideran animal population pet) that is modeledby the equation

dP 2-==0.0004P(P- 150)==0.0004P-0.06P.dt

We want to find pet) if (a) P(O)==200;(b) P(O)== 100.)

(14))

Solution To solve the equation in (14),we separate the variablesand integrate.We get

fdP

==f 0.0004dt,PcP- 150))

-\037f (

\037 - 1

) dP ==f 0.0004dt[partialfractions],150 P P - 150)

In IPI -InIP- 1501==-0.06t+ C,)

P==::i:eCe-0.06t==Be-0.06t

P - 150) [whereB ==::i:ec].) (15))))


p)

M)

FIGURE1.7.6.Typicalsolutioncurves for the explosion/extinctionequation P'= kP(P-M).)

_ Problems)

(a) Substitution of t ==0 and P ==200into (15)gives B ==4. With this value of Bwe solve Eq.(15)for)

600e-O.06t

pet) == .4e-O.06t -1)

(16))

Note that, as t increasesand approachesT == In(4)/0.06\037 23.105,the positivedenominatoron the right in (16)decreasesand approachesO.Consequentlypet) \037

+00as t --+T-.This is a doomsdaysituation-areal populationexplosion.(b) Substitution of t ==0 and P == 100into (15)gives B ==-2.With this value ofB we solve Eq.(15)for)

300e-O.06t 300Pet ) --2e-O.06t + 1-

2 +eO.06t .) (17))

Note that, as t increaseswithout bound, the positive denominator on the right in

(16)approaches+00.Consequently,pet) --+0 as t --++00.This is an (eventual)extinctionsituation. .)

t)

Thus the population in Example7 either explodesor is an endangeredspeciesthreatened with extinction, dependingon whether or not its initial sizeexceedsthethreshold population M == 150.An approximation to this phenomenon is some-times observedwith animal populations, such as the alligator population in certainareas of the southern United States.

Figure 1.7.6showstypical solution curves that illustrate the two possibilitiesfor a population pet) satisfying Eq. (13).If Po == M (exactly!),then the popula-tion remains constant. However, this equilibrium situation is very unstable. If PoexceedsM (even slightly), then pet) rapidly increaseswithout bound, whereas ifthe initial (positive)population is lessthan M (howeverslightly), then it decreases(more gradually) toward zero as t --++00.SeeProblem 33.)

Separatevariablesand usepartial fractionsto solvethe initial

value problems in Problems1-8.Useeither the exactsolutionora computer-generatedslopefieldto sketch the graphs ofsev-eralsolutions of the given differential equation, and highlightthe indicated particular solution.)

dx1.dt

= x -x2, X(0)= 2

dx3.dt

= 1-x2, x(O)= 3

dx5. -=3x(5-x),x(0)=8dtdx6.-= 3x(x-5),x(O)= 2dtdx7.dt

= 4x(7-x),x(O)= 11dx8.dt = 7x(x - 13),x(O)= 17)

dx2.-= lOx-x2,x(0)= 1dtdx4. -= 9-4x2, x(O)= 0dt)

the population numbers 100rabbits and is increasing at

the rate of 20 rabbits per month. How many rabbits will

there beoneyear later?

10.Supposethat the fish population pet) in a lake is attacked

by a diseaseat time t = 0, with the result that the fish

ceaseto reproduce(sothat the birth rate is f3= 0) and the

death rate 8 (deaths per weekper fish) is thereafter propor-tional to Ij-JP.If there were initially 900fish in the lakeand 441were left after 6 weeks,how long did it take all

the fish in the lake to die?

11.Supposethat when a certain lake is stockedwith fish, the

birth and death rates f3 and 8 are both inversely propor-tional to -JP.(a) Show that)

pet) =Okt + \037r')

9.Thetime rate of change of a rabbit population P is pro-portional to the square root of P.At time t = 0 (months))

where k is a constant. (b) If Po= 100and after 6 months

there are169fish in the lake, how many will there beafter

1 year?)))

12.The time rate of change of an alligator population P in

a swamp is proportional to the square of P.The swampcontained a dozenalligators in 1988,two dozenin 1998.When will there be four dozenalligators in the swamp?What happens thereafter?

13.Considera prolific breedof rabbits whose birth and death

rates, f3 and 8, areeachproportional to the rabbit popula-tion P = pet),with f3 > 8. (a) Show that)

PoP (t ) - k constant.-1- k Pot'

Note that P (t) \037 +00as t \037 1/(kPo).This is dooms-day. (b) Supposethat Po = 6 and that there are ninerabbits after ten months. When doesdoomsday occur?

14.Repeatpart (a) of Problem 13 in the case f3 < 8. Whatnow happens to the rabbit population in the long run?

15.Considera population P (t) satisfying the logistic equa-tion dP/dt = aP- bP2, where B = aP is the time rateat which births occurand D = bp2 is the rate at whichdeaths occur.If the initial population is P(0) = Po, and

Bo births per month and Do deaths per month areoccur-ring at time t = 0, show that the limiting population isM = BoPo/Do.

16.Considera rabbit population P (t) satisfying the logisticequation as in Problem 15.If the initial population is 120rabbits and there are 8 births per month and 6 deaths permonth occurring at time t = 0, how many months doesit

take for P (t) to reach95%of the limiting population M?17.Considera rabbit population pet) satisfying the logistic

equation as in Problem 15.If the initial population is 240rabbits and there are9 births per month and 12deaths permonth occurring at time t = 0, how many months doesit

take for P (t) to reach105%of the limiting population M?18.Considera population P (t) satisfying the extinction-

explosionequation dP/dt = ap2-bP,where B = ap2is the time rate at which births occurand D = bP isthe rate at which deaths occur. If the initial populationis P(0) = Po and Bo births per month and Do deaths permonth areoccurring at time t = 0, show that the threshold

population is M = DoPolBo.19.Consideran alligator population P (t) satisfying the

extinction/explosionequation as in Problem 18.If the ini-tial population is 100alligators and there are10births permonth and 9 deaths per months occurring at time t = 0,how many months doesit take for P (t) to reach10timesthe threshold population M?

20. Consideran alligator population P (t) satisfying the

extinction/ explosionequation as in Problem 18.If the ini-tial population is 110alligators and there are11births permonth and 12deaths per month occurring at time t = 0,how many months doesit take for P (t) to reach10%ofthe threshold population M?

21.Supposethat the population P(t) ofa country satisfiesthedifferential equation dP/dt = k P(200- P) with k con-stant. Its population in 1940was 100million and was then)


growing at the rate of 1million per year. Predict this coun-try's population for the year 2000.

22.Supposethat at time t = 0, half of a \"logistic\" popula-tion of 100,000personshave heard a certain rumor, and

that the number of those who have heard it is then increas-ing at the rate of 1000personsper day. How long will it

take for this rumor to spreadto 80%of the population?(Suggestion: Find the value ofk by substituting P(O)and

P'(0) in the logisticequation, Eq. (3).)23.As the salt KN03 dissolvesin methanol, the number x(t)

of grams of the salt in a solution after t secondssatisfiesthe differential equation dx/dt = 0.8x-0.004x2.(a) What is the maximum amount of the salt that will ever

dissolvein the methanol?

(b) If x = 50 when t = 0, how long will it take for an

additional 50g of salt to dissolve?24. Supposethat a community contains 15,000peoplewho

aresusceptibleto Michaud's syndrome, a contagious dis-ease.At time t = 0 the number N (t) ofpeoplewho have

developedMichaud's syndrome is 5000and is increasingat the rate of 500per day. Assume that N '

(t) is propor-tional to the product of the numbers of those who have

caught the diseaseand of those who have not. How longwill it take for another 5000peopleto developMichaud's

syndrome?25.Thedata in the table in Fig. 1.7.7aregiven for a certain

population P (t) that satisfies the logistic equation in (3).(a) What is the limiting population M? (Suggestion: Usethe approximation)

I pet + h) - P(t -h)P (t) \037

2h)

with h = 1 to estimate the values of P'(t) when P25.00and when P = 47.54.Then substitute these valuesin the logistic equation and solvefor k and M.) (b) Usethe values ofk and M found in part (a) to determine whenP = 75. (Suggestion: Take t = 0 to correspondto the

year 1925.))

Year P (millions)

1924 24.631925 25.001926 25.38)

197419751976)

47.0447.5448.04)

FIGURE1.7.7.Population data for Problem 25.

26.A population P (t) of small rodents has birth rate f3

(O.OOI)P(births per month per rodent) and constant deathrate 8. If P(0) = 100and P'(0) = 8, how long (in)))

months) will it take this population to double to 200ro-dents? (Suggestion: First find the value of8.)

27.Consideran animal population P (t) with constant deathrate 8 = 0.01(deaths per animal per month) and with

birth rate f3 proportional to P.Supposethat P(0) = 200and P'(O)= 2.(a) When is P = 1000?(h) When doesdoomsday occur?

28.Supposethat the number x(t)(with t in months) ofalliga-tors in a swamp satisfiesthe differential equation dx/dt =0.0001x2 -O.Olx.(a) If initially there are25 alligators in the swamp, solve

this differential equation to determine what happensto the alligator population in the long run.

(h) Repeatpart (a),exceptwith 150alligators initially.

29.During the periodfrom 1790to 1930,the U.S.populationpet) (t in years) grew from 3.9million to 123.2million.Throughout this period,P (t) remained closeto the solu-tion of the initial value problem

dPdt

= 0.03135P-0.0001489p2, P(O)= 3.9.(a) What 1930population doesthis logistic equation pre-

dict?(h) What limiting population doesit predict?(c) Hasthis logistic equation continued since1930to ac-

curately model the U.S.population?

[Thisproblem is basedon a computation by Verhulst, whoin 1845usedthe 1790-1840U.S.population data to pre-dict accurately the U.S.population through the year 1930(long after his own death, ofcourse).]

30.A tumor may be regarded as a population of multiplyingcells.It is found empirically that the \"birth rate\" of thecellsin a tumor decreasesexponentially with time, so that

f3(t) = f3oe-at(where a and f30 are positive constants),

and hence)

dPdt

= f3oe-at P, P(O)= Po.

Solvethis initial value problem for

pet)= Poexp(\037o

(1-e-at

)).

Observethat P (t) approachesthe finite limiting popula-tion Poexp(f3o/a) as t \037 +00.

31.For the tumor of Problem 30, supposethat at time t = 0there are Po = 106 cellsand that P (t) is then increasingat the rate of 3 x 105 cellsper month. After 6 months thetumor has doubled (in sizeand in number ofcells).Solvenumerically for a, and then find the limiting population ofthe tumor.

32.Derivethe solution

pet)= MPo

Po+ (M - Po)e-kMt

of the logistic initial value problem P' = kP(M- P),P(0)= Po.Make it clearhow your derivation dependsonwhether 0 < Po < M or Po > M.)

33.(a) Derivethe solution)

MPopet)=

Po + (M - Po)ekMt)

of the extinction-explosion initial value problem P'=kP(P-M),P(O)= Po.

(h) How doesthe behavior of P (t) as t increasesdependon whether 0 < Po < M or Po > M?

34. If P (t) satisfiesthe logistic equation in (3),usethe chainrule to show that)

PII(t) = 2k2PcP- 4M)(p-M).)

Concludethat P\" > 0 if 0 0if P > M. In particular, it follows that any solutioncurve that crossesthe line P = 4M has an inflection pointwhere it crossesthat line, and therefore resemblesoneofthe lowerS-shapedcurves in Fig. 1.7.3.

35.Considertwo population functions PI(t) and P2(t), bothof which satisfy the logisticequation with the same limit-

ing population M but with different values k I and k2 of theconstant k in Eq.(3).Assume that k I < k2. Which pop-ulation approachesM the most rapidly? You can reasongeometrically by examining slopefields (especiallyif ap-propriate software is available), symbolically by analyzingthe solution given in Eq. (7), or numerically by substitut-

ing successivevalues of t.

36.Tosolvethe two equations in (10)for the values of k and

M, begin by solving the first equation for the quantityx = e-50kM and the secondequation for x2 = e-IOOkM .Upon equating the two resulting expressionsfor x2 in

terms of M, you get an equation that is readily solvedforM. With M now known, either of the original equationsis readily solvedfor k. This technique can beused to \"fit\"

the logisticequation to any three population values Po,PI,and P2 correspondingto equally spacedtimes to = 0, tI,and t2 = 2tI .

37.Usethe method of Problem 36 to fit the logisticequationto the actual U.S.population data (Fig. 1.7.4)for the years1850,1900,and 1950.Solvethe resulting logistic equa-tion and comparethe predictedand actual populations forthe years 1990and 2000.

38.Fit the logistic equation to the actual U.S.population data

(Fig. 1.7.4)for the years 1900,1930,and 1960.Solvethe

resulting logisticequation, then comparethe predictedandactual populations for the years 1980,1990,and 2000.

39.Birth and death rates of animal populations typically arenot constant; instead, they vary periodically with the pas-sageof seasons.Find P (t) if the population P satisfiesthe differential equation)

dP-= (k + bcos2Jrt)P,dt)))

1.8Acceleration-VelocityModels 85)

where t is in years and k and b arepositive constants.Thusthe growth-rate function r(t) = k + b cos21ft varies pe-riodically about its mean value k. Construct a graph that

contrasts the growth of this population with one that has)

the sameinitial value Po but satisfiesthe natural growth

equation P'= kP (sameconstant k). How would the two

populations compareafter the passageof many years?)

_ Accel\037ration\037Yelocitr...\037\037cl.\037l\037)

In Section1.2we discussedverticalmotion of a massm near the surfaceof the earth

under the influenceof constant gravitational acceleration.If we neglect any effectsof air resistance,then Newton'ssecondlaw (F ==ma)impliesthat the velocityv ofthe massm satisfies the equation)

dvm-== \037 Gdt ') (1))

where FG == -mgis the (downward-directed)force of gravity, where the gravita-tional accelerationis g \037 9.8m/s2 (in mks units; g \037 32 ft/s 2 in fps units).)

Example1)

v _ \037 .. \037 \"\"_ \037 ........\" ..... ...\037 _................ \"',,\"n\" ..'A\" 'n\".', \037,.\"....\" ...\" 'n_\" \"\" \"\"...... ...... 'A'''''' ............,.\".............. ...... ................ \037 .......\037...\"\" '\"\" ...........

Supposethat a crossbowbolt is shot straight upward from the ground (Yo ==0) with

initial velocity Vo ==49 (m/s).Then Eq.(1)with g ==9.8gives)

dv-==-9.8, so vet) ==-(9.8)t+ Vo ==-(9.8)t+49.dt)

Hencethe bolt'sheight function yet) is given by)

y(t) = f [-(9.8)t+49]dt = -(4.9)t2 +49t + Yo = -(4.9)t2 +49t.)

The bolt reachesits maximum height when v == -(9.8)t+49 == 0,hence when

t ==5 (s).Thus its maximum height is)

Ymax ==y(5) ==-(4.9)(52) + (49)(5)== 122.5(m).)

The bolt returns to the ground when y == -(4.9)t(t- 10)== 0,and thus after 10secondsaloft. .)

Now we want to take account of air resistancein a problem like Example1.The force FR exertedby air resistanceon the moving massm must be addedin

Eq.(1),so now)

dvm-== FG + FR.dt)

(2))

Newton showedin his PrincipiaMathematica that certain simplephysicalassump-tions imply that FR is proportional to the squareof the velocity: FR == kv 2. But

empirical investigationsindicate that the actual dependenceof air resistanceon ve-locity can be quite complicated.For many purposesit sufficesto assumethat)

FR ==kv P,)

where 1 < p < 2 and the value of k dependson the sizeand shapeof the body,aswell as the density and viscosityof the air. Generallyspeaking,p == 1 for relatively)))

I

Y

FR)

m)

(Note:FR acts upward whenthe body is falling.)m

INet force F =FR + FG)

FG)

Ground level)

FIGURE1.8.1.Vertical motionwith air resistance.)

low speedsand p = 2 for high speeds,whereas1 <p <2 for intermediatespeeds.But how slow\"low speed\"and how fast \"high speed\"are dependon the samefactorsthat determine the value of the coefficientk.

Thus air resistanceis a complicatedphysical phenomenon. But the simplify-ing assumptionthat FR isexactlyof the form given here,with either p = 1or p = 2,yieldsa tractable mathematical modelthat exhibits the most important qualitativefeatures of motion with resistance.)

ResistanceProportionaltoVelocityLet us first considerthe vertical motion of a body with massm near the surfaceof the earth, subjectto two forces:a downwardgravitational force FG and a forceFR of air resistancethat is proportional to velocity (sothat p = 1)and of coursedirectedoppositethe direction of motion of the body. If we set up a coordinatesystemwith the positive y-direction upward and with y = 0 at ground level, then

FG = -mgand)

\037) (3))FR = -kv,)

where k is a positive constant and v = dyjdt is the velocity of the body.Note that

the minus signin Eq. (3)makesFR positive (an upward force) if the body is falling

(v is negative)and makesFR negative(a downwardforce) if the body is rising (v ispositive).As indicated in Fig.1.8.1,the net force acting on the body is then)

F = FR + FG = -kv -mg,)

and Newton'slaw of motion F = m(dv/dt) yieldsthe equation)

dvm-= -kv -mg.dt)

Thus)

\037)

dv-= -pv-g,dt)

(4))

where p = kjm >O.You should verify for yourself that if the positivey-axisweredirecteddownward, then Eq.(4)would take the form dv/dt = -pv+g.

Equation (4)is a separablefirst-orderdifferentialequation, and its solution is)

v(t) =(v

o +\037 )e-pt -

\037

.) (5))

Here,Va = v (0)is the initial velocityof the body.Note that)

v, = lim vet) = -g.t\037oo P)

(6))

Thus the speedof a body falling with air resistancedoesnot increaseindefinitely;instead,it approachesafinite limiting speed,or terminalspeed,)

g mgIv I

----,- - .p k)

(7))))


This fact is what makesa parachute a practical invention; it even helpsexplainthe occasionalsurvival of peoplewho fall without parachutesfrom high-flying air-planes.

We now rewriteEq.(5)in the form)

dy _ -pt-- (va - v,)e + V,.dt)

(8))

Integrationgives)1 t

yet) = --(vo-v,)e-P + v,t +C.p

We substitute 0 for t and let Yo = y(O)denote the initial height of the body. Thuswe find that C = Yo + (va - v, )Ip,and so)

1 ptyet) = Yo + V,t + -(va-v,)(l-e- ).p)

(9))

Equations (8)and (9)give the velocity v and height y of a body moving ver-tically under the influence of gravity and air resistance.The formulas dependonthe initial height Yo of the body, its initial velocity Va, and the drag coefficientp,the constant such that the accelerationdue to air resistanceis aR = -pV.The two

equations alsoinvolve the terminal velocity v, defined in Eq.(6).For a persondescendingwith the aid of a parachute, a typical value of p is

1.5,which correspondsto a terminal speedof lv, I\037 21.3ftls, or about 14.5mi/h.

With an unbuttoned overcoatflapping in the wind in placeof a parachute,an unlucky

skydiver might increasep to perhapsas much as0.5,which gives a terminal speedof

I v, I\037 65 ftl s, about 44mi/h.SeeProblems10and 11for someparachute-jump

computations.)

Example2)........... .......... ...... ........... ................. ........ . ..... ........... ........................................................

We again considera bolt shot straight upward with initial velocity Va = 49 m/sfrom a crossbowat ground level. But now we take air resistanceinto account,with

p = 0.04in Eq. (4). We ask how the resulting maximum height and time aloftcomparewith the values found in Example1.)

Solution We substitute Yo = 0,Va = 49, and v, = -g/p= -245in Eqs.(5)and (9),andobtain)

vet) = 294e-t/25 -245,

yet) = 7350-245t -7350e-t/25.)

To find the time requiredfor the bolt to reach its maximum height (when V = 0),we solve the equation

vet) = 294e-t/25 -245 = 0for tm = 25 In (294/245)\037 4.558(s).Its maximum height is then Ymax = v(t m) \037

108.280meters (asopposedto 122.5meters without air resistance).To find when

the bolt strikesthe ground, we must solve the equation)

yet) = 7350-245t -7350e-t/25 = O.)

UsingNewton'smethod, we can begin with the initial guessto = 10and carry out

the iteration tn+l = tn- y(tn)ly' (tn ) to generate successiveapproximationsto the

root.Or we can simply use the Solvecommandon a calculatoror computer.We)))

find that the bolt is in the air for tf \037 9.411seconds(as opposedto 10secondswithout air resistance).It hits the ground with a reducedspeedof IV(tf)1 \037 43.227m/s (asopposedto its initial velocity of 49m/s).

Thus the effect of air resistanceis to decreasethe bolt'smaximum height, thetotal time spent aloft, and its final impact speed.Note alsothat the bolt now spendsmore time in descent(tf - tm \037 4.853s) than in ascent(tm \037 4.558s). .)

ResistanceProportionaltoSquareofVelocityNow we assumethat the force of air resistanceis proportional to the squareof thevelocity:)

\037) FR = ::I::kv2,) (10))

with k > O. The choiceof signshere dependson the direction of motion, whichthe force of resistancealways opposes.Taking the positive y-direction as upward,FR < 0 for upward motion (when v > 0) while FR > 0 for downward motion(when v < 0). Thus the signof FR is always oppositethat of v, so we can rewriteEq.(10)as)

FR = -kvlvl.) (10'))

Then Newton'ssecondlaw gives

dvm-= FG + FR = -mg- kvlvl;dt)

that is,)

dv

dt= -g- pvlvl,) (11))

where p = k/m > O. We must discussthe casesof upward and downward motion

separately.)

UPWARD MOTION: Supposethat a projectileis launched straight upward fromthe initial position Yo with initial velocity Vo > O. Then Eq.(11)with v > 0 givesthe differentialequation)

\037 \037

= -g-Pv2 = -g (1+:v2

) .

In Problem 13we ask you to make the substitution u = v-Jpig and apply thefamiliar integral)

(12))

f1

2 du = tan-1 u +cl+uto derive the projectile'svelocity function

V(f)=!!tan(C,-f,Jlii)with Cl = tan- 1

(v0/r) . (13)

Becausef tan u du = - InIcosu

I +C, a second integration (seeProblem 14)yieldsthe position function)

1 cos(C1- t-JfJi)

yet) = Yo + -Inp cosC

1)

(14))))

Example3)


DOWNWARD MOTION:Suppose that a projectileis launched (or dropped)straight downward from the initial position Yo with initial velocity Vo

< O. ThenEq.(11)with v <0 gives the differentialequation)

dv 2

(P 2

)dt =-g+pv=-g 1-g

V .) (15))

In Problem 15we ask you to make the substitution u = vv'pjg and apply the

integral)

f1

2 du = tanh- 1 u +cl-uto derive the projectile'svelocity function)

v(t) = f!tanh (C2- t..[jig)with) C2 = tanh- 1

(voID.) (16))

Becausef tanh u du = InIcoshu

I +C,another integration (Problem16)yieldsthe

position function)

1 cosh(C2- t y'Pg)yet) = Yo

-- Inp coshC2)

(17))

(Note the analogy betweenEqs. (16)and (17)and Eqs.(13)and (14)for upward

motion.)If Vo = 0,then C2 = 0,so vet) = -y'gjp tanh (t y'Pg).Because)

sinh x !(eX -e-X )lim tanh x = lim = lim 2 = 1,

x\037oo x\037oo coshx x\037oo !(eX +e-X))

it follows that in the caseof downward motion the body approachesthe terminal

speed)

lv, I= f!) (18))

(ascomparedwithI Vi I

= gjp in the caseof downward motion with linearresistancedescribedby Eq.(4)).)

\" OA ....... .. \037 '\" .\037

We consideroncemore a bolt shot straight upward with initial velocityVo = 49 mlsfrom a crossbowat ground level,as in Example2.But now we assumeair resistanceproportional to the squareof the velocity,with p = 0.0011in Eqs.(12)and (15).InProblems17and 18we askyou to verify the entries in the last line of the followingtable.)

j!i\037\";\\[\".!.\037.,.'.\037i\037;.'I;!f

,.,

..'.:.;.:.:.:::];.;.'.!J:,]trll!_\037lgl!.!.!;;\037;;.!;.iti.,;.i:\"i[.;i.;:j'Iil\037.<';.;.;.,:

11\037*iillfJiilj:iiilj!\037\037!lillllli!'il'ii!'il\037iiiiii\\II;I\0371I\037\\il\\i

0.0 122.5 10(0.04)v 108.28 9.41(0.0011)v2 108.47 9.41)

, ',--,,-'------- -- .,..c;::;,jA;seel1t

):']1\037i\037\037(sJ

54.564.61)

DescentTil11\037 (s)

54.854.80)

ImpactSpeed\037f\037s)

4943.2343.49)))


y)

12010080604020)

Withoutresistance)

1 2 3 4 5 6 7 8 9 10 t)

FIGURE1.8.2.Theheight functions in Example 1 (without air

resistance),Example 2 (with linear air resistance),and Example 3(with air resistanceproportional to the squareof the velocity) areall

plotted. Thegraphs of the latter two arevisually indistinguishable.)

Comparisonof the last two lines of data here indicateslittle difference-forthemotion of our crossbowbolt-betweenlinear air resistanceand air resistancepro-portional to the squareof the velocity. And in Fig.1.8.2,where the correspondingheight functions are graphed,the difference is hardly visible.However,the differ-encebetweenlinear and nonlinear resistancecan be significant in more complexsituations-suchas, for instance, the atmospheric reentry and descentof a spacevehicle. .)VariableGravitationalAcceleration)Unlessa projectilein verticalmotion remains in the immediatevicinity of the earth'ssurface, the gravitational acceleration acting on it is not constant. According toNewton'slaw of gravitation, the gravitational force of attraction betweentwo pointmassesM and m locatedat a distancer apart is given by)

\037)

GMmF= r2)(19))

where G is a certain empirical constant (G \037 6.6726x 10-11N.(m/kg)2in mks

units).The formula is alsovalid if eitheror both of the two massesare homogeneousspheres;in this case,the distancer is measuredbetweenthe centersof the spheres.

The following exampleis similar to Example2 in Section1.2,but now wetake account of lunar gravity.)

A lunar lander is free-falling toward the moon, and at an altitude of 53 kilometersabove the lunar surface its downwardvelocity is measuredat 1477kmIh. Its retro-rockets,when fired in free space,provide a decelerationof T = 4 mls2. At what

height above the lunar surface should the retrorocketsbe activatedto ensurea \"soft

touchdown\" (v = 0at impact)?)

Solution Let r(t) denote the lander'sdistancefrom the center of the moon at time t

(Fig.1.8.3).When we combinethe (constant) thrust acceleration T and the (neg-ative) lunar accelerationF/m = GM/r2 of Eq. (19),we get the (acceleration)differentialequation)

d2r GM-= T --,dt2 r2)(20))

where M = 7.35X 1022(kg) is the mass of the moon, which has a radius of

R = 1.74x 106 meters (or1740kIn, a little over a quarter of the earth'sradius).)))

FIGURE1.8.3.Thelunar lander

descendingto the surfaceof themoon.)


Noting that this second-orderdifferentialequationdoesnot involve the independentvariable t, we substitute)

drv =-,

dt)

d2r dv dv dr dv-----.--v-dt2 - dt

-dr dt

-dr)

(as in Eq. (36)of Section1.6)and obtain the first-orderequation)

dv GMv- = T --dr r2)

with the new independent variable r. Integration with respectto r now yields the

equation)

1 2 GM-v =Tr+-+C2 r)

(21))

that we can apply both before ignition (T = 0 ) and after ignition (T = 4).

Beforeignition:Substitution of T = 0 in (21)gives the equation)

1 2 GM-v =-+C12 r)(21a))

where the constant is given by Cl = v5/2 -GM/rowith)

kIn m 1hVo = -1477h x 1000

kmx 3600s)

14770m36 s)

and ro = (1.74x 106)+53,000= 1.793x 106 m (from the initial velocity-positionmeasurement).)

After ignition:Substitution of T = 4 and v = 0,r = R (at touchdown) into (21)gIves)

1 2 GM-v = 4r +-+C22 r)(21b))

where the constantC2 = -4R-GM/R isobtainedby substituting the valuesv = 0,r = R at touchdown.)

At the instant of ignition the lunar lander'sposition and velocitysatisfy both

(21a)and (21b).Therefore we can find its desiredheight h above the lunar surfaceat ignition by equating the right-hand sides in (21a)and (21b).This gives r =\037(Cl

-C2) = 1.78187x 106 and finally h = r-R = 41,870meters (that is,41.87kilometers-justover 26miles).Moreover, substitution of this value of r in (21a)gives the velocity v = -450mls at the instant of ignition. .)))


Ir

ml\037)r(t))

EscapeVelocityIn his novel From the Earth to the Moon (1865),JulesVerne raisedthe questionof the initial velocity necesaryfor a projectilefired from the surface of the earthto reach the moon.Similarly, we can ask what initial velocity Va is necessaryforthe projectileto escapefrom the earth altogether. This will be so if its velocityv = drldtremainspositive for all t >0,so it continuesforeverto moveaway fromthe earth. With r(t) denoting the projectile'sdistancefrom the earth'scenter at timet (Fig.1.8.4),we have the equation)

dv d2r GMdt

=dt2 = --;:2') (22))

FIGURE1.8.4.A mass m at agreat distancefrom the earth.)

similar to Eq. (20),but with T = 0 (no thrust) and with M = 5.975X 1024(kg)

denoting the massof the earth, which has an equatorial radius of R = 6.378X 106(m).Substitution of the chain rule expressiondvldt = v(dvldr) as in Example4gIves)

dv GMv- ---dr

-r2 .

Then integration of both sideswith respectto r yields)

1 2 GM-v =-+c.2 r)

Now v = Va and r = R when t = 0,so C = !v5-GMIR,and hencesolution forv2

gives)

2 2

(1 1

)v = va +2GM r-

R.) (23))

This implicit solutionof Eq.(22)determinesthe projectile'svelocity v asa function

of its distancer from the earth'scenter.In particular,)

2 2 2GMv > va

-R

')

sov will remainpositiveprovided that v5 >2GMIR.Therefore,the escapevelocityfrom the earth is given by)

Va =j2\037M

.) (24))

In Problem 27 we ask you to show that, if the projectile'sinitial velocity exceeds-J2GMIR, then r(t) \037 00 as t \037 00,so it does,indeed,\"escape\"from the

earth. With the given values of G and the earth'smassM and radius R, this givesVa \037 11,180(m/s)(about36,680ft/s, about 6.95mils,about 25,000mi/h).)

Remark:Equation (24)gives the escapevelocity for any other (spherical)planetary body when we use its massand radius. For instance, when we use themassM and radius R for the moon given in Example4, we find that escapevelocityfrom the lunar surface is Va \037 2375m/s.This is just over one-fifth of the escapevelocity from the earth'ssurface,a fact that greatly facilitates the return trip (\"Fromthe Moonto the Earth\.") .)))

11I1Problems)1.Theaccelerationof a Maserati is proportional to the dif-

ferencebetween 250kmjh and the velocity of this sportscar. If this machine can acceleratefrom rest to 100kmjhin lOs,how long will it take for the carto acceleratefromrest to 200kmjh?2.Supposethat a body moves through a resisting mediumwith resistanceproportional to its velocity v, so that

dvjdt = -kv. (a) Show that its velocity and positionat time t aregiven by

v(t) = voe-kt)

and)

xU)= xo+ (;) (1-e-kt).(b) Concludethat the body travels only a finite distance,and find that distance.

3.Supposethat a motorboat is moving at 40 ftj s when itsmotor suddenly quits, and that 10s later the boat hasslowedto 20 ftj s.Assume, as in Problem 2, that the re-sistanceit encounterswhile coasting is proportional to its

velocity. How far will the boat coastin all?4. Considera body that moves horizontally through a

medium whoseresistanceis proportional to the squareofthe velocity v, sothat dvjdt = -kv2.Show that

Vov(t) =

1+ vokt)

and that)1

x(t)= Xo +k

In(l + vokt).

Note that, in contrast with the result ofProblem2,x(t)--*+00as t --* +00.Which offers lessresistancewhen the

body is moving fairly slowly-themedium in this prob-lem or the one in Problem 2? Doesyour answer seemconsistent with the observedbehaviors ofx(t)as t --* oo?

5.Assuming resistanceproportional to the squareof the ve-locity (as in Problem 4), how far doesthe motorboat ofProblem 3 coastin the first minute after its motor quits?

6. Assume that a body moving with velocity v encountersresistanceof the form dvjdt = -kV 3/2.Show that

4vov(t) =

2(kt0JO+ 2))

and that

xU)= xo+ \0370JO (1_ 2

).

k kt0JO+2Concludethat under a

\037-powerresistancea body coasts

only a finite distancebeforecoming to a stop.7. Supposethat a carstarts from rest, its engineproviding an

accelerationof 10ftj S2,while air resistanceprovides0.1ft/ S2 of decelerationfor eachfoot per secondof the car'svelocity. (a) Find the car'smaximum possible(limiting)velocity. (b) Find how long it takes the carto attain 90%of its limiting velocity, and how far it travels while doingso.)


8.Rework both parts of Problem7, with the soledifference

that the decelerationdue to air resistancenow is (0.001)v2

ftj S2 when the car'svelocity is v feet per second.9.A motorboat weighs 32,000Ib and its motor provides a

thrust of 5000lb. Assume that the water resistanceis 100pounds for eachfootpersecondof the speedv of the boat.Then)

dv1000-= 5000- 100v.dt

If the boat starts from rest, what is the maximum velocitythat it can attain?

10.A woman bails out of an airplane at an altitude of 10,000ft, falls freely for 20 s, then opensher parachute. How

long will it take her to reachthe ground? Assume lin-

ear air resistancepv ftj s2, taking p = 0.15without the

parachute and p = 1.5with the parachute. (Suggestion:First determine her height abovethe ground and velocitywhen the parachute opens.)

11.According to a newspaperaccount,a paratrooper surviveda training jump from 1200ft when his parachute failed to

openbut provided someresistanceby flapping unopenedin the wind. Allegedly he hit the ground at 100milh after

falling for 8 s.Testthe accuracyof this account.(Sugges-tion: Find p in Eq.(4)by assuming a terminal velocityof 100mijh. Then calculatethe time required to fall 1200ft.)

12.It is proposedto disposeofnuclear wastes-indrums with

weight W = 640Ib and volume 8 ft 3-bydropping them

into the ocean(vo = 0). Theforceequation for a drum

falling through water is)

dvm-=-W+B+FR ,dt)

where the buoyant forceB is equal to the weight (at 62.5Ibjft

3) of the volume of water displacedby the drum

(Archimedes' principle) and FR is the forceof water re-sistance,found empirically to be 1 Ib for eachfoot persecondof the velocity of a drum. If the drums are likelyto burst upon an impact of more than 75 ftjs, what is themaximum depth to which they canbedroppedin the oceanwithout likelihood ofbursting?

13.Separatevariables in Eq.(12)and substitute u = v,Jpigto obtain the upward-motion velocity function given in

Eq.(13)with initial condition v(O) = vo.14.Integrate the velocity function in Eq.(13)to obtain the

upward-motion position function given in Eq. (14)with

initial condition y (0)= Yo.

15.Separatevariables in Eq.(15)and substitute u = v,Jpigto obtain the downward-motion velocity function given in

Eq.(16)with initial condition v(O) = Vo.

16.Integrate the velocity function in Eq.(16)to obtain the

downward-motion position function given in Eq.(17)with

initial condition y(O)= Yo.)))

17.Considerthe crossbowbolt of Example 3, shot straightupward from the ground (y = 0) at time t = 0 with initial

velocity Vo = 49mjs.Takeg = 9.8mjs2 and p = 0.0011in Eq.(12).Then use Eqs.(13)and (14)to show thatthe bolt reachesits maximum height of about 108.47m in

about 4.61s.18.Continuing Problem 17,supposethat the bolt is now

dropped(vo = 0) from a height of Yo = 108.47m. ThenuseEqs.(16)and (17)to show that it hits the ground about4.80s later with an impact speedof about 43.49mjs.19.A motorboat starts from rest (initial velocity v (0) = Vo =0). Its motor provides a constant accelerationof 4 ftj s2,but water resistancecausesa decelerationof v 2j400ftj S2.Find v when t = lOs,and also find the limiting velocityas t \037 +00 (that is,the maximum possiblespeedof the

boat).20.An arrow is shot straight upward from the ground with aninitial velocity of 160ftj s.It experiencesboth the decel-eration of gravity and decelerationv 2j800due to air resis-tance.How high in the air doesit go?21.If a ball is projectedupward from the ground with initial

velocity Vo and resistanceproportional to v 2, deducefromEq.(14)that the maximum height it attains is

1

(PV5

)Ymax =-In 1+- .

2p g22.Supposethat p = 0.075(in fps units, with g = 32 ftjs2)

in Eq.(15)for a paratrooper falling with parachute open.If he jumps from an altitude of 10,000ft and openshis

parachute immediately, what will be his terminal speed?How long will it take him to reachthe ground?23.Supposethat the paratrooperofProblem22falls freely for30s with p = 0.00075beforeopening his parachute. How

long will it now take him to reachthe ground?24.Themassof the sun is 329,320times that of the earth and

its radius is 109times the radius of the earth. (a) Towhatradius (in meters) would the earth have to becompressedin order for it to becomea blackhole-theescapevelocityfrom its surfaceequal to the velocity c = 3 x 108 mj s oflight? (b) Repeatpart (a) with the sun in placeof theearth.

25.(a) Show that if a projectileis launched straight upwardfrom the surface of the earth w ith initial velocity Vo lessthan escapevelocity y'2GMjR,then the maximum dis-tance from the centerof the earth attained by the projectileIS)

2GMRr -

max - 2GM_ Rv 2 'o

where M and R are the mass and radius of the earth, re-spectively. (b) With what initial velocity Vo must such aprojectilebe launched to yield a maximum altitude of 100kilometers abovethe surfaceof the earth? (c) Find themaximum distancefrom the centerof the earth, expressedin terms of earth radii, attained by a projectilelaunchedfrom the surfaceof the earth with 90%ofescapevelocity.26.Supposethat you are stranded-your rocket engine hasfailed-onan asteroid of diameter 3 miles, with densityequal to that of the earth with radius 3960miles. If you)

have enough spring in your legsto jump 4 feetstraight upon earth while wearing your spacesuit, can you blast offfrom this asteroidusing legpoweralone?

27.(a) Supposea projectileis launched vertically from thesurfacer = R of the earth with initial velocity Vo =y'2GMjR sov5 = k2jR wherek2 = 2GM.Thensolvethe differential equation drjdt = kjy'r (from Eq.(23)in this section)explicitly to deducethat r(t) \037 00ast \037 00.

(b) If the projectile is lau nched vertically with initial ve-locity Vo > y'2GMjR,deducethat

dr = /k2+a > \037.dt r y'r

Why doesit again follow that r(t) \037 00as t \037 oo?28.(a) Supposethat a body is dropped(vo = 0) from a dis-

tance ro > R from the earth's center, so its accelerationis dvjdt = -GMjr2. Ignoring air resistance,show that it

reachesthe height r < ro at time

t = \037(Jrro-r2+ ro cos-I rr

).

VWM v\037

(Suggestion: Substitute r ro cos2 () to evaluate

f y'rj(ro - r)dr.) (b) If a body is droppedfrom a heightof 1000km abovethe earth's surface and air resistanceis neglected,how long doesit take to fall and with what

speedwill it strike the earth's surface?29.Supposethat a projectileis fired straight upward from the

surfaceof the earth with initial velocity Vo < y'2GMjR.Then its height y (t) abovethe surface satisfiesthe initial

value problem)

d2y

dt 2)

GM(y + R)2

') y(O)= 0, y' (0)= Vo.)

Substitute dvjdt = v(dvjdy) and then integrate to obtain

2GMyv

2 = v5-

R(R + y)

for the velocity v of the projectileat height y. What maxi-mum altitude doesit reachif its initial velocity is 1kmjs?30.In Jules Verne's original problem, the projectilelaunchedfrom the surfaceof the earth is attracted by both the earthand the moon, so its distance r (t) from the centerof theearth satisfiesthe initial value problem

d2r _ GMe GMm . ,----+ , r (0)= R, r (0)= Vodt 2 r2 (S-r)2

where Me and Mm denotethe massesof the earth andthe moon, respectively; R is the radius of the earth andS = 384,400km is the distancebetween the centersofthe earth and the moon. Toreachthe moon, the projectilemust only just passthe point between the moon and earthwhere its net accelerationvanishes.Thereafterit is \"under

the control\" of the moon, and falls from there to the lunar

surface.Find the minimal launch velocity Vo that sufficesfor the projectileto make it \"From the Earth to the Moon.\)

1.8Application)y)

tv)

F

f)

III\037

c)

FIGURE1.8.5.An ascendingrocket.)


RocketPropulsion)Supposethat the rocketof Fig.1.8.5blastsoff straight upward from the surfaceofthe earth at time t = O.We want to calculate its height y and velocityv = dyjdt at

time t. The rocketis propelledby exhaust gasesthat exit (rearward)with constantspeedc (relative to the rocket).Becauseof the combustion of its fuel, the massm = m (t) of the rocketis variable.

To derive the equation of motion of the rocket,we useNewton'ssecondlaw

in the form)

dP-=Fdt)

(1))

where P ismomentum (the productof massand velocity)and F denotesnet externalforce (gravity, air resistance,etc.).If the massm of the rocketisconstantsom' (t) =O-whenits rocketsare turned off or burned out, for instance-thenEq.(1)gives

d(mv) dv dm dvF= =m-+-v=m-dt dt dt dt 'which (with dvjdt = a) is the more familiar form F = ma of Newton'ssecondlaw.

But here m is not constant. Supposem changesto m + tlm and v to v + L\\ v

during the short time interval from t to t + tlt.Then the change in the momentum

of the rocketitself is

tlP\037 (m + tlm)(v + \037v)-mv = m tlv + v tlm + tlm tlv.)

But the system also includesthe exhaust gasesexpelledduring this time interval,with mass -\037m and approximate velocity v-c.Hencethe total change in mo-mentum during the time interval tlt is)

tlP \037 (m tlv + v tlm + tlm tlv) + (-tlm)(v-c)= m tlv +c tlm + tlm tlv.

Now we divide by tlt and take the limit as tlt \037 0,so tlm \037 0,assumingcontinuity of met). The substitution of the resulting expressionfor dPjdt in (1)yieldsthe rocketpropulsionequation

dv dmm-+c-= F. (2)dt dtIf F = FG + FR , where FG = -mgis a constant force of gravity and FR = -kv isa force of air resistanceproportional to velocity,then Eq.(2)finally gives

dv dmm-+c-= -mg-kv. (3)dt dt)

ConstantThrust)

Now supposethat the rocketfuel is consumedat the constant \"bum rate\" fJ duringthe time interval [0,tl],during which time the massof the rocketdecreasesfrom moto mI.Thus)

m(O)= mo,

met) = mo - fJt,)

m(tl)= ml,dm-= -fJ for t < tl,dt)

(4))

with burnout occurringat time t =tl.)))


PROBLEM1 Substitutethe expressionsin (4)into Eq. (3)to obtain the differentialequation)

dv(m -fJt)-+kv = fJe- (mo - fJt)g.

dt)(5))

Solvethis linear equation for)

v(t) = voMk/{J +\037c

(1-Mk/{J) + ::ok(1-Mk/{J),) (6))

where Vo = v(O)and)

met) mo -fJtM---- -

mo mo)

denotesthe rocket'sfractionalmassat time t.)

No Resistance)PROBLEM2 For the caseof no air resistance,setk = 0 in Eq. (5)and integrateto obtain)

movet) = Vo

- gt +eIn .mo -

fJt)(7))

Becausemo -fJtl = ml,it follows that the velocityof the rocketat burnout (t = tl)IS)

moVI = V(tl) = Vo

-gtl +eln-.ml)

(8))

PROBLEM3 Start with Eq. (7)and integrate to obtain)

1 2 e moy(t) = (vo +c)t- -gt - -(mo- ,Bt) In

,B2 fJ mo- t)

(9))

It follows that the rocket'saltitude at burnout is)

1 2 em1 moYI = y(tl) = (vo +e)tl- -gtl --In-.

2 fJ ml)(10))

PROBLEM4 The V-2 rocketthat was usedto attack London in World War IIhadan initial massof 12,850kg, of which 68.5%was fuel. This fuel burned uniformlyfor 70secondswith an exhaust velocity of 2 kmIs.Assume it encountersair resis-tance of 1.45N permls of velocity.Then find the velocity and altitude of the V-2 atburnout under the assumption that it is launchedverticallyupward from reston theground.)))


PROBLEM5 Actually, our basicdifferentialequation in (3)applieswithout qual-ification only when the rocketis already in motion. However, when a rocket issitting on its launch pad stand and its enginesare turned on initially, it is observedthat a certain time interval passesbefore the rocketactually \"blasts off\" and beginsto ascend.The reasonis that if v = 0 in (3),then the resulting initial acceleration)

dv e dm-=---gdt m dt)

of the rocketmay be negative. But the rocketdoesnot descendinto the ground;it just \"sits there\" while (becausem is decreasing)this calculated accelerationin-creasesuntil it reaches0 and (thereafter)positive values so the rocket can begin toascend.With the notation introduced to describedthe constant-thrust case,showthat the rocketinitially just \"sits there\" if the exhaust velocitye is lessthan mog/f3,and that the time tB which then elapsesbefore actual blastoff is given by)

mog - f3etB = .

fJg)

FreeSpaceSupposefinally that the rocketis accelerating in free space,where there is neithergravity nor resistance,sog = k = O.With g = 0 in Eq.(8)we seethat, as the massof the rocketdecreasesfrom mo to m I,its increasein velocity is)

mo\037 v = v} - Vo = eIn -.

mi)(11))

Note that \037v dependsonly on the exhaust gasspeede and the initial-to-tinal massratio molm},but doesnot dependon the bum rate fJ. For example,if the rocketblastsoff from rest (vo = 0)and e = 5 km/s and molm}= 20,then its velocityat

burnout is VI = 5ln20\037 15km/s.Thus if a rocketinitially consistspredominantlyof fuel, then it can attain velocities significantly greater than the (relative)velocityof its exhaustgases.)

\037 U\037 \037\037!Ymm___mmm'\037m'\037 \037m'\037'm\037m'mm' \"m_'\037'm\037_'_\037'\037m_\"__\" '\"\037\037m\"\"_ ____ \"_m'\"m_'_\037\"\037____ _m____\037__._________\037_\037_____\037_)

In this chapter we have discussedapplicationsof and solution methodsfor severalimportant types of first-orderdifferentialequations, including those that are separa-ble(Section1.4),linear (Section1.5),or exact (Section1.6).In Section1.6we alsodiscussedsubstitution techniques that can sometimesbe usedto transform a givenfirst-orderdifferentialequation into one that is either separable,linear,or exact.

Lest it appearthat thesemethods constitute a \"grab bag\" of specialand unre-lated techniques,it is important to note that they are all versions of a singleidea.Given a differentialequation)

f(x,y, y') = 0,) (1))

we attempt to write it in the form)

ddx[G(x,y)] = o.) (2))))


It ispreciselyto obtain the form in Eq.(2)that we multiply the terms in Eq.(1)by an

appropriate integrating factor (even if all we are doing is separating the variables).But oncewe have found a function G(x,y) such that Eqs.(1)and (2)are equivalent,a general solution is defined implicitly by means of the equation)

G(x,y) = C) (3))

that one obtains by integrating Eq.(2).Given a specificfirst-orderdifferentialequation to be solved,we can attack it

by means of the followingsteps:). Isit separable?If so,separatethe variablesand integrate (Section1.4).. Isit linear?That is, can it be written in the form)

dy-+ P(x)y = Q(x)?dx)

If so,multiply by the integrating factor p = exp(f Pdx) of Section1.5.. Isit exact?That is,when the equationis written in the form Mdx+N dy = 0,is aM/ay = aN/ax(Section1.6)?. If the equation as it stands is not separable,linear,or exact,is there a plausiblesubstitution that will make it so? For instance, is it homogeneous(Section1.6)?)

Many first-orderdifferentialequations succumbto the line of attack outlinedhere.Nevertheless,many more do not. Becauseof the wideavailability of com-puters,numerical techniquesare commonly used to approximate the solutions ofdifferential equations that cannot be solved readily or explicitlyby the methods ofthis chapter. Indeed,most of the solution curves shown in figures in this chapterwereplotted using numerical approximations rather than exact solutions.Severalnumericalmethods for the appropriatesolutionof differentialequationswill bedis-cussedin Chapter 6.)

Chapter1) ReviewProbleIns)

Find generalsolutionsof the differentialequationsin Problems1through 30.Primesdenote derivatives with respecttox.)

1.x3 + 3y -xy'= 02.xy2 + 3y2 -x2y' = 03.xy + y2

-x2y' = 04. 2xy

3 + eX + (3x2y2 + sin y)y' = 05. 3y + x4y' = 2xy6.2xy

2 + x2y' = y2

7. 2x2y + x3y' = 18.2xy + x2y' = y2

9.xy' + 2y = 6x2,JY

10.y' = 1+X2+ y2 +X2y211.x2y' = xy + 3y2

12.6xy3 + 2y

4 + (9x2y2+ 8xy3)y' = 0)

13.4xy2 + y' = 5x4

y214.x3y' = x2

y_

y3

15.y' + 3y = 3x2e-3x

16.y' = x2 -2xy + y217.eX + yexy + (eY + xeYX)y' = 018.2x2y-x3y' = y3

19.3x5y2 + x3y' = 2y

2

20.xy' + 3y = 3X-3/2

21.(x2 - l)y' + (x - l)y = 122.xy' = 6y + 12x4y2/323.eY + ycosx+ (xeY + sinx)y'= 024. 9x2

y2 + X3/2y' = y225.2y + (x + l)y' = 3x + 3)))

26.9X I/2y4/3 -12xI/5y3/2 + (8X3/2yI/3 - 15x6/5yI/2)y' = 0

27.3y + x3y4 + 3xy' = 0

28.y + xy'= 2e2x

29.(2x+ l)y' + y = (2x+ 1)3/230.y' = -Jx + y)

Each of the differential equations in Problems31through 36is of two different types consideredin this chapter-separable,linear, homogeneous, Bernoulli, exact,etc.Hence,derive gen-)


eralsolutions for eachoftheseequations in two different ways;then reconcileyour results.)

dy dy31.-=3(y+7)x2 32.-= xy3 -xydx dx

33.dy 3x2 + 2y2

dy x + 3y- 34. -=dx 4xy dx y

-3x

35.dy 2xy + 2x36.

dY y'Y-y-dx x2 + 1 dx tan x)))

LinearEquationsof HigherOrder)

IfIIm I!ltroduction:Second-Order\037inearEquations)

In Chapter 1 we investigated first-order differential equations.We now turn to

equations of higher ordern >2,beginning in this chapter with equations that arelinear. The general theory of linear differentialequationsparallelsthe second-ordercase(n = 2),which we outline in this initial section.

Recall that a second-orderdifferential equation in the (unknown) function

y(x) is one of the form)

G(x,y, y', y\") = O.) (1))

This differentialequation is saidto be linearprovided that G is linear in the depen-dent variable y and its derivativesy' and y\". Thus a linear second-orderequationtakes(orcan be written in) the form)

\037) A(x)y\" + B(x)y'+C(x)y= F(x).) (2))

Unlessotherwisenoted, we will always assumethat the (known) coefficientfunc-tions A(x),B(x),C(x),and F(x)are continuouson someopeninterval I (perhapsunbounded) on which we wish to solve this differential equation, but we do not

require that they be linear functionsof x.Thus the differentialequation)

eXy\" + (cosx)y'+ (1+ -JX)y = tan-1 x)

is linear becausethe dependentvariable y and its derivativesy' and y\" appear lin-

early. By contrast, the equations)

y\" = yy' and y\" +3(y')2+4y3 = 0)

are not linear becauseproductsand powersof y or its derivativesappear.)

100)))

Spring Mass) Dashpot)

x(t)x=o x>oEquilibrium

position)

FIGURE2.1.1.A mass-spring-dashpot system.)

m)

Fs) II() FR)

x, v > 0)

FIGURE2.1.2.Directionsof theforcesacting on m.)

2.1Introduction:Second-OrderLinearEquations 101)

If the function F(x)on the right-hand sideof Eq. (2)vanishes identically onI,then we call Eq. (2)a homogeneouslinear equation; otherwise, it is nonhomo-geneous.For example,the second-orderequation)

x2y\" +2xy'+3y = cosx)

is nonhomogeneous;its associatedhomogeneousequation is)

x2y\" +2xy'+3y = O.)

In general,the homogeneouslinear equationassociatedwith Eq.(2)is)

\037) A(x)y\" + B(x)y'+C(x)y= O.) (3))

In casethe differential equation in (2)modelsa physical system, the nonhomoge-neousterm F(x)frequently correspondsto someexternal influenceon the system.

Remark:Note that the meaning of the term \"homogeneous\"for a second-orderlinear differentialequation is quite differentfrom its meaning for a first-orderdifferential equation (as in Section1.6).Of course,it is not unusual-eitherinmathematicsor in the Englishlanguagemore generally-forthe sameword to havedifferentmeanings in differentcontexts.)

m.'...l..'.:.

f:\

A TypicalApplicationLinear differentialequations frequently appearasmathematicalmodelsof mechan-ical systemsand electricalcircuits.For example,supposethat a massm is attachedboth to a spring that exertson it a force Fs and to a dashpot (shockabsorber)that

exertsa force FR on the mass(Fig.2.1.1).Assume that the restoringforce Fs ofthe spring is proportional to the displacementx of the massfrom its equilibriumpositionand acts oppositeto the directionof displacement.Then)

Fs= -kx (with k > 0))

so Fs < 0 if x > 0 (springstretched)while Fs > 0 if x < 0 (spring compressed).We assumethat the dashpot force FR is proportional to the velocityv = dxjdt ofthe massand acts oppositeto the directionof motion. Then)

dxFR = -cv= -c- (with c >0)dt)

so FR <0 if v >0 (motion to the right) while FR >0 if v <0 (motion to the left).If FRand Fs are the only forcesacting on the massm and its resulting accel-

eration is a = dvjdt, then Newton'slaw F = ma gives)

mx\" = Fs + FR ;) (4))

that is,)

\037)

d2x dxm-+c-+kx = O.dt2 dt)

(5))

Thus we have a differential equation satisfiedby the position function x(t) of the

massm.This homogeneoussecond-orderlinearequationgovernsthefreevibrationsof the mass;we will return to this problemin detail in Section2.4.)))

102 Chapter2 LinearEquationsof HigherOrder)

If, in addition to FsandFR, the massm is acted on by an external forceF(t)-whichmust then be addedto the right-hand side in Eq. (4)-theresulting

equation is)

\037)

d2x dxm\037 +c-+kx ==F(t).dt dt)

(6))

This nonhomogeneouslinear differentialequation governs the forcedvibrations ofthe massunder the influenceof the external force F(t).)

HomogeneousSecond-OrderLinearEquationsConsiderthe general second-orderlinear equation)

A(x)y\" + B(x)y'+C(x)y== F(x),) (7))

where the coefficientfunctions A, B,C,and F are continuouson the openinterval

I.Herewe assumein addition that A (x) i= 0 at eachpoint of I, so we can divideeachterm in Eq.(7)by A(x) and write it in the form)

\037) y\" +p(x)y'+q(x)y ==f(x).) (8))

We will discussfirst the associatedhomogeneousequation)

\037) y\" +p(x)y'+q(x)y ==O.) (9))

A particularly usefulpropertyof this homogeneouslinearequationis the fact that the

sum of any two solutions of Eq. (9)is again a solution, as is any constant multipleof a solution. This is the central ideaof the following theorem.)

THEOREM1 PrincipleofSuperpositionforHomogeneousEquations

Let Yl and Y2 be two solutions of the homogeneouslinear equation in (9)on theinterval I.If Cl and C2are constants, then the linear combination)

y ==CIYl + C2Y2) (10))

is alsoa solution of Eq.(9)on I.)

Proof:The conclusionfollows almost immediatelyfrom the linearity of the

operation of differentiation,which gives)

, , , d \" \" \"

Y ==CIYl +C2Y2 an Y ==CIYl +C2Y2.)

Then)

y\" + PY' +qy == (CIYl +C2Y2)\" +P(CIYl+C2Y2)'+q(CIYl +C2Y2)==

(CIY\037' +C2Y;)+ p(CIY\037 + C2Y\037) +q(CIYl +C2Y2)

(\" ,

) (\" ,

)==Cl Yl + PYI +qYl +C2 Y2 + PY2 +qY2

==Cl .0 +C2.0 ==0)

becauseYl and Y2 are solutions.Thus Y ==ClYl +C2Y2 is alsoa solution. .)))

2.1Introduction:Second-OrderLinearEquations 103):N \"\" ............__\037 \037 \037...

Example1) We can seeby inspectionthat)

Yl (x)= cosx and Y2 (x)= sin x)

are two solutionsof the equation)

\" 0Y +y= .)

Theorem 1 tellsus that any linear combinationof thesesolutions,such as)

y(x) = 3Yl(X) - 2Y2(X) = 3cosx-2sinx,)

is alsoa solution. We will seelater that, conversely,every solutionof y\" + y = 0is a linear combinationof thesetwo particular solutions Yl and Y2. Thus a generalsolutionof y\" + y = 0 is given by)

y(x) = Clcosx+C2sinx.)

It is important to understand that this singleformula for the generalsolution encom-passesa \"twofold infinity\" of particular solutions, becausethe two coefficientsCland C2can be selectedindependently.Figures2.1.3through 2.1.5illustrate someofthe possibilities,with either Cl or C2setequal to zero,or with both nonzero. .)

8 8 106 6 8

4 4 64

2 2 2\037 0 \037 0 \037 0

-2 -2 -2-4 -4 -4

-6-6 -6 -8-8 0 231:

-8 -10-31: 31: -31: 0 31: 231: -31: 0)

y = 6cosx -2 sin xy = 3cosx + 4 sin x

/)

x) x)

31:

X)

231:) 331:)

FIGURE2.1.3.Solutions

y(x) = Clcosxof y\" + y = o.)FIGURE2.1.4.Solutions

y(x) = C2sin x of y\" + y = o.)FIGURE2.1.5.Solutions ofy\" + y = 0 with Cl and C2bothnonzero.)

Earlier in this sectionwe gave the linear equation mx\" +CX' +kx = F(t) asa mathematicalmodelof the motion of the massshown in Fig.2.1.1.Physicalcon-siderationssuggestthat the motion of the massshould be determinedby its initial

positionand initial velocity.Hence,given any preassignedvaluesof x(0)and x'(0),Eq. (6)ought to have a unique solution satisfying these initial conditions. Moregenerally, in orderto be a \"good\" mathematicalmodel of a deterministicphysicalsituation, a differential equation must have unique solutions satisfying any appro-priate initial conditions.The followingexistenceand uniquenesstheorem(provedin the Appendix)givesus this assurancefor the general second-orderequation.)))

1043

2

1;;..\"

0

-1)

Chapter2 LinearEquationsof HigherOrder)

-2-1 0 1 2 3 4 5x)

FIGURE2.1.6.Solutions ofy\" + 3y'+ 2y = 0 with the sameinitial value y (0)= 1but differentinitial slopes.)

54321

;;..\" 0-1-2-3-4-\0372 -1)

y(O) = 3

y(O) = 0)

o 1 234x)

FIGURE2.1.7.Solutions ofy\" + 3y'+ 2y = 0 with the sameinitial slopey' (0)= 1but differentinitial values.)

Example1)

Continued)

THEOREM2 ExistenceandUniquenessforLinearEquationsSupposethat the functions p,q,and I are continuous on the open interval Icontaining the point a.Then, given any two numbers bo and bl , the equation)

y\" +p(x)y'+q(x)y= I(x)) (8))

has a unique (that is,one and only one)solution on the entire interval I that

satisfiesthe initial conditions)

yea) = bo, y'(a) = bl .) (11))

Remark 1:Equation (8)and the conditions in (11)constitute a second-orderlinear initial value problem.Theorem 2 tellsus that any such initial valueproblemhas a unique solution on the whole interval I where the coefficientfunc-tions in (8)are continuous. Recall from Section1.3that a nonlinear differentialequation generally has a unique solutionon only a smaller interval.

Remark 2: Whereas afirst-orderdifferentialequation dy/dx = F(x,y)generally admits only a singlesolution curve y = y(x) passingthrough a giveninitial point (a,b), Theorem 2 impliesthat the second-orderequation in (8)hasinfinitely many solution curves passingthrough the point (a, bo)-namely,one foreach(real number) value of the initial slopey'(a) = b l . That is, instead of therebeingonly oneline through (a,bo) tangent to a solution curve, every nonverticalstraight line through (a,bo) is tangent to somesolution curve of Eq. (8). Figure2.1.6showsa number of solution curves of the equation y\" + 3y' + 2y = 0 allhaving the sameinitial value y (0)= 1,while Fig.2.1.7showsa number of solutioncurves all having the sameinitial slopey'(0) = 1.The application at the end ofthis sectionsuggestshow to construct such families of solution curves for a givenhomogeneoussecond-orderlinear differentialequation. .)

\037,... ... '\" \037 .... \037...\" \037\037\"\"\037\"\"\037\"\",.. \037\"\"\037...\"\"\"\"\037... \"\"\037\"\".\037 \"'\" \"'\"....\"\"\037\"\"..........,,\037\"\"\037....\037 \037\"\"\"'-\037\"\"\037\"'\037\"\"\037\"\"\"\"'\037\"\"\"'-\"''' \037\"\"\" ...\"\037\",,,.....,............ .... ....\"\"\", ,\"\",\"\",\"\"\"\"......\"'....\" \",...\"\",,,,,,,,,,\037,,,,, ...\"\"\",.....,,,\"\"\"'... ''''',''''\037....,......''''',..\",.....\"......,..... \"'... \",..... \"'.........\"\" \"'... ._ \"'...... ......u \037\"\" '\" \037....-u ....\" u ... ...,..... .... ....,.........'.,....v\"',...\"\",\"\"......-...c<,... ...'\" \"\"''''''''\"'''\037''''.....-.. '\" ...

We saw in the first part of Example1 that y (x)= 3cosx-2 sin x is a solution(onthe entire real line) of y\" + y = O. It has the initial values y (0)= 3, y'(0)= -2.Theorem 2 tells us that this is the only solution with these initial values. Moregenerally, the solution

y(x) = bocosx+b l sin xsatisfiesthe arbitrary initial conditions y(O) = bo, y'(0) = b l ; this illustrates theexistenceof such a solution,alsoas guaranteedby Theorem2. .

Example1 suggestshow, given a homogeneoussecond-orderlinear equation,we might actually find the solution y(x) whoseexistenceis assuredby Theorem2.First, we find two \"essentiallydifferent\" solutions YI and Y2; second,we attempt toimposeon the general solution)

y = ClYI + C2Y2) (12))

the initial conditions yea)simultaneousequations)

bo, y'(a) = b l . That is, we attempt to solve the)

ClYI (a)+c2Y2(a)= bo,

ClY\037 (a)+ c2y\037(a)= b i)

(13))

for the coefficientsCI and C2.)))

108642

\037 0-2-4-6-8-10-2)

Example2)


Verify that the functions)

YI(x)==ex and Y2(x)==xex)

are solutionsof the differentialequation

y\"-2y' +Y ==0,

and then find a solution satisfying the initial conditionsy(O)==3, y'(0)== 1.)

Solution The verificationis routine; we omit it. We imposethe given initial conditionson the

general solution)

-1) ox)

1)

FIGURE2.1.8.Differentsolutions y (x) = 3ex + C2XeX ofy\"

-2y' + y = 0 with the sameinitial value y(O)= 3.)

Example3)

y(x) ==clex +C2xex,)

for which)

y'(x)== (CI +c2)eX +C2xex,to obtain the simultaneousequations)

y(O) ==CI ==3,y'(O)==CI +C2 == 1.)

2)

The resulting solution is Cl ==3, C2==-2.Hencethe solutionof the original initial

value problemisy(x) ==3ex -2xex .

Figure 2.1.8showsseveraladditionalsolutionsof y\"-2y' + y ==0,all having the

sameinitial value y(O) ==3. .)In orderfor the procedureof Example2 to succeed,the two solutionsYI and

Y2 must have the elusiveproperty that the equations in (13)can alwaysbesolvedforCI and C2,no matter what the initial conditions bo and b l might be.The followingdefinition tellspreciselyhow different the two functions Yl and Y2 must be.)

DEFINITION LinearIndependenceof Two FunctionsTwofunctions definedon an openinterval I are saidto belinearlyindependenton I provided that neither is a constant multiple of the other.)

Twofunctionsare saidto belinearlydependenton an openinterval providedthat they are not linearly independentthere; that is,one of them is a constantmulti-

pleof the other. We can alwaysdeterminewhether two given functions f and g arelinearly dependenton an interval I by noting at a glancewhether either of the two

quotients fjg or gjf is a constant-valuedfunction on I.)

Thus it is clearthat the followingpairsof functions are linearly independenton the

entire real line:)

sInx and cosx;eX and e-2x.,eX and xex .,

x+l and x2.,x and Ix I.)))


That is, neither sin xIcosx = tan x nor cosxIsin x = cotx is a constant-valuedfunction; neither eXle-2x = e3x nor e-2xlex is a constant-valuedfunction; and soforth. But the identically zero function I(x) = 0 and any other function g arelinearly dependenton every interval, because0 .g(x) = 0 = I(x).Also, thefunctions

I(x)= sin2x and g(x)= sin x cosxare linearlydependenton any interval becauseI (x)= 2g(x)is the familiar trigono-metric identity sin 2x= 2 sin xcosx. .)GeneralSolutionsBut doesthe homogeneousequation y\" + PY' + qy = 0 always have two linearlyindependentsolutions?Theorem 2 saysyes!We needonly chooseYl and Y2 so that)

Yl (a) = 1,Y\037 (a) = 0 and Y2(a)= 0, y\037(a)

= 1.)

It is then impossiblethat either Yl = kY2 or Y2 = kYl becausek .0 i= 1 for any

constant k. Theorem 2 tellsus that two such linearly independent solutions exist;actually finding them is a crucial matter that we will discussbriefly at the end ofthis section,and in greater detail beginning in Section2.3.

We want to show,finally, that given any two linearly independentsolutionsYland Y2 of the homogeneousequation)

y\"(x) +p(x)y'(x)+q(x)y(x)= 0,) (9))

every solution y of Eq.(9)can beexpressedas a linear combination)

y = CIYl + C2Y2) (12))

of Yl and Y2. This means that the function in (12)is a generalsolutionofEq.(9)-itprovidesall possiblesolutionsof the differentialequation.

As suggestedby the equations in (13),the determinationof the constants Cland C2 in (12)dependson a certain 2 x 2 determinant of values of Yl, Y2, andtheir derivatives. Given two functions I and g, the Wronskian of I and g is thedeterminant)

\037) w= II')

g = Ig'-I'g.g')

We write either W(/,g) or W(x), dependingon whether we wish to emphasizethetwo functionsor the point x at which the Wronskian is to beevaluated.For example,)

. cosx SIn x 2 \302\267 2W(cosx,sIn x)=. = cosx + sIn x = 1-sInx cosx)

and)X

W(eX,xeX ) = exe)xeX

eX +xex)= e2x

.)

Theseare examplesof linearly independentpairsof solutions of differentialequa-tions (seeExamples1 and 2).Note that in both casesthe Wronskian is everywherenonzero.)))


On the other hand, if the functions f and g are linearly dependent,with f =kg (for example),then)

kg gW(f, g) =

kg' g' = kgg'-kg'g = O.)

Thus the Wronskian of two linearly dependentfunctions is identically zero.In

Section2.2we will prove that, if the two functions Yl and Y2 are solutionsof ahomogeneoussecond-orderlinear equation, then the strong conversestated in part

(b)of Theorem 3 holds.)

THEOREM3WronskiansofSolutions)

SupposetliatYlandY2aretwo.solutionsof the homogeneoussecond-orderlinear

equation(]3q.(9)))y\"+ p(x)y'+q(x)y= 0

on allopen.Interval1on which p and q are continuous.

(a)i.I\037;)Il\0374\0372areli\037earlyd

7pendent, then W (Yl, Y2) = 0on I.(b)'lfYlandYi{arelinearly independent,then W (YI, Y2) 1:-0 at eachpoint of I.)

Thus, given two solutions of Eq. (9), there are just two possibilities:TheWronskian W is identicallyzero if the solutionsare linearly dependent;the Wron-skian is never zero if the solutions are linearly independent.The latter fact is what

we needto show that Y = ClYl +C2Y2 is the general solutionof Eq.(9)if Yl and Y2are linearly independentsolutions.)

THEOREAA4. GeneralSolutionsof HomogeneousEquationsLetYlandY2betwolinearlyindependentsolutionsof the homogeneousequation(]3q.(9)))

y\"+p(x)y'+q(x)y= 0)

withjJan.dqcontinuousonthe openinterval I.If Y is any solutionwhatsoeverof Eq.(9)on 1,thenthere existnumbers Cl and C2such that)

Y(x)= CIYl(X) +C2Y2(X))

for.....allix..........inil...)

In essence,Theorem 4 tells us that when we have found two linearly inde-pendent solutions of the second-orderhomogeneousequation in (9),then we havefound allof its solutions.We thereforecall the linear combinationf = ClYl + C2Y2a generalsolution of the differentialequation.

ProofofTheorem 4:Choosea point a of I, and considerthe simultaneousequations)

CIYl (a)+c2Y2(a)= f (a),CIY\037 (a)+ c2y\037(a)

= f'ea).)(14))

The determinant of the coefficients in this system of linear equations in the un-knowns Cl and C2 is simply the Wronskian W (Yl, Y2) evaluated at x = a. By)))


Example4)

Theorem 3,this determinant is nonzero, soby elementaryalgebra it follows that the

equations in (14)can be solved for Cl and C2. With thesevalues of Cl and C2,wedefine the solution)

G(x)= CIYl(X) +C2Y2(X))

of Eq.(9);then)

G(a)==CIYl (a)+c2Y2(a)== yea))

and)

G'(a)==CIY\037 (a)+ c2y\037(a)

= Y'(a).Thus the two solutions Y and G have the sameinitial values at a;likewise,sodo Y'

and G'.By the uniquenessof a solutiondeterminedby such initial values (Theorem2),it follows that Y and G agreeon I.Thus we seethat)

Y(x)= G(x)= CIYl (x)+C2Y2(X),)

as desired.) .)............. ...... ......

If Yl (x)= e2x and Y2 (x)= e-2x , then)

Y\037'

= (2)(2)e2x = 4e2x = 4Yl and Y; = (-2)(_2)e-2x = 4e-2x = 4Y2.)

Therefore, Yl and Y2 are linearly independentsolutionsof)

y\"-4y = O.) (15))

But Y3(X) = cosh2xand Y4(X) = sinh 2xare alsosolutionsof Eq.(15),because)

d2 d\037(cosh2x)= -(2sinh2x)= 4cosh2xdx dx)

and, similarly, (sinh 2x)\" = 4 sinh 2x.It therefore follows from Theorem4 that the

functions cosh2xand sinh 2xcan be expressedas linear combinationsof Yl (x) ==

e2x and Y2(X) = e-2x.Of course,this is no surprise,because)

cosh2x= 4e2x+ 4e-2x and sinh 2x== 4e2x- 4e-2x)

by the definitionsof the hyperboliccosineand hyperbolic sine.) .)Remark:Becausee2x , e-2x and coshx, sinh x are two different pairs of

linearly independent solutions of the equation y\"- 4y == 0 in (15),Theorem 4

impliesthat every particular solution Y (x) of this equation can be written both in

the form)

Y(x)==cle2x +C2e-2x)

and in the form)

Y(x)==a coshx +b sinhx.Thus thesetwo different linear combinations (with arbitrary constant coefficients)provide two differentdescriptionsof the setof all solutionsof the samedifferentialequation y\"

-4Y == O. Henceeachof thesetwo linear combinations is a generalsolution of the equation. Indeed,this is why it is accurate to refer to a specificsuchlinear combinationas \"a general solution\" rather than as \"the general solution.\" .)))


LinearSecond-OrderEquationswith ConstantCoefficientsAs an illustration of the general theory introduced in this section,we discussthe

homogeneoussecond-orderlinear differentialequation)

\037) ay\" +by' +cy ==0) (16))

with constant coefficientsa,b, and c.We first lookfor a singlesolution of Eq.(16)and beginwith the observation that)

(erx ), ==rerx and (erx)\" ==r 2erx

,) (17))

so any derivative of erx is a constant multiple of erx . Hence,if we substituted

y == erx in Eq. (16),then each term would be a constant multiple of erx , with the

constant coefficientsdependenton r and the coefficientsa, b, and c.This suggeststhat we try to find a value of r so that thesemultiples of erx will have sum zero.Ifwesucceed,then y ==erx will be a solutionofEq.(16).

For example,if we substitute y ==erx in the equation)

y\"-5y' +6y ==0,)

we obtain)

r 2erx -5rerx +6erx ==o.)

Thus)

(r2-5r+6)erx ==0; (r -2)(r -3)erx ==o.Hencey == erx will be a solution if either r == 2 or r == 3.So,in searchingfor asinglesolution,we actually have found two solutions:YI (x)==e2x and Y2(x)== e3x.

To carry out this procedurein the general case,we substitute y == erx in

Eq.(16).With the aid of the equations in (17),we find the result to be)

ar2erx +brerx +cerx ==o.)

Becauseerx is never zero,we concludethat y(x) ==erx will satisfy the differential

equation in (16)preciselywhen r is a root of the algebraicequation)

\037) ar2 +br +c ==o.) (18))

This quadratic equation is calledthe characteristicequationof the homogeneouslinear differentialequation)

\037) ay\" +by' +cy ==o.) (16))

If Eq. (18)has two distinct (unequal) roots rl and r2, then the correspondingsolu-tions YI (x) ==er1x and Y2(X) ==er2x of (16)are linearly independent. (Why?) Thisgives the following result.)

THEOREM5 DistinctRealRoots)

If the roots rl and r2 of the characteristicequation in (18)are real and distinct,then)

\037) y(x) ==cler1x +C2er2x) (19))

is a general solutionof Eq. (16).)))


Example5) Find the general solution of)

2y\"-7y' +3y ==o.)

Solution We can solve the characteristicequation)

Example6)

3

2

1

\037

0

-1-2-0.5 0 0.5 1 1.5 2

x)


y(x) = 1+ C2e-2xof y\" + 2y' = 0with different values of C2.)

2r2-7r +3 ==0)

by factoring:)(2r- 1)(r- 3) ==O.

The roots rl == 1and r2 == 3 are real and distinct, so Theorem 5 yieldsthe generalsolution)

y(x) ==Clex/2 +C2e3x.) .)

\037 , '\" \" \037\037 \037 .u\"\"

The differentialequation y\" +2y' ==0 has characteristicequation)

r 2 +2r ==r(r +2) ==0)

with distinct real roots rl == 0 and r2 ==-2.BecauseeO.x = 1,we get the generalsolution)

y(x) ==Cl +C2e-2x.)

Figure 2.1.9showsseveral different solution curves with CI == 1,all appearing to

approach the solution curve y(x) = 1 (with C2 ==0)asx \037 +00. .Remark:Note that Theorem 5 changesa problem involving a differential

equation into one involving only the solutionof an algebraicequation. .)If the characteristic equation in (18)has equal roots rl == r2, we get (at first)

only the singlesolution Yl (x) == ertx of Eq. (16).The problem in this caseis to

producethe \"missing\" secondsolution of the differentialequation.A doubleroot r == r} will occurpreciselywhen the characteristicequation is

a constant multiple of the equation)

(r - rl)2== r 2-2rlr+ rf ==O.)

Any differentialequation with this characteristicequation is equivalent to)

\" 2 ' 2 0Y- rly+r1y==.) (20))

But it is easyto verify by direct substitution that y ==xertx is a secondsolution ofEq. (20).It isclear(but you should verify) that)

Yl (x) ==er1x and Y2 (x)==xertx)

are linearly independent functions, so the general solution of the differentialequa-tion in (20)is)

y(x) ==cler1x + c2xerlx.)))

Example7)

10)

5)

;;::....)

o)

-5-4 -3-2-1 0 1 2 3 4 5x)


y(x) == Cle-X + 2xe-x ofy\" + 2y/ + y == 0 with differentvalues ofCl.)

_ Problems)


THEOREM6 RepeatedRootsIf the characteristic equation in (18)has equal (necessarilyreal) roots r} = r2,then)

\037) y(x) = (CI+c2x)er1X) (21))

is a general solutionof Eq.(16).)

To solve the initial value problem)

y\" +2y' + y = 0;y(O)= 5, y'(0)= -3,)

we note first that the characteristicequation)

r 2 +2r+ 1 = (r + 1)2= 0)

has equal roots r} = r2 = -1.Hencethe general solution providedby Theorem6IS)

y(x) = cle-x +C2xe-x.)

Differentiationyields)

'( )

-x+ -x -xy X =-Cle C2e -C2xe ,)

so the initial conditionsyield the equations)

y(O) = C} 5,y'(O)= -Cl+C2 = -3,)

which imply that C} = 5 and C2 = 2.Thus the desiredparticular solution of the

initial value problem is)

y(x) = 5e-x +2xe-x .This particular solution, together with several others of the form y(x) = cle-x +2xe-x , is illustrated in Fig.2.1.10. .

The characteristicequation in (18)may haveeither real or complexroots.Thecaseof complexroots will be discussedin Section2.3.)

In Problems 1through 16,a homogeneoussecond-orderlin-eardifferential equation, two functions Yl and Y2, and a pairof initial conditions are given. First verify that Yl and Y2 aresolutions of the differential equation. Then find a particularsolution of the form Y == ClYl + C2Y2 that satisfies the giveninitial conditions.Primes denotederivatives with respectto x.1.y\" - Y == 0; Yl == eX, Y2 == e-x ; Y(0) == 0, y/ (0) == 5)

2.y\" -9y== 0; YI == e3x , Y2 == e-3x ; y(O) == -1,y/(O) == 153.y\" + 4y == 0; Yl == cos2x,Y2 == sin2x;y(O) == 3,

y/(O) == 84. y\" + 25y == 0; Yl == COS5x, Y2 == sin 5x; y(O) == 10,

y/(O) == -105. y\"

-3y'+2y == 0; Yl == eX, Y2 == e2x ; y(O) == 1,y/(O) == 06.y\" + y/

- 6y == 0; YI == e2x , Y2 == e-3x ; y(O) == 7,y/ (0) == -1)))


7. y\" + y/ = 0; Yl = 1,Y2 = e-x ; y(O)= -2,y/(O) = 88.y\"

-3y' = 0; Yl = 1,Y2 = e3x ; y(O)= 4, y/(O) = -29.y\" + 2y' + y = 0; Yl = e-x , Y2 = xe-x ; y(O) = 2,

y/(O) = -110.y\"

- 10y'+ 25y = 0; YI = e5x , Y2 = xe5x ; y(O) = 3,y/(O) = 13

11.y\" -2y'+2y = 0; Yl = eX cosx,Y2 = eX sinx;y(O)= 0,y/(O) = 5

12.y\" + 6y' + 13y= 0; Yl = e-3x cos2x, Y2 = e-3x sin 2x;y(O) = 2, y/(O) = 0

13.x2y\"

- 2xy' + 2y = 0; Yl = X, Y2 = x2; y(l) = 3,y/ (1)= 1

14.x2y\" + 2xy'- 6y = 0; Yl = x2, Y2 = x-3; y(2) = 10,

y'(2)= 1515.x2

y\"- xy/ + y = 0; Yl = X, Y2 = x lnx;y(l) = 7,

y'(I)= 216.x2

y\" + xy/ + y = 0; Yl = cos(lnx),Y2 = sin(lnx);y(l) = 2,y'(I)= 3)

Thefollowing three problems illustrate the fact that the super-position principle doesnot generally holdfor nonlinear equa-tions.)

17.Show that y = 1/x is a solution of y/ + y2 = 0, but that ifc i= 0 and c i= 1,then y = c/xis not a solution.

18.Show that y = x3 is a solution of yy\" = 6x4, but that ifc2

i= 1,then y = cx3 is not a solution.19.Show that Yl = 1 and Y2 = -JX are solutions of yy\" +

(y/)2 = 0, but that their sum y = Yl + Y2 is not a solution.)

Determine whether the pairs of functions in Problems 20through 26are linearly independent or linearly dependent onthe real line.

20. f(x) = JT, g(x)= cos2 x + sin 2 x21.f(x)=x3,g(x)=x2

lxl22. f(x)= 1+x,g(x)= 1+ Ixl23.f(x)=xex , g(x)= Ixlex

24. f(x)= sin 2x,g(x)= 1-cos2x25.f(x) = eX sinx,g(x)= eX cosx26.f(x) = 2cosx+ 3 sinx,g(x)= 3cosx-2sinx27.Let y p be a particular solution of the nonhomogeneous

equation y\" + py/ + qy = f(x) and let Yc be a solu-tion of its associatedhomogeneous equation. Show that

y = Yc + y p is a solution of the given nonhomogeneousequation.28.With y p = 1 and Yc = ClCOSx + C2sin x in the notationofProblem 27,find a solution of y\" + y = 1 satisfying theinitial conditions y (0)= -1= y/ (0).

29. Show that YI = x2 and Y2 = x3 are two different solu-tions ofx2

y\"-4xy' + 6y = 0, both satisfying the initial

conditions y(O) = 0 = y/ (0).Explain why thesefacts donot contradict Theorem2 (with respectto the guaranteeduniqueness ).

30.(a) Show that Yl = x3 and Y2 = Ix3

1

are linearly

independent solutions on the real line of the equationx2

y\"-3xy/ + 3y = O. (b) Verify that W (Yl , Y2) is iden-

tically zero. Why do these facts not contradict Theorem3?)

31.Show that Yl = sinx 2 and Y2 = cosx2 are linearly in-

dependent functions, but that their Wronskian vanishes at

x = O. Why doesthis imply that there is no differential

equation of the form y\" + p(x)y/+ q(x)y= 0, with both

p and q continuous everywhere, having both Yl and Y2 assolutions?

32.Let Yl and Y2 be two solutions of A (x)y\" + B(x)y/ +C(x)y = 0 on an open interval I where A, B, and Care continuous and A (x) is never zero. (a) Let W =W (Yl , Y2). Show that)

dW \" \"A(x)-= (Yl)(AY2)- (Y2)(AYl).dx)

Then substitute forAy\037

andAy\037'

from the original differ-

ential equation to show that)

dWA(x)-= -B(x)W(x).dx)

(b)Solvethis first-order equation to deduceAbel'sfor-mula)

W(x) = K exp(-f

B(x)dX

),

A(x)

where K is a constant. (c) Why does Abel's formula

imply that the Wronskian W (Yl , Y2) is either zeroevery-where or nonzero everywhere (asstated in Theorem3)?)

Apply Theorems5 and 6 to find generalsolutions of the dif-

ferential equations given in Problems 33 through 42. Primesdenotederivatives with respectto x.)

33.y\"-3y' + 2y = 0

35.y\" + 5y/ = 037.2y\"

-y/-

Y = 039.4y\" + 4y' + y = 041.6y\"

-7y' - 20y = 0)

34. y\" + 2y'- 15y= 036.2y\" + 3y' = 038. 4y\" + 8y'+ 3y = 040. 9y\"

- 12y'+ 4y = 042. 35y\"

-y/- 12y = 0)

Each of Problems 43 through 48 gives a general solution

y (x) of a homogeneous second-order'differentialequation

ay\" + by' + cy = 0 with constant coefficients.Find such an

equation.

43. y(x) = Cl + C2e-lOx 44. y(x) = clelOx + C2e-lOx45. y(x) = cle-lOx + C2xe-lOx46. y(x) = clelOx + C2elOOx 47. y(x) = CI+ C2X

48. y(x) = eX (c1exh+ C2e

-xh))

Problems49and 50dealwith the solution curvesofy\" + 3y/ +2y = 0 shown in Figs. 2.1.6and 2.1.7.49. Find the highest point on the solution curve with y (0)= 1

and y/ (0)= 6 in Fig. 2.1.6.50.Find the third-quadrant point of intersection of the solu-

tion curves shown in Fig. 2.1.7.)))

2.2GeneralSolutionsof LinearEquations 113)

51.A second-orderEuler equationis oneof the form

ax2y// + bxy' + cy = 0 (22)

where a, b, c are constants. (a) Show that if x > 0,then the substitution v = In x transforms Eq.(22) intothe constant-coefficientlinear equation

d2y dya

dv 2 + (b -a)dv

+ cy = 0 (23)

with independent variable v. (b) If the roots rl and r2 ofthe characteristicequation ofEq.(23)arereal and distinct,)

concludethat a general solution of the Euler equation in

(22)is y(x) = CIX rl + C2Xr2.)

Make the substitution v = In x ofProblem 51to find generalsolutions (for x > 0) of the Euler equations in Problems 52-56.)

52.x2y\" + xy/

-Y = 054. 4x2

y\" + 8xy'-3y = 0

56.x2y//

-3xy'+ 4y = 0)

53. x2y\" + 2xy'- 12y=0

55. x2y\" + xy/ = 0)

IIfIJ GeneralSoluti()ns()f Lineargq\037\037!\037_()!!\037)

We now show that our discussionin Section2.1of second-orderlinear equationsgeneralizesin a very natural way to the generalnth-orderlineardifferential equa-tion of the form)

Po(x)y(n)+ Pl(x)y(n-l)+...+ Pn-l(x)y'+ Pn(X)Y ==F(x). (1))

Unlessotherwisenoted, we will always assumethat the coefficientfunctions Pi (X)and F(x)are continuous on someopeninterval I (perhapsunbounded)wherewewish to solve the equation. Underthe additionalassumptionthat Po(x)i-0 at eachpoint of I, we can divide each term in Eq. (1)by Po(x) to obtain an equationwith

leading coefficient1,of the form)

\037) yen) +PI (x)y(n-l)+ ...+Pn-l(x)y'+ Pn (x)y == j(x).) (2))

The homogeneouslinear equationassociatedwith Eq.(2)is)

\037) yen) +PI(X)y(n-l)+...+Pn-I(X)y'+Pn(x)y==O.) (3))

Just as in the second-ordercase,a homogeneousnth-order linear differential equa-tion has the valuableproperty that any superposition,or linearcombination,of so-lutions of the equation is again a solution. The proof of the following theoremisessentiallythe same-aroutine verification-asthat of Theorem 1 of Section2.1.)

THEOREM1 Principleof SuperpositionforHomogeneousEquations

Let Yl, yz, ..., Yn be n solutions of the homogeneouslinear equation in (3)onthe interval I.If CI,Cz,..., Cn are constants, then the linear combination)

\037) y ==CIYl +czyz+ ...+ CnYn) (4))


Example1)

\".v, \037 ........ .................... '\" .-.< .-.' '\" '\"

It is easyto verify that the three functions)

YI (x)==e-3x , yz(x)==cos2x, and Y3(X) ==sin 2x)))


\037O)

6

4

2)

-2-4-6)

o 2 4 6 8 10x)

FIGURE2.2.1.Theparticularsolution y (x) =-3e-3x + 3cos2x -2 sin 2x.)

Example1)

Continued)

are all solutionsof the homogeneousthird-orderequation

y(3) + 3y\" +4y' + 12y= 0)

on the entire real line. Theorem 1 tells us that any linear combination of thesesolutions,such as)

y(x) = -3Yl(X)+3yz(x)- 2Y3(X) = -3e-3x +3cos2x-2sin2x,)

is alsoa solution on the entire real line.We will seethat, conversely,every solu-tion of the differentialequation of this exampleis a linear combinationof the threeparticular solutions Yl, yz, and Y3. Thus a general solution is given by)

y(x) = cle-3x +czcos2x+c3sin2x.) .)ExistenceandUniquenessofSolutionsWe saw in Section2.1that a particular solutionof a second-orderlinear differential

equation is determinedby two initial conditions.Similarly, a particular solution ofan nth-order linear differential equation is determined by n initial conditions.Thefollowingtheorem,proved in the Appendix,is the natural generalizationof Theorem2 of Section2.1.)

THEOREM2 ExistenceandUniquenessforLinearEquationsSupposethat the functions PI,pz, ...,Pn, and I are continuous on the openinterval I containing the point a.Then, given n numbers bo, bI, ..., bn-I, thenth-order linear equation (Eq.(2)))

yen) +PI(x)y(n-l)+...+ Pn-l(X)y'+ Pn(X)Y = I(x))

has a unique (that is,one and only one)solution on the entire interval I that

satisfies the n initial conditions)

yea) = bo, y'(a) = bI ,) y(n-I)(a)= bn-1 .) (5))... ,)

Equation (2)and the conditions in (5)constitute an nth-order initial valueproblem.Theorem 2 tellsus that any such initial value problemhas a unique so-lution on the whole interval I where the coefficientfunctions in (2)are continuous.It tellsus nothing, however,about how to find this solution. In Section2.3we will

seehow to construct explicitsolutions of initial value problemsin the consta\037t-coefficientcasethat occursoften in applications.)

We saw earlier that)

y(x) = _3e-3x +3cos2x -2 sin 2x)

is a solution of)

y(3) + 3y\" +4y' + 12y= 0

on the entire real line.This particularsolutionhas initial valuesy (0)= 0,y'(0)= 5,and y\" (0)= -39,and Theorem 2 impliesthat there is no other solution with thesesameinitial values. Note that its graph (in Fig.2.2.1)looksperiodicon the right.

Indeed,becauseof the negative exponent, we seethat y (x) \037 3cos2x - 2 sin 2xfor large positivex. .)))

54321

\037 0-1-2-3-4-5 -1)


Remark:Becauseits general solutioninvolvesthe three arbitrary constantsCI, C2, and C3, the third-order equation in Example1 has a \"threefold infinity\" ofsolutions,including three families of especiallysimplesolutions:)

. y(x) = cIe-3x (obtainedfrom the general solution with C2 = C3= 0),. y(x) = C2cos2x (with Cl = C3 = 0),and. y(x) = C3sin2x(with Cl = C2 = 0).)

Alternatively, Theorem 2 suggestsa threefold infinity of particularsolutionscorre-spondingto independentchoicesof the three initial values y (0)= bo, y'(0)= b l ,and y\" (0)= b2.Figures2.2.2through 2.2.4illustrate three correspondingfamiliesof solutions-foreach of which, two of thesethree initial valuesare zero. .)

3) 1

0.80.60.40.2

\037 0-0.2-0.4-0.6-0.8-1-1)

y'(O) = 3/)2) y\"(0)=3

/)1)

\037 0)

-1)

'\"y'(O) = -3) y\"(O) = -3)-2)

-3-2 -1)o) 2 3 4 5)1) o 1 2 3x)

o 1 2 345x)

4)

x)

FIGURE2.2.2.Solutions ofy(3) + 3y\" + 4y' + 12y= 0 with

y/ (0) = y\" (0)= 0 but withdifferent values for y(O).)


y (0)= y\" (0)= 0 but with

different values for y/ (0).)


y(O)= y/ (0)= 0 but with

different values for y\" (0).)

Note that Theorem 2 impliesthat the trivial solution y(x) = 0 is the onlysolution of the homogeneousequation)

yen) +PI (x)y(n-I)+ ...+Pn-l(x)y'+Pn(x)y= 0) (3))

that satisfiesthe trivial initial conditions)

yea) = y'(a)=...= y(n-l)(a) = o.)

- .. -

Example2) It is easyto verify that)

YI (x)= x2 and Y2(X) = x 3)

are two different solutionsof)

x2y\"

-4xy' +6y = 0,)

and that both satisfy the initial conditions y (0) = y'(0) = O. Why doesthis not

contradict the uniquenesspart of Theorem2?It isbecausethe leadingcoefficientinthis differentialequationvanishesat x = 0,so this equationcannotbewritten in theform ofEq.(3)with coefficientfunctions continuouson an openinterval containingthe point x = O. .)))


LinearlyIndependentSolutionsOn the basisof our knowledgeof general solutionsof second-orderlinearequations,we anticipate that a general solutionof the homogeneousnth-order linear equation)

\037) yen) +PI (x)y(n-1)+ ...+Pn-1(x)y'+ Pn(X)Y ==0) (3))

will be a linear combination)

\037) Y ==C1Y1 +C2Y2 + ...+Cn Yn ,) (4))

where Y1, Y2, ..., Yn are particular solutionsof Eq.(3).But thesen particularsolu-tions must be \"sufficiently independent\" that we can alwayschoosethe coefficientsC1,C2,..., Cn in (4) to satisfy arbitrary initial conditions of the form in (5). Thequestion is this: What should be meant by independenceof three or more func-tions?

Recall that two functions 11and 12are linearly dependentif one is a constantmultiple of the other; that is, if either 11==kl2 or 12==kl1 for someconstantk. Ifwe write theseequations as)

(1)/1+ (-k)/2==0 or (k)/1+ (-1)/2==0,)

we seethat the linear dependenceof 11and 12impliesthat there existtwo constantsC1and C2not both zero such that)

c1/1+c2/2==o.) (6))

Conversely,if C1and C2are not both zero, then Eq.(6)certainly impliesthat 11and12are linearly dependent.

In analogy with Eq. (6),we say that n functions 11,12,...,In are linearlydependentprovided that somenontrivial linear combination)

c1/1+c2/2+...+cnln)

of them vanishes identically;nontrivial means that not allof the coefficientsCI,C2,..., Cn are zero (although someof them may be zero).)

DEFINITION LinearDependenceofFunctionsThe n functions 11,12,..., In are saidto be linearlydependenton the intervalI provided that there existconstantsCl,C2,..., Cn not all zero such that)

C1 11+c2/2+ ...+Cn In ==0) (7))

on I;that is,)

cl/1(x)+c2/2(X)+...+cnln(x)==0)

for all x in I.)

If not all the coefficients in Eq. (7)are zero,then clearly we can solve for at leastone of the functions as a linear combinationof the others, and conversely.Thus thefunctions 11,12,..., In are linearly dependentif and only if at least one of them isa linear combinationof the others.)))

Example3)


The functions)

fl (X) = sin 2x, f2(x)= sin x cosx, and f3(x)= eX

are linearly dependenton the real line because)

(l)fI+ (-2)f2+ (0)f3= 0)

(by the familiar trigonometric identity sin 2x = 2 sin x cosx).) .)The n functionsfI,f2, ..., fn are calledlinearlyindependenton the interval

I provided that they are not linearly dependentthere. Equivalently, they are linearly

independenton I provided that the identity)

C1 fl +C2f2 + ...+Cn fn = 0

holdson I only in the trivial case)

(7))

CI = C2 = ...= Cn = 0;)

that is, no nontrivial linear combination of thesefunctions vanisheson I. Put yetanother way, the functions fI,f2, ..., fn are linearly independentif no one of themis a linear combinationof the others.(Why?)

Sometimesone can show that n given functions are linearly dependentby

finding, as in Example3, nontrivial values of the coefficientsso that Eq.(7)holds.But in orderto show that n given functions are linearly independent,we must provethat nontrivial values of the coefficientscannot be found, and this is seldomeasy todo in any direct or obviousmanner.

Fortunately, in the caseof n solutionsof a homogeneousnth-orderlinearequa-tion, there is a tool that makesthe determinationof their linear dependenceor inde-pendencea routine matter in many examples.This tool is the Wronskian determi-nant, which we introduced (for the casen = 2) in Section2.1.Supposethat the n

functions fI,f2, ..., fn are each n - 1 times differentiable.Then their Wronskianis the n x n determinant)

fl f2 fn

\037 W= f{ f\037 f\037

(8)

fI(n-I) fin-I) f\037n-l))

We write W(fI, f2,..., fn) or W(x), dependingon whether we wish to empha-size the functions or the point x at which their Wronskian is to be evaluated. TheWronskian is named after the Polishmathematician J.M.H.Wronski (1778-1853).

We saw in Section2.1that the Wronskian of two linearly dependentfunctions

vanishes identically. Moregenerally, the Wronskian of n linearly dependentfunc-tions fl,f2, ..., fn is identicallyzero.To prove this, assumethat Eq.(7)holdsonthe interval I for somechoiceof the constants CI,C2,..., Cn not all zero.We then

differentiate this equation n - 1 times in succession,obtaining the n equations)

clf!(x)+c2f2(X)

clf{(x)+ c2f\037(x))

+...++...+)

cnfn(x)= 0,cnf\037(x)

= 0,)

(9))

CIfI(n-I)(x)+c2fin-I )(x)+...+ Cn f\037n-l) (x)=

0,)))


which hold for all x in I. We recallfrom linear algebra that a system of n lin-ear homogeneousequations in n unknowns has a nontrivial solution if and onlyif the determinant of coefficientsvanishes. In Eq. (9)the unknowns are the con-stants Cl, C2,...,Cn and the determinant of coefficients is simply the WronskianW (11,12,... , In) evaluatedat the typical point x of I.Becausewe know that theCi are not all zero,it follows that W (x)= 0,as we wanted to prove.

Therefore, to show that the functions 11,12,..., In are linearly independenton the interval I, it sufficesto show that their Wronskian is nonzeroat justone pointof I.)

Example4)... .\037.

Showthat the functions Yl (x) = e-3x, Y2(X)

Example1)are linearly independent.)cos2x,and Y3(X))

.....

sin 2x (of)

Solution Their Wronskian is)

-3xe) cos2x) sin 2x)

W= -3e-3x -2sin2x) 2cos2x)

ge-3x -4cos2x -4sin 2x)

-3x=e)-2sin 2x) 2 cos2x +3e-3x)

cos2x) sin 2x)

-4cos2x -4sin 2x) -4cos2x -4sin 2x)

+ge-3x)cos2x) sin 2x = 26e-3x

i= O.)-2sin 2x 2 cos2x)

BecauseW i= 0 everywhere, it follows that Yl, Y2, and Y3 are linearly independenton any openinterval (including the entire real line). .)

Example5)......

Showfirst that the three solutions)

Yl(X)=X, Y2(x)=xlnx, and Y3(X)=X2)

of the third-orderequation

x3y(3) -x2

y\" +2xy'-2y = 0) (10))

are linearly independenton the openinterval x >O.Then find a particular solutionof Eq. (10)that satisfiesthe initial conditions)

y(l) = 3, y'(I) = 2, y\"(I)= 1.) (11))

Solution Note that for x > 0,we coulddivide each term in (10)by x3 to obtain a homoge-neouslinear equationof the standardform in (3).When we compute the Wronskianof the three given solutions,we find that)

x x In x x2

W= 1 1 + In x 2x = x.1

0 2x)))


Thus W (x) i= 0 for x >0,soYl, Y2, and Y3 are linearly independenton the interval

x > O. To find the desiredparticular solution, we imposethe initial conditionsin

(11)on)

y(x) = ClX + C2X In x) 2+ C3X ,)

y'(x) = Cl + c2(1+ In x) + 2C3X,)

y\"(x) = 0)C2+-x)

+ 2C3.)

This yieldsthe simultaneousequations)

y(I) = Cl + C3 = 3,y'(I) = Cl + C2 + 2C3= 2,)

y\" (1)=) C2 + 2C3= 1;)

we solve to find Clquestion is)

1,C2 = -3,and C3 = 2. Thus the particular solution in)

y(x) = x-3xIn x+2x2.) .)

Provided that W (Yl, Y2, ... , Yn) i= 0,it turns out (Theorem4) that we canalways find valuesof the coefficientsin the linear combination)

y = ClYl +C2Y2 + ...+ CnYn)

that satisfy any given initial conditions of the form in (5).Theorem3 providesthe

necessarynonvanishing of W in the caseof linearly independentsolutions.)

THEOREM3 Wronskiansof Solutions)

Supposethat YI, Y2, ..., Yn are n solutionsof the homogeneousnth-orderlinearequation)

y(n) +PI (x)y(n-l)+ ...+Pn-l(X)y' +Pn(x)y= 0 (3))

on an openinterval I,where eachPi is continuous. Let)

w = W (YI, Y2, ..., Yn).)

(a) If YI, Y2, ..., Yn are linearly dependent,then W = 0 on I.(b) If YI, Y2, ..., Yn are linearly independent,then W i= 0at each point of I.Thus there are just two possibilities:Either W = 0 everywhereon I,or W f.= 0everywhereon I.)

Proof:We have already proven part (a). To prove part (b), it is sufficient

to assumethat W (a) = 0 at somepoint of I,and show this implies that the solu-tions YI, Y2, ..., Yn are linearly dependent.But W (a) is simply the determinant of)))


coefficientsof the systemof n homogeneouslinear equations)

CIYl (a) +

CIY\037 (a) +)

C2Y2 (a) + ...+c2y\037(a) +...+)

CnYn(a) = 0,cny\037(a)

= 0,)

(12))

clyin-1)(a) + c2yi

n-l)(a) + ...+ cny\037n-l) (a) = 0

in the n unknowns Cl, C2, ..., Cn . BecauseW (a) = 0,the basicfact from linearalgebraquoted just after (9) impliesthat the equations in (12)have a nontrivial

solution.That is, the numbers Cl,C2,..., Cn are not all zero.We now usethesevalues to define the particular solution)

Y (x)= C1 Yl (X) +C2Y2 (X) + ...+Cn Yn (X)) (13))

of Eq. (3). The equations in (12)then imply that Y satisfies the trivial initial

conditionsyea) = Y/(a) = ...= y(n-l)(a)= O.

Theorem 2 (uniqueness)therefore impliesthat Y(x) = 0 on I. In view of (13)and the fact that Cl, C2, ..., Cn are not all zero,this is the desiredconclusionthat

the solutions Yl, Y2, ...,Yn are linearly dependent.This completesthe proof ofTheorem 3. .)GeneralSolutions)We can now show that, given any fixed set of n linearly independent solutions ofa homogeneousnth-order equation, every (other) solution of the equation can beexpressedas a linear combination of those n particular solutions.Usingthe factfrom Theorem 3 that the Wronskian of n linearly independentsolutions is nonzero,the proof of the following theorem is essentiallythe sameas the proof of Theorem4 of Section2.1(the casen = 2).)

THEOREM4 GeneralSolutionsof HomogeneousEquationsLetYl, Y2, ...,Yn be n linearly independentsolutionsof the homogeneousequa-tion)

yen) +Pl(X)y(n-l)+...+Pn-l(X)Y'+Pn(x)y= 0) (3))

on an openinterval I where the Pi are continuous. If Y is any solutionwhatsoeverofEq.(3),then there existnumbers Cl,C2,...,Cn such that)

Y(x)= CIYl (x)+C2Y2(X)+...+ cnYn(x))

for all x in I.)

Thus every solution of a homogeneousnth-order linear differential equationis a linear combination)

\037) Y = CIYl +C2Y2 + ...+ CnYn)

of any n given linearly independent solutions.On this basiswe call such a linearcombination a generalsolutionof the differentialequation.)))

Example6)

2.2GeneralSolutionsof LinearEquations 121)\"\" \"

According to Example4, the particular solutions YI (x) == e-3x, Y2 (x) == cos2x,and Y3(X) ==sin 2xof the linear differentialequationy(3) +3y\"+4y' +12y== 0 arelinearly independent.Now Theorem2 saysthat-givenbo, b I , and b2-thereexistsa particular solution y(x) satisfying the initial conditions y(O) == bo, y/ (0) == b I ,and y\" (0) == b2. HenceTheorem 4 impliesthat this particular solution is a linearcombination of YI, Y2, and Y3. That is, there existcoefficientsCI,C2,and C3 suchthat)

y(x) ==Cle-3x +C2cos2x+C3sin 2x.

Uponsuccessivedifferentiation and substitution of x ==0,we discover that to find

thesecoefficients,we needonly solve the three linear equations)

CI + C2 ==bo,

-3CI + 2C3==b I ,

9CI- 4C2 ==b2.)

(Seethe applicationfor this section.)) .)

NonhomogeneousEquationsWe now considerthe nonhomogeneousnth-order linear differential equation)

\037) yen) + PI(x)y(n-I)+ ...+Pn-I(x)y/+Pn(x)y==I(x)) (2))

with associatedhomogeneousequation)

\037) yen) +PI(x)y(n-I)+ ...+Pn-I(x)y/+ Pn(x)y==O.) (3))

Supposethat a singlefixed particular solution yp of the nonhomogeneousequation in (2)is known, and that Y is any other solutionof Eq.(2).If Yc == Y -Yp,then subsitution of Yc in the differentialequationgives (using the linearity of differ-entiation))

(n) + (n-I)+ + / +Yc PlYc ... Pn-IYc PnYc

== [(y(n) +PI yen-I) + ...+Pn-Iy/ + Pn y]

[( (n) (n-I) /]-

yp +PIYp +...+Pn-IYp + PnYp

==I(x)- I(x)==o.)

Thus Yc == Y -Yp isa solutionof the associatedhomogeneousequationin (3).Then)

\037) (14))y==Yc+YP')

and it follows from Theorem4 that)

\037) (15))Yc ==CIYI +C2Y2 + ...+ CnYn,)

where YI, Y2, ...,Yn are linearly independent solutions of the associatedhomo-geneousequation. We call Yc a complementaryfunction of the nonhomogeneousequationand have thus proved that a generalsolutionof the nonhomogeneousequa-tion in (2)is the sum of its complementaryfunction Yc and a singleparticular solu-tion Yp of Eq. (2).)))


THEOREM5 SolutionsofNonhomogeneousEquationsLet Yp be a particular solution of the nonhomogeneous equation in (2)on an

openinterval I where the functions Pi and f are continuous. Let Yl, Y2, ..., Yn

be linearly independentsolutionsof the associatedhomogeneousequation in (3).If Y is any solution whatsoeverof Eq. (2)on I,then there existnumbers CI,C2,..., Cn suchthat)

Y(x)==CIYl(x)+C2Y2(X)+...+ cnYn(x) +Yp(x) (16))

for all x in I.)

Example7) It is evident that Yp==3xis a particular solutionof the equation)

y\" +4y == 12x,) (17))

and that Yc(x) ==Clcas2x+C2sin 2xis its complementarysolution. Find a solutionof Eq.(17)that satisfiesthe initial conditionsy(O) ==5,y'(0)==7.)

Solution The general solution of Eq.(17)is)

Now)

y(x) == CICOS2x+C2sin 2x+ 3x.)

y'(x) ==-2CIsin 2x+2C2cas2x+3.Hencethe initial conditionsgive)

y(O) == Cl ==5,

y'(O)==2C2+ 3 ==7.)

We find that Cl ==5 and C2 ==2.Thus the desiredsolution is)

IIf!JProblems)

In Problems1through 6, show directly that the givenfunctionsare linearly dependent on the real line. That is, find a non-trivial linear combination of the given functions that vanishes

identically.

1.f(x)= 2x,g(x)= 3x2, hex) = 5x- 8x2

2.f(x)= 5,g(x)= 2- 3x2, hex) = 10+ 15x23.f(x)= 0,g(x)= sin x,hex) = eX

4. f(x)= 17,g(x)= 2 sin 2x,hex) = 3cos2 X

5. f(x)= 17,g(x)= cos2x,hex) = cos2x6.f(x)= eX, g(x)= coshx,hex) = sinh x)

In Problems7 through 12,usethe Wronskian to prove that the

given functions are linearly independent on the indicated in-terval.)

y(x) ==5cos2x+2sin2x+ 3x.) .)

7. f(x) = 1,g(x)=x,hex) = x2; the real line

8.f(x) = eX, g(x)= e2x , hex) = e3x ; the real line

9.f(x) = eX, g(x)= cosx,hex) = sin x;the real line

10.f(x) = eX, g(x)= x-2, hex) =x-2 1n x;x > 011.f(x) =x,g(x)= xeX , hex) = x2ex ; the real line

12.f(x) = x,g(x)= cos(1nx),hex) = sin(1nx);x > 0)

In Problems13through 20,a third-order homogeneous linearequation and three linearly independent solutions are given.Find a particular solution satisfying the given initial condi-tions.)

13.y(3) + 2y\"-

y/-2y = 0; y(O)= 1,y/(O) = 2, y\"(O) = 0;

X -x -2xYl = e , Y2 = e , Y3 = e)))

14.y(3) - 6y\" + IIy'- 6y = 0; y(O) = 0, y'(O) = 0,y\" (0)= 3; YI = eX, Y2 = e2x , Y3 = e3x

15.y(3) -3y\" + 3y'-y = 0; y(O)= 2,y' (0)= 0, y\" (0)= 0;YI = eX, Y2 = XeX, Y3 = X2eX

16.y(3) -5y\"+8y'-4y = 0;y(O)= 1,y'(O)= 4, y\"(O) = 0;YI = eX, Y2 = e2x , Y3 = Xe2x

17.y(3) + 9y' = 0; y(O)= 3,y'(O)= -1,y\"(O) = 2; YI = 1,Y2 = cos3x, Y3 = sin 3x

18.y(3) -3y\"+4y'-2y= 0;y(O)= 1,y'(O)= 0, y\"(O) = 0;YI = eX, Y2 = eX cosx,Y3 = eX sinx.

19.x3y(3) - 3x2

y\" + 6xy'- 6y = 0; y(l) = 6, y'(I)= 14,y\"(I) = 22;YI = x,Y2 = x2, Y3 = x3

20.x3y(3) + 6x2

y\" + 4xy' - 4y = 0; y(l) = 1,y'(I)= 5,y\"(I) = -11;YI = x,Y2 = x-2, Y3 =x-2 lnx)

In Problems 21through 24, a nonhomogeneous differentialequation, a complementary solution Yo and a particular so-lution y p are given. Find a solution satisfying the given initial

conditions.)

21.y\" + y = 3x;y(O)= 2, y'(O)= -2;Yc = CIcosx+ C2sin x;YP = 3x

22.y\"-4y = 12;y(O)= 0, y' (0)= 10;

Yc = cle2x + C2e-2x ; YP = -323.y\" -2y'- 3y = 6; y(O)= 3,y'(O)= 11;

Yc = cle-x + C2e3x; YP = -224. y\"

-2y' + 2y = 2x;y(O)= 4, y' (0)= 8;Yc = CIeX cosx + C2exsin x;y p = x + 1

25.LetLy = y\" + py'+ qy. Supposethat YI and Y2 are twofunctions such that)

LYI = I(x) and LY2 = g(x).)

Show that their sum y = YI + Y2 satisfiesthe nonhomoge-neous equation Ly = I(x)+ g(x).26.(a)Find by inspectionparticular solutions of the two non-

homogeneousequations

y\"+2y=4 and y\"+2y=6x.(b) Usethe method of Problem 25 to find a particular so-lution of the differential equation y\" + 2y = 6x + 4.

27.Prove directly that the functions

II(x)= 1, 12(X) = x, and 13(X) = x 2)


30.Verify that YI = x and Y2 = x2 are linearly independentsolutions on the entire real line of the equation)

x2y\"

-2xy'+ 2y = 0,)

but that W (x, x2) vanishes at x = O.Why do theseobser-vations not contradict part (b) ofTheorem3?

31.This problem indicates why we can imposeonly n initial

conditions on a solution of an nth-order linear differential

equation. (a)Given the equation)\", 0y + py + qy =,)

explain why the value ofy\"(a) is determined by the valuesof y (a) and y'(a). (b) Prove that the equation)

y\"-2y'- 5y = 0)

has a solution satisfying the conditions)

y(O)= 1, y'(O)= 0, and y\"(O) = C)

32.)if and only if C = 5.Prove that an nth-order homogeneous linear differential

equation satisfying the hypotheses ofTheorem2 has n lin-

early independent solutions YI, Y2, ..., Yn. (Suggestion:LetYi be the unique solution such that)

(i-I)( ) - 1Yi a-) and Yi(k) (a) = 0 if k i= i-I.))33.Supposethat the three numbers rl, r2, and r3 are dis-

tinct. Show that the three functions exp(rlx),exp(r2x),and exp(r3x)are linearly independent by showing that

their Wronskian)

1 1 1)

W = exp[(rl+ r2 + r3)x].rl r2 r3)

r2 r2 r2I 2 3)

is nonzero for all x.34. Assume asknown that the Vandermonde determinant)

are linearly independent on the whole real line. (Sugges-tion: Assume that CI + C2X + C3X2 = O.Differentiate this 1 1 1equation twice, and concludefrom the equations you get

rl r2 rnthat CI = C2 = C3 = 0.)28.Generalizethe method ofProblem27 to provedirectly that v= r2 r2 r2I 2 n

the functions)

10(x) = 1, II(x)= x, 12(x) = X2, ..., In (X) = X n)

are linearly independent on the real line.29.Usethe result of Problem 28 and the definition of linear

independenceto provedirectly that, for any constant r, thefunctions)

lo(x)= erx,) II(x) = xerx

,) ...,) In (x) = xnerx)

are linearly independent on the whole real line.)

n-I n-Ir l r2)

n-Irn)

is nonzero if the numbers rl , r2, ..., rn aredistinct. Prove

by the method of Problem 33 that the functions)

fi(x) = exp(rix),) 1 < i < n- -)

are linearly independent.)))

35.According to Problem 32 of Section2.1,the WronskianW (YI , Y2) of two solutions of the second-orderequation

y\" + PI (x)y/ + P2(X)Y = 0

is given by Abel's's formula

W(x) = K exp(-f PI(x)dx)

for someconstant K. It can be shown that the Wronskianofn solutions YI, Y2, ..., Yn of the nth-order equation

y(n) + PI(X)y(n-I) +...+ Pn-I(X)y/ + Pn(X)Y = 0

satisfiesthe sameidentity. Prove this for the casen = 3as follows: (a) Thederivative of a determinant of func-tions is the sum of the determinants obtained by separatelydifferentiating the rows of the original determinant. Con-cludethat)

YI) Y2) Y3)

w/ =Y\037 Y\037 Y\037

(3) (3) (3)Y1 Y2 Y3

(b) Substitute for Y?), yi3), and yj3) from the equation

y(3) + PIy\" + P2Y/ + P3Y = 0,

and then show that W/ = -PI W. Integration now givesAbel's formula.

36.Supposethat one solution YI (x) of the homogeneoussecond-orderlinear differential equation)

y\" + p(x)y/+ q(x)y= 0) (18))

is known (on an interval I where P and q are continuous

functions). The method of reduction of orderconsistsof substituting Y2(X) = V(X)YI (x) in (18)and attemptingto determine the function v (x) so that Y2 (x) is a secondlinearly independent solution of (18).After substitutingY = V(X)YI (x) in Eq.(18),use the fact that YI (x) is asolution to deducethat)

YI v\" + (2y\037 + PYI )v/ = o.) (19))

If YI (x) is known, then (19)is a separableequation thatis readily solvedfor the derivative v/ (x) of v(x). Integra-tion of v/ (x) then gives the desired(nonconstant) functionv (x).)

37.Before applying Eq. (19)with a given homogeneoussecond-orderlinear differential equation and a known so-lution YI (x),the equation must first bewritten in the form

of (18)with leading coefficient 1 in order to correctlydetermine the coefficientfunction P(x).Frequently it ismore convenient to simply substitute Y = v (x)YI (x) in

the given differential equation and then proceeddirectlyto find v (x).Thus, starting with the readily verified solu-tion YI (x) = x3 of the equation)

x2y\"

-5xy'+ 9y = 0 (x > 0),)

substitute Y = vx 3 and deducethat x v\" + v/ = O.Thencesolvefor vex) = CIn x,and thereby obtain (with C = 1)the secondsolution Y2 (x) = x3 In x.)

In eachofProblems38 through 42,a differential equation andonesolution YI aregiven. Usethe method ofreduction oforderas in Problem37 to find a secondlinearly independent solution

Y2.

38.x2y\" + xy/ -9y = 0 (x > 0); YI (x) = x3

39.4y\" -4y' + Y = 0; YI (x) = exI2

40. x2y\"

-x(x+ 2)y'+ (x + 2)y = 0 (x > 0); YI (x) = x41.(x + l)y\"

- (x + 2)y' + Y = 0 (x > -1);YI(X) = eX

42. (1-x2)y\" + 2xy'-2y = 0 (-1< x < 1);YI (x) = x

43. First note that YI (x) = x is one solution of Legendre'sequation oforder 1,)

(1-x2)y\" -2xy'+ 2y = O.)

Then use the method of reduction of order to derive the

secondsolution)

x 1+xY2(X) = 1--In (for-l< x < 1).2 I-x)

44. First verify by substitution that YI (x) = X-1/2cosx is onesolution (for x > 0)ofBessel'sequation oforder

\037,)

x2y\" + xy/ + (x2 -

\037)y= O.)

Then derive by reduction of order the secondsolution

Y2(X) = X-I/2 sinx.)

_ I-I0mogeneousEq1Jatiolls with C()nstantCoeffic\037ents)

In Section2.2we saw that a general solution of an nth-order homogeneouslinearequation is a linear combinationof n linearly independentparticular solutions, but

we said little about how actually to find even a singlesolution. The solution of alinear differential equation with variable coefficientsordinarily requiresnumericalmethods (Chapter 6)or infinite seriesmethods (Chapter 3). But we can now showhow to find, explicitly and in a rather straightforward way, n linearly independentsolutionsof a given nth-order linear equation if it has constant coefficients. The)))

2.3HomogeneousEquationswith ConstantCoefficients125)

general such equation may be written in the form)

\037) any(n) +an_Iy(n-l) + ...+ a2Y// +alY/ +aoy ==0,) (1))

where the coefficientsao, aI,a2, ..., an are real constantswith an f=. O.)

The CharacteristicEquationWe first lookfor a singlesolutionof Eq.(1),and begin with the observationthat)

dk_ (erx) ==rkerx

dxk ') (2))

so any derivative of erx is a constant multiple of erx . Hence,if we substituted

y ==erx in Eq.(1),each term would be a constant multiple of erx , with the constantcoefficientsdependingon r and the coefficientsak.This suggeststhat we try to find

r so that all thesemultiples of erx will have sum zero,in which casey ==erx will bea solutionof Eq.(1).

For example,in Section2.1we substituted y ==erx in the second-orderequa-)tion)

// b / 0ay + y +cy ==)

to derive the characteristicequation)

ar2 +br +c ==0)

that r must satisfy.Tocarryout this technique in the generalcase,we substitute y ==erx in Eq.(1),

and with the aid of Eq.(2)we find the result to be)

a rnerx +a rn-1erx +...+a r 2erx +a rerx +a erx ==o.n n-l 2 I 0 ,)

that is,)

erx(anr

n +an_Ir n-1 + ...+a2r2 +aIr+ao) ==O.

Becauseerx is neverzero,we seethat y ==erx will bea solutionof Eq.(1)preciselywhen r is a root of the equation)

\037) anrn +an_Ir n-1 + ...+a2r2 +aIr+ao ==o.) (3))

This equation is calledthe characteristicequationor auxiliary equationof the

differential equation in (1).Our problem,then, is 'reducedto the solution of this

purely algebraicequation.According to the fundamental theorem of algebra, every nth-degree poly-

nomial-suchas the one in Eq. (3)-hasn zeros,though not necessarilydistinct

and not necessarilyreal.Finding the exact values of thesezerosmay be difficult

or even impossible;the quadratic formula is sufficient for second-degreeequations,but for equations of higher degreewe may need either to spot a fortuitous fac-torization or to apply a numerical technique such as Newton'smethod (or useacalculator/computersolvecommand).)))


DistinctRealRoots)

Whatever the method we use,let us supposethat we have solvedthe characteristicequation. Then we can always write a general solution of the differentialequation.The situation is slightly more complicatedin the caseof repeatedroots or complexroots of Eq.(3),so let us first examine the simplestcase-inwhich the characteristicequation has n distinct (no two equal) realroots rl,r2, ..., rn. Then the functions)

ert x er2x ernx, ...,)are all solutionsof Eq.(1),and (by Problem 34 of Section2.2)thesen solutionsarelinearly independenton the entire real line.In summary, we haveprovedTheorem 1.)

THEOREM1 DistinctRealRoots)

If the roots rl,r2, ..., rn of the characteristicequation in (3)are real and distinct,then)

\037) y(x) = Clertx +C2er2x + ...+cnernx) (4))

is a general solution of Eq.(1).)

Example1)

...........

Solvethe initial valueproblem

y(3) + 3y\"- 10y'= 0;

y(O)= 7, y/ (0)= 0, y\" (0)= 70.)

Solution The characteristicequation of the given differentialequation is

r 3 +3r2- lOr = O.)

We solve by factoring:)

r(r2 +3r- 10)= r(r +5)(r-2)= 0,)

and so the characteristic equation has the three distinct real roots r = 0,r = -5,and r = 2.BecauseeO= 1,Theorem 1 gives the general solution

( ) + -5x + 2xY X = Cl C2e C3e .)

Then the given initial conditionsyield the linear equations)

y(O)= Cl +y/ (0)=

y\" (0)=)

C2+ C3 = 7,5C2+ 2C3= 0,

25c2+ 4C3= 70)

in the coefficientsCl, C2, and C3. The last two equations give y\" (0)- 2y/ (0)35c2 = 70,so C2 = 2.Then the secondequation gives C3 = 5,and finally the first

equation gives Cl = O.Thus the desiredparticular solution is)

y(x) = 2e-5x +5e2x.) .)))


PolynomialDifferentialOperatorsIf the roots of the characteristicequation in (3)are not distinct-thereare repeatedroots-thenwe cannot producen linearly independent solutionsof Eq. (1)by themethod of Theorem 1.For example,if the roots are 1,2,2,and 2,we obtain onlythe two functions eX and e2x.The problem,then, is to producethe missing linearly

independentsolutions.For this purpose,it isconvenientto adopt \"operatornotation\"

and write Eq.(1)in the form Ly = 0,where the operator)

dn dn-1 d2 dL=an-+an-l +...+a2-+aI-+aO (5)dxn dxn-I dx2 dx)

operateson the n-times differentiablefunction y(x) to producethe linearcombina-tion)

Ly = anyCn) +an_1yCn-l)+ ...+a2y(2)+aIY/ +aoy

of y and its first n derivatives. We also denote by D = d/dx the operationofdifferentiationwith respectto x,so that)

Dy = y/, D2y = y\", D3y = y(3),)

and soon.In terms of D,the operator L in (5)may be written)

L = anDn +an_1Dn-I +...+a2D2+aID+ao,) (6))

and we will find it useful to think of the right-hand sidein Eq.(6)asa (formal) nth-

degreepolynomial in the \"variable\" D;it is a polynomialdifferential operator.A first-degree polynomial operator with leading coefficient 1 has the form

D -a, where a is a real number. It operateson a function y = y(x) to produce)

(D-a)y = Dy -ay = y/-ay.)

The important fact about such operators is that any two of them commute:)

(D-a)(D-b)y = (D-b)(D-a)y) (7))

for any twice differentiable function y = y(x). The proof of the formula in (7) isthe followingcomputation:)

(D-a)(D-b)y = (D-a)(y/ -by)= D(y'-by) -aCy'-by)= y\"

- (b+a)y/ +aby = y\"- (a+b)y' +bay

= D(y'-ay) -b(y' -ay)= (D-b)(y'-ay) = (D-b)(D-a)y.)

We seehere also that (D-a)(D-b) = D2 - (a +b)D+ab. Similarly, it canbe shown by induction on the number of factors that an operator product of the

form (D-al)(D- a2) ...(D- an) expands-bymultiplying out and collectingcoefficients-inthe sameway as does an ordinary product (x- aI)(x- a2) ...(x-an) of linear factors, with xdenotinga real variable. Consequently,the algebraof polynomial differentialoperators closelyresemblesthe algebra of ordinary realpolynomials.)))


RepeatedRealRootsLetus now considerthe possibilitythat the characteristicequation)

anrn +an_lr n-1 +...+a2r2 +aIr+ao ==0) (3))

has repeatedroots.For example,supposethat Eq.(3)has only two distinct roots, roof multiplicity 1 and rl of multiplicity k == n - 1 > 1.Then (after dividing by an)Eq.(3)can be rewritten in the form)

(r - rl)k(r- ro) == (r - ro)(r- rl)k ==o.) (8))

Similarly, the correspondingoperator L in (6)can be written as)

L == (D- rl)k(D- ro) == (D- ro)(D- rl)k,) (9))

the orderof the factors making no differencebecauseof the formula in (7).Two solutions of the differentialequation Ly ==0 are certainly Yo ==erox and

Yl == ert x. This is, however, not sufficient; we needk + 1 linearly independentsolutions in orderto construct a general solution, becausethe equation is of orderk + 1.To find the missingk - 1 solutions,we note that)

Ly == (D- ro)[(D- rl)k y ] ==O.)

Consequently,everysolution of the kth-order equation)

(D- rl)ky ==0) (10))

will also be a solution of the original equation Ly == O. Henceour problem isreducedto that of finding a general solution of the differentialequation in (10).

The fact that Yl == erl x is one solution of Eq. (10)suggeststhat we try thesubstitution)

y(x) ==U(X)YI(X) ==u(x)erIX,) (11))

where u (x)is a function yet to bedetermined.Observethat

(D- rl) [uerIX] == (Du)erIX +u(rlerIX ) - rl(uerIX ) == (Du)er1x. (12)

Uponk applicationsof this fact, it follows that)

(D- rI)k [uerIX] == (Dku )erlX) (13))

for any sufficiently differentiablefunction u (x).Hencey ==uerl x will bea solutionof Eq. (10)if and only if Dku ==u Ck ) ==O.But this is so if and only if

u(x)==Cl +C2X +C3x2 + ...+Ckxk-l

,)

a polynomialof degreeat most k -1.Henceour desiredsolutionof Eq.(10)is

y(x) ==uer1x == (CI+C2X +C3x2 +...+Ckxk-I)er\\x.)

In particular,we seehere the additional solutionsxerlX , x2erlX , ..., xk-Ier1x of the

original differentialequation Ly ==O.The precedinganalysis can be carriedout with the operator D - rl replaced

with an arbitrary polynomialoperator. When this isdone,the result is a proof of the

following theorem.)))


THEOREM2 RepeatedRootsIf the characteristic equation in (3)has a repeatedroot r of multiplicity k, then

the part of a general solution of the differentialequation in (1)correspondingtor is of the form)

>-) (CI +C2X +C3x2 + ...+Ckxk-l )erx

.) (14))

We may observethat, accordingto Problem 29 of Section2.2,the k func-tions erx , xerx , x2erx , ..., and xk-1erx involved in (14)are linearly independentonthe real line. Thus a root of multiplicity k correspondsto k linearly independentsolutionsof the differentialequation.)

Example2).\037..

Find a general solutionof the fifth-orderdifferentialequation

9y(5) _ 6y

(4) + y(3) ==o.)

Solution The characteristicequation is

9r5 -6r4 + r 3 ==r 3(9r2-6r+ 1)==r 3(3r- 1)2==o.It has the triple root r == 0 and the doubleroot r == 1.The triple root r == 0contributes

CleO.x +C2XeO.x +c3x2eO.x ==Cl +C2X +C3x2

to the solution, while the doubleroot r == 1contributesC4ex/3+C5xex/3.Henceageneral solutionof the given differentialequation is)

y(x) ==Cl +C2X +C3x2 +C4ex/3+C5xex/3.) .)

Complex-ValuedFunctionsandEuler'sFormulaBecausewe have assumedthat the coefficientsof the differential equation and its

characteristicequation are real, any complex(nonreal) roots will occurin complexconjugatepairsa ::f::bi wherea and b are real and i ==H.This raisesthe questionof what might be meant by an exponentialsuch as eCa+bi)x.

To answer this question, we recallfrom elementary calculus the Taylor(orMacLaurin)seriesfor the exponentialfunction)

00tn t 2 t3 t 4

et ==L- == 1 + t +-+-+-+ ... .n=0 n! 2! 3! 4!)

If we substitute t == i () in this seriesand recall that i2 ==-1,i3 ==-i,i4 == 1,andsoon, we get)

ei(J = >;(io)n

n=On!)

()2 i()3 ()4 i ()5== 1 + i() ----+-+--...2! 3! 4! 5!)

== (1 _ 02+ 04

_ ...) + i (0 _ 03+ 05 -...

) .2! 4! 3! 5!)))


Becausethe two real seriesin the last line are the Taylorseriesfor cos() and sin (),respectively,this impliesthat)

>-) eie ==cos() + i sin ().) (15))

This result is known as Euler'sformula.Becauseof it, we define the exponentialfunction eZ , for z ==x + iy an arbitrary complexnumber, to be)

eZ ==ex+iy ==eX eiy ==eX (cosy + i sin y).) (16))

Thus it appearsthat complexroots of the characteristicequation will lead tocomplex-valuedsolutionsof the differentialequation. A complex-valuedfunctionF of the real variablex associateswith eachreal number x (in its domainof defini-tion) the complexnumber)

F(x)==f(x) + ig(x).) (17))

The real-valuedfunctions f and g are calledthe realand imaginaryparts, respec-tively, of F.If they are differentiable,we define the derivative F'of F by)

F'(x) ==f'(x)+ ig'(x).) (18))

Thus we simply differentiate the real and imaginary parts of F separately.We say that the complex-valuedfunction F(x)satisfiesthe homogeneouslin-

ear differential equation L[F(x)]== 0 provided that its real and imaginary parts in

(17)separately satisfy this equation-soL[F(x)]==L[f(x)]+ iL[g(x)]==O.The particular complex-valued functions of interest here are of the form

F(x) ==erx , where r ==a ::i:bi.We note from Euler'sformula that)

e(a+bi)x==eaxeibx ==eax (cosbx + i sin bx)) (19a))

and)

e(a-bi)x==eaxe-ibx ==eax (cosbx - i sin bx).) (19b))

The most important property of erx is that)

Dx(erx) ==rerx

,) (20))

if r is a complexnumber. The proof of this assertionis a straightforward computa-tion basedon the definitionsand formulas given earlier:)

Dx(erx ) == Dx(eax cosbx)+ iDx(eaxsinbx)== (ae

ax cosbx -beax sin bx)+ i (aeax sin bx +beax cosbx)

== (a+bi)(eax cosbx + ieax sin bx)==rerx.)))

Example3)

Example4)


ComplexRootsIt follows from Eq.(20)that when r is complexGust aswhen r is real),erx will beasolutionof the differentialequation in (1)if and only if r isa rootof its characteristicequation. If the complexconjugate pair of roots rl == a + bi and r2 ==a -bi aresimple(nonrepeated),then the correspondingpart of a general solution of Eq.(1)is)

y(x) ==Cler1x +C2er2x ==Cle(a+bi)x+C2e(a-bi)x==Cleax(cosbx+ i sinbx)+C2eax(cosbx- i sinbx)

y(x) == (CI+C2)eax cosbx+ i(CI-C2)eax

sinbx,)

where the arbitrary constants CI and C2 can be complex.For instance, the choiceCI == C2 == 4 gives the real-valued solution YI (x) == eax cosbx,while the choiceCI ==-4i , C2 == 4i gives the independentreal-valuedsolutionY2(X) ==eax sinbx.This yieldsthe followingresult.)

THEOREM3 ComplexRoots)

If the characteristicequation in (3)has an unrepeatedpair of complexconjugateroots a ::f::bi (with b =f. 0), then the correspondingpart of a general solution ofEq.(1)has the fonn)

\037) eax(CIcosbx +C2sin bx).) (21))

The characteristicequation of)

y\" +b2y ==0 (b > 0))

is r 2 + b2 ==0,with roots r == ::f::bi.SoTheorem 3 (with a ==0)gives the generalsolution)

y(x) ==CICOSbx +C2sin bx.) .)

Find the particular solutionof)

y\"-4y' +5y ==0)

for which y(O) == 1 and y'(0)==5.)

Solution Completionof the square in the characteristicequationyields)

r 2-4r +5 == (r -2)2+ 1 ==0,)

sor -2 == ::f::H ==::f::i.Thus we obtain the complexconjugateroots2::f::i (whichcouldalsobe found directly using the quadratic formula). HenceTheorem3 with

a ==2 and b == 1 gives the general solution)

y(x) ==e2x(CIcosx+C2sin x).)))


y)

(x,y))

FIGURE2.3.1.Modulus and

argument of the complexnumberx + iy.)

Example5)

Then)

y'(x) ==2e2x(CICOSX+C2sin x) +e2x(-clsin x +C2cosx),)

so the initial conditions give)

y(O) ==CI == 1 and y'(0)==2CI+C2==5.)

It follows that C2 ==3, and so the desiredparticular solution is)

y(x) ==e2x(cosx+3 sin x).) .)In Example5 belowwe employ the polarform)

z ==x + iy ==reie)

(22))

of the complexnumber z.This form follows from Euler'sformula upon writing)

z = r (;+ i\037 )

= r(cose+ i sine)= rei6)

x)

in terms of the modulus r ==Jx2 +y2 > 0 of the number z and its argument()indicated in Fig.2.3.1.For instance, the imaginary number i has modulus 1 and

argument Jr/2,so i ==eiJr /2.Similarly, -i==e3Jr /2.Another consequenceis the factthat the nonzero complexnumber z ==reie has the two squareroots)

v'Z == ::f::(reie)I/2==::f::,Jreie/2,) (23))

where ,Jrdenotes(asusual for a positive real number) the positive square root ofthe modulus of z.)

Find a general solution of y(4) +4y ==O.)

Solution The characteristicequation is)

r 4 -4 == (r2)2 - (2i)2== (r2 +2i)(r2-2i) ==0,)

and its four roots are ::f::,J::f::2i.Sincei ==eiJr /2 and -i==ei3Jr/2, we find that)

.../2i= (2eiJf /2)1/2= heiJf /4 = h (cos:+ i sin:)

= 1 + i)

and)

,J 2i = (2ei3Jf /2f/

2 = hei3Jf /4 = h (cos3:+ i sin 3:) = -1+ i.)

Thus the four (distinct)roots of the characteristicequationare r == ::f::(::f::1+i).Thesetwo pairsof complexconjugate roots,1 ::f::i and -1::f::i,give a general solution)

y(x) ==eX (CIcosx+C2sinx)+e-x (C3cosx+C4sinx))

of the differentialequation y(4) +4y ==O.) .)))

Example6)


RepeatedComplexRootsTheorem 2 holdsfor repeatedcomplexroots.If the conjugatepair a ::f::bi has mul-

tiplicity k, then the correspondingpart of the general solutionhas the form)

(AI + A 2x+...+ Akxk-l)eCa+bi)x + (Bl + B2X +...+ Bkxk-l)eCa-bi)xk-l

==LxPeax(Ci cosbx +di sin bx). (24)

p=o)

It can be shown that the 2k functions)

xPeax cosbx,) xPeax sin bx,) 0 <p <k-1)that appearin Eq.(24)are linearly independent.)

Find a general solutionof (D2 +6D+ 13)2y==O.)

Solution By completing the square,we seethat the characteristicequation)

Example7)

(r2 +6r+ 13)2== [(r+3)2+4]2 ==0)

has as its roots the conjugate pair -3::f::2i of multiplicity k == 2.HenceEq.(24)gives the general solution)

y(x) ==e-3x(Clcos2x +dl sin 2x)+xe-3x

(C2cos2x +d2sin 2x). .)In applications we are seldompresentedin advance with a factorizationas

convenientas the one in Example6.Often the most difficult part of solving a homo-geneouslinear equation is finding the roots of its characteristicequation. Example7 illustratesan approach that may succeedwhen a root of the characteristicequationcan be found by inspection.)

\037 \" co'''C\"' C\"'\" .-. C\"' \037. _N.. ..,_u.'u

The characteristicequationof the differentialequation)

y(3) +y' - 10y==0)

is the cubicequation)r 3 +r-10==0.

By a standard theorem of elementary algebra, the only possiblerational roots arethe factors ::f::1,::f::2,::f::5,and ::f::10of the constant term 10.By trial and error (if not

by inspection)we discoverthe root 2. The factor theorem of elementaryalgebraimpliesthat r -2 is a factor of r 3 +r -10,and divisionof the former into the latter

producesas quotient the quadraticpolynomial)

r 2 +2r+5 == (r + 1)2+4.)

The roots of this quotient are the complexconjugates-1::f::2i.The three roots wehave found now yield the general solution)

y(x) ==Cle2x +e-x (C2cos2x +C3sin 2x).) .)))

Example8)\037 \037 _ A

The roots of the characteristicequation of a certain differentialequation are 3, -5,0,0,0,0,-5,2 ::f::3i,and 2 ::f::3i.Write a general solution of this homogeneousdifferentialequation.)

Solution The solution can be read directly from the list of roots.It is)

y(X) ==CI +C2X +C3x2 +C4x3 +cse3x +C6e-sx +C7xe-sx

+e2x(cscos3x +Cg sin 3x)+xe2x(CIOcos3x +CII sin 3x). .)

IfI1Problemsu)

Find the generalsolutions ofthe differential equations in Prob-lems 1through 20.)

1.y\"-4y == 0

3.y\" + 3y'- 10y == 05. y\" + 6y' + 9y == 07. 4y\"

- 12y'+ 9y == 09.y\" + 8y'+ 25y == 011.y(4) - 8y(3)+ 16y\" == 012.y(4) - 3y(3)+ 3y\"

- y' == 013.9y(3)+ 12y\" + 4y' == 0 14.y(4) + 3y\"

-4y == 015.y(4) - 8y\" + 16y== 0 16.y(4) + 18y\" + 81y == 017.6y(4)+ lly\" + 4y == 0 18.y(4) == 16y19.y (3)+ y\"

-y' - y == 020.y(4) + 2y(3) + 3y\" + 2y' + y == 0 (Suggestion: Expand

(r2 +r+l)2.))

2. 2y\"- 3y' == 0

4. 2y\"-7y' + 3y == 0

6. y\" + 5y' + 5y == 08. y\"

-6y' + 13y == 010.5y(4) + 3y(3) == 0)

Solvethe initial value problems given in Problems21through26.

21.y\"-4y' + 3y == 0; y(O) == 7, y'(O)== 11

22.9y\" + 6y' + 4y == 0; y(O) == 3, y'(O)== 423.y\"

-6y' + 25y == 0; y(O) == 3, y'(O)== 124. 2y(3) - 3y\"

-2y' == 0; y(O) == 1,y'(O)== -1,y\"(O) == 325.3y(3) + 2y\" == 0; y(O) == -1,y'(O)== 0, y\"(O) == 126.y(3) + 10y\" + 25y' == 0; y(O) == 3, y' (0) == 4, y\" (0) == 5)

Find generalsolutions ofthe equations in Problems27 through32. First find a small integral root of the characteristicequa-tion by inspection;then factor by division.

27.y(3) + 3y\"-4y == 0

28.2y(3) -y\" -5y'-2y == 029.y(3) + 27y == 030.y(4)

- y(3) + y\"-3y'- 6y == 0

31.y(3) + 3y\" + 4y' - 8y == 032.y(4) + y(3) - 3y\"

-5y'- 2y == 0)

In Problems 33 through 36, one solution of the differentialequation is given. Find the generalsolution.

33.y(3) + 3y\"-54y == 0; y == e3x

34. 3y(3)- 2y\" + 12y'- 8y == 0; y == e2x /3

35.6y(4)+ 5y(3) + 25y\" + 20y'+ 4y == 0; y == cos2x36.9y(3)+ lly\" + 4y' - 14y == 0; y == e-X sin x)

37.Find a function y(x) such that y(4)(x) == y(3)(X) for all xand y(O) == 18,y'(O)== 12,y\"(O) == 13,and y(3)(0) == 7.

38.Solvethe initial value problem)

y(3) _5y\" + 100y'-500y== 0;

y(O) == 0, y' (0) == 10, y\" (0) == 250)

given that YI (x) == e5x is one particular solution of the

differential equation.)

In Problems39 through 42, find a linear homogeneousconstant-coefficient equation with the given generalsolution.

39.y(x) == (A + Bx + Cx2)e2x

40. y(x) == Ae2x + B cos2x + Csin 2x41.y(x) == A cos2x + B sin 2x + Ccosh2x + D sinh 2x42. y(x) == (A+Bx+Cx2)cos2x+(D+Ex+Fx2)sin2x)

Problems 43 through 47pertain to the solution ofdifferential

equations with complexcoefficients.

43. (a) UseEuler'sformula to show that every complexnum-

ber can be written in the form rei(), where r > 0 and

-Jr < e < Jr. (b) Expressthe numbers 4, -2,3i,1+ i,and -1+ i,J3in the form rei(). (c) Thetwo squareroots of rei() are :!::.-Jiei()/2.Find the square roots of the

numbers 2-2i,J3and -2+ 2i,J3.44. Usethe quadratic formula to solvethe following equa-

tions. Note in eachcase that the roots are not complexconjugates.(a) x2 + ix + 2 == 0 (b) x2 -2ix + 3 == 0

45. Find a general solution of y\"-2iy' + 3y == O.

46. Find a general solution of y\"- iy' + 6y == 0

47. Find a general solution of y\" == (-2+ 2i,J3)y.48. Solvethe initial value problem)

y(3) == y; y(O) == 1, y' (0) == y\" (0) == O.)

(Suggestion: Imposethe given initial conditions on the

general solution)

y(x) == Aex + Beax + Ce/3x,)))

where a and fJ arethe complexconjugate rootsofr 3-1=0, to discoverthat)

1

(x\037

)y(x) =\"3

eX + 2e-x /2cos2)is a solution.)

49. Solvethe initial value problem)

y(4) = y(3) + y\" + y' + 2y;

y(O)= y' (0) = y\" (0)= 0, 2y(3)(0)= 30.)

50.Thedifferential equation)

y\" + (sgnx)y= 0) (25))

has the discontinuous coefficientfunction)

{+1

sgnx=-1)

if x > 0,if x < o.)

Show that Eq.(25)neverthelesshas two linearly indepen-dent solutions YI (x) and Y2 (x)defined for all x such that. Each satisfiesEq.(25)at eachpoint x i= 0;. Each has a continuous derivative at x = 0;. YI (0)=

y\037(O)= 1 and Y2(0) = y\037 (0)= O.

(Suggestion: Each Yi (x)will be defined by one formulafor x < 0 and by another for x > 0.)Thegraphs of thesetwo solutions areshown in Fig. 2.3.2.)

Ell) MechanicalVibrations)

m c)

I

I

Equilibrium

position)

FIGURE2.4.1.A mass-spring-dashpot system.)

2.4MechanicalVibrations 135)

Y2(x))

YI (x))Y

4)

-4)

51.)

FIGURE2.3.2.Graphs of YI (x) and

Y2 (x) in Problem 50.

According to Problem 51in Section2.1,the substitutionv = In x (x > 0) transforms the second-orderEuler equa-tion ax2

y\" + bxy' + cy = 0 to a constant-coefficient ho-mogeneouslinear equation. Show similarly that this samesubstitution transforms the third-order Euler equation)

311I b 211 ' d 0ax y + x y + cxy + y =)

(where a, b, c, d are constants) into the constant-coefficientequation)

d3y d2

y dya-+ (b -3a)-+ (c-b + 2a)-+ dy =O.dv 3 dv 2 dv)

Make the substitution v = In x ofProblem 51to find generalsolutions (for x > 0) of the Euler equations in Problems 52throug h 58.

52.x2y\" + xy' + 9y = 053.x2y\" + 7xy' + 25y = 054. X3

ylll + 6x2y\" + 4xy' = 0

55.X3y

lll -x2y\" + xy' = 0

56.x3y'\" + 3x2

y\" + xy' = 057.X3

ylll - 3x2

y\" + xy' = 058.X3

ylll + 6x2

y\" + 7xy' + y = 0)

The motion of a massattached to a spring servesas a relatively simple exampleof the vibrations that occurin more complexmechanical systems.For many suchsystems,the analysisof thesevibrations is a problemin the solutionof lineardiffer-ential equations with constant coefficients.

We considera body of massm attached to one end of an ordinary spring that

resistscompressionas well as stretching; the other end of the spring is attachedtoa fixed wall, as shown in Fig.2.4.1.Assume that the body restson a frictionlesshorizontal plane, so that it can move only backand forth as the spring compressesand stretches.Denoteby x the distanceof the body from its equilibriumposition-its position when the spring is unstretched. We take x > 0 when the spring isstretched,and thus x <0 when it is compressed.

According to Hooke'slaw, the restorative force Fs that the spring exerts onthe massis proportionalto the distancex that the spring has beenstretchedor com-pressed.Becausethis is the sameas the displacementx of the mass m from its)))

$j

i) Unstretched

spnng)

Static

equilibrium)

Systemin motion)

y)

FIGURE2.4.2.A masssuspendedvertically from a spring.)

equilibrium position,it follows that)

Fs==-kx.) (1))

The positive constant of proportionalityk is calledthe springconstant.Note that

Fsand x have oppositesigns:Fs<0 when x >0,Fs >0 when x <o.Figure 2.4.1showsthe massattached to a dashpot-adevice,like a shock

absorber,that providesa force directedoppositeto the instantaneous direction ofmotion of the massm.We assumethe dashpot is so designedthat this force FR isproportional to the velocity v ==dxjdtof the mass;that is,)

dxFR == -ev==-e-.dt)

(2))

The positive constant e is the damping constantof the dashpot.Moregenerally,we may regard Eq. (2)as specifying frictional forcesin our system (including airresistanceto the motion of m).

If, in addition to the forcesFsand FR , the massis subjectedto a given exter-nal forceFE == F(t),then the total force acting on the massis F == Fs+FR +FE.UsingNewton'slaw)

d2x \"F ==ma ==m-==mxdt2 '

we obtain the second-orderlinear differentialequation)

\037) mx\" +ex'+kx == F(t)) (3))

that governs the motion of the mass.If there is no dashpot (and we ignore all frictional forces),then we sete ==0

in Eq. (3)and call the motion undamped;it is dampedmotion if e >O.If there isno external force, we replaceF(t) with 0 in Eq. (3).We refer to the motion asfreein this caseand forcedin the caseF(t) =f. O.Thus the homogeneousequation)

\037)\" , k 0mx +ex + x ==) (4))

describesfree motion of a masson a spring with dashpotbut with no externalforcesapplied.We will defer discussionof forced motion until Section2.6.

For an alternative example,we might attach the mass to the lower end of aspring that is suspendedverticallyfrom a fixedsupport,as in Fig.2.4.2.In this casethe weight W == mg of the masswould stretch the spring a distanceSodeterminedby Eq. (1)with Fs==-Wand x ==So.That is, mg ==kso, so that So ==mgjk. Thisgives the static equilibrium position of the mass. If y denotesthe displacementofthe massin motion, measureddownward from its static equilibrium position, then

we askyou to show in Problem 9 that y satisfiesEq.(3);specifically,that)

my\" +ey' +ky == F(t)) (5))

if we includedamping and externalforces(meaningthoseother than gravity).)))

o)

FIGURE2.4.3.Thesimplependulum.)


TheSimplePendulumThe importance of the differential equation that appearsin Eqs.(3)and (5) stemsfrom the fact that it describesthe motion of many other simplemechanicalsystems.For example,a simplependulumconsistsof a massm swinging backand forth onthe end of a string (or better, a masslessrod)of length L, as shown in Fig.2.4.3.We may specify the position of the massat time t by giving the counterclockwiseangle () == e(t) that the string or rod makeswith the vertical at time t. To analyzethe motion of the massm, we will apply the law of the conservationof mechanicalenergy, accordingto which the sum of the kinetic energy and the potential energyof m remains constant.

The distancealong the circular arc from 0 to m is s == L(),so the velocityofthe massis v ==dsjdt ==L(d()jdt),and therefore its kinetic energy is)

T == \037mv2 == \037m (dS

)2

= \037mL2 (de

)2.2 2 dt 2 dt)

We next chooseas referencepoint the lowestpoint 0 reachedby the mass (seeFig.2.4.3).Then its potential energy V is the product of its weight mg and its

vertical height h ==L(1-cos()) above 0,so)

V ==mgL(I-cos()).)

The fact that the sum of T and V is a constant C thereforegives)

1

(d()

)2

-mL2 - +mgL(I-cose)==c.2 dt)

We differentiateboth sidesof this identity with respectto t to obtain)

(d()

)(d2()

). d()

mL2 -\037 +mgL(sIn())-==0,

dt dt dt)

so)

d2() g.-+-SIn () ==0dt2 L)

(6))

after removal of the common factor mL2(d()ldt).This differential equation canbe derived in a seemingly more elementary manner using the familiar secondlaw

F == ma of Newton (appliedto tangential components of the accelerationof the

mass and the force acting on it). However, derivations of differential equationsbasedon conservationof energy are often seenin more complexsituations whereNewton'slaw isnot sodirectlyapplicable,and it may be instructive to seethe energymethod in a simplerapplicationlikethe pendulum.

Now recallthat if () is small, then sin () \037 () (this approximationobtainedby

retaining just the first term in the Taylorseriesfor sin e).In fact, sin () and e agreeto two decimalplaceswhen I()I is at most ]T/12(that is, 15\302\260).

In a typical pendulumclock,for example,e would never exceed15\302\260. It therefore seemsreasonableto

simplify our mathematicalmodel of the simplependulum by replacingsin e with ein Eq.(6).If we alsoinsert a term C()'to account for the frictional resistanceof the

surrounding medium, the result is an equation in the form of Eq.(4):)

()\" +C()'+k() ==0,) (7))))

A)

FIGURE2.4.4.Theangle Ci.)

where k == g/L. Note that this equation is independent of the massm on the endof the rod.We might, however,expectthe effects of the discrepancybetween () andsin () to accumulateover a periodof time, so that Eq.(7)will probably not describeaccurately the actual motion of the pendulum over a long periodof time.

In the remainder of this section,we first analyze free undamped motion andthen free dampedmotion.)

FreeUndampedMotionIf we have only a masson a spring,with neither damping nor external force, then

Eq.(3)takesthe simplerform)

mx\" +kx ==O.) (8))

It is convenientto define)

Wo = [!;) (9))

and rewrite Eq. (8)as)

\" 2 0X +wax == .) (8'))

The general solution of Eq.(8')is)

x(t) == A coswot+ B sinwot.) (10))

To analyze the motion describedby this solution, we chooseconstants C anda so that)

Acosa ==-,C)

. BsIna ==-,

C)(11))C == viA2 + B2,) and)

as indicated in Fig.2.4.4.Note that, although tan a == BIA, the anglea is not givenby the principal branch of the inverse tangent function (which gives values only in

the interval -n/2< x < n/2). Instead,a is the angle between 0 and 2n whoseB cosineand sinehave the signsgiven in (11),where either A or B or both may be

negative. Thus)

a==)tan- l (B/A)n + tan- l (B/A)2n + tan- l

(BIA))

if A >0,B >0 (first quadrant),if A <0 (secondor third quadrant),if A >0,B <0 (fourth quadrant),)

where tan- l (B/A)is the angle in (-n/2,n/2) given by a calculatoror computer.In any event, from (10)and (11)we get)

(A B.

). .

)x(t) ==CC

cosw ot +C slnwot ==C(cosacoswot + sIn a slnwot .)

With the aid of the cosineaddition formula, we find that)

x(t) ==Ccos(wot-a).) (12))

Thus the massoscillatesto and from about its equilibrium position with)))

x)

x(t) = Ccos(root- a))

I

I

II--I

I

I-C --1-----I. T

FIGURE2.4.5.Simpleharmonic motion.)

I

I

I

I

I

I

I

I

-I)

Example1)


1.Amplitude2.Circularfrequency3.Phaseangle)

C,(Va, and

a.)

Suchmotion is calledsimpleharmonicmotion.If time t is measuredin seconds,the circular frequency (Va has dimensionsof

radians per second(rad/s).The periodof the motion is the time required for the

system to completeone full oscillation,so is given by)

2nT=-(Va)

(13))

seconds;its frequencyis)1 (Va

v ----- -T 2n)

(14))

t)

in hertz (Hz),which measuresthe number of completecyclesper second.Notethat frequency is measuredin cyclespersecond,whereascircularfrequencyhas the

dimensionsof radians persecond.A typical graph of a simpleharmonicposition function)

x(t) = Ccos(wat -ex) = Ccos(Wa (t-

:a))= Ccos(Wo(t- 8\302\273)

is shown in Fig.2.4.5,where the geometric significanceof the amplitude C, the

periodT, and the time lag) a8=-(Va)

are indicated.If the initial position x(0) = Xa and initial velocityx'(0) = Va of the mass

are given,we first determine the valuesof the coefficientsA and B in Eq.(10),then

find the amplitude C and phaseangle a by carrying out the transformation of x(t)to the form in Eq.(12),as indicatedpreviously.)

..... ...... ..... ............................ ..... ....,. ...... ...... ..\"..

A body with massm = 4 kilogram (kg) is attached to the end of a spring that isstretched2 meters (m) by a force of 100newtons(N).It is set in motion with initial

position Xa = 1 (m) and initial velocity Va = -5(rn/s). (Note that these initial

conditions indicate that the body is displacedto the right and is moving to the leftat time t = 0.) Find the position function of the body as well as the amplitude,frequency,periodof oscillation,and time lag of its motion.)

Solution The spring constant is k = (100N)/(2m) = 50(N/m),so Eq. (8)yields \037X\"+

50x= 0;that is,)x\" + 100x= O.

Consequently,the circularfrequencyof the resulting simpleharmonic motion of the

body will be (Va = v'100= 10(rad/s).Henceit will oscillatewith period

2n 2nT =-=- \037 0.6283s

(Va 10)

and with frequency)

1 (Va 10v = - =-=- \037 1.5915Hz.

T 2n 2n)))

We now imposethe initial conditionsx(0)== 1 and x'(0)==-5on the positionfunction)

x(t) == A coslOt + B sin lOt with X'(t) == -lOA sin lOt + lOBcoslOt.)

It follows readily that A == 1 and B ==-4, so the position function of the body is)

1 .0x(t) ==coslOt --SIn 1 t.

2)

Henceits amplitudeof motion is)

c = J(1)2+ (-i)2= i.J5m.)

To find the time lag, we write)

0(

2 1

)0

x(t) = 2 0coslOt -0sin lOt = 2cos(10t- ex),)

where the phaseangle a satisfies)

2cosex= 0>0 and)

1sin a ==--<o.o)

Hencea is the fourth-quadrant angle)

(-1/0

)ex = 2:rr + tan- 1

2/0 = 2:rr - tan- 1 (i) \037 5.8195,)

and the time lag of the motion is)

a8 ==- \037 0.5820s.

(Va)

With the amplitudeand approximatephaseangle shown explicitly,the positionfunc-tion of the body takesthe form)

x(t) \037 !.J5cos(10t-5.8195),)

and its graph is shown in Fig.2.4.6.) .)))

(0,Xo))

\037)

o)

o)

t)

FIGURE2.4.7.Overdampedmotion: x(t)= cIer1t + C2er2t with

rl < 0 and r2 < O.Solution curvesare graphed with the sameinitial

position Xo and different initialvelocities.)


x)

8) T)I-)

-I)

0.5) c)

-0.5)

.5)

-1)

FIGURE2.4.6.Graph of the position function

x(t)= Ccos(wot-a) in Example 1,with amplitudeC \037 1.118,periodT \037 0.628,and time lag 8 \037 0.582.)

FreeDampedMotionWith damping but no externalforce, the differentialequationwe have beenstudyingtakesthe form mx\" +cx'+kx ==0;alternatively,)

II 2 ' 2 0X + px +wax ==,) (15))

where Wo == ,Jk/mis the correspondingundampedcircular frequencyand)

CP ==->o.

2m

The characteristicequation r 2 +2pr+w6 ==0 ofEq.(15)has roots

rl, r2 ==-p ::i:(p2-(6)1/2)

(16))

(17))

that dependon the signof)

2 2 c2 kp -w ==---a 4m2 m)

C2-4km

4m 2)(18))

The criticaldampingCer is given by Cer == ,J4km , and we distinguish three cases,accordingas C >Cer' C ==Cer' or C < Cere)

OVERDAMPED CASE:C > Ccr(C2 > 4km). BecauseC is relatively large in this

case,we are dealing with a strong resistancein comparisonwith a relatively weak

spring or a small mass.Then (17)gives distinct real roots rl and r2, both of which

are negative.The position function has the form)

x(t) ==CIerI!+C2er2!.) (19))

It iseasyto seethat x(t) \037 0 as t \037 +00and that the body settlesto its equilibriumposition without any oscillations(Problem 29). Figure 2.4.7shows sometypicalgraphs of the position function for the overdampedcase;wechoseXo a fixedpositivenumber and illustrated the effects of changing the initial velocityva. In everycasethe would-beoscillationsare dampedout.)))

(0,XO))

\037)

o)

o)

t)

FIGURE2.4.8.Criticallydamped motion:x(t) = (c]+ c2t)e-ptwith p > o.Solution curves aregraphed withthe same initial position Xo anddifferent initial velocities.)

a)

,WI I)

\" X = Ce-pt cos(w It -a).......(X =+Ce-pt/',--.I)

\037 0)

o)

t)

FIGURE2.4.9.Underdampedoscillations:x(t) = Ce-pt cos(w]t-a).)

CRITICALLYDAMPEDCASE: C = Ccr(c2 = 4km). In this case,(17)givesequal roots rl = r2 = -p of the characteristicequation, so the general solution is)

x(t) = e-pt (CI+C2t ).) (20))

Becausee-pt > 0 and CI +C2t has at most one positive zero,the body passesthrough its equilibrium positionat most once,and it is clearthat x(t) --+ 0 ast --+ +00.Somegraphs of the motion in the critically dampedcaseappear in

Fig.2.4.8,and they resemblethose of the overdamped case(Fig.2.4.7).In the

critically dampedcase,the resistanceof the dashpot is just large enough to dampout any oscillations,but even a slight reduction in resistancewill bring us to the

remaining case,the one that showsthe most dramaticbehavior.)

UNDERDAMPED CASE:C < Ccr(c2 < 4km). The characteristic equation nowhas two complexconjugate roots -p ::i:iJwB

-p2,and the general solution is)

\037) x(t) = e-pt (A coswIt + B sin wIt),) (21))

where)

_ J 2 2 _ .J4km -c2WI - Wa

-p -2m)

(22))

Usingthe cosineaddition formula as in the derivation of Eq.(12),we may rewriteEq.(20)as)

x(t) = Ce-pt

( \037

COswlt+\037

sinWit)

,)

so)

\037) x(t) = Ce-pt COS(WIt-a)) (23))

where)

\037) C = JA2 + B2,)Acosa= -,C)

. BSlna= -.

C)and)

The solution in (22)representsexponentiallydampedoscillationsof the bodyaround its equilibrium position. The graph of x(t) liesbetweenthe \"amplitudeenvelope\"curves x = -Ce-pt and x = Ce-pt and touches them when wIt -a isan integral multiple of n.The motion is not actually periodic,but it is neverthelessuseful to call WI its circularfrequency(moreproperly,its pseudofrequency),Tl =2nIWI its pseudoperiodof oscillation,and Ce-pt its time-varying amplitude.Most of thesequantities are shown in the typical graph of underdamped motionin Fig.2.4.9.Note from Eq. (21)that in this caseWI is lessthan the undampedcircular frequency Wa, so TI is larger than the periodT of oscillation of the samemasswithout damping on the samespring. Thus the action of the dashpot has atleasttwo effects:)

1.It exponentially damps the oscillations,in accord with the time-varyingamplitude.

2. It slowsthe motion; that is, the dashpotdecreasesthe frequencyof the motion.)))

As the followingexampleillustrates,damping typically alsodelaysthe motionfurther-that is, increasesthe time lag-ascomparedwith undampedmotion with

the sameinitial conditions.)

Example2)-

The massand spring of Example1 are now attached also to a dashpot that pro-vides 1 N of resistancefor each meter per secondof velocity. The massis set in

motion with the sameinitial position x(O)== 1 and initial velocityx'(0) == -5asin Example1.Now find the position function of the mass,its new frequencyand

pseudoperiodof motion, its new time lag, and the times of its first four passagesthrough the initial positionx == O.)

Solution Rather than memorizingthe variousformulasgiven in the precedingdiscussion,it isbetter practicein a particular caseto setup the differentialequationand then solveit directly. Recall that m == 1and k == 50;we are now given c == 1 in mks units.

HenceEq. (4)is 1X\" +x' +50x==0;that is,

x\" +2X' + 100x==O.)

The characteristic equation r 2 + 2r + 100== (r + 1)2+ 99r2 ==-1::i:,J99i,so the general solution is)

o has roots rI,)

x(t) ==e-t (A cos,J99t + Bsin,J99t).) (24))

Consequently,the new circular (pseudo)frequencyis WI ==,J99\037 9.9499(ascom-paredwith Wa == 10in Example1).The new (pseudo)periodand frequencyare)

2n 2nTI ==-==

fAi\\\037 0.6315S

WI V 99)

and)

1 WI ,J99VI ==-==-==-\037 1.5836Hz

TI 2n 2n

(ascomparedwith T \037 0.6283< TI and V \037 1.5915> VI in Example1).We now imposethe initial conditionsx(0)== 1 and x'(0)==-5 on the position

function in (23)and the resulting velocity function)

x'(t) ==-e-t (A cos,J99t +Bsin,J99t) +,J99e-t (-Asin,J99t +B cosJ99t).)

It follows that)

x(O)== A == 1 and x'(O)==-A+B,J99== -5,)

whence we find that A == 1 and B == -4/,J99.Thus the new position function ofthe body is

x(t) = e-t

(cas,J99t -

\037sin,J99t)

.Henceits time-varying amplitude of motion is)

C -tIe ==) (4

)2 {115(1)2+ -

,J99e-t =

V 99e-t .)))

We therefore write)

vTIS(

\037 4

)x(t) =\037

e-tvTIS

cosJ99t -vTIS

sin J99t99 115 115

(ill=V 9ge-t cos(J99t -al),)

where the phaseangle al satisfies)

\037cosal=vTIS

>0 and115)

4sinal= -

vTIS< O.)

Henceal is the fourth-quadrant angle)

(-4/vTIS

) (4

)al = 2][+ tan-l

\037vTIS= 2][- tan-l

IN>\037 5.9009,

99/ 115 v99)

and the time lag of the motion is)

al81 = -\0370.5931S

WI)

(as comparedwith 8 \037 0.5820< 81 in Example1).With the time-varying am-plitude and approximatephaseangle shown explicitly, the position function of themasstakesthe form)

f{l15x(t) \037 -e-t cos(J99t -5.9009),

99)(25))

and its graph is the dampedexponential that is shown in Fig.2.4.10(in comparisonwith the undamped oscillationsof Example1).)

x)

x = Cte-t)

\ \.......)

FIGURE2.4.10.Graphs of the position function

x(t)= C]e-t cos(w]t-a])ofExample 2 (dampedoscillations),the position function x(t)= Ccos(wot-a) ofExample 1 (undamped oscillations),and the envelopecurvesx(t)= ::l:C1e-t

.)))

From (24)we seethat the masspassesthrough its equilibrium positionx ==0when COS(WIt-aI)==0,and thus when)

3nwIt -al ==--,

2)

n2')

n2')

3n2') ...,)

that is, when)

3nt ==81

--,2WI)

n81--,

2WI)

n81 +-,

2w})

3n81 +-,

2WI)

We seesimilarly that the undampedmassof Example1 passesthrough equilibriumwhen)

3n n n 3nt ==80 --, 80--, 80 +-, 80 +-,

2wa 2wa 2wa 2wa

The following table comparesthe first four values tI, t2, t3, t4 we calculatefor the

undamped and dampedcases,respectively.)

n 1 2 3 4tn (undamped) 0.11070.42490.73901.0532tn (damped) 0.11950.43520.7509 1.0667)

Accordingly, in Fig.2.4.11(where only the first three equilibrium passagesareshown)we seethe dampedoscillationslagging slightly behind the undamped ones.

.)x)

x(t) =Ccos(Wo t - a))x =Cle-t)

- -) ---)

- --)- --)

-1)

x(t) =Cle-t cos( wit - a\037)

FIGURE2.4.11.Graphs on the interval 0 < t < 0.8illustrating the additional delay associatedwith damping.)

1IIl\037.!'!\037\037 \037_!: In\037..\037.\037 . _)

1.Determine the periodand frequency of the simple har-monic motion of a 4-kgmass on the end of a spring with

spring constant 16NIm.2.Determine the periodand frequency of the simple har-

monic motion of a body of mass 0.75kg on the end ofa spring with spring constant 48NIm.

3.A mass of 3 kg is attached to the end of a spring that is)

stretched 20cm by a forceof 15N. It is set in motion with

initial position Xo = 0 and initial velocity Vo = -10m/s.Find the amplitude, period,and frequency of the resultingmotion.)

4. A body with mass 250g is attached to the endofa springthat is stretched 25 cm by a forceof 9 N. At time t = 0)))

the body is pulled 1 m to the right, stretching the spring,and set in motion with an initial velocity of 5 mjs to theleft. (a) Find x(t)in the form Ccos(wot-a).(b) Findthe amplitude and periodof motion of the body.)

In Problems5 through 8, assumethat the differential equationofa simple pendulum of length L is LO\" + gO = 0, where

g = GMjR 2 is the gravitational accelerationat the locationofthe pendulum (at distanceR from the centerofthe earth; Mdenotesthe mass of the earth).

5.Twopendulums areof lengths LI and L2 and-whenlo-catedat the respectivedistancesR I and R2 from the centerof the earth-have periodsPI and P2.Show that)

PI RI-JL;P2 R2-JL;.)

6.A certain pendulum keepsperfecttime in Paris, where theradius of the earth is R = 3956(mi). But this clockloses2 min 40 s per day at a location on the equator. Usetheresult of Problem 5 to find the amount of the equatorialbulge of the earth.)

7. A pendulum of length 100.10in., locatedat a point atsealevel where the radius of the earth is R = 3960(mi),has the sameperiodas doesa pendulum of length 100.00in. atop a nearby mountain. Usethe result ofProblem5 tofind the height of the mountain.)

8.Most grandfather clockshave pendulums with adjustablelengths. One such clockloses10min per day when the

length of its pendulum is 30 in. With what length pendu-lum will this clockkeepperfecttime?)

9.DeriveEq.(5)describingthe motion ofa mass attached tothe bottom of a vertically suspendedspring. (Suggestion:First denote by x(t) the displacementof the mass belowthe unstretched position of the spring; setup the differen-tial equation for x. Then substitute y = x - Xo in thisdifferential equation.))

10.Considera floating cylindrical buoy with radius r, heighth, and uniform density p < 0.5(recallthat the densityof water is 1 gjcm3). Thebuoy is initially suspendedatrest with its bottom at the top surface of the water andis releasedat time t = O. Thereafterit is actedon bytwo forces: a downward gravitational forceequal to its

weight mg = n r2hg and (by Archimedes' principle of

buoyancy) an upward forceequal to the weight nr2xg ofwater displaced,where x = x(t)is the depth of the bot-tom of the buoy beneath the surfaceat time t (Fig.2.4.12).Concludethat the buoy undergoes simple harmonic mo-tion around its equilibrium position Xe = ph with periodP = 2n,Jphjg .Compute P and the amplitude of the mo-tion if p = 0.5gjcm3, h = 200cm, and g = 980cmjs2.)

FIGURE2.4.12.Thebuoy ofProblem 10.11.A cylindrical buoy weighing 100Ib (thus of mass m

3.125slugs in ft-Ib-s (fps) units) floats in water with its

axis vertical (as in Problem 10).When depressedslightlyand released,it oscillatesup and down four times every10s.Assume that friction is negligible.Find the radius ofthe buoy.12.Assume that the earth is a solid sphereofuniform density,with mass M and radius R = 3960(mi). For a particleofmass m within the earth at distance r from the center ofthe earth, the gravitational forceattracting m toward the

centeris Fr = -GMrmjr2, where Mr is the mass of the

part ofthe earth within a sphereofradius r. (a) Show that

Fr = -GMmrjR3. (b) Now supposethat a small hole isdrilled straight through the centerof the earth, thus con-necting two antipodal points on its surface.Leta particleofmass m bedroppedat time t = 0 into this hole with ini-

tial speedzero,and let r(t)be its distance from the centerof the earth at time t (Fig. 2.4.13).Concludefrom New-ton's secondlaw and part (a) that r\" (t) = -k2r(t),where

k2 = GMjR3 = gjR.)

FIGURE2.4.13.A mass m falling downa hole through the centerof the earth

(Problem 12).(c) Takeg = 32.2ftjs2, and concludefrom part (b) that

the particle undergoes simple harmonic motion back and

forth between the endsof the hole,with a period of about

84min. (d) Lookup (or derive) the period of a satellite

that just skims the surfaceof the earth; comparewith the

result in part (c).How do you explain the coincidence?Or is it a coincidence? (e) With what speed(in miles)))

per hour) doesthe particle passthrough the centerof theearth? (f) Lookup (or derive) the orbital velocity of asatellite that just skims the surface of the earth; comparewith the result in part (e).How doyou explain the coinci-dence?Or is ita coincidence?

13.Supposethat the mass in a mass-spring-dashpotsystemwith m = 10,c = 9, and k = 2 is set in motion with

x(O)= 0 and x'(0) = 5. (a) Find the position func-tion x(t) and show that its graph looksas indicated in

Fig. 2.4.14.(b) Find how far the mass moves to the

right beforestarting backtoward the origin.)

5

4

3

2\037

1

0

-1-2 50 10 15 20

t)

FIGURE2.4.14.Theposition function

x(t)ofProblem 13.14.Supposethat the mass in a mass-spring-dashpotsystem

with m = 25, c = 10,and k = 226 is set in motionwith x(O)= 20 and x'(0) = 41.(a) Find the positionfunction x(t) and show that its graph looksas indicated in

Fig. 2.4.15.(b) Find the pseudoperiodof the oscillationsand the equations of the \"envelope curves\" that aredashedin the figure.)

20\\

10 \"

\037 0

-10//-20/

0 5) 10t)

15) 20)

FIGURE2.4.15.Theposition function

x(t)ofProblem 14.)

Theremaining problems in this sectiondealwith free dampedmotion. In Problems15through 21,a mass m is attachedto both a spring (with given spring constant k) and a dash-pot (with given damping constant c). The mass is set in

motion with initial position Xo and initial velocity Vo. Findthe position function x(t) and determine whether the mo-tion is overdamped, critically damped, or underdamped. Ifit is underdamped, write the position function in the formx(t) = C1e-pt

COS(WI t -al).Also, find the undamped position)


function u(t) = Cocos(wot-ao) that would result if the masson the spring wereset in motion with the same initial positionand velocity, but with the dashpot disconnected(soc = 0).Finally, construct afigure that illustrates the effect ofdamping

by comparing the graphs ofx(t)and u(t).

15.m =\037,

c = 3,k = 4;Xo = 2, Vo = 016.m = 3,c = 30,k = 63;Xo = 2, Va = 217.m = 1,c = 8,k = 16;Xo = 5, Vo = -1018.m = 2,c = 12,k = 50;Xo = 0, Vo = -819.m = 4,c = 20,k = 169;Xo = 4, Vo = 1620.m = 2,c = 16,k = 40;Xo = 5, Vo = 421.m = 1,c = 10,k = 125;Xo = 6, Vo = 5022.A 12-lbweight (mass m = 0.375 slugs in fps units)

is attached both to a vertically suspended spring that it

stretches 6 in. and to a dashpot that provides 3 lb of re-sistancefor every foot per secondof velocity. (a) If the

weight is pulled down 1 ft below its static equilibrium po-sition and then releasedfrom rest at time t = 0, find its po-sition function x(t). (b) Find the frequency, time-varying

amplitude, and phaseangleof the motion.

23.This problem dealswith a highly simplified model ofa carof weight 3200lb (massm = 100slugs in fps units). As-sume that the suspensionsystem actslike a single springand its shock absorberslike a single dashpot, so that its

vertical vibrations satisfy Eq.(4)with appropriate values

of the coefficients. (a) Find the stiffness coefficient k

of the spring if the carundergoesfreevibrations at 80cy-clesper minute (cycles/min)when its shockabsorbersaredisconnected.(b) With the shock absorbersconnected,the car is set into vibration by driving it overa bump, and

the resulting damped vibrations have a frequency of 78cycles/min.After how long will the time-varying ampli-tude be 1% of its initial value?)

Problems24 through 34dealwith a mass-spring-dashpotsys-tem having position function x (t) satisfying Eq. (4). We write

Xo = x (0) and Vo = x'(0) and recall that p = c/(2m),w6 = k/m, and wi = w6

- p2. The system is critically

damped, overdamped, or un derdamped, as specifiedin eachproblem.

24. (Critically damped) Show in this casethat)

x(t)= (xo + vot + pxot)e-pt.)

25.(Critically damped) Deducefrom Problem 24 that the

mass passesthrough x = 0 at someinstant t > 0 if and

only if Xo and Va + pXo have oppositesigns.26.(Critically damped) Deducefrom Problem 24that x(t) has

a localmaximum or minimum at someinstant t > 0 if and

only if Vo and Vo + pXo have the same sign.27.(Overdamped)Show in this casethat)

x(t)= \037 [(vo- r2xO)erjl - (vo-r,xo)er21 ] ,2y)

where rl , r2 = -p ::l:Jp2 -wB and y = (rl -r2)/2 > O.)))


28.(Overdamped)If Xo = 0,deducefrom Problem27 that)

X(t) = Vo e-pt sinh yt.y)

29. (Overdamped) Prove that in this casethe mass can passthrough its equilibrium position x = 0 at most once.

30.(Underdamped) Show that in this case)

_

(Vo + pXo.

)x(t)= e pI xOcosw)t+

WIsmw)t.)

31.(Underdamped) If the damping constant c is small in com-parison with v'8mk , apply the binomial seriesto show that)

WI \037 WO (1- \037).

8mk)

32.(Underdamped) Show that the localmaxima and minimaof)

X(t) = Ce-pt COS(Wlt - a))

occurwhere)p

tan(wlt -a) =--.WI

Concludethat t2 - tl = 2njWI if two consecutivemaximaoccurat times tl and t2.

33.(Underdamped) Let Xl and X2 be two consecutivelocalmaximum values ofx(t).Deducefrom the result ofProb-lem 32 that)

InXl = 2npX2 WI

The constant \037 = 2npjWI is called the logarithmicdecrementof the oscillation.Note alsothat c = mWI \037jn

becausep = cj(2m).)

Note: The result ofProblem 33providesan accuratemethodfor measuring the viscosity ofa fluid, which is an importantparameter in fluid dynamics but is not easyto measuredirectly.According to Stokes'sdrag law, a sphericalbody of radius amoving at a (relatively slow)speedthrough a fluid ofviscosityf.L experiencesa resistive force FR = 6n f.La v. Thus ifa spheri-calmasson a spring is immersed in the fluid and set in motion,this drag resistancedamps its oscillationswith damping con-stant c = 6naf.L. Thefrequency WI and logarithmic decrement\037 of the oscillationscan be measuredby directobservation.Thefinal formula in Problem 33 then gives c and hencethe

viscosity ofthe fluid.)

34. (Underdamped) A body weighing 100 Ib (mass m =3.125slugs in fps units) is oscillating attached to a springand a dashpot. Its first two maximum displacementsof6.73in. and 1.46in. areobservedto occurat times 0.34sand 1.17s, respectively. Compute the damping con-stant (in pound-secondsper foot) and spring constant (in

pounds perfoot).)

Differential Equationsand DeterminismGiven a mass m, a dashpot constant c, and a spring constant

k, Theorem 2 ofSection2.1implies that the equation)

mx// + cx/+ kx = 0) (26))

has a unique solution for t > 0 satisfying given initial condi-tions X(0) = Xo, x/(0)= vo. Thus the future motion ofan idealmass-spring-dashpotsystem is completely determined by the

differential equation and the initial conditions. Ofcoursein

a realphysical system it is impossibleto measure the param-etersm, c, and k precisely.Problems 35 through 38 explorethe resulting uncertainty in predicting the future behavior ofaphysical system.

35.Supposethat m = 1,c = 2, and k = 1 in Eq.(26).Showthat the solution with X (0)= 0 and x/(0)= 1 is)

Xl (t) = te-t.36.Supposethat m = 1 and c = 2 but k = 1- 10-2n.Show

that the solution ofEq.(26)with x(O)= 0 and x/ (0) = 1IS

X2(t) = 10ne-t sinh 10-n t.

37.Supposethat m = 1 and c = 2 but that k = 1 + 10-2n .Show that the solution of Eq.(26) with x(O) = 0 and

x/CO)= 1 is)

X3(t) = 10ne-t sin 10-n t.)

38.Whereas the graphs of Xl (t) and X2 (t) resemblethoseshown in Figs.2.4.7and 2.4.8,the graph ofX3(t) exhibits

damped oscillationslike those illustrated in Fig.2.4.9,but

with a very long pseudoperiod.Nevertheless,show that

for eachfixed t > 0 it is true that)

lim X2(t) = lim X3(t) = Xl (t).n\037oo n\037oo)

Concludethat on a given finite time interval the three solu-tions are in \"practical\" agreement if n is sufficiently large.)

_ NonhomogeneousEqt:iationsand UndeterminedC()efficients)We learned in Section2.3how to solvehomogeneouslinearequationswith constantcoefficients,but we saw in Section2.4that an external force in a simplemechanicalsystemcontributesa nonhomogeneousterm to its differentialequation. The generalnonhomogeneousnth-order linear equation with constant coefficientshas the form)

\037) any(n) +an_Iy(n-l) + ...+alY'+aoy = f(x).) (1))))

Example1)

2.5NonhomogeneousEquationsand UndeterminedCoefficients149)

By Theorem 5 of Section2.2,a general solutionof Eq.(1)has the form)

\037) y = Yc + Yp) (2))

where the complementaryfunction yc(x)is a general solutionof the associatedho-mogeneousequation)

an y(n) + an -1y

(n -1)+ ...+a 1y' +aOY = 0,) (3))

and y p (x)is a particular solutionof Eq.(1).Thus our remaining task is to find y p.The methodof undeterminedcoefficients is a straightforward way of doing

this when the given function f (x) in Eq.(1)is sufficiently simplethat we can makean intelligent guessas to the general form of y p. For example,supposethat f (x)is a polynomial of degreem. Then, becausethe derivativesof a polynomial arethemselves polynomials of lower degree,it is reasonableto suspecta particularsolution

yp(x)= Amxm + Am_1xm-1 +...+A1x + Ao

that is alsoa polynomialof degreem, but with asyet undeterminedcoefficients.We

may, therefore, substitute this expressionfor y pinto Eq.(1),and then-by equatingcoefficientsof likepowersof x on the two sidesof the resulting equation-attemptto determine the coefficientsAo, AI, ..., Am so that y p will, indeed,bea particularsolution of Eq.(1).

Similarly, supposethat)

f (x)= a coskx +b sin kx .)

Then it is reasonableto expecta particular solutionof the same form:)

yp(x)= A coskx+ B sinkx,)

a linear combination with undetermined coefficientsA and B. The reason is that

any derivativeof such a linear combinationof coskx and sin kx has the sameform.We may therefore substitute this form of yp in Eq.(1),and then-by equating co-efficientsof coskx and sin kx on both sidesof the resulting equation-attempttodetermine the coefficientsA and B so that y p will, indeed,bea particular solution.

It turns out that this approach doessucceedwhenever all the derivatives off (x)have the sameform as f (x) itself. Beforedescribingthe method in full gen-erality, we illustrate it with severalpreliminary examples.)

Find a particular solutionof y\" +3y' +4y = 3x+2.)

Solution Heref (x)= 3x+2 is a polynomialof degree1,soour guessis that)

yp(x)= Ax + B.)

Theny\037

= A andy\037

= 0,so yp will satisfy the differentialequationprovidedthat)

(0)+3(A) +4(Ax+ B)= 3x +2,)

that is,)

(4A)x+ (3A +4B)= 3x+ 2)))


for all x. This will be true if the x-termsand constant terms on the two sidesof this

equation agree.It therefore sufficesfor A and B to satisfy the two linear equations4A ==3 and 3A +4B ==2 that we readily solve for A ==

\037

and B ==-/6 .Thus we

have found the particular solution)

yp(x) ==\037x

-/6 .) .)

\"'''',h _,'_'\" ,

Example2)..,. ,..\"....\" .....,.,......\037. .... \037 \037 ,..\" ... ...\" ,.. \037 ... \" \",.. \037'.....'-nn'.n.vu.....-\"\"..

Find a particular solution of y/l-4y ==2e3x

.)

Solution Any derivativeof e3x is a constant multiple of e3x, so it is reasonableto try)

yp(x) == Ae3x.)

Theny\037

==9Ae3x, so the given differentialequation will be satisfiedprovided that)

9Ae3x -4(Ae3x) ==2e3x

;)

that is,5A ==2,so that A ==\037.

Thus our particular solution is y p (x)==\037

e3x. .)

Example3) Find a particular solutionof 3y\" +y' -2y ==2cosx.)

Solution A first guessmight be yp(x) == A cosx,but the presenceof y' on the left-handsidesignalsthat we probably needa term involving sin x as well.Sowe try)

yp(x) == Acosx+ Bsinx;y\037(x)

== -Asin x + Bcosx,y\037

(x) ==-A cosx - B sin x.)

Then substitution of y p and its derivativesinto the given differentialequationyields)

3(-A cosx-B sin x) +(-A sin x + Bcosx)- 2(A cosx + B sin x) ==2cosx,)

that is (collectingcoefficientson the left),)

(-5A+ B)cosx+(-A-5B)sin x ==2cosx.)

This will be true for all x provided that the cosineand sineterms on the two sidesof this equation agree.It therefore suffices for A and B to satisfy the two linearequations)

-5A + B ==2,-A- 5B ==0)

with readily found solution A ==-t3 ' B ==

/3 .Hencea particular solution is)

( )5 1 .

Yp x ==-13cosx + 13SIn x.) .)The following example,which superficiallyresemblesExample2,indicates

that the method of undetermined coefficients is not always quite so simpleas wehave made it appear.)

Example4).,..\".\"......,..\".\"..'.. .n.nnn'\",,,', \" n \"... \" .0-' ....... \037 _ .,...,..\",,_ ...., \037 ...... .......0.......

Find a particular solution of y\"-4y ==2e2x

.)))


Solution If we try y p (x)= Ae2x, we find that)

y;-

4yp = 4Ae2x -4Ae2x = 0 =1= 2e2x.)

Thus, no matter how A is chosen,Ae2x cannot satisfy the given nonhomogeneousequation. In fact, the precedingcomputation showsthat Ae2x satisfiesinstead the

associatedhomogeneousequation. Therefore,we shouldbegin with a trial function

yp(x) whosederivativeinvolvesboth e2x and somethingelsethat can canceluponsubstitution into the differential equation to leave the e2x term that we need. A

reasonableguessis)

yp(x)= Axe2x,)

for which)

y\037(x)=Ae2x+2Axe2xand y;(x)=4Ae2x +4Axe2X .Substitution into the original differentialequationyields

(4Ae2x +4Axe2x) -4(Axe2x

) = 2e2x.)

The terms involving xe2x obligingly cancel,leaving only 4Ae2x

A =\037.

Consequently,a particular solution is)

2e2x, so that)

yp(x)=\037xe2x.)

.)

The GeneralApproachOur initial difficulty in Example4 resultedfrom the fact that f (x)= 2e2x satisfiesthe associatedhomogeneousequation. Rule 1,given shortly, tells what to do when

we do not have this difficulty, and Rule 2 tells what to do when we do have it.The method of undeterminedcoefficientsapplieswheneverthe function f (x)

in Eq. (1)is a linear combinationof (finite) products of functions of the followingthree types:

1.A polynomial in x;2.An exponential function erx ;3.coskx or sin kx .)

(4))

Any such function-forexample,

f(x) = (3 -4x2)e5x -4x3 coslOx,)

has the crucial property that only finitely many linearly independentfunctions ap-pearas terms (summands) in f(x) and its derivativesof all orders.In Rules 1 and2 we assumethat Ly = f(x) is a nonhomogeneouslinear equation with constantcoefficientsand that f (x)is a function of this kind.)

RULE 1 Methodof UndeterminedCoefficientsSupposethat no term appearingeither in f (x)or in any of its derivativessatisfiesthe associatedhomogeneousequation Ly = O. Then take as a trial solution for

yp a linear combinationof all linearly independent such terms and their deriva-tives. Then determine the coefficientsby substitution of this trial solution into

the nonhomogeneousequation Ly = f(x).)))


Note that this rule is not a theorem requiring proof; it is merelya proceduretobe followed in searchingfor a particular solution y p. If we succeedin finding y p,then nothing more needbe said.(It can beproved,however,that this procedurewill

always succeedunder the conditions specifiedhere.)In practicewe checkthe suppositionmade in Rule 1 by first using the charac-

teristic equation to find the complementary function Yc, and then write a list of allthe terms appearing in f (x) and its successivederivatives. If none of the terms in

this list duplicatesa term in Yc, then we proceedwith Rule 1.)

Example5)\037

Find a particular solution of)

y\" +4y ==3x3.) (5))

Solution The (familiar) complementarysolutionof Eq.(5)is)

Yc(x) ==Clcos2x+C2sin 2x.)

The function f (x) == 3x3 and its derivativesare constant multiples of the linearly

independentfunctionsx3, x2, x,and 1.Becausenone of theseappearsin Yc, we try)

yp == Ax 3 + Bx2+Cx+ D,

y\037==3Ax 2 +2Bx+C,

y; ==6Ax +2B.)

Substitution in Eq.(5)gives)

y; +4yp == (6Ax +2B)+4(Ax3 + Bx2+Cx+ D)==4Ax3 +4Bx2 + (6A +4C)x+ (2B+D) ==3x3

.)

We equate coefficientsof likepowersof x in the last equation to get)

4A ==3,6A +4C ==0,)

4B ==0,2B + D ==0)

with solution A ==\037,

B ==0,C ==-\037,

and D == O.Hencea particular solution ofEq.(5)is)

( )3 3 9

yp X == 4x -gX.) .)

Example6) Solvethe initial value problem)

y\"-3y' +2y ==3e-x -10cos3x;

y(O) == 1, y'(0)==2.)(6))

Solution The characteristic equation r 2 - 3r + 2 == 0 has roots r

complementaryfunction is)

1 and r == 2,so the)

Yc(x) == Ctex+c2e2x.)))

.....

Example7)


The terms involved in f (x) ==3e-x - lOcos3x and its derivativesaree-x , cos3x,and sin 3x.Becausenone of theseappearsin Yc, we try)

yp== Ae-x + Bcos3x+Csin3x,y\037

==-Ae-x - 3Bsin3x+ 3Ccos3x,y\037

== Ae-x - 9Bcos3x - 9Csin 3x.)

After we substitute theseexpressionsinto the differentialequation in (6)and collectcoefficients,we get)

y; - 3y\037+2yp ==6Ae-x + (-7B-9C)cos3x + (9B-7C)sin3x

==3e-x -10cos3x.)

We equate the coefficientsof the terms involving e-x , those involving cos3x,and

those involving sin 3x.The result is the system)

6A == 3,-7B-9C==-10,

9B - 7C == 0)

with solution A == 4, B ==}73 , and C ==

i3 .This gives the particularsolution)

( )} -x 7 3 9.3ypx ==\"2e +TIcosx+TIslnx,)

which, however,doesnot have the requiredinitial values in (6).To satisfy thoseinitial conditions,we begin with the generalsolution)

y(x) ==Yc(x) + yp(x))x 2x}-x 7 3 9.3==c}e+C2e +\"2e +TIcosx+TIslnx,)

with derivative)

y'(x)==c}ex +2c2e2x- 4e-x - ij sin3x+i\037

cos3x.)

The initial conditions in (6)leadto the equations)

y (0)==c}+C2+ 4 + ?3== 1,

y'(0)==c}+2C2- 4 +i\037

==2)

with solutionc}==-4, C2 ==}63.The desiredparticular solutionis therefore)

( )} x + 6 2x + } -x+ 7 3 + 9 .

3y x == -\"2e TIe \"2

e TIcos x TI SIn x.) .)

Find the general form of a particular solutionof)

y(3) +9y' ==xsin x +x2e2x.) (7))))


Solution The characteristic equation r 3 +9r ==0 has roots r ==0,r ==-3i,and r ==3i.Sothe complementaryfunction is)

Yc(x) ==Cl +Czcos3x +C3sin 3x.)

The derivativesof the right-hand sidein Eq.(7)involve the terms)

cosx, SIn x, xcosx, xSIn x,

eZx, xezx

, and xZezx.)

Becausethere is no duplication with the terms of the complementary function, the

trial solution takesthe form)

Yp(x) == A cosx+ B sin x +Cxcosx+ Dxsin x + Eezx + Fxezx +Gxzezx.)

Uponsubstituting Yp in Eq.(7)and equatingcoefficientsof liketerms, we get sevenequations determining the sevencoefficientsA, B,C,D,E,F,and G. .)

The CaseofDuplicationNow we turn our attention to the situation in which Rule 1 doesnot apply:Someofthe terms involved in f(x) and its derivatives satisfy the associatedhomogeneousequation. For instance, supposethat we want to find a particular solution of thedifferentialequation)

(D- r)3y == (2x-3)erx .) (8))

Proceedingas in Rule 1,our first guesswould be)

yp(x) == Aerx + Bxerx .) (9))

This form of yp(x) will not be adequate becausethe complementary function ofEq. (8)is)

Yc(x) ==clerx +czxerx +c3xzerx,) (10))

so substitution of (9) in the left-hand side of (8) would yield zero rather than

(2x-3)erx .To seehow to amend our first guess,we observethat)

(D- r)z[(2x-3)erx ] == [Dz(2x-3)]erx ==0)

by Eq. (13)of Section2.3.If y(x) is any solution of Eq. (8)and we apply the

operator (D- r)z to both sides,we seethat y (x) is alsoa solution of the equation(D- r)5y ==O.The general solution of this homogeneousequation can be written

as)

y(x) ==Cterx +czxerx +c3xzerx+Ax 3erx + Bx4erx .\\. \037 \\.. j

V V

Yc yp

Thus every solution of our original equation in (8)is the sum of a complementaryfunction and a particularsolution of the form)

yp(x) == Ax 3erx + Bx4erx.) (11))))


Note that the right-hand sidein Eq.(11)can be obtained by multiplying each termof our first guessin (9)by the least positive integral powerof x (in this case,x3)that sufficesto eliminateduplicationbetween the terms of the resulting trial solution

yp(x)and the complementaryfunction Yc(x) given in (10).This proceduresucceedsin the general case.

To simplify the general statement of Rule2,we observethat to find a particularsolutionof the nonhomogeneouslinear differentialequation)

Ly = II(x)+ Iz(x),) (12))

it suffices to find separatelyparticular solutions Y I (x) and Yz (x)of the two equa-tions)

Ly = II(x) and Ly = Iz(x),) (13))

respectively.For linearity then gives)

L[YI + Yz] = LY1 +LYz = II(x)+ Iz(x),)

and therefore yp = Y I + Yz is a particular solution of Eq.(12).(This is a type of\"superpositionprinciple\" for nonhomogeneouslinear equations.)

Now our problemis to find a particular solution of the equation Ly = f(x),whereI(x)is a linear combinationof productsof the elementaryfunctions listedin

(4).Thus I(x)can be written as a sum of terms each of the form)

Pm (x)erX coskx or Pm (x)erXsinkx,) (14))

where Pm (x) is a polynomial in x of degreem. Note that any derivative of sucha term is of the sameform but with both sinesand cosinesappearing. The proce-dure by which we arrivedearlier at the particular solution in (11)of Eq.(8)can begeneralizedto show that the followingprocedureis always successful.)

RULE 2 MethodofUndeterminedCoefficientsIf the function I(x)is of either form in (14),take as the trial solution)

yp(x)= xS[(Ao +Alx + Azxz +...+Amxm)e

rX coskx+ (Bo+B1x+ Bzxz +...+ Bmxm)e

rX sinkx], (15))

where s is the smallestnonnegativeinteger such that such that no term in yp du-plicatesa term in the complementaryfunction Yc. Then determinethe coefficientsin Eq.(15)by substituting yp into the nonhomogeneousequation.)

In practicewe seldomneedto deal with a function I(x) exhibiting the full

generality in (14).The table in Fig.2.5.1lists the form of y p in various commoncases,correspondingto the possibilitiesm = 0,r = 0,and k = O.

On the other hand, it is common to have)

I(x)= II(x)+ Iz(x),)

where II(x) and Iz(x) are different functions of the sort listed in the table in

Fig.2.5.1.In this event we take as yp the sum of the trial solutions for II(x)andIz(x),choosings separatelyfor each part to eliminateduplicationwith the comple-mentary function. This procedureis illustrated in Examples8 through 10.)))


I{x},)

YP)

Pm = bo+ b1x+ b2x2 + ...+ bmxm

a coskx + b sin kx

erx (acoskx + b sin kx)

Pm (x)erx

Pm (x)(acoskx + b sin kx))

xS(Ao+ A1x + A2x2+...+ Amxm)

XS(A coskx + B sin kx)

XSerx(A coskx + B sin kx)

XS(Ao + A1x + A2x2 + ...+ Amxm)e

rx

xS[(Ao + A1x +...+ Amxm) coskx+ (Bo+ B1x+...+ Bmxm) sinkx])

FIGURE2.5.1.Substitutions in the method of undetermined coefficients.)

\037'N' ..:-: ....N 'u.,.... '''\"N N..N..U..,......,.... N N \037.-N\037..... Nn

Example8)\"\037..\037.,,\037.,,,..,...'...'\037nN''''.''\"''A-'....N..,._..,_. \037 ,\037 ,... ....

Find a particular solution of)

y(3) + y\" ==3ex +4x2.) (16))

Solution The characteristicequation r 3 + r 2 ==0 has roots rl == r2 ==0 and r3 ==-1,so the

complementaryfunction is)

Yc(X) ==CI +C2X +C3e-x.)

As a first steptoward our particular solution,we form the sum)

(AeX) + (B+Cx+Dx2

).)

The part AeX correspondingto 3ex doesnot duplicate any part of the complemen-tary function, but the part B + Cx+ Dx2 must be multiplied by x2 to eliminateduplication.Hencewe take)

Yp== Aex + Bx2 +Cx3 + Dx4,

y\037== Aex +2Bx+3Cx2 +4Dx3,

Y; == Aex +2B +6Cx+ 12Dx2, and

y\0373)== Aex +6C+24Dx.)

Substitution of thesederivativesin Eq.(16)yields)

2Aex + (2B+6C)+ (6C+24D)x+ 12Dx2 ==3ex +4x2.)

The systemof equations)

2A ==3,6C+24D ==0,)

2B +6C==0,12D==4)

has the solution A ==\037,

B ==4, C ==-\037,

and D == 1.Hencethe desiredparticularsolution is)

Yp(x) ==\037eX

+4x2-\037x3

+ 1x4.) .)))


Example9)

........ .Determine the appropriateform for a particular solutionof)

y\" +6y' + 13y= e-3x cos2x.)

Solution The characteristicequation r 2+6r+13= 0 has roots -3::f::2i,so the complemen-tary function is)

yc(x)= e-3x(CIcos2x+C2sin 2x).

This is the sameform as a first attempt e-3x(A cos2x+ B sin 2x)at a particular

solution, sowe must multiply by x to eliminateduplication.Hencewe would take)

yp(x)= e-3X(Axcos2x+ Bxsin2x).) .)

Example10)N,\"NN.\",'N,'NN.'n\037\"'N,\"\",... ,-';\"\" '\" \"'''..........nN,.''...... ....;\".\"...... .........\"....n.....,..\".... .......\"\"....\"\" \" \",..\" \"''''''' ...........'A...........\" \"\"\"\"\"'......... ;\"\".,..... ... \037 .... '\" ,..... ........ \037 .... \037._.\", .....,_\"....... _., ....... \037v

Determine the appropriateform for a particular solutionof the fifth-order equation)

(D-2)3(D2 +9)y = x2e2x +xsin3x.)

Solution The characteristicequation (r-2)3(r2 +9)= 0 has roots r = 2,2,2, 3i,and -3i,so the complementaryfunction is)

yc(x)= Cte2x+C2xe2x+c3x2e2x+C4cos3x +C5sin 3x.)

As a first steptoward the form of a particular solution,we examine the sum)

[(A +Bx+Cx2)e2x] + [(D+ Ex)cos3x+ (F+Gx)sin 3x].)

Toeliminateduplicationwith terms of Yc (x),the first part-correspondingto x2e2x_must bemultiplied by x3, and the secondpart-correspondingto xsin 3x-mustbemultiplied by x. Hencewe would take)

yp(x)= (Ax3 + Bx4 +Cx5)e2x + (Dx+ Ex2)cos3x+ (Fx+Gx2

) sin 3x. .)

VariationofParameters)

Finally, let us point out the kind of situation in which the methodof undeterminedcoefficientscannot beused.Consider,for example,the equation)

IIY +y=tanx,) (17))

which at first glancemay appearsimilar to thoseconsideredin the precedingex-amples.Not so;the function f (x)= tan x has infinitely many linearly independentderivatives)

sec2x, 2 sec2x tan x, 4 sec2x tan 2x +2 sec4x,)

Therefore, we do not have available a finite linear combination to use as a trial

solution.)))


We discusshere the method of variation of parameters,which-inprinciple(that is, if the integrals that appearcan beevaluated)-canalwaysbe usedto find a

particular solution of the nonhomogeneouslinear differentialequation)

yen) +Pn_l(X)y(n-l)+...+Pl(X)y'+po(x)y ==f(x), (18))

provided that we already know the general solution)

Yc ==ClYl +czyz+ ...+ CnYn) (19))

of the associatedhomogeneousequation)

yen) +Pn_l(X)y(n-l)+...+ Pl(X)y'+Po(x)y==o.) (20))

Here,in brief, is the basicideaof the method of variation of parameters.Sup-posethat we replacethe constants,orparameters,Cl,Cz,..., Cn in the complemen-tary function in Eq. (19)with variables: functions U 1,Uz, ...,Un of x. We askwhether it is possibleto choosethesefunctions in such a way that the combination)

yp(x) ==Ul (X)Yl (x)+UZ(X)yZ(X) + ...+ Un (X)Yn (X) (21))

is a particular solution of the nonhomogeneousequation in (18).It turns out that

this is alwayspossible.The method is essentiallythe samefor all ordersn >2,but we will describe

it in detail only for the casen ==2.Sowe beginwith the second-ordernonhomoge-neousequation)

\037) L[y] == y\" + P(X)y'+ Q(x)y == f(x)) (22))

with complementary function)

Yc(x) ==ClYl(X) +czyz(x)) (23))

on someopeninterval I where the functions P and Q are continuous. We want to

find functions uland Uz such that)

\037) yp(x) ==Ul(X)Yl(X) +uz(x)yz(x)) (24))

is a particular solutionof Eq.(22).Onecondition on the two functions Ul and Uz is that L[yp]==f(x).Because

two conditions are requiredto determine two functions, we are free to imposean

additional condition of our choice.We will do so in a way that simplifiesthe com-putations as much aspossible.But first, to imposethe condition L[yp]==f(x),wemust compute the derivatives

y\037and

y\037.The product rule gives)

1(

1 I) (

1 I)yp

==UlYl + uzyz + U1Yl + uzyz .)

Toavoid the appearanceof the secondderivativesu '{andu\037,

the additionalconditionthat we now imposeis that the secondsum here must vanish:)

1 1 0U1Yl + uzYz == .) (25))))


Then)

, , ,Yp

==UIYI +U2Y2') (26))

and the product rule gives)

\"(

\" \") (

\" \"

)Yp==

UIYI +U2Y2 + UIYI + U2Y2 .) (27))

But both YI and Y2 satisfy the homogeneousequation)

y\" + Py' + Qy ==0)

associatedwith the nonhomogeneousequation in (22),so)

\" P 'QYi

==-Yi-

Yi) (28))

for i == 1,2.It therefore follows from Eq.(27)that)

Y\037

==(u'lY\037 + u;Y\037)

-P .(u IY\037

+ U2Y\037)- Q . (u IYI +U2Y2).)

In view of Eqs.(24)and (26),this means that)

\"(

\" , ') P '

Q .Yp

== uIYI + u 2Y2-

Yp-

YP')

hence)

L[ ]\" \"

YP ==uIYI +u2Y2.) (29))

The requirement that Yp satisfy the nonhomogeneousequation in (22)-thatis, that

L[yp]==f(x)-thereforeimpliesthat)

\" \" f( )uIYI + u 2Y2 == x.) (30))

Finally, Eqs. (25)and (30)determine the functions U I and U2 that we need.Collectingtheseequations,we obtain a system)

\037)

, , 0UIYI + U2Y2 == ,\" \" f( )uIYI + u 2Y2 == x)

(31))

of two linear equations in the two derivatives u'l and u;.Note that the determinantof coefficients in (31)is simply the Wronskian W (YI, Y2). Oncewe have solvedthe equations in (31)for the derivativesu'l and u;,we integrateeach to obtain the

functions uland U2 such that)

\037)Yp

== UIYI + U2Y2) (32))

is the desiredparticular solution of Eq. (22). In Problem 63 we ask you to carryout this processexplicitlyand thereby verify the formula for Yp (x) in the followingtheorem.)))


THEOREM1 Variationof ParametersIf the nonhomogeneous equation y\" + P(x)y'+ Q(x)y == f(x) has comple-mentary function Yc(x) == CIYl(x)+ C2Y2(X), then a particular solution is givenby)

( ) -- ( ) fY2(x)f(x)d + ( ) f

Yl(x)f(x)dYp x - Yl X X Y2 X x,W(x) W(x))

(33))

where W == W(Yl, Y2) is the Wronskian of the two independentsolutionsYl andY2 of the associatedhomogeneousequation.)

Example11)..\037... ...\" . ... ..,. \037 ,........... \037

Find a particular solution of the equation y\" +Y == tan x.)

Solution The complementary function is Yc(x) == Clcosx+ C2sinx,and we couldsimplysubstitute directly in Eq. (33).But it is more instructive to setup the equations in

(31)and solve for u; and u;,sowe beginwith)

Yl ==cosx,, .

Yl==-SIn x,)

Y2 ==sInx,,

Y2 ==COSx.)

Hencethe equations in (31)are)

(U'l)(cosx) + (u;)(sinx)==0,(u;)(-sinx)+ (u;)(cosx)== tanx.)

We easilysolve theseequations for),. sin2xu

1==-SInxtan x ==- ==cosx-secx,cosx)

, .U2 ==COSx tan x ==SIn x.)

Hencewe take)

Ul = f (eosx- seex)dx= sinx-InIseex+ tanxl)

and)

U2 = f sinxdx= -eosx.

(Do you seewhy we choosethe constants of integration to be zero?)Thus our

particular solution is)

Yp(x) == U 1 (X)Yl (x)+ U2(x)Y2 (x)== (sinx- In Isecx + tan xI)cosx +(-cosx)(sinx);)

that is,)

Yp(x) ==-(cosx)In Isecx+ tanxl.) .)))


_Pro\037lem\037)In Problems 1through 20,find a particular solution y p of the

given equation. In all theseproblems,primes denotederiva-tives with respectto x.1.y\" + 16y== e3x 2. y\" -y' -2y == 3x + 43.y\" - y/ -6y == 2 sin 3x 4. 4y\" + 4y/ + y == 3xeX5. y\" + y/ + y == sin 2x 6. 2y\" + 4y/ + 7y == x2

7. y\" -4y == sinh x 8. y\" -4y == cosh2x9.y// +2y' - 3y == 1+xex

10.y\" + 9y == 2cos3x + 3 sin 3x11.y(3) + 4y' == 3x - 1 12.y(3) + y/ == 2- sinx13.y\" + 2y' + 5y == eX sin x 14.y(4) - 2y\" + y == xeX

15.y(5) + 5y(4)- y == 17 16.y\" + 9y == 2x2e3x + 517.y\" + y == sinx+ x cosx18.y(4) -5y\" + 4y == eX -xe2x

19.y(5) + 2y(3) + 2y\" == 3x2 - 120.y(3) -Y == eX + 7)

In Problems21through 30, set up the appropriate form ofaparticular solution YP' but do not determine the values of the

coefficients.21.y\" -2y' + 2y == eX sin x22.y(5) - y(3) == eX + 2x2-523.y'/ + 4y == 3x cos2x24. y(3) - y\" - 12y'== X -2xe-3x

25.y// + 3y' + 2y == x(e-X -e-2x )26.y\" -6y' + 13y== xe3x sin 2x27.y(4) + 5y\" + 4y == sin x + cos2x28.y(4) + 9y\" == (x2 + 1)sin 3x29.(D- 1)3(D2 -4)y == xeX + e2x + e-2x

30.y(4) - 2y\" + y == x2 cosX)

Solvethe initial value problems in Problems31through 40.

31.y\" + 4y == 2x;y(O) == 1,y'(O)== 232.y\" + 3y' + 2y == eX; y(O) == 0, y'(O)== 333.y\" + 9y == sin 2x;y(O) == 1,y/(O) == 034. y\" + y == cosx;y(O) == 1,y/(O) == -135.y\"

-2y' + 2y == x + 1;y(O) == 3,y/(O) == 036.y(4) -4y\" == x2; y(O) == y'(O)== 1,y\"(O) == y(3)(0) == -137.y(3) - 2y\" + y/ == 1+ xex; y(O) == y/(O) == 0, y\"(O) == 138.y\" + 2y' + 2y == sin 3x;y(O) == 2, y/(O) == 039.y(3) + y\" == x + e-x; y(O) == 1,y' (0) == 0, y\" (0) == 140. y(4) -Y == 5; y(O) == y' (0) == y\" (0) == y(3)(0) == 041.Find a particular solution of the equation

y(4) _ y(3) _y\" _ y' _ 2y == 8x5.

42. Find the solution of the initial value problem consistingof the differential equation of Problem 41and the initial

conditions)

y (0) == y/ (0) == y\" (0) == y(3)(0) == O.)

43. (a) Write

cos3x+ i sin 3x == e3ix== (cosx+ i sinx)3

by Euler'sformula, expand,and equatereal and imag-inary parts to derive the identities

cos3x ==\037

cosx+ icos3x,sin 3x ==

\037

sin x - i sin 3x.)

(b) Usethe result ofpart (a) to find a general solution of)

y\" + 4y == cos3x.)

Use trigonometric identities to find generalsolutions of the

equations in Problems44 through 46.

44. y\" + y' + y == sinxsin 3x45. y\" + 9y == sin 4 x46. y\" + y == x cos3 X)

In Problems47 through 56,use the method ofvariation ofpa-rameters to find a particular solution of the given differential

equation.

47. y\" + 3y' + 2y == 4ex 48. y\"-2y'- 8y == 3e-2x

49. y\"-4y' + 4y == 2e2x 50. y\" -4y == sinh 2x

51.y// + 4y == cos3x 52. y\" + 9y == sin 3x53.y\" + 9y == 2sec3x 54. y\" + y == csc2x55.y\" + 4y == sin 2x 56. y\"

-4y == xeX

57.You can verify by substitution that Yc == CIX + C2X-1 is acomplementary function for the nonhomogeneous second-order equation)

x2y\" + xy/

-y == 72x5

.)

But beforeapplying the method of variation of parame-ters, you must first divide this equation by its leading co-efficient x2 to rewrite it in the standard form)

\" 1, 1 3Y + -

y --y == 72x .x x 2)

Thus f(x) == 72x3 in Eq. (22).Now proceedto solvethe

equations in (31)and thereby derive the particular solution

Yp== 3x5

.)

In Problems 58 through 62,a nonhomogeneous second-orderlinear equation and a complementary function Yc are given.Apply the method ofProblem 57 to find a particular solution

of the equation.

58.x2y\"

-4xy'+ 6y == x3; Yc == CIX2 + C2x3

59.x2y/'

-3xy'+ 4y == x4; Yc == X2(Cl + c2lnx )

60.4x2y\"

-4xy'+ 3y == 8x4/3; Yc == CIX + C2x3/4

61.x2y\" + xy/ + y == In x;Yc == CIcos(1nx)+ C2sin(lnx)

62.(x2 -l)y\"

-2xy'+ 2y == x2 -1;Yc == CIX + c2(1+ x2)63.Carry out the solution processindicated in the text to

derive the variation of parameters formula in (33)from

Eqs.(31)and (32).64. Apply the variation of parameters formula in (33)to find

the particular solution y p(x)== -xcosx of the nonhomo-

geneousequation y\" + y == 2sin x.)))

162 Chapter2 LinearEquationsof HigherOrder)_ ForcedOscillationsand Resonance)

Equilibrium

positionI

I

k l)

FIGURE2.6.1.Thecart-with-flywheel system.)

In Section2.4we derived the differentialequation)

\037) mx\" +ex'+kx = F(t)) (1))

that governs the one-dimensionalmotion of a massm that is attached to a spring(with constant k) and a dashpot (with constantc)and is alsoacted on by an externalforce F(t).Machineswith rotating components commonly involve mass-springsystems(or their equivalents)in which the external force is simpleharmonic:)

F(t) = Focoswt or F(t) = Fosinwt,) (2))

where the constant Fo is the amplitude of the periodicforce and w is its circularfrequency.

For an exampleof how a rotating machine component can provide a sim-ple harmonic force, considerthe cart with a rotating vertical flywheel shown in

Fig.2.6.1.The cart has massm -mo, not including the flywheel of massmo. Thecentroid of the flywheel is off center at a distancea from its center, and its angular

speedis w radians per second.The cart is attached to a spring (with constant k)as shown.Assume that the centroid of the cart itself is directly beneath the centerof the flywheel, and denoteby x(t) its displacementfrom its equilibrium position(wherethe spring is unstretched).Figure 2.6.1helpsus to seethat the displacementx of the centroid of the combinedcart plusflywheel is given by)

_ (m -mo)x +mo(x +a coswt) moax = = x+-coswt.m m)

Let us ignore friction and apply Newton'ssecondlaw mx\" = -kx,becausethe

force exertedby the spring is -kx.We substitute for x in the last equation to obtain)

mx\" -moaw2coswt = -kx;)

that is,)mx\" +kx = moaw2coswt .) (3))

Thus the cart with its rotating flywheel acts likea masson a spring under the in-fluence of a simpleharmonic external force with amplitude Fo = moaw2. Suchasystem is a reasonablemodel of a front-loading washing machine with the clothesbeingwashedloadedoff center.This illustrates the practical importanceof analyz-ing solutionsof Eq.(1)with external forcesas in (2).)

UndampedForcedOscillationsTo study undamped oscillationsunder the influence of the external force F(t)Focoswt, we setc = 0 in Eq.(1),and therebybeginwith the equation)

\037) mx\" +kx = Fo coswt) (4))

whosecomplementaryfunction is Xc = Clcoswot +C2sin wot. Here)

\037) wo=(f)))

2.6ForcedOscillationsand Resonance 163)

(as in Eq.(9)of Section2.4)is the (circular)natural frequencyof the mass-springsystem.The fact that the angle wot is measuredin (dimensionless)radiansremindsus that if t is measuredin seconds(s),then Wo is measuredin radianspersecond-that is, in inverse seconds(S-I).Also recallfrom Eq. (14)in Section2.4that di-vision of a circular frequency w by the number 2n of radians in a cyclegives the

corresponding(ordinary)frequencyv ==w/2n in Hz (hertz ==cyclesper second).Let us assumeinitially that the external and natural frequenciesare unequal:

w -1= WOo We substitute xp == A coswt in Eq. (4) to find a particular solution. (Nosineterm is neededin xp becausethere is no term involving x' on the left-hand sidein Eq.(4).)This gives)

-mw2A coswt +kA coswt == Focoswt,)

so)Fo

A==k -mw 2)

Fo/m

w 2-w 2 'o)

(5))

and thus)Fo/m

xp(t)== 2 2 coswt.Wo -w) (6))

Therefore, the general solutionx ==Xc +xp is given by)

Fo/mx(t) ==Clcoswot +C2sin wot + 2 2 coswt,w -w

o)

(7))

where the constants Cl and C2 are determined by the initial valuesx (0)and x'(0).Equivalently, as in Eq.(12)of Section2.4,we can rewriteEq.(7)as)

Fo/mx(t)==Ccos(wot-a)+2 2 coswt,w -w

o)

(8))

so we seethat the resulting motion is a superpositionof two oscillations,one with

natural circular frequencyWo, the other with the frequencyw of the externalforce.)

Example1)

\"\"N,'N'.'nn_'n\037n\037n\037,-\" '\" '\" '\" .. ,,,_n.. N \037 \" '\" '\" N \"\"\"\"\" \"'\037\"\" A.' \037 .....,\037_ \037..... \037 \"\"'\"' '\" \037 \037\"N\"\"\"\" \037\"\"\" N '\" '\" \"\"\"\", A.' \037\037N' \037\037 N' \"\"\"\", ..,\"\"\",......\".,,,'\" '\" \"\"'\"' \"'N\037 '\" '\" ... '\" \". \"'.. ..... ... \037

Supposethat m == 1,k ==9, Fo == 80, and w ==5,so the differential equationin (4)IS)

x\" +9x==80cos5t.)

Find x(t) if x(O)==x'(0)== O.)

Solution Herethe natural frequency Wo == 3 and the frequency w == 5 of the externalforceare unequal, as in the precedingdiscussion.First we substitute xp == A cos5t in

the differential equation and find that -25A+ 9A == 80,so that A == -5.Thus a

particular solution isxp (t) ==-5cos5t.

The complementaryfunction is Xc ==Clcos3t +C2sin 3t, so the generalsolution ofthe given nonhomogeneousequation is)

x(t) ==Clcos3t +C2sin 3t -5 cos5t,)

with derivative)

x'(t) ==-3Clsin 3t +3C2cos3t + 25 sin 5t.)))


15

10

5)

Period = 2nI- -II I

I I

I I

I I)

\037 0)

-5-10-15

o n 2n 3n 4n 5n 6nt)

FIGURE2.6.2.Theresponsex(t) = 5cos3t -5cos5t in

Example 1.)

Example2)

1.5x = sin 5t

1.0I

0.5I

\037 0.0I \\

-0.5 \\

\\

\\I-1.0 \\../ \\.

x = sin 5t sin 50t)

\\ I)

\\ I)

I \\

\\)

-1.50.0 0.5 1.0 1.5 2.0 2.5 3.0t)

FIGURE2.6.3.Thephenomenon of beats.)

The initial conditionsx(0)==0 and x'(0)==0 now yield Cl ==5 and C2 ==0, so the

desiredparticular solution is)

x(t) ==5 cos3t -5cos5t.)

As indicated in Fig.2.6.2,the periodof x(t) is the least commonmultiple 2n of the

periods2n13and 2n15of the two cosineterms. .)

Beats)If we imposethe initial conditionsx(0)==x'(0)==0 on the solution in (7),we find

that)

FoC1 ==- and C2 ==0,

m(w6-(2)

so the particular solution is)

Fox(t) == 2 2 (coswt -coswot).

m(wo -w )

The trigonometric identity 2 sin A sin B ==cos(A-B)-cos(A+ B),appliedwith

A ==\037(wo +w)t and B ==

\037(wo-w)t, enablesus to rewriteEq.(9)in the form)

(9))

2Fo 1 1x(t) == 2 2 sin 2(wO-w)t sin 2(wO+w)t.

m (wo -w))

(10))

Supposenow that w \037 Wo, so that Wo+w is very large in comparisonwith Iwo -wi.Then sin

\037(wo + w)t is a rapidly varying function, whereassin\037(wo

- w)t is aslowly varying function. We may therefore interpret Eq. (10)as a rapid oscillationwith circular frequency \037(wo +w),)

x(t) == A(t) sin\037

(wo +w)t,)

but with a slowly varying amplitude

2Fo . 1A (t) == 2 2 sIn 2 (wo -w) t .m (wo -w

))

With m ==0.1,Fo ==50,Wo ==55,and w ==45,Eq.(10)gives

x(t) ==sin 5t sin 50t.

Figure 2.6.3showsthe correspondingoscillationof frequency\037

(wo +w) ==50 that

is \"modulated\" by the amplitudefunction A(t) ==sin 5t of frequency \037(wo-w)==5.

.)

I)

A rapid oscillationwith a (comparatively)slowly varying periodicamplitudeexhibitsthe phenomenon of beats. For example,if two horns not exactly attuned

to one another simultaneouslyplay their middleC, one at wo/(2n)== 258 Hz andthe other at w/(2n)== 254Hz, then one hears a beat-anaudible variation in the

amplitude of the combinedsound-witha frequencyof)

(wo -w)/22n)

258-254==2 (Hz).

2)))

\037 0.0)

-0.5)

-1.0)

1.5)....)

1.0) ....)

....)

0.5)

.......)-1.5 .......

0.000.250.500.751.001.251.50t)

FIGURE2.6.4.Thephenomenon ofresonance.)

Example3)


ResonanceLookingat Eq.(6),we seethat the amplitude A of xp is large when the natural andexternal frequenciesWo and wareapproximatelyequal. It is sometimesuseful torewrite Eq.(5)in the form)

FoA==

k -mw 2)

Folk ==::i:pFo1 - (wlwO)2 k

') (11))

where Folk is the static displacementof a spring with constantk due to a constantforce Fo, and the amplification factor p is defined to be)

1p--11- (wlwo)21.)

(12))

It is clearthat p \037 +00as w ----+ wo. This is the phenomenonof resonance-the increasewithout bound (as w ----+ wo) in the amplitude of oscillationsof an

undamped system with natural frequency Wo in responseto an external force with

frequencyw \037 wo.We have beenassuming that w i= WOo What sort of catastrophe should one

expectif wand Wo are,preciselyequal?Then Eq.(4),upon divisionof each term bym, becomes)

/I 2 Fox +wox ==-coswot.m)

(13))

Becausecoswot is a term of the complementaryfunction, the method of undeter-mined coefficientscallsfor us to try)

Xp(t) ==teA coswot + B sin wot).)

We substitute this in Eq. (13)and thereby find that A == 0 and BHencethe particular solution is)

Fo/(2mw o).)

Fo .xp(t)== t SIn wot.

2mwo)(14))

The graph of xp(t)in Fig.2.6.4(in which m == 1,Fo == 100,and Wo ==50)showsvividly how the amplitude of the oscillation theoretically would increasewithout

bound in this caseof pure resonance,w == wo. We may interpret this phenomenonas reinforcement of the natural vibrations of the system by externally impressedvibrations at the samefrequency.)

Supposethat m ==5 kg and k ==500N/m in the cart with the flywheel of Fig.2.6.1.Then the natural frequency is Wo == -Jklm == 10rad/s; that is, 10/(2n)\037 1.59Hz.We would therefore expectoscillationsof very large amplitude to occur if the

flywheel revolvesat about (1.59)(60)\037 95 revolutionsperminute (rpm). .In practice,a mechanicalsystem with very little damping can bedestroyedby

resonancevibrations. A spectacularexamplecan occurwhen a column of soldiersmarches in stepover a bridge.Any complicatedstructure suchasa bridgehas manynatural frequenciesof vibration. If the frequencyof the soldiers'cadenceisapprox-imately equal to one of the natural frequenciesof the structure, then-justas in oursimpleexampleof a masson a spring-resonancewill occur.Indeed,the resulting)))


Example4)

resonancevibrations can be of such large amplitude that the bridgewill collapse.This has actually happened-forexample,the collapseof Broughton BridgenearManchester,England, in 1831-andit is the reasonfor the now-standardpracticeof breakingcadencewhen crossinga bridge. Resonancemay have been involvedin the 1981KansasCity disasterin which a hotel balcony (calleda skywalk) col-lapsedwith dancerson it. The collapseof a building in an earthquakeis sometimesdue to resonancevibrations causedby the ground oscillating at one of the natural

frequenciesof the structure; this happenedto many buildings in the MexicoCityearthquakeof September19,1985.On occasionan airplane has crashedbecauseofresonant wing oscillationscausedby vibrations of the engines.It is reported that

for someof the first commercialjet aircraft, the natural frequency of the verticalvibrations of the airplane during turbulence was almost exactly that of the mass-springsystemconsistingof the pilot'shead (mass)and spine(spring).Resonanceoccurred,causingpilotsto have difficulty in reading the instruments. Largemodemcommercialjetshave different natural frequencies,so that this resonanceproblemno longer occurs.)

ModelingMechanicalSystems)The avoidance of destructiveresonancevibrations is an ever-presentconsiderationin the designof mechanical structures and systemsof all types. Often the mostimportant step in determining the natural frequency of vibration of a system is theformulation of its differential equation. In addition to Newton'slaw F == ma, the

principleof conservation of energy is sometimesuseful for this purpose(as in thederivation of the pendulum equation in Section2.4).The following kinetic andpotential energy formulas are often useful.

1.Kineticenergy:T == 1mv2for translation of a massm with velocity v;

2.Kinetic energy:T == 1/w 2 for rotation of a body of a moment of inertia /with angular velocity w;

3.Potential energy: V == 1kx2for a spring with constant k stretched or com-presseda distancex;

4. Potential energy:V = mgh for the gravitational potential energy of a massmat height h above the referencelevel (the level at which V = 0),provided that

g may be regardedas essentiallyconstant.)

Find the natural frequency of a massm on a spring with constant k if, instead ofslidingwithout friction, it is a uniform diskof radius a that rollswithout slipping,as shown in Fig 2.6.5.)

Solution With the precedingnotation, the principleof conservationof energy gives

1mv2+ 1/w2+ 4kx2 == E

where E is a constant (the total mechanical energy of the system).We note that

v == aw and recallthat / == ma2/2 for a uniform circular disk. Then we maysimplify the last equation to)

Equilibrium

positionI

I

I)

\037mv2 +\037kx2

==E.Becausethe right-hand sideof this equation is constant,differentiation with respect

x =0 to t (with v ==x' and v' = x\") now gives

G 3 III k I 0FI URE 2.6.5.Therolling disk. 2mx x + xx ==.)))


We divide eachterm by \037mx'to obtain

\" 2kx +-x==o.3m

Thus the natural frequency of horizontal back-and-forth oscillation of our rollingdisk is y'2k/3m,which is y'2/3\037 0.8165times the familiar natural frequencyy'k/m of a masson a spring that is slidingwithout friction rather than rolling without

sliding.It is interesting(andperhapssurprising) that this natural frequencydoesnot

dependon the radiusof the disk. It couldbe either a dime or a large disk with aradius of one meter (but of the samemass). .)

Example5) Supposethat a car oscillatesverticallyas if it werea massm == 800 kg on a singlespring (with constantk ==7 X 104 N/m),attached to a singledashpot (with constantc == 3000Nos/m).Supposethat this car with the dashpot disconnectedis driven

along a washboard road surface with an amplitude of 5 cm and a wavelength ofL == 10m (Fig.2.6.6).At what car speedwill resonancevibrations occur?)

m)

m)\037}\037)

Surface)

2nsy=acosT)

c)

_=Iy=-)

s=o)Equilibrium

position)

FIGURE2.6.6.Thewashboardroad surfaceofExample 5.)

FIGURE2.6.7.The\"unicyclemodel\" ofa car.)

Solution We think of the car as a unicycle, as pictured in Fig.2.6.7.Let x(t) denote the

upward displacementof the massm from its equilibrium position; we ignore the

force of gravity, becauseit merely displacesthe equilibrium positionas in Problem9 of Section2.4.We write the equationof the road surfaceas

27TSy ==a cos- (a ==0.05m, L == 10m).

L)(15))

When the car is in motion, the spring is stretchedby the amount x-y, soNewton'ssecondlaw, F ==ma, gives)

mx\" ==-k(x- y);)

that is,)

mx\" +kx ==ky) (16))

If the velocityof the car is v, then s == vt in Eq.(15),soEq.(16)takes the form)

, 27TVtmx'+kx ==ka cos .L)

(16'))))

This is the differential equation that governs the vertical oscillationsof the car. In

comparing it with Eq. (4), we seethat we have forced oscillationswith circularfrequency w == 2nvjL.Resonancewill occurwhen w == Wo == ,Jklm. We useour

numericaldata to find the speedof the car at resonance:)

L If 10\037X 104V ==- - ==- \037 14.89(m/s);2n m 2n 800)

that is, about 33.3mijh(using the conversionfactor of 2.237mijh perm/s). .)

DampedForcedOscillationsIn real physicalsystemsthere is always somedamping, from frictional effects ifnothing else.The complementaryfunction Xc of the equation)

\037)1/ I k 17mx +cx + X == rocoswt) (17))

is given by Eq.(19),(20),or (21)of Section2.4,dependingon whether c > Cer =,J4km , C == Cer' or C < Cere The specificform is not important here. What isimportant is that, in any case,theseformulas show that xc(t) \037 0 as t \037 +00.Thus Xc is a transientsolutionof Eq.(17)-onethat diesout with the passageoftime, leaving only the particular solutionXp.

The method of undeterminedcoefficientsindicatesthat we should substitute)

x(t) == A coswt + B sin wt)

in Eq. (17).When we do this, collectterms, and equate coefficientsof coswt andsin wt, we obtain the two equations)

(k -m(2)A +cwB == Fo,) -cwA+ (k -m(2)B ==0) (18))

that we solve without difficulty for)

(k -m(2)FoA==

(k -m(2)2+ (cw)2')

cwFoB== .(k -m(2)2+ (cw)2)

(19))

If we write)

A coswt + B sin wt ==C(coswt cosa + sin wt sin a) ==Ccos(wt -a))

as usual, we seethat the resulting steadyperiodicoscillation)

xp(t)==Ccos(wt-a)) (20))

has amplitude)

c = .JA2 + B2 = Fa

.J(k -m(2)2+ (cw)2)(21))

Now (19)impliesthat sin a == BjC>0,so it follows that the phaseangle a lies in

the first or secondquadrant. Thus)

B cwtan a ==-== with 0 <a <n,

A k -mw 2)(22))))

so)

ot=)

cwtan -1

k -mw 2cw

1T+ tan- 1

2k-mw)

if k >mw 2,)

if k <mw 2)

(whereasot = 1T12if k = m(2).Note that if c > 0,then the \"forcedamplitude\"-definedas a function C(w)

by (21)-alwaysremains finite, in contrast with the caseof resonancein the un-

dampedcasewhen the forcing frequency w equalsthe critical frequency Wo =-Jklm. But the forced amplitude may attain a maximum for somevalue of w, in

which casewe speakof practicalresonance.To seeif and when practical res-onance occurs,we need only graph C as a function of wand look for a globalmaximum. It can be shown (Problem 27)that C is a steadily decreasingfunction

of w if c > y'2km. But if c < y'2km, then the amplitude of C attains a maxi-mum value-andso practical resonanceoccurs-atsomevalue of w lessthan wo,and then approacheszero as w \037 +00.It follows that an underdampedsystemtypically will undergoforced oscillationswhoseamplitude is). Largeif w iscloseto the critical resonancefrequency;. Closeto Folk if w is very small;. Very small if w is very large.)

Example6).... ........... ............

Find the transient motion and steady periodicoscillationsof a damped mass-and-spring system with m = 1,c = 2,and k = 26 under the influenceof an externalforce F(t) = 82cos4t with x(O)= 6 and x'(0)= O.Also investigatethe possibilityof practical resonancefor this system.)

Solution The resulting motion x(t) = Xtr(t) + xsp(t)of the masssatisfies the initial value

problem)

x\" +2X' +26x= 82cos4t; x(O)= 6, x'(0)= O. (23))

Insteadof applying the general formulasderivedearlier in this section,it isbetter in

a concreteproblemto work it directly.The roots of the characteristicequation)

r 2 +2r+26 = (r + 1)2+25 = 0)

are r = -1::l::5i,so the complementaryfunction is)

xc(t)= e-t (Clcos5t +C2sin 5t).)

When we substitute the trial solution)

x(t) = A cos4t+ B sin4t)

in the given equation,collectliketerms, and equate coefficientsof cos4t and sin 4t,we get the equations)

lOA + 8B = 82,-8A+ lOB= 0)))


with solution A ==5,B ==4.Hencethe general solutionof the equation in (23)is)

X(t) ==e-t(C1cos5t +C2sin 5t) +5 cos4t +4 sin 4t.)

At this point we imposethe initial conditions x(O) == 6,x'(0) == 0 and find that

C1 == 1 and C2 == -3.Therefore, the transient motion and the steady periodicoscillationof the massare given by)

Xtr(t) ==e-t (cos5t - 3 sin 5t))

and)

.(

5 4 .)xsp(t)= 5 cos4t +4 sm 4t =.J41

.J4Tcos4t +

.J4Tsm 4t

==,J41cos(4t-a))

where a == tan- 1

(\037)\037 0.6747.

Figure 2.6.8showsgraphs of the solutionx(t) ==Xtr(t) +xsp(t)of the initial

value problem)

x\" + 2X' +26x==82cos4t,x(O)==Xo, x'(O)==0 (24))

for the different values Xo == -20,-10,0,10,and 20of the initial position.Herewe seeclearly what it means for the transient solution Xtr(t) to \"die out with the

passageof time,\" leaving only the steady periodicmotion xsp(t).Indeed,becauseXtr(t) --+ 0 exponentially, within a very few cyclesthe full solution x(t) and the

steady periodicsolution xsp(t)are virtually indistinguishable(whatever the initial

positionxo).)

x)

-10)

10)

t)

-20)

FIGURE2.6.8.Solutions of the initial value problem in (24)with Xo = -20,-10,0, 10,and 20.)))

To investigate the possibilityof practical resonancein the given system, wesubstitute the values m = 1,c = 2,and k = 26 in (21)and find that the forcedamplitude at frequencyw is)

The graph of C(w)is shown in Fig.2.6.9.The maximum amplitude occurswhen)

109876

u 54321

00)

Practical resonance)


82C(w)= .

,J676-48w2 +w 4)

-164w(w2-24) = o.(676-48w2 +(4)3/2)

10 15w)

20) I -41(4w3 -96w)C (w) --(676-48w2 +(4)3/2)

5)

FIGURE2.6.9.Plot ofamplitude C versus external

frequency w.)

Thus practical resonanceoccurswhen the externalfrequencyisw = J24(a bit lessthan the mass-and-spring'sundampedcritical frequencyof Wo = ,Jk/m = ,J26)..)

IfD Problems)

In Problems1through 6, expressthe solution of the given ini-tial value problem as a sum of two oscillationsas in Eq. (8).Throughout, primes denotederivatives with respectto time t.In Problems 1-4,graph the solution function x (t) in such away that you can identify and label(as in Fig. 2.6.2)its pe-riod.)

1.x\" + 9x = 10cos2t; x(O)=x'(0)= 02.x\" + 4x = 5 sin 3t; x(O)= x'(O)= 03.x\" + 100x = 225cos5t + 300sin 5t; x(O) 375,

x'(O)= 04. x\" + 25x= 90cos4t; x (0)= 0,x'(0)= 905. mx\" + kx = Focoswt with w i= Wo;x (0)= Xo,x'(0)= 06.mx\" +kx = Focoswt with w = Wo;x(O)= 0,x'(O)= Vo)

In eachofProblems 7 through 10,find the steadyperiodicso-lution xsp(t) = Ccos(wt-a) of the given equation mx\" +cx'+ kx = F(t) with periodicforcing function F(t)offre-quency w. Then graph xsp(t) together with (for comparison)the adjustedforcing function F](t) = F(t)jmw.7. x\" + 4x'+ 4x = lOcos3t8.x\" + 3x'+ 5x = -4cos5t9.2X\" + 2x'+ x = 3 sin lOt10.x\" + 3x'+ 3x = 8coslOt + 6 sin lOt)

In each of Problems 11through 14,find and plot both the

steady periodicsolution xsp(t) = Ccos(wt-a) of the givendifferential equation and the transient solution Xtr(t) that sat-isfies the given initial conditions.

11.x\" + 4x'+ 5x = lOcos3t;x(O)= x'(O)= 012.x\" + 6x'+ 13x= 10sin 5t; x(O)= x'(0)= 013.x\" + 2x'+ 26x= 600coslOt;x(O)= 10,x'(O)= 014.x\" + 8x'+ 25x = 200cost + 520sin t; x(O) = -30,

x'(O)= -10)

Eachof Problems 15through 18gives the parametersfor aforcedmass-spring-dashpotsystem with equation mx\" +cx'+kx = Focoswt. Investigate the possibility ofpracticalreso-nanceof this system. In particular, find the amplitude C(w)ofsteadyperiodicforcedoscillationswithfrequency w. Sketch

the graph of C(w) and find the practicalresonancefrequencyw (if any).

15.m = 1,c = 2, k = 2, Fo = 2)

16.m=1,c=4,k=5,Fo =10)

17.m=1,c=6,k=45,Fo =50)

18.m = 1,c = 10,k = 650,Fo = 100)

19.A mass weighing 100lb (mass m = 3.125slugs in fpsunits) is attached to the end of a spring that is stretched

1 in. by a forceof 100lb. A forceFocoswt acts on the

mass.At what frequency (in hertz) will resonanceoscilla-tions occur?Neglectdamping.)

20.A front -loading washing machine is mounted on a thick

rubber pad that acts like a spring; the weight W = mg(with g = 9.8mjs2) of the machine depressesthe pad ex-actly 0.5cm.When its rotor spins at w radians persecond,the rotor exertsa vertical forceFocoswt newtons on the

machine. At what speed(in revolutions per minute) will

resonancevibrations occur?Neglectfriction.)

21.Figure 2.6.10shows a mass m on the end of a pendulum(of length L) also attached to a horizontal spring (with

constant k). Assume small oscillationsof m so that the

spring remains essentially horizontal and neglectdamp-ing. Find the natural circular frequency Wo of motion ofthe mass in terms of L, k, m, and the gravitational con-stant

g.)))

FIGURE2.6.10.Thependulum-and-spring system ofProblem21.

22.A mass m hangs on the end of a cordaround a pulley ofradius a and moment of inertia I,as shown in Fig. 2.6.11.Therim of the pulley is attached to a spring (with constant

k). Assume small oscillationsso that the spring remainsessentially horizontal and neglectfriction. Find the natu-ral circular frequency of the system in terms ofm, a, k, I,and g.)

FIGURE2.6.11.Themass-spring-pulley system ofProblem22.

23.A building consistsof two floors. The first floor is at-tached rigidly to the ground, and the secondfloor is ofmass m = 1000slugs (fps units) and weighs 16tons

(32,000lb). Theelasticframe of the building behavesasaspring that resistshorizontal displacementsof the secondfloor; it requiresa horizontal forceof5 tons to displacethesecondfloor a distance of 1 ft. Assume that in an earth-

quake the ground oscillateshorizontally with amplitudeAo and circular frequency w, resulting in an external hor-izontal forceF(t) = mAow2 sin wt on the secondfloor.(a) What is the natural frequency (in hertz) ofoscillationsof the secondfloor? (b) If the ground undergoes oneoscillation every 2.25s with an amplitude of 3 in., whatis the amplitude of the resulting forcedoscillationsof thesecondfloor?

24. A mass on a spring without damping is actedon by theexternal forceF (t) = Focos3 wt. Show that there are twovalues of w for which resonanceoccurs,and find both.

25.Derivethe steady periodicsolution of)

\" ' k 17.mx + ex + x = 1'0SIn wt.)

In particular, show that it is what onewould expect-thesameas the formula in (20)with the samevalues ofC and

w, exceptwith sin(wt -a) in placeofcos(wt-a).26.Given the differential equation)

\" ' k E D'mx + ex + x = 0 coswt + 1'0SIn wt)

-with both cosineand sine forcing terms-derive the

steady periodicsolution)

J \302\2432 + F.2

xsp(t) = 0 0cos(wt-a - fJ),J(k -m(2)2+ (ew)2)

where a is defined in Eq.(22) and fJ = tan- 1(Fo/Eo).(Suggestion:Add the steady periodicsolutions separatelycorresponding to Eocoswt and Fo sin wt (seeProblem

25).)27.According to Eq.(21),the amplitude of forcedsteady

periodicoscillations for the system mx\" + ex'+ kx -Focoswt is given by)

FoC(w)= .J(k -m(2)2+ (ew)2)

(a) If e > ecr/\037' where ecr = -vi 4km , show that Csteadily decreasesas w increases. (b) If e < ecr/\037,show that C attains a maximum value (practical reso-nance) when)

W = Wm = J \037 - \037 < Wo = fk.m 2m2 V ;;)

28.) As indicated by the cart-with-flywheel examplediscussedin this section,an unbalanced rotating machine part typ-ically results in a forcehaving amplitude proportional tothe squareof the frequency w. (a) Show that the am-

plitude of the steady periodicsolution of the differential

equation)

mx\" + ex'+ kx = mAw 2 coswt)

(with a forcing term similar to that in Eq.(17))is given by)

mAw 2

C(w)= .J(k -m(2)2+ (ew)2)

(b) Supposethat e2 < 2mk. Show that the maximum

amplitude occursat the frequency Wm given by)

w m =)k

(2mk

)m 2mk -e2 .)

Thus the resonancefrequency in this caseis larger (incontrast with the result ofProblem27) than the natural fre-quency Wo = -vlk/m . (Suggestion: Maximize the squareof C.))))

2.7ElectricalCircuits 173)

Automobile VibrationsProblems 29 and 30 deal further with the car of Example5. Its upward displacementfunction satisfies the equationmx\" + cx'+ kx = cy' + ky when the shockabsorberis con-nected(sothat c > 0).With y = a sin wt for the roadsurface,this differential equation becomes

mx\" + cx'+ kx = Eocoswt + Fo sin wt)

amplitude slightly over 5 cm.Maximum resonancevibra-tions with amplitude about 14cm occuraround 32mi/h,but then subside to more tolerable levels at high speeds.Verify thesegraphically basedconclusionsby analyzingthe function C(w). In particular, find the practical reso-nance frequency and the corresponding amplitude.)

c = aJP+ (cw)2

J(k -m(2)2+ (cw)2Becausew = 2nv/L when the caris moving with velocityv, this gives C asa function of v.

30.Figure 2.6.12shows the graph of the amplitude function

C(w) using the numerical data given in Example 5 (in-cluding c = 3000N.s/m). It indicates that, as the caracceleratesgradually from rest, it initially oscillateswith)

15

12\037

eu 9'-'\"

(1)\"'0

::::s

.\037 6\037e<r:

3

00 20 40 60 80 100Velocity (mi/h))

where Eo = cwaand Fo = ka.29. Apply the result ofProblem26to show that the amplitude

C of the resulting steady periodicoscillationfor the carisgiven by)

FIGURE2.6.12.Amplitude ofvibrations ofthe caron a washboard surface.)

\037ElectricalCircuits\037\"..UN< m\", _\".\" _,_\"_\",\"_\037,, __\037, ,,__..,_m_)

oSwitch)

Herewe examine the RLCcircuit that is a basicbuilding blockin morecomplicatedelectricalcircuits and networks.As shown in Fig.2.7.1,it consistsof)

L) A resistorwith a resistanceof R ohms,An inductorwith an inductanceof L henries,andA capacitorwith a capacitanceof Cfarads)

R)

<;ir\037ui\037

ElemeDt),..YQJ\037ftgl,>

::,lJi!Op..<

dlL-dt

RI

1C Q)

in serieswith a sourceof electromotive force (suchas a battery or a generator)that suppliesa voltage of E(t) volts at time t. If the switch shown in the circuitof Fig.2.7.1is closed,this resultsin a current of I(t) amperesin the circuit anda charge of Q(t) coulombson the capacitor at time t. The relation between the

functions Q and I is)

FIGURE2.7.1.TheseriesRLCcircuit.)

\037;= I(t).) (1))

Inductor) We will always usemks electricunits, in which time is measured in seconds.According to elementary principlesof electricity, the voltage drops across

the three circuit elementsare thoseshown in the table in Fig.2.7.2.We can analyzethe behavior of the seriescircuit of Fig.2.7.1with the aid of this table and one ofKirchhoff's laws:

The (algebraic)sum of the voltagedropsacrossthe elements in asimpleloopof an electricalcircuit is equal to the appliedvoltage.

As a consequence,the current and charge in the simpleRLC circuit of Fig.2.7.1satisfy the basiccircuit equation)

Resistor)

Capacitor)

FIGURE2.7.2.Tableof voltagedrops.)

\037)

dI 1L-+RI+-Q=E(t).dt C)(2))))


If we substitute (1)in Eq.(2),we get the second-orderlinear differentialequation)

1LQ\"+ RQ'+ -Q= E(t)

C)(3))

for the charge Q(t),under the assumption that the voltage E(t) is known.In most practical problemsit is the current I rather than the charge Q that is

of primary interest, so we differentiateboth sidesof Eq.(3)and substitute I for Q'to obtain)

\037)

1LI\"+RI'+-I= E'(t).

C)(4))

We donot assumehere a prior familiarity with electricalcircuits.It sufficestoregard the resistor,inductor, and capacitor in an electricalcircuit as \"black boxes\"that are calibratedby the constants R, L, and C.A battery or generator isdescribedby the voltage E(t) that it supplies.When the switch isopen,no current flows in the

circuit; when the switch is closed,there is a current I(t) in the circuit and a chargeQ(t)on the capacitor.All we needto know about theseconstants and functions isthat they satisfy Eqs.(1)through (4),our mathematicalmodel for the RLC circuit.We can then learn a gooddealaboutelectricityby studying this mathematicalmodel.)

TheMechanical-ElectricalAnalogyIt is striking that Eqs.(3)and (4)have preciselythe sameform as the equation)

\037) mx\" +ex'+kx = F(t)) (5))

of a mass-spring-dashpotsystem with external force F(t).The table in Fig.2.7.3detailsthis important mechanical-electricalanalogy.As a consequence,most ofthe resultsderived in Section2.6for mechanical systemscan be appliedat oncetoelectricalcircuits.The fact that the samedifferentialequationservesasa mathemat-ical modelfor such differentphysical systemsis a powerful illustration of the unify-

ing roleof mathematics in the investigation of natural phenomena.Moreconcretely,the correspondencesin Fig.2.7.3can be usedto construct an electricalmodel of agiven mechanical system,using inexpensiveand readily availablecircuit elements.The performance of the mechanical system can then be predictedby means of ac-curate but simplemeasurements in the electricalmodel.This is especiallyusefulwhen the actual mechanical system would be expensiveto construct or when mea-surements of displacementsand velocities would be inconvenient, inaccurate, oreven dangerous.This ideais the basisof analogeomputers--electricalmodelsofmechanical systems.Analog computersmodeledthe first nuclear reactors for com-mercial powerand submarinepropulsionbefore the reactors themselveswerebuilt.)

':'!\"li_{II_\037{\"\":tt.)Me\037II.a.rdc\037....$\037$\037gdt..!....i?!.!..........;.......)

Massm

Damping constant cSpring constant k

Position xForceF)

Inductance LResistanceR

Reciprocalcapacitance1/CCharge Q (using (3)(or current I using (4)))Electromotive forceE (or its derivative E'))

FIGURE2.7.3.Mechanical-electricalanalogies.)))

In the typical caseof an alternating current voltage E(t) = Eosinwt,Eq.(4)takesthe form)

1LI\"+ RI'+-I= wEocoswt.

C)(6))

As in a mass-spring-dashpotsystem with a simpleharmonic external force, thesolution of Eq. (6)is the sum of a transientcurrent Itr that approacheszero ast \037 +00(under the assumption that the coefficients in Eq. (6)are all positive,so the roots of the characteristic equation have negative real parts), and a steadyperiodiccurrentIsp; thus)

I = Itr + Isp.) (7))

Recall from Section2.6(Eqs.(19)through (22)there) that the steadyperiodicsolu-tion of Eq.(5)with F(t) = Focoswt is

Fo cos(wt-a)xsp(t)= ,J(k -m(2)2+ (cw)2)

where)

-1 cwa = tan 2 ' 0 <a < Jr.k-mw)If we make the substitutions L for m, R for c, IIC for k, and wEo for Fo, we getthe steadyperiodiccurrent)

Eocos(wt-a)Isp(t)=

JR2 +(WL

-W\037 r)

(8))

with the phaseangle)

-1 wRCa=tan 2 '1-LCw)o <a <Jr.- -) (9))

ReactanceandImpedanceThe quantity in the denominator in (8),)

z = JR2 +(WL

-W\037 Y

(ohms),) (10))

is calledthe impedanceof the circuit. Then the steadyperiodiccurrent)

EoIsp(t)= -cos(wt-a)

Z)(11))

has amplitude)

Eo10= Z') (12))

reminiscentof Ohm'slaw, I = EIR.)))


R)

-5)

5)

R)

D=a-!!2)

FIGURE2.7.4.Reactanceand

delay angle.)

-1r-\037I I

I I

I I

I)

FIGURE2.7.5.Time lag ofcurrent behind imposedvoltage.)

Example1)

Equation (11)gives the steady periodiccurrent as a cosinefunction, whereasthe input voltage E(t) == Eo sin wt was a sine function. To convert Isp to a sinefunction, we first introduce the reactance)

1S ==wL --.

wC)(13))

Then Z == ,JR2 + S2,and we seefrom Eq.(9)that a is as in Fig.2.7.4,with delayangle 8 ==a - !n.Equation (11)now yields)

lsp(t)=Eo

(cosa coswt + sin a sin wt)Z)

Eo

(SR.

)==- --coswt+-SlnwtZ Z Z)

= Eo(cos8 sin wt - sin 8coswt).

Z)

Therefore,)

Eo .Isp(t)==-sln(wt -8),

Z)(14))

where)

S LCw2- 18 ==tan- 1 -==tan- 1

R wRC)(15))

This finally gives the time lag8/w (in seconds)of the steady periodiccurrent Ispbehind the input voltage (Fig.2.7.5).)

InitialValue Problems)When we want to find the transient current, we are usually given the initial valuesI(0)and Q(O).SOwe must first find I'(0).To doso,we substitute t ==0 in Eq.(2)to obtain the equation)

1LI'(0)+ RI(0)+-Q(O)== E(O)C)

(16))

to determine I'(0)in terms of the initial values of current, charge, and voltage.)

Consideran RLC circuit with R == 50 ohms (Q),L == 0.1henry (H),and C5 x 10-4 farad (F).At time t ==0,when both I(0)and Q(0)are zero, the circuit isconnectedto a 110-V, 60-Hz alternating current generator. Find the current in thecircuit and the time lag of the steadyperiodiccurrent behind the voltage.)

Solution A frequencyof 60Hz means that w == (2n)(60)rad/s,approximately377 rad/s.Sowe take E(t) == 110sin 377t and useequality in placeof the symbolfor approximateequality in this discussion.The differentialequation in (6)takes the form)

(0.1)1\"+SOl'+20001== (377)(110)cos3771.)))

Example2)


We substitute the given values of R, L, C,and w == 377 in Eq.(10)to find that the

impedanceis Z ==59.58Q, so the steadyperiodicamplitude is)

110(volts)10== == 1.846amperes(A).59.58(ohms))

With the samedata, Eq.(15)gives the sinephaseangle:)

8 ==tan- 1 (0.648)==0.575.)

Thus the time lag of current behind voltage is)

8) 0.575==0.0015s,

377)w)

and the steadyperiodiccurrent is)

Isp ==(1.846)sin(377t-0.575).)

The characteristicequation (0.1)r2 +50r+2000==0 has the two roots'l\037

-44and r2 \037 -456.With theseapproximations,the general solution is)

I(t)==cle-44t +c2e-456t+ (1.846)sin(377t-0.575),)

with derivative)

I'(t) ==-44cle-44t -456c2e-456t+696cos(377t-0.575).)

Because1(0)== Q(O) == 0,Eq. (16)gives I'(0) == 0 as well. With these initial

values substituted, we obtain the equations)

1(0)==Cl +C2- 1.004==0,I'(0)==-44cl-456c2+584==0;)

their solution is Cl ==-0.307,C2== 1.311.Thus the transient solution is)

Itr(t) ==(-0.307)e-44t + (1.311)e-456t.)

The observation that after one-fifth of a secondwe haveIItr (0.2)I < 0.000047A

(comparableto the current in a singlehuman nerve fiber)indicatesthat the transient

solutiondiesout very rapidly, indeed. .)

Supposethat the RLC circuit of Example1,still with I(0) == Q(0) == 0, is con-nectedat time t == 0 to a battery supplying a constant 110V. Now find the current

in the circuit.)))


Solution We now have E(t) = 110,soEq.(16)gives)

10)

EO10=

-YR2 + (w L-

w1c )2)

Wm W)

FIGURE2.7.6.Theeffectoffrequency on 10.)

, E(0) 110I (0)= =-= 1100(A/s),L 0.1)

and the differentialequation is)

(0.1)/\"+SOl'+20001= E'(t) = O.)

Its general solution is the complementaryfunction we found in Example1:)

I(t) = cle-44t +c2e-456t.)

We solve the equations)

1(0)= Cl +C2= 0,1'(0)= -44cl-456c2= 1100)

for Cl = -C2= 2.670.Therefore,)

I(t) = (2.670)(e-44t -e-456t).)

Note that I(t) --+0 as t --++00even though the voltage is constant.) .)

ElectricalResonance)Consideragain the current differentialequation in (6)correspondingto a sinusoidalinput voltage E(t) = Eo sin wt. We have seenthat the amplitude of its steadyperiodiccurrent is)

Eo Eo10= -=z

/R2 + (wL -w\037 Y)

(17))

For typicalvaluesof the constants R, L, C,and Eo,the graph of 10asa function of (J)

resemblesthe one shown in Fig.2.7.6.It reachesa maximum valueat Wm = 1jv'LCand then approacheszero as w --++00;the critical frequency Wm is the resonancefrequencyof the circuit.

In Section2.6we emphasizedthe importanceof avoiding resonancein mostmechanical systems(the cellois an exampleof a mechanicalsystem in which reso-nance is sought).By contrast, many common electricaldevicescouldnot function

properly without taking advantageof the phenomenonof resonance.The radio is afamiliar example.A highly simplifiedmodel of its tuning circuit is the RLCcircuitwe have discussed.Its inductance L and resistanceR are constant, but its capaci-tance C is varied as one operatesthe tuning dial.

Supposethat we wanted to pick up a particular radio station that is broad-casting at frequency w, and thereby (in effect) provides an input voltage E(t) =Eo sin wt to the tuning circuit of the radio.The resulting steadyperiodiccurrent Ispin the tuning circuit drives its amplifier, and in turn its loudspeaker,with the volumeof sound we hear roughly proportional to the amplitude 10of Isp.To hear our pre-ferred station (of frequency w) the loudest-andsimultaneously tune out stationsbroadcastingat other frequencies-wetherefore want to chooseC to maximize 10.)))


But examineEq.(17),thinking of w as a constant with C the only variable.We seeat a glance-nocalculusrequired-that10is maximal when)

that is, when)

1wL --==o.

wC ')

1C ==-.

Lw 2)(18))

Sowe merely turn the dial to setthe capacitanceto this value.This is the way that oldcrystal radiosworked,but modem AM radioshave a

more sophisticateddesign.A pairof variablecapacitorsare used.The first controlsthe frequency selectedas describedearlier;the secondcontrols the frequencyof asignal that the radio itself generates,kept closeto 455 kilohertz (kHz)above the

desiredfrequency.The resulting beat frequencyof 455kHz,known as the interme-diatefrequency, is then amplifiedin severalstages.This techniquehas the advantagethat the severalRLCcircuits usedin the amplificationstageseasilycan bedesignedto resonateat 455kHzand rejectother frequencies,resulting in far more selectivityof the receiveras well asbetter amplificationof the desiredsignal.)

_ Problems\"_)

Problems1through 6 deal with the RL circuit ofFig. 2.7.7,aseriescircuit containing an inductor with an inductance of Lhenries,a resistorwith a resistanceofR ohms, and a sourceofelectromotive force(emf), but no capacitor.In this caseEq. (2)reducesto the linearfirst-orderequation)

LI'+ RI = E(t).)

L)

R)

FIGURE2.7.7.Thecircuit forProblems 1 through 6.)

1.In the circuit ofFig. 2.7.7,supposethat L = 5 H, R = 25Q, and the sourceE of emf is a battery supplying 100Vto the circuit. Supposealsothat the switch has beenin po-sition 1 for a long time, so that a steady current of 4 A isflowing in the circuit. At time t = 0, the switch is thrownto position 2, so that 1(0)= 4 and E = 0 for t > O.Find

I(t).2.Given the samecircuit as in Problem 1,supposethat theswitch is initially in position 2,but is thrown to position 1at time t = 0, so that 1(0)= 0 and E = 100for t > O.Find I(t) and show that I(t) \037 4 as t \037 +00.)

3.Supposethat the battery in Problem 2 is replacedwith

an alternating-current generator that suppliesa voltage ofE(t) = 100cos60t volts. With everything elsethe same,now find I(t).

4. In the circuit of Fig. 2.7.7,with the switch in position 1,supposethat L = 2, R = 40, E (t) = 1OOe-lOt , and

1(0)= O. Find the maximum current in the circuit fort > O.

5. In the circuit of Fig. 2.7.7,with the switch in position 1,supposethat E(t) = 1OOe-lOt cos60t, R = 20,L = 2,and 1(0)= O.Find I(t).

6.In the circuit of Fig. 2.7.7,with the switch in position 1,take L = 1,R = 10,and E(t) = 30cos60t+ 40sin60t.(a) Substitute Isp(t) = A cos60t + B sin 60t and thendetermine A and B to find the steady-state current Isp in

the circuit. (b) Write the solution in the form Isp(t) =Ccos(wt-a).)

Problems 7 through 10deal with the RCcircuit in Fig. 2.7.8,containing a resistor(R ohms), a capacitor(Cfarads), aswitch, a sourceofemf, but no inductor. Substitution ofL = 0in Eq. (3)gives the linearfirst-orderdifferential equation)

dQ 1R-+-Q=E(t)dt C)

for the chargeQ = Q(t) on the capacitorat time t. Note that

I(t) = Q/(t).)))


SWitch1C)

R)

FIGURE2.7.8.Thecircuit forProblems7 through 10.)

7. (a) Find the charge Q(t) and current I(t) in the RCcircuitif E(t)= Eo(a constant voltage suppliedby a battery) andthe switch is closedat time t = 0, so that Q(0) = O. (b)Show that)

lim Q(t) = EoC and that lim I(t) = O.t\037+oo t\037+oo)

8.Supposethat in the circuit ofFig. 2.7.8,we have R = 10,C = 0.02,Q(O)= 0, and E(t) = 100e-5t (volts). (a)Find Q(t) and I(t). (b) What is the maximum charge onthe capacitorfor t > 0 and when doesit occur?

9.Supposethat in the circuit of Fig. 2.7.8,R = 200,C = 2.5X 10-4, Q(O)= 0, and E(t) = 100cos120t.(a) Find Q(t) and I(t). (b) What is the amplitude of the

steady-statecurrent?10.An emf of voltage E(t) = Eocoswt is applied to the RC

circuit ofFig. 2.7.8at time t = 0 (with the switch closed),and Q(O)= O.Substitute Qsp(t)= A coswt + B sin wt in

the differential equation to show that the steady periodicchargeon the capacitoris

EoCQsp(t)= cos(wt-fJ)-J1+ w 2R2C2

where fJ = tan-](wRC).)

In Problems 11through 16,the parametersofan RLCcircuitwith input voltage E(t) are given. Substitute

Isp(t) = A coswt + B sin wt

in Eq. (4), using the appropriate value ofw, to find the steadyperiodiccurrent in the form Isp(t) = 10sin(wt - 8).)

11.R = 30Q, L = 10H, C = 0.02F; E(t) = 50sin2tV12.R = 200Q, L = 5 H, C =0.001F;

E(t) = 100sin lOt V

13.R = 20Q, L = 10H, C = 0.01F;E(t) = 200cos5tV

14.R = 50Q, L = 5 H, C = 0.005F;E(t) = 300coslOOt + 400sin lOOt V

15.R = 100Q,L = 2 H, C = 5 X 10-6 F;E(t) = 110sin 601(tV

16.R = 25Q, L = 0.2H, C = 5 X 10-4 F;E(t) = 120cos377t V)

In Problems 17through 22, an RLCcircuit with input volt-

age E(t) is described.Find the current I(t) using the giveninitial current (in amperes)and chargeon the capacitor(incoulombs).)

17.R = 16Q, L = 2 H, C = 0.02F;E(t) = 100V; 1(0)= 0, Q(O)= 5

18.R = 60Q, L = 2 H, C = 0.0025F;E(t) = 100e-t

V; 1(0)= 0, Q(O)= 019.R = 60Q, L = 2 H, C = 0.0025F;

E(t) = 100e-lOt V; 1(0)= 0, Q(O)= 1)

In eachofProblems 20 through 22,plot both the steady peri-odiccurrent Isp(t) and the total current I(t) = Isp(t) + Itr(t).

20.Thecircuit and input voltage ofProblem 11with 1(0)= 0and Q(O)= 0



23.Consideran LCcircuit-that is, an RLCcircuit with R =O-with input voltage E(t) = Eosin wt. Show that un-bounded oscillationsof current occurfor a certain reso-nance frequency; expressthis frequency in terms ofLandC.

24. It was stated in the text that, if R, L, and C arepositive,then any solution of LI\" + RI'+ 1/C = 0 is a transient

solution-itapproacheszeroas t \037 +00.Prove this.

25.Prove that the amplitude 10of the steady periodicsolutionofEq.(6)is maximal at frequency w = 1/-JLC.)

_ Endp()intProblemsand Eig\037nvalues)

You are now familiar with the fact that a solutionof a second-orderlineardifferentialequation is uniquely determined by two initial conditions.In particular, the onlysolution of the initial value problem)

y\" +p(x)y'+q(x)y==0; yea) ==0, y'(a) ==0) (1))

is the trivial solution y (x)= O.Mostof Chapter 2 has beenbased,directly or indi-

rectly,on the uniquenessof solutionsof linear initial valueproblems(asguaranteedby Theorem 2 of Section2.2).)))

Example1)

y4321)

-1-2-3-4)

FIGURE2.8.1.Various possiblesolutions y (x) = B sin x.J3of the

endpoint value problem in

Example 1.For no B i= 0 doesthesolution hit the target value y = 0for x = T(.)

Example2)

y4321)

-1-2-3-4) B=-4)

FIGURE2.8.2.Various possiblesolutions y (x) = B sin 2x of theendoint value problem in Example2.No matter what the coefficientB is, the solution automaticallyhits the target value y = 0 forX=T(.)

2.8Endpoint Problemsand Eigenvalues 181)

In this sectionwe will seethat the situation is radicallydifferent for a problemsuch as)

y\" +p(x)y'+q(x)y ==0; yea) ==0, y(b) ==O.) (2))

The differencebetween the problemsin Eqs.(1)and (2)is that in (2)the two con-ditions are imposedat two different points a and b with (say)a < b.In (2)we areto find a solution of the differentialequation on the interval (a,b) that satisfiesthe

conditions yea) ==0 and y(b) ==0 at the endpointsof the interval. Sucha problemiscalledan endpointor boundary value problem.Examples1 and 2 illustrate the

sortsof complicationsthat can arisein endpointproblems.)

Considerthe endpointproblem)

y\" +3y ==0; y(O) ==0, yen) ==o.) (3))

The general solutionof the differentialequation is)

y(x) == A cosx-J3+ Bsinx-J3.Now y(O) == A, so the condition y(O) ==0 impliesthat A ==O.Thereforethe only

possiblesolutionsare of the form y (x)== B sin x,J3.But then)

yen) == Bsinn-J3\037 -0.7458B,)so the other condition yen) == 0 requiresthat B == 0 also.Graphically,Fig.2.8.1illustrates the fact that no possiblesolution y (x) == B sin x,J3with B i= 0 hits the

desiredtarget value y == 0 when x == n. Thus the only solution of the endpointvalueproblemin (3)is the trivial solutiony(x) = 0 (which probablyisno surprise)..)

..\037.. d..\037

Considerthe endpoint problem)

y\" +4y ==0; y(O) ==0, yen) ==O.) (4))

The general solutionof the differentialequation is)

y (x)== A cos2x+ B sin 2x.Again, y(O) == A, so the conditiony(O) ==0 impliesthat A ==O.Thereforethe only

possiblesolutionsare of the form y(x) == B sin 2x.But now yen) == Bsin 2n = 0no matter what the value of the coefficient B is.Hence,as illustrated graphicallyin Fig.2.8.2,everypossiblesolution y (x)== B sin 2xhits automatically the desiredtarget value y ==0 when x ==n (whateverthe value of B).Thus the endpoint value

problemin (4)has infinitely many different nontrivial solutions.Perhaps this doesseema bit surprising. .

Remark 1:Note that the bigdifferencein the resultsof Examples1 and 2stemsfrom the seeminglysmall differencebetweenthe differential equationsin (3)and (4),with the coefficient3 in one replacedby the coefficient4 in the other. Inmathematicsaselsewhere,sometimes\"big doorsturn on small hinges.\"

Remark 2: The \"shooting\" terminologyusedin Examples1 and 2 is of-ten useful in discussingendpoint value problems.We considera possiblesolutionwhich starts at the left endpoint value and askwhether it hits the \"target\" specifiedby the right endpoint value.)))


EigenvalueProblemsRather than being the exceptionalcases,Examples1 and 2 illustrate the typicalsituation for an endpoint problemas in (2):It may have no nontrivial solutions,orit may have infinitely many nontrivial solutions.Note that the problemsin (3)and(4)both can be written in the form)

\037) y\" +p(x)y'+Aq(X)y = 0; yea) = 0, y(b) = 0,) (5))

with p(x) = 0,q(x) = 1,a = 0,and b = n. The number A is a parameter in

the problem(nothing to do with the parameters that werevaried in Section2.5).Ifwe take A = 3,we get the equations in (3);with A = 4, we obtain the equations in

(4).Examples1 and 2 show that the situation in an endpoint problemcontaining aparameter can (and generally will) dependstrongly on the specificnumericalvalueof the parameter.

An endpoint value problemsuch as the problemin (5)-onethat contains an

unspecifiedparameterA-iscalledan eigenvalue problem.The questionwe askin

an eigenvalueproblemis this: For what values of the parameter A doesthere exista nontrivial (i.e.,nonzero) solution of the endpoint value problem?Sucha valueof A is calledan eigenvalueof the problem.Onemight think of such a value as a\"proper\" valueof A for which there existproper(nonzero)solutionsof the problem.Indeed,the prefixeigenis a German word that (in somecontexts)may be translatedas the Englishwordproper,so eigenvaluesare sometimescalledpropervalues (orcharacteristicvalues).

Thus we saw in Example2 that A = 4 isan eigenvalueof the endpointproblem)

y\" +Ay = 0, y(O)= 0, yen) = 0,) (6))

whereasExample1 showsthat A = 3 is not an eigenvalueof this problem.Supposethat A* is an eigenvalue of the problemin (5)and that y*(x) is a

nontrivial solution of the endpoint problemthat resultswhen the parameterA in (5)is replacedby the specificnumericalvalue A*, so)

y:+ p(x)y\037 +A*q(X)y* = 0 and y*(a) = 0, y*(b) = O.)

Then we call y* an eigenfunctionassociatedwith the eigenvalueA*. Thus we sawin Example2 that y*(x)= sin 2xis an eigenfunctionassociatedwith the eigenvalueA* = 4, as is any constant multiple of sin 2x.

Moregenerally,note that the problemin (5)is homogeneousin the sensethat

any constant multiple of an eigenfunction is again an eigenfunction-indeed,oneassociatedwith the sameeigenvalue. That is, if y = y*(x)satisfies the problem in

(5)with A = A*, then sodoesany constant multiple cy*(x).It can beproved (undermild restrictions on the coefficientfunctions p and q) that any two eigenfunctionsassociatedwith the sameeigenvaluemust be linearly dependent.)

Example3) Determine the eigenvaluesand associatedeigenfunctionsfor the endpointproblem)

y\" +Ay = 0; y(O)= 0, y(L) = 0 (L >0).) (7))

Solution We must considerall possible(real)values of A-positive,zero,and negative.If A = 0,then the equation is simply y\" = 0 and its general solution is)

y(x) = Ax + B.)))

FIGURE2.8.3.Thehyperbolicsine and cosinegraphs.)

y)

n=1) n=3)

-I)

FIGURE2.8.4.The.

f .(

. nn xeigen unctions Yn x)= SIn-

Lforn = 1,2,3,4.)


Then the endpointconditionsy(O) ==0 ==y(L) immediatelyimply that A ==B == 0,so the only solution in this caseis the trivial function y (x)= O.Therefore,A == 0 isnot an eigenvalueof the problemin (7).

If A < 0,let us then write A == -a2(with a > 0) to be specific.Then the

differentialequation takesthe form)

\" 2 0Y -ay==,)and its general solution is)

y(x) ==cleax +C2e-ax== A coshax + B sinh ax,)

where A == CI + C2 and B == CI -C2.(Recall that coshax == (eax +e-ax )j2and

that sinh ax == (eax -e-ax)j2.)The conditiony(O) ==0 then gives)

y(O) == A cosh0 + B sinh 0 == A ==0,)

x)

so that y(x) == B sinh ax.But now the secondendpoint condition,y(L) ==0, givesy (L) == B sinh aL == O. This impliesthat B == 0,becausea i= 0,and sinh x == 0only for x == 0 (examine the graphs of y == sinh x and y == coshx in Fig.2.8.3.)Thus the only solutionof the problemin (7)in the caseA < 0 is the trivial solution

y = 0,and we may thereforeconcludethat the problemhas no negativeeigenvalues.The only remaining possibilityis that A ==a2 >0 with a >O.In this casethe

differentialequation is)

y\" +a2y ==0,)

with general solution)

y(x) ==Acosax+Bsinax.The condition y (0) == 0 impliesthat A == 0,so y (x) == B sin ax.The conditiony(L) ==0 then gives)

y(L) == BsinaL==O.

Can this occurif B i= O? Yes,but only provided that aL is a (positive)integral

multiple of n:)aL ==n,) 2n,) 3n,) ...,) nn,) ...,)

that is, if)

n 2 4n2 9n2 n 2n 2A ==a2 ==-

L2' L2' L2' ..., L2'Thus we have discoveredthat the problem in (7)has an infinite sequenceof positiveeigenvalues)

n 2n 2An ==

L2 ' n == 1,2,3,....) (8))Lx)

With B == 1,the eigenfunctionassociatedwith the eigenvalueAn is)

nnxYn (x)==sin-, n == 1,2,3,... .

L)(9))

Figure 2.8.4showsgraphs of the first severalof theseeigenfunctions. We seevis-ibly how the endpoint conditions y (0) == y (L) == 0 serve to selectjust those sinefunctions that start a periodat x ==0 and wind up at x ==L preciselyat the end of ahalf-period. .)))

Example4)

Example3 illustrates the general situation. According to a theorem whoseprecisestatementwe will defer until Section9.1,under the assumption that q (x)>o on the interval [a,b], any eigenvalueof the form in (5)has a divergentincreasingsequence)

A I < A2 < A3 < ...< An < ... ---+ +00of eigenvalues,eachwith an associatedeigenfunction. This is also true of the fol-lowing more general type of eigenvalueproblem,in which the endpoint conditionsinvolve values of the derivativey' aswell as values of y:)

y\" +p(x)y'+Aq(X)Y ==0;alyea)+a2Y'(a) ==0, b i y(b) +b2Y'(b) ==0,)

(10))

whereaI,a2, b I , and b2 are given constants.With al == 1 ==b2 and a2 ==0 ==b I , weget the problemof Example4 (in which p(x) = 0 and q (x)= 1,as in the previousexample).)

Determine the eigenvaluesand eigenfunctionsof the problem)

y\" +AY ==0; y(O) ==0, y'(L)==O.) (11))

Solution Virtually the sameargument as that usedin Example3 showsthat the only possibleeigenvaluesare positive, so we take A == a2 > 0 (a > 0) to be specific.Then the

differentialequation is)

y)

n=2) n=I)

-1)

FIGURE2.8.5.Theeigenfunctions

. (2n - l)nxYn (x) = SIn

2Lforn = 1,2,3,4.)

y\" +a2y ==0,)


y(x) == A cosax+ B sin ax.The condition y(O) ==0 immediatelygives A ==0,so)

y(x) == B sin ax and y'(x)==Bacosax.)

The secondendpoint condition y'(L)==0 now gives)

y'(L)==BacosaL ==O.)

This will hold with B =1= 0 provided that aL is an oddpositive integral multiple ofn /2:)

naL ==-2')

3n2') ...,)

(2n - l)n2)

... ,)

that is, if)

n 2 9n2 (2n - 1)2n2A==-4L2' 4L2' ..., 4L2Thus the nth eigenvalueAn and associatedeigenfunctionof the problem in (11)aregiven by)

(2n - 1)2n2A -

n - 4L2)(2n - l)nx

and Yn (x)==sin2L)

(12))

for n == 1,2,3, ....Figure 2.8.5showsgraphs of the first severalof theseeigen-functions. We seevisibly how the endpoint conditions y (0) == y'(L) == 0 serve toselectjust those sinefunctions that start a periodat x == 0 but wind up at x == Lpreciselyin the middleof a half-period. .)))


A general procedurefor determining the eigenvaluesof the problem in (10)can be outlined as follows. We first write the general solution of the differential

equation in the form)

Y == A Y 1 (x,A) + BY2 (x,A).

We write Yi (x,A) becauseYI and Y2 will dependon A, as in Examples3 and 4, in

which)

YI(X) ==cosax==cosx-JIand Y2(X) ==sin ax ==sinx-JI.)

Then we imposethe two endpoint conditions,noting that each is linear in Y and y',and hencealso linear in A and B.When we collectcoefficientsof A and B in the

resulting pair of equations,we thereforeget a systemof the form)

al (A)A + fJI (A)B ==0,a2(A)A + fJ2(A)B ==O.)

(13))

Now A isan eigenvalueif and only if the system in (13)has a nontrivial solution (onewith A and B not both zero).But such a homogeneoussystem of linearequationshas a nontrivial solution if and only if the determinant of its coefficientsvanishes.We therefore concludethat the eigenvalues of the problem in (10)are the (real)solutionsof the equation)

D(A) ==a I (A)fJ2 (A) -a2(A) fJ I (A) ==O.) (14))

To illustrate Eq. (14)in a concreteproblem,let'srevisit the eigenvalueprob-lem of Example3. If A > 0,then the differential equation y\" + Ay == 0 hasthe general solution y(x) == A cos(,J):x)+ Bsin(,J):x).The endpointconditionsy(O) ==0 and y(L) ==0 then yield the equations)

Y (0) == A .1 + B .0 ==0,y(L) == A cos(,J):L)+ Bsin(,J):L)==0)

(in the unknowns A and B) which correspondto the equations in (13).The de-terminant equation D(A) == 0 correspondingto (14)is then simply the equationsin(,J):L)==0,which impliesthat,J):L==nn, or A ==n 2n 2jL2for n == 1,2,3,...(aswe saw in Example3).

For more generalproblems,the solutionof the equation D(A) ==0 in (14)maypresentformidabledifficultiesand require a numericalapproximationmethod (suchasNewton'smethod) or recourseto a computeralgebra system.

Mostof the interest in eigenvalueproblemsis due to their very diversephysi-cal applications.The remainderof this sectionisdevotedto three suchapplications.Numerous additional applicationsare included in Chapters 8 and 9 (on partial dif-ferential equations and boundary valueproblems).)

TheWhirlingStringWho of us has not wonderedabout the shapeof a quickly spinning jump rope?Letus considerthe shapeassumedby a tightly stretchedflexiblestring of length Landconstant lineardensity p (massperunit length) if it is rotatedor whirled (like ajumprope)with constant angular speed()) (in radians persecond)around its equilibriumposition along the x-axis.We assumethat the portion of the string to one sideofany point exertsa constant tension force T on the portion of the string to the other)))


y)

T T)

x=O x=LEquilibrium position

(a))

x=O) x=L)

Whirling string

(b))

I Vertical component:(J I Tsin (J__ __J)

String)

(c)

FIGURE2.8.6.Thewhirlingstring.)

y)T)

//)

x X x + L\\x x)

FIGURE2.8.7.Forceson a short

segment of the whirling string.)

x)

sideof the point, with the direction of T tangential to the string at that point. Wefurther assumethat, as the string whirls around the x-axis,eachpoint moves in acirclecenteredat that point'sequilibrium position on the x-axis.Thus the string iselastic,so that as it whirls it also stretchesto assumea curved shape. Denote by

y (x)the displacementof the string from the point x on the axisof rotation. Finally,we assumethat the deflectionof the string is so slight that sin e \037 tan e ==y'(x) in

Fig.2.8.6(c).We plan to derive a differentialequation for y(x) by applicationof Newton's

law F == ma to the pieceof string of massp \037x correspondingto the interval

[x,x + \037x]. The only forcesacting on this pieceare the tension forcesat its twoends.From Fig.2.8.7we seethat the net vertical force in the positivey-direction is)x)

F == T sinCe+ \037e)- T sin e \037 T tan(e + \037e)

- T tan e,)

so that)

F \037 Ty'(x + \037x)-Ty'(x).) (15))

Next we recallfrom elementary calculusor physicsthe formula a == rw 2 for the(inward) centripetalaccelerationof a body in uniform circularmotion (r is the radiusof the circleand w is the angular velocity of the body). Herewe have r == y, sothe vertical accelerationof our pieceof string is a == -w2

y, the minus sign becausethe inward direction is the negativey-direction.Becausem ==p \037x, substitution ofthis and (15)in F ==ma yields

Ty'(x + \037x)-Ty'(x) \037 -pw2

y \037x,)

so that)

y'(x+ \037x)-y'(x) 2T . \037 -pw y.

\037x

We now take the limit as \037x --+0 to get the differentialequation of motion of thestring:)

T \" 2 0Y + pw y == .) (16))

If we write)

pw2

A==-T)

(17))

and imposethe condition that the endsof the string are fixed, we finally get theeigenvalueproblem)

y\" + Ay ==0; y(O) ==0, y(L) ==0) (7))

that we consideredin Example3.We found there that the eigenvaluesof the problemin (7)are)

n 2n 2An == , n == 1,2,3, ...,

L2)(8))

with the eigenfunctionYn(x) ==sin(nnxjL)associatedwith An.But what doesall this mean in terms of the whirling string? It means that un-

lessA in (17)is one of the eigenvaluesin (8),then the only solutionof the problem)))

/jlil!!f!!l\">'''''''''''''''''.''''''_\" \"70,%Ci;;;ry\037.t\,!:") ;;.r;.i.:.)FIGURE2.8.8.Distortion ofahorizontal beam.)

L)

\037\037\037-____---\037 x)

Positive

y-values)

FIGURE2.8.9.Thedeflectioncurve.)


in (7)is the trivial solution y(x) = O. In this casethe string remains in its equilib-rium position with zero deflection. But if we equate (17)and (8)and solve for thevalue Wn correspondingto An,)

Wn = /A\037T

= n;\037)(18))

for n == 1,2,3,..., we get a sequenceof criticalspeedsof angular rotation. Onlyat thesecritical angular speedscan the string whirl up out of its equilibrium position.At angular speedW it assumesa shapeof the form Yn == Cn sin(nnx/L) illustrated

in Fig.2.8.4(where Cn = 1).Our mathematicalmodel is not sufficiently complete(or realistic)to determine the coefficientCn , but it assumesmuch smallerdeflectionsthan thoseobservedin Fig.2.8.4,so the numericalvalue of Cn would necessarilybesignificantly smaller than 1.

Supposethat we start the string rotating at speed)

W <WI =\037 \037

,)

then gradually increaseits speedof rotation. Solong as W < WI, the string remainsin its undeflectedposition y = O.But when W ==WI, the string popsinto a whirling

position y == Cl sin(nx/L).And when W is increasedstill further, the string popsbackinto its undeflectedposition along the axisof rotation!)

The Deflectionofa UniformBeam)

We includenow an exampleof the useof a relatively simpleendpointvalue problemto explain a complicatedphysicalphenomenon-theshapeof a horizontal beamonwhich a vertical force is acting.

Considerthe horizontal beam shown in Fig.2.8.8,uniform both in crosssec-tion and in material. If it is supportedonly at its ends, then the force of its own

weight distorts its longitudinal axisof symmetry into the curve shown as a dashedline in the figure. We want to investigate the shapey == y (x) of this curve, the

deflectioncurve of the beam. We will use the coordinate system indicated in

Fig.2.8.9,with the positivey-axisdirecteddownward.A consequenceof the theory of elasticityis that for relatively small deflections

of such a beam (sosmall that [y'(x)]2is negligiblein comparison with unity), an

adequate mathematicalmodel of the deflectioncurve is the fourth-orderdifferential

equation)

Ely(4)== F(x),) (19))

where). E is a constant known as the Young's modulus of the material of the beam,. I denotesthe moment of inertia of the crosssectionof the beam around ahorizontal line through the centroid of the crosssection,and. F(x)denotesthe density of downward force acting vertically on the beam at

the point x.)))

x=oIi....

:...<,:;.;.;-:)

x=L

11)Simply supported or hinged)

x=o

-t)

x=L-t)

Built in)

FIGURE2.8.10.Twoways ofsupporting a beam.)

Density of force?Yes;this means that the force acting downward on a veryshort segment[x,x + \037x] of the beam is approximately F(x)\037x. The units ofF(x)are thoseof force per unit length, such as poundsper foot. We will considerhere the casein which the only force distributed along the beam is its own weight,w poundsperfoot, so that F(x)= w. Then Eq.(19)takesthe form)

Ely(4)== W) (20))

where E,I,and wareall constant.

Note: We assumeno previousfamiliarity with the theory of elasticityor with

Eq.(19)or (20)here.It is important to be ableto begin with a differentialequationthat arises in a specificapplieddisciplineand then analyze its implications; thus

we develop an understanding of the equation by examining its solutions.Observethat, in essence,Eq. (20)impliesthat the fourth derivative y(4) is proportional tothe weight density w. This proportionality involves, however, two constants:E,which dependsonly on the material of the beam,and I,which dependsonly on the

shapeof the crosssectionof the beam.Values of the Young'smodulus E of variousmaterialscan be found in handbooksof physical constants; I ==

\037Jl'a4for a circular

crosssectionof radius a.)

Although Eq.(20)is a fourth-orderdifferentialequation, its solution involves

only the solution of simplefirst-orderequations by successivesimpleintegrations.Oneintegration of Eq.(20)yields)

Ely(3)== wx + C1 ;)

a secondyields)EI \" 1 2 C C .

y == 2wX + IX + 2,)

another yields)

Ely'==\037wx3 + \037CIx2 + C2X + C3;

a final integration gives)

Ely ==2\037

wx4 + \037CIx3 + \037C2x2 + C3x +C4,)

where C1 , C2, C3, and C4 are arbitrary constants.Thus we obtain a solution ofEq.(20)of the form)

wy(x)== x2 +Ax3 +Bx2 +Cx+D,24EI) (21))

where A, B,C,and D are constants resulting from the four integrations.Theselast four constantsare determinedby the way in which the beam is sup-

ported at its ends,wherex ==0 and x ==L.Figure2.8.10showstwo commontypesof support.A beam might alsobe supportedone way at one end but another way at

the other end.For instance,Fig.2.8.11showsa cantilever-abeam firmly fastenedat x ==0 butfree (no support whatsoever)at x ==L.The followingtable showsthe

boundary or endpointconditionscorrespondingto the three most commoncases.We will seethat theseconditions are appliedreadily in beam problems,although adiscussionhere of their origin would take us too far afield.)))

Cantilever)

FIGURE2.8.11.Thecantilever.)

Example5)


(;:\",\";\037;Sr.PP\037f.ta:) EndpointCondition)

Simply supported

Built -in or fixed endFreeend)

y = y\" = 0

y = y' = 0

y\" = y(3) = 0)

For example,the deflection curve of the cantilever in Fig.2.8.11would begiven by Eq.(21),with the coefficientsA, B,C,and D determinedby the conditions)

y(O)= y'(O)= 0 and y\"(L)= y(3)(L)= 0,) (22))

correspondingto the fixedend at x = 0 and the free end at x = L.The conditionsin (22)together with the differentialequation in (21)constitutean endpointvalueproblem.)

....-. ...........

Determine the shapeof the deflectioncurve of a uniform horizontalbeam of lengthL and weight w perunit length and simply supportedat each end.)

Solution We have the endpoint conditions)

y(O)= y\"(O) = 0 = y(L) = y\"(L).)

Rather than imposing then directly on Eq. (21),let us begin with the differential

equation E1y(4) = wand determine the constants as we proceedwith the four

successiveintegrations.The first two integrationsyield)

Ely(3)= wx + A; Ely\"= !wx2 + Ax + B.)

Hencey\"(O) = 0 impliesthat B = 0,and then y\"(L)= 0 gives

0= !wL2 +AL.)

It follows that A = -wLj2and thus that)

EI \" 1 2 1 Ly = 2x - 2w x.)

Then two more integrationsgive)

Ely'= iWx3-

\037wLx2 + C,)

and finally,)

Ely(x) =2\037

wx4 - /2wLx3 +Cx+ D.) (23))

Now y(O)= 0 impliesthat D = 0;then, becausey(L) = 0,)

0=2\037

wL4-l\037

wL4+CL.)

It follows that C = wL3j24.Hencefrom Eq.(23)we obtain)

wy(x) = (x4 -2Lx3 +L3x)24EI) (24))))


Example6)

as the shapeof the simply supportedbeam.It is apparent from symmetry (seealsoProblem 17)that the maximum deflection Ymax of the beam occursat its midpointx = L/2,and thus has the value)

(L

)w

(1 4 2 4 1 4

)Ymax=Y\"2

= 24E/ 16L-

8L +

2L ;)

that is,)

5wL4Ymax =

384E/.) (25).)

\037n \037 \037

For instance, supposethat we want to calculate the maximum deflectionof a simplysupportedsteelrod 20 ft long with a circularcrosssection1 in. in diameter.From ahandbook we find that typical steelhas density 8 = 7.75g/cm3 and that its Young'smodulus is E = 2 X 1012g/cm.s2, so it will be more convenient to work in cgsunits. Thus our rod has)

length: L = (20ft)(30.48e;) = 609.60em)

and)

.(1.

)(cm

)radIus: a = - In. 2.54-=-- = 1.27cm.2 In.)

Its linearmassdensity (that is, its massperunit length) is)

p = na28 = n(1.27)2(7.75)\037 39.27\037,cm)

so)

w = pg = (39.27\037) (980

e\037 )\037 38484.6dyn .

cm s cm

The area moment of inertia of a circular diskof radius a around a diameter is / ==

\037na4,so)

1/ = -n(1.27)4\037 2.04cm4.

4)

Therefore Eq.(25)yields)

\037(5)(38484.6)(609.60)4

\037 16.96emYmax(384)(2 x 1012)(2.04)

,)

about 6.68in.,as the maximum deflectionof the rod at its midpoint. It is interestingto note that Ymax is proportional to L4, so if the rod wereonly 10ftlong, its maxi-mum deflectionwould beonly one-sixteenthasmuch-onlyabout0.42in. Because/ =

\037na4,we seefrom Eq. (25)that the samereduction in maximum deflection

couldbe achievedby doubling the radius a of the rod. .)))

y)y =y(x))

p) p)

x=O)I

Ix=L)

FIGURE2.8.12.Thebuckledrod.)


TheBuckledRod)

x)

Figure 2.8.12showsa uniform rod of length L, hinged at each end, that has been\"buckled\" by an axial force of compressionP appliedat one end. We assumethis buckling to be so slight that the deflection curve y == y (x) of the rod mayberegardedas definedon the interval 0 <x <L.

In the theory of elasticity the linear endpointboundary value problem)

Ely\"+ Py ==0, y(O) ==y(L) ==0) (26))

is usedto model the actual (nonlinear)behaviorof the rod. As in our discussionofthe deflectionof a uniform beam, E denotesthe Young's modulus of the materialof the beam and I denotesthe moment of inertia of each crosssectionof the beamaround a horizontal line through its centroid.

If we write)

PA==-EI') (27))

then the problem in (26)becomesthe eigenvalueproblem)

y\" +AY ==0; y(O) ==y(L) ==0) (7))

that we consideredin Example3.We found that its eigenvalues{An} are given by)

n 2JT2An== L2

' n==I,2,3,...) (8))

with the eigenfunctionYn ==sin(nJTxjL)associatedwith An. (Thus whirling stringsand buckledrodsleadto the sameeigenvaluesand eigenfunctions.)

To interpret this result in terms of the buckledrod, recall from Eg.(27)that

P ==AE I.The forces)

n 2n 2EIPn ==An E1==

L2)n==I,2,3,...) (28))

are the criticalbuckling forcesof the rod. Only when the compressiveforce P isone of thesecritical forcesshould the rod \"buckle\" out of its straight (undeflected)shape.The smallestcompressiveforce for which this occursis)

JT2EIP1 == .L2)

(29))

This smallestcritical force PI is calledthe Eulerbuckling forcefor the rod; it is the

upperbound for thosecompressiveforcesto which the rod can safelybesubjectedwithout buckling.(In practicea rod may fail at a significantly smallerforcedue to acontribution of factors not taken into accountby the mathematicalmodeldiscussedhere.

))))

Example7)

...... \037 ,.... ...............

For instance, supposethat we want to compute the Euler buckling force for a steelrod 10ftlong having a circular crosssection1 in. in diameter. In cgsunits we have)

E = 2 X 1012gjcm.s2,

L = (10ft) (30.48e;) = 304.8em, and

I = JT

[<0.5in.)(2.54\037m )r \037 2.04em4.4 In.)

Uponsubstituting thesevalues in Eq.(29)we find that the critical force for this rodIS)

PI \037 4.34X 108 dyn \037 976lb,

using the conversionfactor 4.448x 105 dynjlb.)

m Problems)The eigenvalues in Problems 1through 5 are all nonnegative.First determine whether 'A = 0 is an eigenvalue; then find the

positive eigenvalues and associatedeigenfunctions.)

1.y\" + 'Ay = 0; y/(O) = 0, y(l) = 02.y\" + 'Ay = 0; y/(O) = 0, y/(n) = 03.y\" + 'Ay = 0; y(-n)= 0, y(n) = 04. y\" + 'Ay = 0; y/( -n)= 0, y/(n) = 05. y\" + 'Ay = 0; y(-2) = 0, y/ (2) = 06.Considerthe eigenvalue problem)

y\" + 'Ay = 0; y/(O) = 0, y(I) + y'(I)= O.)

All the eigenvalues are nonnegative, so write 'A = a2

where a > 0. (a) Show that 'A = 0 is not an eigen-value. (b) Show that y = A cosax+ B sin ax satisfiesthe endpoint conditions if and only if B = 0 and a isa positive root of the equation tan z = I/z. Theseroots{an}r arethe abscissasof the points of intersection of thecurves y = tan z and y = 1/z, as indicated in Fig. 2.8.13.Thus the eigenvalues and eigenfunctions of this problemare the numbers

{a\037} r and the functions {cosanx}r, re-spectively.)

z)

FIGURE2.8.13.Theeigenvaluesaredetermined bythe intersections of the graphs of y = tan z and y = I/z(Problem 6).)

7. Considerthe eigenvalue problem)

y\" + 'Ay = 0; y(O) = 0, y(l) + y/ (I)= 0;)

.)

all its eigenvaluesarenonnegative. (a) Show that 'A = 0is not an eigenvalue. (b) Show that the eigenfunctionsare the functions {sin anx}f,where an is the nth positiveroot of the equation tan z = -z. (c) Draw a sketch indi-

cating the roots {an}fas the points of intersection of the

curves y = tan z and y = -z. Deducefrom this sketch

that an \037 (2n - I)n/2 when n is large.8.Considerthe eigenvalue problem)

y\" + 'Ay = 0; y(O)= 0, y(l) = y/ (1);)

all its eigenvaluesarenonnegative. (a) Show that 'A = 0is an eigenvalue with associatedeigenfunction Yo (x) =x.(b) Show that the remaining eigenfunctions aregiven by

Yn (x)= sin fJnx, where fJn is the nth positive root of the

equation tan z = z. Draw a sketch showing theseroots.Deducefrom this sketch that fJn

\037 (2n + I)n/2when n islarge.

9.Prove that the eigenvalue problem of Example 4 has no

negative eigenvalues.10.Prove that the eigenvalue problem)

y\" + 'Ay = 0; y(O)= 0, y(l) + y/(I)= 0)

has no negative eigenvalues. (Suggestion: Show graph-

ically that the only root of the equation tanh z = -z isz = 0.)

11.Usea method similar to that suggestedin Problem 10toshow that the eigenvalue problem in Problem 6has no neg-ative eigenvalues.

12.Considerthe eigenvalue problem)

y\" + 'Ay = 0; y(-n) = y(n), y/(-n) = y/(n),)

which is not of the type in (10)becausethe two endpointconditions arenot \"separated\" betweenthe two endpoints.(a) Show that 'Ao = 0 is an eigenvalue with associated)))

eigenfunction Yo (x) = 1.(b) Show that there areno neg-ative eigenvalues.(c) Show that the nth positive eigen-value is n 2 and that it has two linearly independent associ-ated eigenfunctions, cosnx and sin nx.

13.Considerthe eigenvalue problem)

y\" + 2y' + AY = 0; y(O)= y(l) = O.)

(a) Show that A = 1 is not an eigenvalue. (b) Showthat there is no eigenvalue A such that A < 1.(c) Showthat the nth positive eigenvalue is An = n 2n 2 + 1,with

associatedeigenfunction Yn (x) = e-X sin nn x.14.Considerthe eigenvalue problem)

y\"+2y'+AY=0; y(O)=0, y'(I)=O.)Show that the eigenvaluesareall positive and that the nth

positive eigenvalue is An =a\037

+ 1 with associatedeigen-function Yn (x)= e-X sin an x, where an is the nth positiveroot of tan z = z.

15.(a) A uniform cantilever beam is fixed at x = 0 and freeat its other end, wherex = L. Show that its shapeis givenby)

wy(x) = (x4 -4Lx3 + 6L2X

2).24\302\2431

(b) Show that y'(x)= 0 only at x = 0, and thus that it fol-lows (why?) that the maximum deflectionof the cantileveris Ymax = y(L) = wL

4/(8\302\243 I).)


16.(a) Supposethat a beam is fixed at its ends x = 0 and

x = L.Show that its shapeis given by)

wy(x) = (x

4 -2Lx3 + L2X2).

24\302\2431

(b) Show that the roots of y'(x)= 0 arex = 0,x = L,and x = L/2, so it follows (why?) that the maximum

deflectionof the beam is)

(L

)WL4

Ymax= Y

2\"

=384\302\2431')

one-fifth that ofa beam with simply supported ends.17.For the simply supported beam whose deflection curve is

given by Eg.(24),show that the only root ofy'(x)= 0 in

[0,L] is x = L/2,so it follows (why?) that the maximum

deflectionis indeedthat given in Eg.(25).18.(a)A beam is fixed at its left end x = 0 but is simply sup-

ported at the other end x = L. Show that its deflection

curve IS)

wy(x) = (2x4 -5Lx3 + 3L2X

2).48\302\2431)

(b) Show that its maximum deflectionoccurswhere x =(15--J33)L/16and is about 41.6%of the maximum de-flection that would occurif the beam were simply sup-

ported at eachend.)))

Pow-erSeriesMethods)

11II_Introductionand Review(:)\037 _\037_?w\037...__\037e

ries)

In Section2.3we saw that solving a homogeneouslinear differential equationwith constant coefficientscan be reducedto the algebraicproblem of finding the

roots of its characteristicequation. There is no similarprocedurefor solving lineardifferentialequations with variable coefficients,at leastnot routinely and in finitely

many steps.With the exceptionof specialtypes,suchas the occasionalequationthat

can be solved by inspection,linear equations with variable coefficientsgenerallyrequire the powerseriestechniquesof this chapter.

Thesetechniquessufficefor many of the nonelementarydifferentialequationsthat appearmost frequently in applications.Perhaps the most important (becauseofits applications in such areasas acoustics,heat flow, and electromagneticradiation)isBessel'sequationof ordern:)

\037) x2y\" +xy' + (x2- n 2

)y = O.)

Legendre'sequationof ordern is important in many applications.It has the form)

\037) (1-x2)y\"

-2xy' -t-n(n + l)y = O.)

In this sectionwe introducethe powerseriesmethod in its simplestform and,along the way, state (without proof) severaltheorems that constitutea reviewof the

basicfacts about powerseries.Recall first that a powerseriesin (powersof) x-ais an infinite seriesof the form)

00

LCn (x-a)n = Co+Cl(x-a) +C2(x-a)2+ ...+Cn (x-a)n + ....(1)n=O)

If a = 0,this is a powerseriesin x:)

\037)

00

Lcnxn = Co+CIX +C2X2 + ...+cnxn + ... .

n=O)

(2))

194)))

3.1Introductionand Reviewof PowerSeries 195)

We will confineour review mainly to powerseriesin x,but everygeneralpropertyof powerseriesin x can beconvertedto a generalpropertyof powerseriesin x-aby replacementof x with x-a.

The powerseriesin (2)convergeson the interval I provided that the limit)

00 N

Lcnxn = lim Lcnxn

N\037oon=O n=O)

(3))

existsfor all x in I.In this casethe sum)

00

f(x) = Lcnxn

n=O)

(4))

is definedon I,and we call the seriesL cnxn a powerseriesrepresentationof thefunction f on I.The followingpowerseriesrepresentationsof elementaryfunctions

should be familiar to you from introductory calculus:)00 xn x2 x3

eX =L,= l+x+,+,+.\";n=O

n. 2. 3.)(5))

00 (_I)nx2n x2 x4cosX = '\"\"' = 1--+-- ....

\037 (2n)' 2'. 4'. 'n=O

.)

(6))

. 00 (_ 1)n X 2n +1 x3 X 5

sInx = L = x--+-- ... .n=O (2n + I)! 3! 5! ') (7))

00x2n x2 x4

coshx = L = 1 +-+-+ ... .n=O (2n)! 2! 4! ') (8))

00 x2n+l x3 x5sinh x = L = x +-+-+ ....

n=O (2n + I)! 3! 5! ') (9))

00 (_I)n+lxn x2 x310(1+x) = L n

= x-2+3-...;n=l)

(10))

1 00= Lxn = 1 +x+x2 +x3 + ... ;I-x

n=O)

(11))

and)

ex a(a- l)x2 a(a- 1)(a -2)x3

(1+x) = 1 +ax + 2! + 3! + .... (12)

In compact summation notation, we observethe usual conventionsthat O!= 1 andthat xO = 1 for all x, including x = O. The seriesin (5)through (9)convergetothe indicated functions for all x. In contrast, the seriesin (10)and (11)convergeifIx I

< 1 but divergeif Ix I> 1.(What if Ix I

= I?)The seriesin (11)is the geometricseries.The seriesin (12),with a an arbitrary real number, is the binomial series.If a is a nonnegativeinteger n, then the seriesin (12)terminatesand the binomialseriesreducesto a polynomial of degreen which convergesfor all x. Otherwise,the seriesis actually infinite and it convergesif Ix I

< 1 and divergesif Ix I> 1;its

behavior for Ix I= 1 dependson the value of a.)))

196 Chapter3 PowerSeriesMethods)

Remark:Power seriessuch as thoselisted in (5) through (12)are oftenderived as Taylorseries.The Taylor serieswith centerx ==a of the function f isthe powerseries)

00fen) (a) f\" (a)

\037 L (x-a)n==f(a)+f'(a)(x-a)+ (x-a)2+... (13)n=O

n! 2!)

in powersof x-a, under the hypothesis that f is infinitely differentiableat x == a(sothat the coefficientsin Eq.(13)are all defined).If a ==0,then the seriesin (13)is the Maclaurinseries)

\037)

00 f en) (0) f \"(O) f (3)(0)L xn ==f (0)+f'(O)x+ x2 + x3 + ... .

n=0 n ! 2! 3!)

(13'))

For example,supposethat f (x) == eX. Then fen) (x) == eX, and hencefen) (0)== 1for all n >O.In this caseEq.(13')reducesto the exponentialseriesin (5). .)PowerSeriesOperationsIf the Taylor seriesof the function f converges to f(x) for all x in someopeninterval containing a, then we say that the function f is analytic at x == a. For

example,). every polynomial function is analytic everywhere;. every rational function is analytic wherever its denominatoris nonzero;. more generally, if the two functions f and g are both analytic at x ==a, then

so are their sum f + g and their product f .g, as is their quotient f/g if

g(a) =1= O.)

For instance, the function h(x) == tan x == (sinx)/(cosx)is analytic at x ==0becausecos0 == 1 =1= 0 and the sineand cosinefunctions are analytic (by virtue

of their convergentpowerseriesrepresentations in Eqs. (6)and (7)). It is rather

awkward to compute the Taylor seriesof the tangent function using Eq. (13)be-causeof the way in which its successivederivativesgrow in complexity (try it!).Fortunately, powerseriesmay be manipulated algebraically in much the samewayaspolynomials.For example,if)

00

f(x) ==Lanxn

n=O)

00and g(x) ==Lbnx

n,

n=O)

(14))

then)

00

f(x) +g(x) ==L(an + bn)xn

n=O)

(15))

and)

00

f(x)g(x)==Lcnxn

n=O)

==aobo+ (aobl +a1bo)x+ (aob2 + a1b1 +a2bo)x2+ ... , (16))))

where Cn = aobn +a1bn-1 +...+anbo.The seriesin (15)is the result of termwiseaddition and the seriesin (16)is the result of formal multiplication-multiplyingeachterm of the first seriesby each term of the secondand then collectingcoef-ficients of likepowersof x. (Thus the processesstrongly resembleaddition andmultiplication of ordinary polynomials.)The seriesin (15)and (16)convergetof(x) + g(x) and f(x)g(x),respectively,on any open interval on which both the

seriesin (14)converge.For example,)

.(

13 15 )( 12 14)sIn xcosx = x-

6x + 120x

-...1 -2x +

24x -...)

= x + (-\037 - \037

)x3 + (\037 + \037 + \037

) x5 + ...6 2 24 12 120)

4 3 16 5= x- -x +-x -...6 120)

1

[(2x)3 (2x)5

]1.= - (2x)- + - ... = -SIn 2x

2 3! 5! 2)

for all x.Similarly, the quotient of two powerseriescan becomputedby long division,

as illustrated by the computation shown in Fig.3.1.1.This division of the Taylorseriesfor cosx into that for sin x yieldsthe first few terms of the series)

1 3 2 5 17 7tan x = x + -x +-x + -x +....3 15 315) (17))

Division of powerseriesis more treacherous than multiplication; the seriesthus

obtained for f/g may fail to convergeat somepoints where the seriesfor f and gboth converge. For example,the sineand cosineseriesconvergefor all x, but the

tangent seriesin (17)convergesonly if Ix I< JT/2.)

ThePowerSeriesMethod)The powerseriesmethodfor solvinga differentialequationconsistsof substitutingthe powerseries)

\037)

00

y = LCnxn

n=O)

(18))

in the differential equation and then attempting to determine what the coefficientsCo, Cl, C2,...must be in order that the powerserieswill satisfy the differential

equation. This is reminiscentof the method of undeterminedcoefficients,but now

we have infinitely many coefficients somehow to determine.This method is not

always successful,but when it is we obtain an infinite seriesrepresentationof asolution, in contrast to the \"closedform\" solutions that our previousmethodshaveyielded.

Beforewe can substitute the powerseriesin (18)in a differential equation,wemust first know what to substitute for the derivativesy', y\", ....The following the-orem (stated without proof) tellsus that the derivativey' of y = L cnxn is obtainedby the simpleprocedureof writing the sum of the derivativesof the individual termsin the seriesfor

y.)))


x3 2x5 17x7x + + + + ...3 15 315

x2 x4 x6

)x3 x5 x7

+ + ...1-2+24

-720+ ... x

6 120 5040

x3 x5 x7X - + + ...

2 24 720

x3 x5 x7+ + ...

3 30 840

x3 x5 x7+ - ...

3 6 72

2x5 4x7+ ...

15 3152x5 x7

+ ...15 15

17x7+ ...

315)

FIGURE3.1.1.Obtaining the seriesfor tan x by division ofseries.)

THEOREM1 TermwiseDifferentiationof PowerSeries)If the powerseriesrepresentation)

\037)

00

f(x) = Lcnxn = Co+ClX +C2x2 +C3x3 + ...n=O)

(19))

of the function f convergeson the openinterval I,then f is differentiableon I,and)

\037)

00

f'(x)= LnCnxn-l= Cl + 2C2X + 3C3X2+...n=l)

(20))

at eachpoint of I.)

For example,differentiationof the geometric series)

1\037 n 2 3

=\037x=1+x+x+x +...I-xn=O)

( 11))))

Example1)


gives)1 00

= \"nXn-l= 1 +2x +3x2 +4x3 + ... .(1-X )2 \037

n=l)

The processof determining the coefficients in the seriesy = L cnxn sothat it will satisfy a given differential equation dependsalsoon Theorem2.Thistheorem-alsostated without proof-tellsus that if two powerseriesrepresentthe

samefunction, then they are the sameseries.In particular, the Taylorseriesin (13)is the only powerseries(in powersof x-a) that representsthe function f.)

THEOREM2 IdentityPrincipleIf)

00 00

Lanxn = Lbnx

n

n=O n=O)

for every point x in someopeninterval I,then an = bn for all n >o.)

In particular, if L anx n = 0 for all x in someopeninterval, it follows fromTheorem 2 that an = 0 for all n >o.)

..... .....

Solvethe equation y' +2y = O.)

Solution We substitute the series)

\037)

00

y = LcnXn

n=O)

00

and y' = Lncnxn-1

,n=l)

and obtain)

00 00

Lncnxn-1 +2Lcnxn = O.

n=l n=O)

(21))

To comparecoefficientshere,we needthe general term in each sum to be the term

containingxn .To accomplishthis, we shift the index of summation in the first sum.To seehow to do this, note that)

00 00

Lncnxn-1 = Cl + 2C2X + 3C3X2+ ...= L(n + I)Cn+lX

n .n=l n=O)

Thus we can replacen with n + 1 if, at the sametime, we start counting one steplower;that is, at n = 0 rather than at n = 1.This is a shift of +1 in the index ofsummation. The result of making this shift in Eq.(21)is the identity)

00 00

L(n+ I)Cn+lXn +2Lcnx

n = 0;n=O n=O)

that is,)00

L[(n+ I)Cn+l +2cn]xn = O.n=O)))


If this equation holdson someinterval, then it follows from the identity principlethat (n + I)Cn+I +2cn = 0 for all n >0;consequently,)

\037)

2cnCn+I = -

n+l) (22))

for all n > O. Equation (22)is a recurrencerelationfrom which we can succes-sivelycomputeCI,C2,C3,...in terms of co;the latter will turn out to be the arbitraryconstant that we expectto find in a general solutionof a first-orderdifferential equa-tion.)

With n = 0,Eq.(22)gives)

2coCI =--.

1)

With n = 1,Eq. (22)gives)

2CI 22co 22coC2 =--=+-=-.

2 1 .2 2!)

With n = 2,Eq.(22)gives)

2C2 23co 23coC3----- ----3- 1.2.3

- 3!.)By now it should beclearthat after n such steps,we will have)

\037)n 2ncoc= (-I) -n , '

n.)n > 1.)

(This is easyto prove by induction on n.)Consequently,our solutiontakes the form)

00 00 2n 00( 2 )

n

L n L n Co n L - X -2xy(x) = Cnx = (-1)-x = Co = coe .n' n'n=O n=O. n=O.)

In the final stepwe haveusedthe familiar exponentialseriesin Eq.(5)to identify ourpowerseriessolution as the samesolution y(x) = coe-2x we couldhave obtainedby the method of separationof variables. .)

ShiftofIndexofSummation)In the solution of Example1 we wrote)

00 00

Lncnxn-I = L(n+ I)Cn+IX

n

n= I n=O)

(23))

by shifting the index of summation by +1 in the serieson the left. That is, wesimultaneously increasedthe index of summation by 1 (replacingn with n + 1,n \037 n + 1)and decreasedthe starting point by 1,from n = 1 to n = 0,therebyobtaining the serieson the right. This procedureis valid becauseeachinfinite seriesin (23)is simply a compact notation for the singleseries)

CI+ 2C2X + 3C3X2+4C4X3+ ....) (24))))

Moregenerally,we can shift the index of summation by k in an infinite seriesby simultaneouslyincreasingthe summation indexby k (n \037 n+k) and decreasingthe starting point by k.For instance, a shift by +2(n \037 n +2) yields)

00 00\037 n-l \037 n+l\037anx

=\037an+2x .

n=3 n=l)

If k is negative,we interpret a \"decreaseby k\" as an increaseby -k= Ikl. Thus ashift by -2(n \037 n -2)in the index of summation yields)

00 00

Lncnxn-l = L(n-2)Cn_2Xn

-3;n= 1 n=3)

we have decreasedthe index of summation by 2 but increasedthe starting point by

2,from n = 1 to n = 3.You shouldcheckthat the summation on the right is merelyanother representationof the seriesin (24).

We know that the powerseriesobtained in Example1 convergesfor all xbecauseit is an exponential series.Morecommonly, a power seriessolution isnot recognizablein terms of the familiar elementary functions. When we get an

unfamiliar powerseriessolution, we need a way of finding where it converges.After all, y = L cnxn is merely an assumedform of the solution. The procedureillustrated in Example1 for finding the coefficients{cn}is merelya formal processand mayormay not be valid. Its validity-in applying Theorem 1 to computey' and applying Theorem 2 to obtain a recurrencerelation for the coefficients-dependson the convergenceof the initially unknown seriesy = L cnxn . Hencethis formal processisjustified only if in the end we can show that the powerserieswe obtain convergeson someopeninterval. If so,it then representsa solution ofthe differential equation on that interval. The following theorem (which we statewithout proof) may beusedfor this purpose.)

THEOREM3 Radiusof ConvergenceGiven the powerseriesL cnxn , supposethat the limit)

\037)

. Cnp= hm -n-+oo Cn+l)

(25))

exists(p is finite) or is infinite (in this casewe will write p = (0).Then

(a) If p = 0,then the seriesdivergesfor all x i= O.

(b) If 0 p.

(c) If p = 00,then the seriesconvergesfor all x.)

The number p in (25)is calledthe radiusof convergenceof the powerseriesL cnxn .For instance, for the powerseriesobtained in Example1,we have)

. (-1)n2nco/n!p= hmn-+oo (_1)n+12n+lco/(n+ I)!)

n+llim = 00,

n-+oo 2)

and consequently the serieswe obtained in Example1 convergesfor all x. Evenif the limit in (25)fails to exist,there always will be a number p such that exactlyone of the three alternatives in Theorem 3 holds. This number may be difficult tofind, but for the powerserieswe will considerin this chapter,Eq.(25)will bequitesufficient for computing the radius of convergence.)))


Example2) Solvethe equation (x-3)y' +2y ==O.)

Solution As before,we substitute)

\037)

00

y ==LcnXn

n=O)

00

and y' ==Lncnxn-1

n=l)

to obtain)00 00

(x- 3)Lncnxn-1 +2Lcnxn ==0,

n= 1 n=O)

so that)00 00 00

Lncnxn - 3Lncnxn-1 +2Lcnxn ==O.

n=l n=l n=O)

In the first sum we can replacen == 1 with n ==0 with no effect on the sum. In the

secondsum we shift the index of summation by +1.This yields)

00 00 00

Lncnxn -3 L(n+ I)Cn+lXn +2Lcnxn ==0;

n=O n=O n=O)

that is,)00

L[ncn - 3(n + I)Cn+l +2cn ] xn ==O.n=O)

The identity principlethen gives)

nCn- 3(n + I)Cn+l +2cn ==0,)

from which we obtain the recurrencerelation)

\037)

n+2C - Cn+l -

3(n + 1)n) for n >O.)

We apply this formula with n ==0,n == 1,and n ==2,in turn, and find that)

2Cl == -co,

3)

3 3C2 ==-Cl==-co3 .2 32 ')

4 4and C3 ==-C2==3cO.3.3 3)

This is almost enough to make the pattern evident; it is not difficult to show byinduction on n that)

\037)

n+lCn ==

3n Co) if n > 1.)

Henceour proposedpowerseriessolution is)

LOOn+ 1ny(x)==co x.3n

n=O)

(26))

Its radius of convergenceis)

p == limn ---+ 00)

Cn) . 3n +33hm == .

n---+oo n + 2)Cn+l)))

Example3)


Thus the seriesin (26)converges if -3< x < 3 and diverges if Ixl > 3.In this

particular examplewe can explain why. An elementary solution (obtainedby sepa-ration of variables)of our differentialequationis y == 1/(3-x)2.If we differentiatetermwise the geometric series)

1

3-x)

1

3 1 00 x n

x =3 L3n '

1-- n=O3)

we get a constant multiple of the seriesin (26).Thus this series(with the arbitraryconstant Co appropriatelychosen)representsthe solution)

1y(x) =

(3 _ X)2)

on the interval -3< x < 3, and the singularity at x == 3 is the reason why theradius of convergenceof the powerseriessolution turned out to bep ==3. .)

Solvethe equation) ==y-x-l.)Solution We make the usual substitutions y ==L cnxn and y' ==L ncnx n-l , which yield)

00 00

x2Lncnxn-1 ==-1-x +Lcnx

n

n=l n=O)

so that)00 00

LnCnxn+1==-1-x +LCnxn.n=l n=O)

Becauseof the presenceof the two terms -1and -xon the right-hand side,weneedto split off the first two terms, Co+Clx,of the serieson the right for comparison.Ifwe also shift the index of summation on the left by -1(replacen == 1 with n == 2and n with n - 1),we get)

00 00

L(n- I)Cn _lXn ==-1-x+Co+CIX +LCnxn .

n=2 n=2)

Becausethe left-hand sidecontains neither a constant term nor a term containingx to the first power, the identity principlenow yieldsCo == 1,C1 == 1,and Cn ==

(n - 1)Cn-l for n >2.It follows that)

C2 == 1 .Cl == I!, C3 ==2 .C2 ==2!, C4==3 .C3 ==3!,)

and, in general,that)

\037) Cn == (n - I)! for n >2.)

Thus we obtain the powerseries)

00

y(x) == 1 +x+L(n-l)!xn .n=2)))


But the radius of convergenceof this seriesis)

. (n - I)! 1p = hm = lim - = 0,

n-+oo n! n-+oo n)

so the seriesconverges only for x = O. What does this mean? Simply that thegiven differentialequationdoesnot have a (convergent)powerseriessolutionof theassumedform y = L cnxn .This exampleservesasa warning that the simpleact ofwriting y = L cnxn involvesan assumption that may befalse. .)

Example4) Solvethe equation y\" + y = O.)

Solution If we assumea solution of the form)

\037)

00

y = LCnxn ,n=O)

we find that)

\037)

00

y' = Lncnxn-}n=})

and)

00\" '\"\"'\" ( 1)

n-2y =

\037 n n - Cnx .n=2)

Substitution for y and y\" in the differentialequation then yields)

00 00

Ln(n -1)cnxn-2 +Lcnxn = O.n=2 n=O)

We shift the index of summation in the first sum by +2(replacen = 2 with n = 0and n with n +2).This gives)

00 00

L(n+2)(n+ l)cn+2xn +Lcnxn = O.

n=O n=O)

The identity (n +2)(n+ l)cn+2 +Cn = 0 now follows from the identity principle,and thus we obtain the recurrencerelation)

\037)

CnC --n+2 -

(n + 1)(n+ 2))(27))

for n >O.It is evident that this formula will determine the coefficientsCn with evensubscriptin terms of Co and those of oddsubscriptin terms of C};Co and C}are not

predeterminedand thus will be the two arbitrary constants we expectto find in ageneral solution of a second-orderequation.

When we apply the recurrencerelation in (27)with n = 0,2,and 4 in turn,we get)

Co CoC2 = -

2!' C4 = 4!'Taking n = 1,3,and 5 in turn, we find that)

and)Co

C6 = --.6!)

C}C3 = -

3!')

C}C--5 - 5!')and)

C}C7 = --.

7!)))

3.1 Introductionand Reviewof PowerSeries 205)

Again, the pattern isclear;we leave it for you to show (by induction) that for k > 1,)

\037) C2k ==)(-1)kco

(2k)!)

and)(-1)kc1

C -2k+l -

(2k+ I)!.)

Thus we get the powerseriessolution)

(x2 x4 x6

) (x3 x5 X

7

)y(x) ==Co 1--+---+...+Cl x--+---+....2! 4! 6! 3! 5! 7! ')

that is, y(x) ==Cocosx+Clsinx.Note that we have no problem with the radius ofconvergencehere;the Taylor seriesfor the sineand cosinefunctions convergefor\037lx. .)

The solution of Example4 can bear further comment. Supposethat we hadneverheard of the sineand cosinefunctions, let alone their Taylorseries.We would

then have discoveredthe two powerseriessolutions)

00 (-1)nX 2n X2 x4

C(x)= L (2n)!= 1- 2!+ 4!

-...n=O)

(28))

and)

00 (-I)nx2n+l x3 x5

S(x)=\037 (2n + I)!

= x-3!+ Sf- ...) (29))

of the differential equation y\" + y == O. Both of thesepowerseriesconvergeforall x. For instance, the ratio test in Theorem 3 impliesconvergencefor all z of the

seriesL(-1)nzn /(2n)!obtained from (28)by writing z ==x2.Henceit followsthat

(28)itself convergesfor all x,asdoes(by a similarploy) the seriesin (29).It isclearthat C(0)== 1 and S(0)==0,and termwisedifferentiation of the two

seriesin (28)and (29)yields)

C'(x)==-Sex) and S'(x)==C(x).) (30))

Consequently,C'(0) == 0 and S'(0) == 1.Thus with the aid of the power seriesmethod (all the while knowing nothing about the sine and cosinefunctions), wehave discoveredthat y ==C(x)is the unique solutionof)

y\" + y ==0)

that satisfies the initial conditions y(O) == 1 and y'(0) == 0,and that y == Sex)is the unique solution that satisfies the initial conditions y (0) == 0 and y'(0)== 1.It follows that C(x)and Sex)are linearly independent, and-recognizingthe im-portance of the differentialequation y\" + y ==O-wecan agreeto call C the cosinefunction and S the sinefunction. Indeed,all the usual propertiesof thesetwo func-tions can beestablished,using only their initial values (at x ==0)and the derivativesin (30);there is no need to refer to triangles or even to angles.(Can you use the

seriesin (28)and (29)to show that [C(x)]2+ [S(x)]2== 1 for all x?)This demon-strates that)

Thecosineand sinefunctions arefully determinedby the differen-tial equationy\" + y = 0 of which they arethe two natural linearlyindependentsolutions.)))

Figures3.1.2and 3.1.3showhow the geometriccharacterof the graphsof cosx and

sin x is revealedby the graphs of the Taylorpolynomialapproximationsthat we getby truncating the infinite seriesin (28)and (29).

This is by no means an uncommon situation. Many important specialfunc-tions of mathematics occurin the first instance as powerseriessolutions of differ-ential equations and thus are in practicedefined by means of thesepowerseries.In the remaining sectionsof this chapter we will seenumerous examplesof suchfunctions.)

y)

n=8) n = 16) n=24)

2)

x)-1-2)

n=6) n = 14) n=22)

FIGURE3.1.2.Taylor polynomial approximations tocosx.)

11IIProblems)

In Problems1through 10,find a powerseriessolution of the

given differential equation. Determine the radius of conver-genceof the resulting series,and use the seriesin Eqs.(5)through (12)to identify the seriessolution in terms offamil-iar elementary functions. (Ofcourse,no onecanprevent youfrom checking your work by also solving the equations by themethods ofearlierchapters!))

1.y/ = Y

3.2y'+ 3y = 05. y' = x2

Y

7. (2x - l)y' + 2y = 09.(x - l)y' + 2y = 0)

2.y/ = 4y

4. y/ + 2xy = 06.(x -2)y'+ y = 08.2(x + l)y' = Y

10.2(x- l)y' = 3y)

In Problems11through 14,usethe method ofExample 4 to findtwo linearly independent powerseriessolutions of the givendifferential equation. Determine the radius ofconvergenceofeachseries,and identify the generalsolution in terms offamil-iar elementary functions.)

11.y\" = y

13.y\" + 9y = 0)

12.y\" = 4y14.y\" + y = x)

Show (asin Example 3) that the powerseriesmethod fails to

yield a powerseriessolution of the form y = L cnxn for the

differential equations in Problems15through 18.)

15.xy/ + y = 017.x2

y/ + y = 0)

16.2xy' = y

18.x3y/ = 2y)

y)

n=5) n = 13) n =21)

2)

-1-2)

n=7) n = 15) n = 23)

FIGURE3.1.3.Taylor polynomial approximations toSIn x .)

In Problems 19through 22,first derive a recurrencerelation

giving Cn for n > 2 in terms of Coor Ct (or both). Then ap-ply the given initial conditions to find the values ofCoand Ct.Next determine Cn (in terms of n, as in the text) and, finally,

identify the particular solution in terms offamiliar elementary

functions.)

19.y\" + 4y = 0; y(O)= 0, y/(O) = 320.y\"

-4y = 0; y(O)= 2, y/(O) = 021.y\"

-2y' + y = 0; y(O) = 0, y/(O) = 122.y\" + y/

-2y = 0; y(O)= 1,y/(O) = -223.Show that the equation)

x2y\" + x2

y' + y = 0)

has no power seriessolution of the form y = L cnxn .

24. Establish the binomial seriesin (12)by means of the fol-

lowing steps. (a) Show that y = (l + x)Q' satisfiesthe

initial value problem (1+x)y' = ay, y(O) = 1.(b) Showthat the power seriesmethod gives the binomial seriesin

(12)as the solution of the initial value problem in part (a),and that this seriesconvergesif Ix I

< 1.(c) Explain whythe validity of the binomial seriesgiven in (12)follows

from parts (a) and (b).25.For the initial value problem)

y\" = y/ + y, y(O)= 0, y(l) =1)))

3.2SeriesSolutionsNearOrdinary Points 207)

derive the power seriessolution

00 F.y(x) = L \037xn

n=l n!)

27.This sectionintroduces the useof infinite seriesto solvedifferential equations. Conversely, differential equationscan sometimes be used to sum infinite series.For exam-ple,considerthe infinite series)

where {Fn}\037o is the sequence0, 1,1,2, 3, 5, 8, 13,...of Fibonaccinumbers defined by Fo = 0, Fl = 1,Fn = Fn-2 + Fn-l for n > 1.

26.(a) Show that the solution of the initial value problem

y' = 1+ y2, Y(0) = 0

is y(x) = tanx. (b) Becausey(x) = tan x is an oddfunction with y' (0) = 1,its Taylor seriesis of the form

y = x + C3X3 + csxs + C7X7 + ... .

Substitute this seriesin y' = 1+ y2 and equate like powersofx to derive the following relations:)

1 1 1 1 11+---+-+---+....I! 2! 3! 4! 5! ')

note the + + -+ + - ...pattern of signs superimposedon the terms of the seriesfor the number e. We couldevaluate this seriesif we could obtain a formula for the

function)

1 2 13 14 1 sf(x) = 1+ x - -x + -x + -x - -x +...2! 3! 4! 5! ')

3C3= 1,7C7= 2cs+ (C3)2,

l1cII= 2cg+ 2C3C7+ (CS)2.

(c) Concludethat)

5cs= 2C3,

9cg = 2C7+ 2C3CS,)

becausethe sum of the numerical seriesin question is sim-

ply f(I). (a) It'spossibleto show that the power seriesgiven hereconvergesfor all x and that termwise differen-tiation is valid. Given thesefacts, show that f (x) satisfies

the initial value problem)

y(3) = y; y(O)= y' (0)= 1, y\" (0)= -1.)1 2 17

tan x = x + -x3 + -xs + _x7

3 15 31562 9 1382 II+ 2835

x + 155925x +....

(d) Would you preferto usethe Maclaurin seriesformulain (13)to derive the tangent seriesin part (c)?Think aboutit!)

(b) Solvethis initial value problem to show that)

1 2

(v'3 .v'3

)f(x) = -ex+ _e-x/2 cos-x+ v'3 sIn -x .

3 3 2 2)

For a suggestion, seeProblem 48ofSection2.3.(c) Eval-uate f (1)to find the sum of the numerical seriesgivenhere.)

_ Se:riesSollltions\037ear Ordina\037Po\037nts)

The powerseriesmethod introduced in Section3.1can be appliedto linear equa-tions of any order(aswell as to certain nonlinearequations),but its most important

applications are to homogeneoussecond-orderlinear differential equationsof the

form)

A(x)y\" + B(x)y'+C(x)y==0,) (1))

where the coefficients A, B,and C are analytic functions of x. Indeed,in mostapplicationsthesecoefficientfunctions are simplepolynomials.

We saw in Example3 of Section3.1that the seriesmethod doesnot alwaysyield a seriessolution. To discoverwhen it doessucceed,we rewriteEq.(1)in the

form)

\037) y\" + P(x)y'+ Q(x)y ==0) (2))

with leading coefficient1,and with P == B/ A and Q == C/ A. Note that P(x) andQ(x)will generally fail to be analytic at points where A(x)vanishes.For instance,considerthe equation)

xy\" +y' +xy ==O.) (3))))


Example1)

Example2)

Example3)

The coefficientfunctions in (3)are continuouseverywhere.But in the form of (2)it

is the equation)

1y\" + -y' + y = 0

x)(4))

with P(x)= 1/x not analytic at x = O.The point x = a is calledan ordinary point of Eq.(2)-andof the equivalent

Eq. (I)-providedthat the functions P(x)and Q(x)are both analytic at x = a.Otherwise,x = a is a singularpoint.Thus the only singular point of Eqs.(3)and(4) is x = O. Recall that a quotient of analytic functions is analytic wherever thedenominator is nonzero.It follows that, if A(a) i-0 in Eq. (1)with analytic coef-ficients, then x = a is an ordinary point. If A(x),B(x),and C(x)are polynomialswith no common factors, then x = a is an ordinary point if and only if A(a) i-O.)

n..,.....u..u..........\".u................... ..... ...\037v ..... .....,..\",..,.... ....\"'...,...\"'.\"_....\037 \037\" ,,,'..,........_,.....'...., \",..........\037\"'\"

The point x = 0 is an ordinary point of the equation)

xy\" + (sinx)y'+x2y = 0,)

despitethe fact that A (x)= x vanishesat x = O.The reasonis that)

sin x 1

(x3 x5

)x2 x4

P(x)=\037

=x x-3!+ Sf

-...= 1 -3!+ Sf- ...)

isneverthelessanalyticat x = 0 becausethe divisionby xyieldsa convergentpowerseries. .)

The point x = 0 is not an ordinary point of the equation)

y\" +x2y' +x1/2y = O.)

For while P(x) = x2 is analytic at the origin, Q(x) = x1/2 is not. The reasonisthat Q(x)is not differentiableat x = 0 and henceis not analytic there. (Theorem1of Section3.1impliesthat an analytic function must be differentiable.) .)

..............

The point x = 0 is an ordinary point of the equation)

(1-x3)y\" + (7x2 +3x5)y' + (5x- I3x4)y = 0)

becausethe coefficient functions A(x), B(x), and C(x) are polynomials with

A(O) i-O. .)Theorem 2 of Section2.1impliesthat Eq. (2)has two linearly independent

solutions on any openinterval where the coefficientfunctions P(x)and Q(x)arecontinuous. The basicfact for our presentpurposeis that near an ordinary point a,thesesolutions will be powerseriesin powersof x-a.A proof of the followingtheorem can be found in Chapter 3 of Coddington, An Introduction to OrdinaryDifferentialEquations (EnglewoodCliffs,N.J.:Prentice Hall,1961).)))


THEOREM1 SolutionsNear anOrdinaryPoint

Supposethat a is an ordinary point of the equation)

A(x)y\" +B(x)y'+C(x)y= 0;) (1))

that is, the functions P = BIA and Q = CIA are analytic at x = a.ThenEq.(1)has two linearly independentsolutions,eachof the form)

\037)

00

y(x) = LCn(X-a)n.

n=O)

(5))

The radius of convergenceof any such seriessolution is at leastas large as the

distancefrom a to the nearest (real or complex)singular point of Eq.(1).Thecoefficientsin the seriesin (5)can bedeterminedby its substitution in Eq.(1).)

Example4)\037 \037 .. '\"

Determine the radius of convergenceguaranteedby Theorem'1 of a seriessolutionof)

(x2 + 9)y\" +xy' +x2y = 0) (6))

in powersof x. Repeat for a seriesin powersof x-4.)

Solution This exampleillustrates the fact that we must take into account complexsingular

points as well as real ones.Because)

xP(x)=

x2 + 9)and)

x2Q(x)=

x2 +9 ')

the only singular points of Eq. (6)are ::l::3i.The distance(in the complexplane)of each of these from 0 is 3, so a seriessolution of the form L cnxn has radiusof convergence at least3. The distanceof each singular point from 4 is 5, so aseriessolution of the form L Cn (x- 4)n has radius of convergenceat least 5 (seeFig.3.2.1). .)

y)

x)

FIGURE3.2.1.Radius ofconvergenceasdistance to nearest singularity.)))


Example5) Find the general solution in powersof x of)

(x2-4)y\" +3xy'+ y = o.) (7))

Then find the particular solution with y(O)= 4, y/ (0)= 1.)

Solution The only singular points of Eq. (7)are ::l::2,so the serieswe get will have radiusof convergenceat least2. (SeeProblem 35 for the exact radius of convergence.)Substitution of)

\037)

00

y = LcnXn

,n=O)

00/

'\"\"\"\" n-1Y =

\037 ncnx ,n=l)

00

and y\" = Ln(n -1)cnxn-2

n=2)

in Eq. (7)yields)

00 00 00 00

Ln(n - l)cnxn -4Ln(n -1)cnxn-2 +3Lncnxn +Lcnxn = O.n=2 n=2 n=l n=O)

We can beginthe first and third summations at n = 0 as well,becauseno nonzeroterms are thereby introduced.We shift the index of summation in the secondsum

by +2,replacingn with n +2 and using the initial value n = O.This gives)

00 00 00 00

Ln(n - l)cnxn -4L(n+2)(n+ 1)cn+2xn +3Lncnxn +Lcnxn = o.n=O n=O n=O n=O)

After collectingcoefficientsof Cn and Cn+2,we obtain)

00

L[(n2 + 2n + l)cn -4(n +2)(n+ 1)cn+2]x

n = O.n=O)

The identity principleyields)

(n + 1)2cn -4(n +2)(n+ 1)cn+2 = 0,)

which leadsto the recurrencerelation)

\037)

(n + l)cnc -n+2 -

4(n + 2))(8))

for n >O.With n = 0,2,and 4 in turn, we get)

Coc --2-4.2')3C2 3co

C4=-=4 .4 42 .2 .4 ') and)

5C4 3 .5coC6=-= .

4 .6 43 .2 .4 .6)

Continuing in this fashion, we evidently would find that)

C2n =)1 .3 .5 ...(2n - 1)

Co.4n .2 .4 ...(2n))

With the common notation)

(2n + I)!(2n + I)!!= 1 .3 .5 ...(2n + 1)= 2n .

n!)))


and the observation that 2 .4 .6 ...(2n)==2n .n!,we finally obtain)

\037)

(2n - I)!!C2n ==

3 Co.2 n .n!)(9))

(We alsousedthe fact that 4n .2n ==23n .)With n == 1,3,and 5 in Eq.(8),we get)

2CIC --3 -4.3 ')

4C3 2 .4cICs ==-==

4 .5 42 .3 .5') and)

6c5 2 .4 .6cIC7 ==-== .

4 .7 43 .3 .5 .7)

It is apparent that the pattern is)

\037)

2 .4 .6 ...(2n) n!C2n+I == CI == CI.4n .1 .3 .5 ...(2n + 1) 2n .(2n + I)!!)

(10))

The formula in (9)gives the coefficientsof even subscriptin terms of Co;the

formula in (10)gives the coefficients of odd subscriptin terms of CI. After weseparatelycollectthe terms of the seriesof even and odddegree,we get the generalsolution)

y(x) ==Co(1 +f (2\037

-1)!!x2n

)+CI (

X +f n!x2n+I

). (11)

n=I2 n .n!

n=I2n .(2n + I)!!)

Alternatively,)

(1234 56

)y(x) ==Co 1 + -x +-x + x +...8 128 1024)

(13 15 17

)+CI x + -x +-x +-x +....6 30 140) (11'))

BecauseY (0) == Coand y/ (0) == CI, the given initial conditions imply that Co == 4and CI == 1.Usingthesevalues in Eq. (11/),the first few terms of the particularsolution satisfying y (0)==4 and y'(0)== 1 are)

1 2 13 3 4 15y(x) ==4+x+2x +

6x +

32x +

30x +.... (12)

.)

Remark:As in Example5, substitution of y ==L cnxn in a linearsecond-orderequation with x ==0 an ordinary point typically leadsto a recurrencerelationthat can be used to expresseach of the successivecoefficientsC2, C3, C4,...interms of the first two, Coand CI.In this event two linearly independentsolutionsareobtained as follows.Let Yo(x) be the solutionobtainedwith Co== 1 and CI ==0,andlet YI (x)be the solutionobtained with Co==0 and CI == 1.Then)

Yo(O) == 1, yb(O) ==0 and YI(O) ==0, y\037(O)== 1,)

so it is clearthat Yo and YI are linearly independent.In Example5, Yo(x) and YI (x)are defined by the two seriesthat appear on the right-hand side in Eg.(11),which

expressesthe general solution in the form Y ==CoYo +CIYI. .)))


TranslatedSeriesSolutions)If in Example5 we had sought a particular solution with given initial values yea)and y/(a),we would have neededthe general solution in the form)

00

y(x) ==LCn(x_a)n;n=O)

(13))

that is,in powersof x - a rather than in powersof x. For only with a solution ofthe form in (13)is it true that the initial conditions)

yea) ==Co and y/(a) ==Cl)

determine the arbitrary constants Coand Cl in terms of the initial valuesof y and y/.Consequently,to solve an initial value problem,we needa seriesexpansionof the

general solution centeredat the point where the initial conditionsare specified.)

Example6) Solvethe initial value problem)

d2y dy

(t2-2t -3)-+ 3(t -1)-+ y ==0; y(l) ==4, y'(I) ==-1. (14)dt2 dt)

Solution We needa general solutionof the form L Cn (t - l)n.But insteadof substituting this

seriesin (14)to determine the coefficients,it simplifiesthe computationsif we first

make the substitution x == t -1,so that we wind up lookingfor a seriesof the form

L cnxn after all.To transform Eq.(14)into one with the new independentvariablex,we note that)

t2-2t - 3 == (x+ 1)2-2(x+ 1)- 3 ==x2-4,

dy dy dx dy /-==--==-==y,dt dxdt dx)

and)

d2y _

[\037

(dY

)]dx _ \037 / _ II

dt2 - dx dx dt-

dx(y ) - y ,)

where primesdenotedifferentiationwith respectto x.Hencewe transform Eq.(14)into)

(x2-4)y\" +3xy'+ y ==0

with initial conditions y ==4 and y/ == 1 at x ==0 (correspondingto t == 1).This isthe initial value problemwe solved in Example5, so the particular solution in (12)is available. We substitute t - 1 for x in Eq. (12)and thereby obtain the desiredparticular solution)

1 1yet) ==4 + (t - 1)+ -(t- 1)2+ -(t- 1)32 6

3 4 1 5+-(t- 1) +-(t- 1) +....32 30)))

This seriesconverges if -1< t < 3.(Why?) A seriessuch as this can beusedtoestimate numericalvalues of the solution. For instance,)

1 1y(0.8)= 4 + (-0.2)+-(-0.2)2+-(-0.2)32 6

3 4 1 5+ -(-0.2)+ -(-0.2)+...32 30

')

so that y(0.8)\037 3.8188.) .)The last computation in Example6 illustrates the fact that seriessolutionsof

differential equations are useful not only for establishinggeneral propertiesof asolution,but alsofor numericalcomputationswhen an expressionof the solution in

terms of elementary functions is unavailable.)

TypesofRecurrenceRelationsThe formula in Eq.(8)is an exampleof a two-termrecurrencerelation; it expresseseach coefficientin the seriesin terms of one of the precedingcoefficients.A many-termrecurrencerelation expresseseach coefficient in the seriesin terms of two ormore precedingcoefficients. In the caseof a many-term recurrence relation, it isgenerally inconvenientor even impossibleto find a formula that gives the typicalcoefficientCn in terms of n. The next exampleshowswhat we sometimescan dowith a three-termrecurrencerelation.)

Example7) Find two linearly independentsolutionsof)

/I I 2 0Y-xy -x y = .) (15))

Solution We make the usual substitution of the powerseriesy = L cnxn .This results in the

equation)00 00 00

Ln(n- l)cnxn-2-LncnXn -LcnX

n+2 = O.n=2 n=l n=O)

We can start the secondsum at n = 0 without changing anything else.To makeeach term includexn in its general term, we shift the indexof summation in the first

sum by +2(replacen with n +2),and we shift it by -2in the third sum (replacenwith n -2).Theseshifts yield)

00 00 00

L(n+2)(n+ l)cn+2xn -LncnXn -LCn-2Xn = O.

n=O n=O n=2)

The common range of thesethree summations is n > 2, so we must separate theterms correspondingto n = 0 and n = 1 in the first two sums before collectingcoefficientsof xn .This gives)

00

2C2+ 6C3X-CIX +L[(n +2)(n+ l)cn+2-nCn-Cn-2]xn = O.

n=2)

The identity principlenow impliesthat 2C2= 0,that C3 = iCl,and the three-termrecurrencerelation)

\037)

nCn +Cn-2Cn+2 =

(n +2)(n+ 1))(16))))


for n >2.In particular,)

C4=) 2C2+Co12)

Cs =) 3C3+Cl20)

4C4+C2C6=

30)(17))

C7 =) 5cs+C3

42)Cg =) 6C6+C4

56)

Thus all values of Cn for n > 4 are given in terms of the arbitrary constantsCo and

Cl becauseC2 = 0 and C3 =\037Cl.

To get our first solution Yl of Eq.(15),we chooseCo= 1 and Cl = 0,so that

C2= C3 = O.Then the formulas in (17)yield)

1C --4 - 12') Cs = 0,)

1C --6- 90') C7 = 0,)

3C - .

g - 1120')thus)

1 4 1 6 3 gYl (x)= 1 + 12x + 90x + 1120x +....) (18))

BecauseCl = C3 = 0,it isclearfrom Eq.(16)that this seriescontainsonly terms ofeven degree.

Toobtaina secondlinearly independentsolutionY2 ofEq.(15),we takeCo= 0and Cl = 1,so that C2= 0 and C3 =

\037.

Then the formulas in (17)yield)

C4= 0,)3

Cs - C6 = 0,=40')

13C7 =

1008')

so that)

1 3 3 s 13 7Y2(X)=X+-X +-x + x +....6 40 1008) (19))

BecauseCo= C2 = 0,it is clearfrom Eq. (16)that this seriescontains only termsof odd degree.The solutions Yl (x) and Y2 (x) are linearly independent becauseYl (0)= 1 and

Y\037(0)= 0,whereasY2 (0)= 0 and

Y\037(0)= 1.A general solution of

Eq. (15)is a linear combinationof the powerseriesin (18)and (19).Equation (15)has no singular points,so the powerseriesrepresenting Yl (x) and Y2 (x)convergefor all x. .)

The LegendreEquationThe Legendreequationof ordera is the second-orderlinear differentialequation)

\037) (1-x2)y\"

-2xy' +a(a+ l)y = 0,) (20))

where the real number a satisfies the inequality a >-1.This differentialequationhas extensive applications,ranging from numerical integration formulas (suchasGaussianquadrature) to the problemof determining the steady-state temperaturewithin a solidsphericalball when the temperatures at points of its boundary areknown. The only singular points of the Legendreequation are at +1 and -1,soit has two linearly independent solutions that can be expressedas powerseriesin)))


powersof x with radius of convergenceat least 1.The substitution y == L cmxm in

Eq.(20)leads(seeProblem31)to the recurrencerelation)

\037)

(a -m)(a+m + 1)C -- Cm+2 -

(m + 1)(m+2) m) (21))

for m >O.We are using m as the indexof summation becausewe have another rolefor n to play.

In terms of the arbitrary constantsCoand CI,Eq.(21)yields)

a(a+ 1)C2 ==- 2! Co,)

(a-l)(a+2)C3 ==- 3! CI,)

a(a-2)(a+ 1)(a +3)C4== Co,4!

(a - 1)(a -3)(a+2)(a+4)CI.

S!)Cs =)

We can show without much trouble that for m > 0,)

maCa-2)(a-4)...(a-2m +2)(a+ 1)(a+3)...(a+2m -1)C2m ==(-1) Co

(2m)!(22))

and)

m (a - 1)(a-3)...(a -2m + 1)(a+2)(a+4)...(a +2m)C2m+! = (-1)

(2m + 1)! C!.(23))

Alternatively,)

C2m = (-I)ma2mcO and C2m+1= (-I)ma2m+lCl,)

where a2m and a2m+1 denote the fractions in Eqs.(22)and (23),respectively.With

this notation, we get two linearly independentpowerseriessolutions)

00

YI(X) = CoL(-I)ma2mx2m

m=O)

00

and Y2(X) = CIL(-I)ma2m+IX2m+lm=O)

(24))

of Legendre'sequation of ordera.Now supposethat a = n, a nonnegative integer. If a == n is even, we see

from Eq. (22)that a2m = 0 when 2m > n. In this case,YI (x) is a polynomialofdegreenand Y2 is a (nonterminating) infinite series.If a == n is an oddpositiveinteger, we seefrom Eq. (23)that a2m+1 == 0 when 2m + 1 > n. In this case,Y2 (x) is a polynomial of degreenand YI is a (nonterminating) infinite series.Thusin either case,one of the two solutions in (24)is a polynomial and the other is anonterminating series.)))

With an appropriatechoice(madeseparatelyfor each n) of the arbitrary con-stants Co(n even) or CI (n odd),the nth-degreepolynomial solution of Legendre'sequation of ordern,)

\037) (1-x2)y\"

-2xy' +n(n + l)y = 0,) (25))

is denotedby Pn (x)and is calledthe Legendrepolynomialof degreen. It is cus-tomary (for a reasonindicated in Problem 32)to choosethe arbitrary constant sothat the coefficientof xn in Pn (x)is (2n)!j [2n (n!)2].It then turns out that)

\037) LN (-1)k(2n-2k)!n-2kp X - Xn( ) -

k=O2nk!(n -k)!(n -2k)!

') (26))

where N = [nj2],the integral part of nj2.The first sixLegendrepolynomialsare)

Po(x)= 1,) PI(x)= x,)

1 2P2(x)= -(3x -1),2

1P4(X)= 8 (35x4 -30x2 + 3),)

1 3P3(x)=2 (5x -3x),

P5(x)= \037(63x5-70x3 + 15x),8)

and their graphs are shown in Fig.3.2.2.)

y)

x)

FIGURE3.2.2.Graphs y = Pn (x) of the Legendrepolynomials forn = 1,2,3,4, and 5.Thegraphs aredistinguished by the fact that all nzerosof Pn (x) lie in the interval -1< x < 1.)

_ Problems)

Find generalsolutions in powersofx of the differential equa-tions in Problems1through 15.State the recurrencerelation

and the guaranteed radius ofconvergencein eachcase.

1.(x2 -l)y\" + 4xy' + 2y = 0

2.(x2 + 2)y\" + 4xy' + 2y = 03.y\" +xy' + y = 04. (x2 + l)y\" + 6xy' + 4y = 05. (x2-3)y\" + 2xy' = 06.(x2 -

l)y\"-6xy' + 12y == 0)

7. (x2 + 3)y\"-7xy' + 16y= 0

8.(2-x2)y\"

-xy' + 16y= 09.(x2 -

l)y\" + 8xy'+ 12y= 010.3y\" +xy'-4y = 011.5y\"

-2xy'+ 10y= 012.y\"

-x2y' -3xy = 013.y\" + x2y' + 2xy = 014.y\" + xy = 0 (an Airy equation)15.y\" + x2

Y = 0)))

Usepowerseriesto solvethe initial value problems in Prob-lems 16and 17.16.(1+ X2

)y\" + 2xy'-2y = 0; y(O)= 0, y'(O)= 117.y\" + xy'- 2y = 0; y(O)= 1,y'(O)= 0)

Solve the initial value problems in Problems 18through 22.First make a substitution of the form t = x -a, then find asolution L cnt

n of the transformed differential equation. Statethe interval ofvalues ofx for which Theorem 1of this sectionguarantees convergence.)

18.y\" + (x - l)y' + y = 0; y(l) = 2,y'(I)= 019.(2x -x2

)y\"-6(x- l)y'-4y = 0; y(l) = 0,y'(I)= 1

20. (x2 -6x + 10)y\" -4(x - 3)y'+ 6y = 0; y(3) = 2,y'(3)= 0

21.(4x2 + 16x+ 17)y\" = 8y; y(-2)= 1,y'(-2)= 022. (x2+6x)y\"+(3x+9)y'-3y= 0;y(-3)= 0,y'(-3)= 2)

In Problems 23 through 26,find a three-term recurrencere-lation for solutions of the form y = L cnx

n. Then find the

first three nonzero terms in eachof two linearly independentsolutions.)

23. y\" + (1+ x)y = 024. (x2 -

l)y\" + 2xy'+ 2xy = 025. y\" + x2y' + x2

y = 026. (1+ x3

)y\" + x4y = 0

27. Solvethe initial value problem)

y\" + xy'+ (2x2 + l)y = 0; y(O)= 1,y'(O)= -1.)Determine sufficiently many terms to compute y (1/2)ac-curate to four decimalplaces.)

In Problems28 through 30,find the first three nonzero termsin each of two linearly independent solutions of the formY == L cnx

n. Substitute known Taylor seriesfor the analyticfunctions and retain enough terms to compute the necessarycoefficients.

28. y\" + e-Xy = 029. (cosx)y\" + y = 030. xy\" + (sin x)y'+ xy = 031.Derivethe recurrencerelation in (21)for the Legendre

equation.32. Follow the stepsoutlined in this problem to establish Ro-

drigues'sformula)

1 dn

Pn(x) = --(x2 - I)nn!2n dx n)

for the nth-degree Legendrepolynomial. (a) Show that

v = (x2 - l)n satisfiesthe differential equation)

(1-x2)v' + 2nxv = O.)

Differentiate eachsideof this equation to obtain)

(1-x2)v\" + 2(n - l)xv'+ 2nv = O.)


(b) Differentiate eachsideof the last equation n times in

successionto obtain)

(1-x2)v(n+2) - 2xv(n+1) + n(n + l)v(n) == O.)

Thus u = v(n) = Dn (x2 - l)n satisfiesLegendre'sequa-tion of order n. (c) Show that the coefficient of xn

in u is (2n)!/n!;then state why this proves Rodrigues'formula. (Note that the coefficient of xn in Pn (x) is(2n)!/ [2

n (n !)2].)33.TheHermiteequationofordera is)

y\"-2xy'+ 2ay == O.)

(a)Derivethe two powerseriessolutions)

\0372ma(a-2)...(a -2m + 2)

YI = 1+ \037(_I)m x2m

m=l (2m)!)

and)

Y2 =x)

\0372m (a - 1)(a -3)...(a -2m + I)+ \037(_I)m x2m+I.

m=I (2m + I)!)

Show that YI is a polynomial if a is an even integer,whereasY2 is a polynomial if a is an odd integer. (b) TheHermitepolynomial of degreen is denotedby Hn (x).Itis the nth-degree polynomial solution of Hermite'sequa-tion, multiplied by a suitable constant so that the coeffi-cient of xn is 2n . Show that the first six Hermite polyno-mials are)

Ho(x)-1,H2(x) = 4x2-2,H4(x) = 16x4 -48x2 + 12,Hs(x)= 32xs -160x3 + 120x.)

HI (x) = 2x,H3(x) == 8x3 - 12x,)

A general formula for the Hermite polynomials is)

2 dn

(2

)Hn(x) = (_I)neX - e-x .dx n)

Verify that this formula does in fact give an nth-degreepolynomial. It is interesting to use a computer alge-bra system to investigate the conjecture that (for eachn) the zerosof the Hermite polynomials Hn and Hn+Iare \"interlaced\" -that is, the n zerosof Hn lie in the nbounded open intervals whose endpoints are successivepairs ofzerosof Hn+1.

34. Thediscussionfollowing Example 4 in Section3.1sug-geststhat the differential equation y\" + y = 0 couldbeusedto introduce and define the familiar sine and cosinefunctions. In a similar fashion, the Airy equation)

\"y =xy)))

servesto introduce two new specialfunctions that appearin applications ranging from radio waves to molecularvi-brations. Derivethe first three or four terms of two dif-ferent power seriessolutions of the Airy equation. Thenverify that your results agreewith the formulas)

y)

\0371 .4 .....(3k -2)

y, (x)= 1+ t:r (3k)!X 3k)

x)

and)

\0372 .5 .....(3k -1)

Y2(X) == x + \037X3k+1

k=l (3k + I)!for the solutions that satisfy the initial conditions YI (0) ==

1,Y\037 (0) == 0 and Y2 (0) == 0,

Y\037 (0) == 1,respectively.Thespecialcombinations)

FIGURE3.2.3.TheAiry function graphsY == Ai(x) and Y == Bi(x).)

35.(a) Todetermine the radius of convergenceof the seriessolution in Example 5, write the seriesof terms of even

degreein Eq.(11)in the form)

A. YI (x)I(X) --

32/3r(\037))

Y2 (x)3113r(

\037))

00 00

Yo(x) == 1+ LC2nX2n == 1+ Lanzn

n=l n=1)and)

Bi(x)_ YI (x) + Y2(X)-31/6r(\037) 3-1/6r(\037)

define the standard Airy functions that appearin math-ematical tables and computer algebra systems. Theirgraphs shown in Fig. 3.2.3exhibit trigonometric-like os-cillatory behavior for x < 0, whereas Ai (x) decreasesexponentially and Bi(x) increases exponentially asx -+ +00.It is interesting to use a computer algebrasystem to investigate how many terms must be retained in

the YI- and Y2-seriesaboveto producea figure that is visu-

ally indistinguishable from Fig. 3.2.3(which is basedon

high-precision approximations to the Airy functions).)

where an == C2n and z == x2. Then apply the recurrencerelation in Eq.(8)and Theorem3 in Section3.1to show

that the radius ofconvergenceof the seriesin z is4.Hencethe radius ofconvergenceof the seriesin x is 2.How doesthis corroborateTheorem1 in this section? (b) Write the

seriesof terms ofodd degreein Eq.(11)in the form)

YI(X) =x(1

+\037C2n+,x2n)

=x(1

+\037bnzn))

to show similarly that its radius of convergence(as apowerseriesin x)is also2.)

IIIJRegularSingularPoints)

We now investigatethe solution of the homogeneoussecond-orderlinear equation)

\037) A(x)y\" + B(x)y'+C(x)y= 0) (1))

near a singular point. Recall that if the functions A, B,and C are polynomialshaving no common factors, then the singular points of Eq. (1)are simply thosepoints where A (x)= O.For instance,x = 0 is the only singular point of the Besselequation of ordern,)

x2y\" +xy' + (x2- n 2

)y = 0,whereasthe Legendreequation of ordern,)

(1-x2)y\"

-2xy' +n(n + l)y = 0,)

has the two singular pointsx = -1and x = 1.It turns out that someof the featuresof the solutionsof suchequationsof the most importancefor applicationsare largelydetermined by their behavior near their singular points.)))

3.3RegularSingularPoints 219)

We will restrict our attention to the casein which x = 0 is a singular pointof Eq.(1).A differentialequation having x = a as a singular point is easily trans-formed by the substitution t = x-a into one having a correspondingsingular pointat O.For example,let us substitute t = x-Iinto the Legendreequationof ordern.Because)

, _ dy _ dy dt _ dyY- -- -- - -,dx dt dx dt)

1/ d2y

[d

(dY

)]dt d2

yY =

dx2=

dt dx dx=

dt2 ')

and 1 -x2 = 1- (t + 1)2= -2t- t 2, we get the equation)

d2y dy-t(t+2)\037-2(t +1)-+n(n + l)y = O.

dt dt)

This new equation has the singular point t = 0 correspondingto x = 1 in theoriginal equation; it has alsothe singular point t = -2correspondingto x = -1.)

TypesofSingularPointsA differential equation having a singular point at 0 ordinarily will not have powerseriessolutionsof the form y(x) = L cnxn , so the straightforward method of Sec-tion 3.2fails in this case.To investigatethe form that a solutionof such an equationmight take, we assumethat Eq. (1)has analytic coefficientfunctions and rewrite it

in the standard form)

\037) y\" + P(x)y'+ Q(x)y = 0,) (2))

where P = BIA and Q = CIA.Recall that x = 0 is an ordinary point (asopposedto a singular point) of Eq. (2)if the functions P(x)and Q(x)are analytic at x ==

0;that is, if P(x) and Q(x) have convergentpowerseriesexpansionsin powersof x on someopen interval containing x = O. Now it can be proved that eachof the functions P(x) and Q(x) either is analytic orapproaches::f::ooas x --+ O.Consequently,x = 0 is a singular point of Eq. (2)provided that either P(x) orQ(x) (orboth) approaches::f::ooas x -+ O. For instance, if we rewrite the Besselequation of ordern in the form)

1/ 1,(

n 2

)y + -y + 1 -- y = 0,X x2)

we seethat P(x)= 11x and Q(x)= 1- (nlx)2both approach infinity asx --+O.We will seepresently that the powerseriesmethodcan begeneralizedto apply

near the singular point x = 0 of Eq.(2),provided that P(x)approachesinfinity nomore rapidly than 1Ix, and Q(x)no more rapidly than 1Ix2, asx -+ O. This is away of saying that P(x)and Q(x)haveonly \"weak\" singularitiesat x = O.Tostatethis more precisely,we rewriteEq.(2)in the form)

\037)II + p(x) , + q(x) _ 0y y \037y- ,x x)

(3))

where)

\037) p(x) = xP(x) and q(x)= x2Q(x).) (4))))


Example1)

DEFINITION RegularSingularPointThe singular point x = 0 of Eq. (3)is a regularsingularpoint if the functions

p(x) and q(x)are both analytic at x = O.Otherwiseit is an irregularsingularpoint.)

In particular, the singular point x = 0 is a regular singular point if p(x)and

q(x) are both polynomials.For instance, we seethat x = 0 is a regular singular

point of Bessel'sequation of ordern by writing that equation in the form

1 22/I , x-n

y + -y + 2 y = 0,x x)

noting that p(x) = 1 and q (x)= x2- n 2 are both polynomials in x.Bycontrast, considerthe equation

2x3y\" + (1+x)y'+3xy = 0,)

which has the singular point x = O.If we write this equation in the form of (3),weget)

\" (1+x)j(2x2), \037 _ 0y + y + 2\"Y- .x x)

Because) l+x 1 1p(x) = =-+- \037 00

2x2 2x2 2xas x \037 0 (although q(x) =

\037

is a polynomial), we seethat x = 0 is an irregular

singular point. We will not discussthe solution of differentialequationsnear irreg-ular singularpoints;this is a considerablymore advancedtopic than the solutionofdifferentialequations near regular singular points.)

Considerthe differentialequation

x2(1+x)y\" +x(4-x2)y' + (2+3x)y = O.

In the standard form y\" + Py' + Qy = 0 it is

\" 4 -x2, 2 +3x

y +x(1+x)y + x2(1+x)y = O.)

Because)4 -x2 2 +3x

P(x)=x(1+x)

and Q(x)= x2(1+x)both approach 00 asx \037 0,we seethat x = 0 is a singular point. To determinethe

nature of this singular point we write the differentialequation in the form of Eq.(3):)

\" (4-x2)j(1+x), (2+3x)j(1+x) 0y + y + 2 y= .X X)

Thus)4 -x2 2 + 3x

p\037)= \037d q\037)= .l+x l+xBecausea quotient of polynomialsis analytic whereverthe denominatoris nonzero,we seethat p(x) and q(x) are both analytic at x = O. Hencex = 0 is a regularsingular point of the given differentialequation. .)))

Example2)


It may happen that when we beginwith a differentialequation in the generalform in Eq.(1)and rewrite it in the form in (3),the functions p(x)and q(x)as givenin (4)are indeterminateforms at x = O. In this casethe situation is determinedbythe limits)

Po = p(O)= lim p(x) = lim xP(x)x---+o x---+o)

(5))

and)

qo = q(O)= lim q(x)= lim x2Q(x).x---+o x---+o)

(6))

If Po = 0 = qo, then x = 0 may be an ordinary point of the differential equationx2y\" +xp(x)y'+q(x)y= 0 in (3).Otherwise:). If both the limits in (5)and (6)existand are finite, then x = 0 is a regular

singularpoint.. If either limit fails to existor is infinite, then x = 0 is an irregular singular

point.)

Remark:The most commoncasein applications,for the differential equa-tion written in the form)

II + p(x) , + q(x) _ 0y y -z-y- ,x x)(3))

is that the functions p(x) and q(x) are polynomials. In this casePo = p(O)andqo = q(O)are simply the constant terms of thesepolynomials,so there is no needto evaluate the limits in Eqs.(5)and (6). .)

To investigatethe nature of the point x = 0 for the differentialequation)

X4

y\" + (x2sinx)y'+(1-cosx)y= 0,)

we first write it in the form in (3):)

II (sinx)/x, (1-cosx)/x2y + y + 2 y = O.x x)

Then I'H6pital'srule gives the values)

sInx cosxPo = lim-= lim

1x---+o x x---+o)

=1)

and)1-cosx sin x 1

qo = lim 2 = lim-= -x---+o X x---+o 2x 2

for the limits in (5)and (6).Sincethey are not both zero, we seethat x = 0 is not

an ordinary point. But both limits are finite, so the singular point x = 0 is regular.Alternatively, we couldwrite)

sin x 1

(x3 x5

)x2 x4

p(x) =\037

=x x-3!+ Sf

-...= 1-3!+ Sf- ...)))


and)

1-cosx 1

[ (X2 X

4X

6

)]q(X) = = - 1- 1--+---+...X 2 X 2 2! 4! 6!)

1 x2 x4----+--...-2! 4! 6! .)

These(convergent)powerseriesshowexplicitly that p(x)and q(x)are analytic andmoreover that Po = p(O)= 1 and qo = q(O)=

\037, thereby verifying directly that

x = 0 is a regular singularpoint. .)

TheMethodofFrobenius)We now approach the task of actually finding solutionsof a second-orderlineardif-ferential equationnear the regular singular point x = O.The simplestsuch equationis the constant-coefficientequidimensionalequation)

2 H , 0X y + poxy + qoY =) (7))

to which Eq.(3)reduceswhen p(x) = Po and q(x)= qo are constants.In this casewe can verify by direct substitution that the simplepowerfunction y(x) = xr is asolution of Eq.(7)if and only if r is a root of the quadratic equation)

r(r - 1)+por +qo = O.) (8))

In the general case,in which p(x) and q(x)are powerseriesrather than con-stants, it is a reasonableconjecture that our differentialequation might have a solu-tion of the form)

\037)

00 00y(x) = xr Lcnxn = Lcnxn+r = coxr +CIXr+1 +C2Xr +2 + ...

n=O n=O)

(9))

-theproduct of xr and a powerseries.This turns out to be a very fruitful con-jecture;accordingto Theorem 1 (soonto be stated formally), every equationof theform in (1)having x = 0 as a regular singular point does,indeed,have at leastonesuch solution. This fact is the basisfor the method of Frobenius,named for the

German mathematicianGeorgFrobenius(1848-1917),who discoveredthe methodin the 1870s.

An infinite seriesof the form in (9)is calleda Frobeniusseries.Note that

a Frobeniusseriesis generally not a powerseries.For instance, with r =-\037

the

seriesin (9)takesthe form)

y = cox-1/2 +cIX 1/2+C2x3/2 +C3x5/2+ ...;)

it is not a seriesin integral powersof x.To investigate the possibleexistenceof Frobenius seriessolutions, we begin

with the equation)

\037) x2yH +xp(x)y'+q(x)y = 0) (10))))


obtained by multiplying the equation in (3)by x2. If x = 0 is a regular singular

point, then p(x) and q(x)are analytic at x = 0,so)

p(x) = Po+PIX +P2x2+P3x3 + ...,q(x)= qo +qlX +q2x2+q3x3 + ...

.)

(11))

Supposethat Eq.(10)has the Frobeniusseriessolution)

\037)

00

y = LcnXn+r .

n=O)

(12))

We may (and always do) assumethat Co i= 0 becausethe seriesmust have a first

nonzero term. Termwisedifferentiation in Eq.(12)leadsto)

\037)

00

y' = Lcn(n+r)xn+r-l

n=O)

(13))

and)

\037)

00

y\" = Lcn(n +r)(n + r - l)xn+r-2.n=O)

(14))

Substitution of the seriesin Eqs.(11)through (14)in Eq.(10)now yields)

[r(r- l)coxr+ (r + l)rclxr+1+ ...]+ [pox+PIX

2 + ...] .[rcoxr-l+ (r + I)CIXr + ...]

+ [qO+q I X + ...] .[coxr +CI X r+ I + ...] = O. (15))

Uponmultiplying initial terms of the two products on the left-hand sidehere andthen collectingcoefficientsof xr , we seethat the lowest-degreeterm in Eq.(15)isco[r(r-1)+por+qo]xr .If Eq.(15)is to besatisfiedidentically,then the coefficientof this term (as well as those of the higher-degreeterms) must vanish. But we areassuming that Coi= 0,so it follows that r must satisfy the quadraticequation)

\037) r(r - 1)+por +qo = 0) (16))

of preciselythe sameform as that obtainedwith the equidimensionalequationin (7).Equation (16)iscalledthe indicialequationof the differentialequationin (10),andits two roots (possiblyequal) are the exponentsof the differential equation (at the

regular singular point x = 0).Our derivationof Eq.(16)showsthat if the Frobeniusseriesy = xr L cnxn is

to bea solutionof the differentialequation in (10),then the exponentr must beoneof the roots rl and r2 of the indicial equation in (16).If rl i= r2, it followsthat thereare two possibleFrobenius seriessolutions, whereas if rl = r2 there is only onepossibleFrobeniusseriessolution; the secondsolutioncannotbea Frobeniusseries.The exponentsrl and r2 in the possibleFrobenius seriessolutionsare determined(usingthe indicial equation) by the values Po = p(O)and qo = q(O) that we havediscussed.In practice,particularly when the coefficientsin the differential equation)))

in the original form in (1)are polynomials,the simplestway of finding Po and qo isoften to write the equation in the form)

Po+ PIX +P2x2 + ... qo +qlX +q2x2 + ...y\" + y' + y ==o. (17)

X x2)

Then inspectionof the seriesthat appearin the two numeratorsreveals the constantsPo and qo.)

Example3) Find the exponentsin the possibleFrobenius seriessolutionsof the equation)

2x2(1+x)y\" +3x(1+x)3y'- (1-x2)y ==O.)

Solution We divide each term by 2x2(1+x) to recastthe differentialequation in the form)

l(l+2x +x2) _1(1-x)

y\" + 2 y' + 2y ==0,

X x2)

and thus seethat Po ==\037

and qo ==-\037.

Hencethe indicial equation is)

r(r - 1)+ \037r

-1==r 2 + 1r -1== (r + l)(r-1)==0,)

with roots rl ==\037

and r2 == -1.The two possibleFrobenius seriessolutions arethen of the forms)

00

Yl (x) ==X 1/2Lanxn

n=O)

00

and Y2(X) ==x-ILbnxn .

n=O)

.)

FrobeniusSeriesSolutions)

Oncethe exponentsrl and r2 are known, the coefficientsin a Frobenius seriesso-lution are determined by substitution of the seriesin Eqs.(12)through (14)in the

differential equation, essentiallythe samemethod as was used to determine coef-ficients in powerseriessolutions in Section3.2.If the exponentsrl and r2 arecomplexconjugates,then there alwaysexisttwo linearly independentFrobeniusse-ries solutions.We will restrict our attention here to the casein which rl and r2 areboth real.We also will seeksolutions only for x > O. Oncesuch a solution hasbeen found, we need only replacexr1 with Ix Ir

1 to obtain a solution for x < o.The following theorem is proved in Chapter 4 of Coddington'sAn Introduction to

Ordinary DifferentialEquations.)))

Example4)


THEOREM1 FrobeniusSeriesSolutions)

Supposethat x = 0 is a regular singular point of the equation)

>-) x2y\" +xp(x)y'+q(x)y = o.) (10))

Letp >0 denote the minimum of the radii of convergenceof the powerseries)

00

p(x) = LPn xn

n=O)

00

and q(x)= Lqnxn.n=O)

Letrl and r2 be the (real)roots,with rl > r2, of the indicial equationr(r-1)+Por +qo = O.Then

(a) For x >0,there existsa solutionof Eq.(10)of the form)

>-)00

Yl (x)= xrt Lanxn

n=O)

(18))(aO i= 0))

correspondingto the larger root rl .(b) If rl - r2 is neither zero nor a positive integer, then there existsa second

linearly independentsolutionfor x >0 of the form)

>-)00

Y2(X) = xr2 Lbnxn

n=O)

(19))(bo i= 0))

correspondingto the smaller root r2.

The radii of convergenceof the powerseriesin Eqs. (18)and (19)are each atleastp. The coefficients in theseseriescan be determined by substituting the

seriesin the differentialequation)

x2y\" +xp(x)Y'+q(x)y= O.)

We have already seenthat if rl = r2, then there can existonly one Frobeniusseriessolution. It turns out that, if rl - r2 is a positive integer, there mayor maynot exista secondFrobeniusseriessolutionof the form in Eq.(19)correspondingtothe smallerroot r2.Theseexceptionalcasesare discussedin Section3.4.Examples4 through 6 illustrate the processof determining the coefficientsin those Frobeniusseriessolutions that are guaranteedby Theorem 1.)

Find the Frobenius seriessolutionsof)

2x2y\" +3xy'- (x2 + l)y = O.) (20))

Solution First we divide each term by 2x2 to put the equation in the form in (17):)3 1 1 2- --- -x

y\" + 1.y' + 2 2Y = O.

X x2)(21))

We now seethat x = 0 is a regular singular point, and that Po =\037

and qo = -\037.

Becausep(x) =\037

and q (x)= -!-!x2 are polynomials,the Frobeniusserieswe)))


obtain will convergefor all x >O.The indicial equation is)

r(r - 1)+ \037r

-\037

= (r - \037)(r + 1)= 0,)

so the exponentsare rl =\037

and r2 = -1.They do not differ by an integer, soTheorem 1 guarantees the existenceof two linearly independent Frobenius seriessolutions.Rather than separately substituting)

00

YI = xl/2Lanxn

n=O)

00

and Y2 = x-ILbnxn

n=O)

in Eq. (20),it is more efficient to beginby substituting Y = xr L cnxn . We will

then get a recurrencerelation that dependson r. With the value rl =\037

it becomesa recurrencerelation for the seriesfor YI, whereaswith r2 = -1it becomesarecurrencerelation for the seriesfor Y2.

When we substitute)

\037)

00

Y = LcnXn+r

,n=O)

00

y' = L(n+r)cnxn+r-l,

n=O)

and)

\037)

00

y\" = L(n+ r)(n + r - 1)cnxn+r-2

n=O)

in Eq.(20)-theoriginal differentialequation, rather than Eq.(21)-weget)

00 00

2 L(n+ r)(n + r - l)cnxn+r + 3 L(n+r)cnxn+r

n=O n=O)

00 00-Lcnxn+r+2-Lcnxn+r = O. (22)n=O n=O)

At this stagethere are several ways to proceed.A goodstandardpracticeis to shift

indicesso that each exponent will be the sameas the smallestone present.In this

example,we shift the index of summation in the third sum by -2 to reduceits

exponent from n + r +2 to n +r.This gives)

00 00

2 L(n+r)(n + r - l)cnxn+r +3 L(n+r)cnxn+r

n=O n=O)

00 00-LCn_2Xn +r -Lcnxn+r = O. (23)n=2 n=O)

The common range of summation is n > 2,so we must treat n = 0 and n = 1

separately.Followingour standard practice,the terms correspondingto n = 0 will

always give the indicial equation)

[2r(r- 1)+3r- l]co= 2 (r2 +

\037r

-\037)

Co= O.)))


The terms correspondingto n = 1 yield)

[2(r+ l)r+3(r+ 1)- I]Cl = (2r2 +5r+2)CI= o.)

Becausethe coefficient2r2 +5r+2 of Cl is nonzero whether r =\037

or r = -1,itfollows that)

Cl = 0) (24))

in either case.The coefficientof xn+r in Eq.(23)is)

2(n +r)(n + r - l)cn +3(n + r)cn-Cn-2-Cn = o.)

We solve for Cn and simplify to obtain the recurrencerelation)

\037)

Cn-2Cn =

2(n + r)2+ (n +r) - 1)

for n >2.) (25))

CASE1:rl =\037.

We now write an in placeof Cn and substitute r =\037

in Eq.(25).This gives the recurrencerelation)

\037)

an-2an =

2n2 + 3n)for n >2.) (26))

With this formula we can determine the coefficientsin the first Frobenius solution

Yl. In view of Eq.(24)we seethat an = 0 whenevern is odd.With n = 2,4, and 6in Eq.(26),we get)

aoa --2- 14')a2 aoa - -4 - 44-616')

a4 aoand a6 = 90=

55,440.)

Hencethe first Frobenius solution is)

(x2 x4 x6

)Yl (x)= aox1/2 1 +-+-+ + ... .14 616 55,440)

CASE2: r2 = -1.We now write bn in placeof Cn and substitute r -1in

Eq.(25).This gives the recurrencerelation)

\037)

bn-2b -

n - 2b2- 3n)for n >2.) (27))

Again, Eq.(24)impliesthat bn = 0 for n odd.With n = 2,4, and 6 in (27),we get)

bob2 = -,

2)b _ b2 _ bo

4 -20- 40')d b _ b4 _ bo

an 6- 54-

2160)

Hencethe secondFrobenius solution is)

(x2 x4 x6

)Y2(X) = box-1 1+-+-+ +....2 40 2160)

.)))


Example5)\037 '-' \" '-\" '\" '\" \"...\037....'\" \",\037\037_..,\037_'\"\"\" \"\"No\"\" \"'..\",_ '\" '\" ....,',.,........'-'\037........ ,....,,'..'..n.'\" \"\"'\" ........... \"'\" ........ ,,\"-\"-\"'''-''' '\" ........ \"......\037 \037\"...... \037.... .....\037 ............. \"....... ..................... .....\037.......\"

Find a Frobeniussolution of Bessel'sequation of orderzero,)

\037) x2y\" +xy' +x2y = O.) (28))

Solution In the form of (17),Eq.(28)becomes)

II 1, x2y + -y + -y = O.

X x2)

Hencex = 0 is a regular singularpoint with p(x) = 1 and q(x)= x2, soour serieswill convergefor all x >O.BecausePo = 1 and qo = 0,the indicial equation is

r(r - 1)+ r = r 2 = O.)

Thus we obtain only the singleexponent r = 0,and so there is only one Frobeniusseriessolution)

00

y(x) = xO Lcnxn

n=O)

of Eq.(28);it is in fact a powerseries.Thus we substitute y = L cnxn in (28);the result is)

00 00 00

Ln(n - l)cnxn +Lncnxn +Lcnxn+2 = O.

n=O n=O n=O)

We combinethe first two sums and shift the index of summation in the third by -2to obtain)

00 00

Ln 2cnxn +LCn_2Xn = O.n=O n=2)

The term correspondingto xO gives0 = 0:no information. The term correspondingto x1 gives CI = 0,and the term for xn yieldsthe recurrencerelation)

\037)

Cn-2Cn =--

n 2)for n >2.) (29))

BecauseCI = 0,we seethat Cn = 0 whenevern is odd. Substituting n = 2,4, and6 in Eq.(29),we get)

CoC2 = -

22 ')

C2 CoC - -4 --

42 - 22 .42 ')

C4 Coand C6= -

62 = -22 .42 .62 .)

Evidently, the pattern is)

\037)

(-l)nco (-l)nco\037n= =.

22 .42 ...(2n)2 22n (n !)2)

The choiceCo = 1 gives us one of the most important specialfunctions in math-

ematics,the Besselfunction of orderzeroof the first kind, denotedby Jo(x).Thus)

\037)

00 (_ 1)n X 2n X 2 x4X6

lo(x)=\037 22n (n!)2

= 1-4+64

-2304

+ ...\302\267) (30))

In this examplewe havenot beenableto find a secondlinearly independentsolutionof Bessel'sequationof orderzero.We will derive that solution in Section3.4;it will

not be a Frobeniusseries. .)))

Example6)


When r1-r2 Isan Integer

Recall that, if rl - r2 is a positive integer, then Theorem 1 guarantees only the

existenceof the Frobeniusseriessolution correspondingto the larger exponentri.Example6 illustrates the fortunate casein which the seriesmethod neverthelessyieldsa secondFrobenius seriessolution. The casein which the secondsolution isnot a Frobeniusserieswill bediscussedin Section3.4.)

.....

Find the Frobeniusseriessolutionsof)

1/ 2 / 0xy + y +xy = .) (31))

Solution In standard form the equationbecomes)

// 2 / x2y + -y + -y = 0,

X x2)

so we seethat xindicial equation)

o is a regular singular point with Po) 2 and qo = O. The)

r(r - 1)+2r = r(r + 1)= 0

has roots rl = 0 and r2 = -1,which differ by an integer. In this casewhen rl - r2is an integer, it is better to depart from the standard procedureof Example4 and

beginour work with the smallerexponent.As you will see,the recurrencerelationwill then tell us whether or not a secondFrobenius seriessolutionexists.If it doesexist,our computations will simultaneouslyyield both Frobenius seriessolutions.If the secondsolutiondoesnot exist,we begin anewwith the largerexponentr = rlto obtain the one Frobeniusseriessolutionguaranteedby Theorem 1.

Hencewe begin by substituting)

00 00

y = x-ILCnxn = LCnxn-1

n=O n=O)

in Eq.(31).This gives)

00 00 00

L(n- 1)(n-2)cnxn-2 +2 L(n- l)cnxn-2 +LcnXn = O.

n=O n=O n=O)

We combinethe first two sumsand shift the index by -2in the third to obtain)

00 00

Ln(n - l)cnxn-2 +LCn_2Xn-2 = O.

n=O n=2)

(32))

The casesn = 0 and n = 1 reduceto)

o .Co= 0 and O.Cl = O.)

Hencewe have two arbitrary constants Coand Cl and thereforecan expectto find ageneral solution incorporatingtwo linearly independentFrobeniusseriessolutions.If, for n = 1,we had obtainedan equationsuch as0 .Cl = 3, which canbesatisfiedfor no choiceof Cl, this would have told us that no secondFrobeniusseriessolutioncouldexist.)))


y)

1)

4n)

FIGURE3.3.1.Thesolutionscosx Slnx

Yl(X) =-and Y2(X) =-x x)

in Example 6.)

Now knowing that all is well,from (32)we read the recurrencerelation)

\037)

Cn-2Cn =-

n(n - 1))(33))for n >2.)

The first few values of n give

1C2= -Nco,)

1C3=-3 .2CI,

1 CICs = --C3= -

5 .4 5!')1

C4= --C24.3)Co4!')

1 CoC6= --C4=--

6 .5 6!'evidently the pattern is)

1 CIC7 = --C6= --.

7 .6 7!')

(-l)nco (-l)nCIC2n = (2n)!' C2n+I =

(2n + I)!for n > 1.Therefore, a general solution of Eq.(31)is)

\037)

00

y(x) = x-ILcnxn

n=O)

= Co

(1 _ x2 + x4 _...) + \037

(X _ x3+ x5 _...

)x 2! 4! x 3! 5!)

Co00 (-1)

nX 2n C1

00 (-1)nX 2n +1

=\037 \037 (2n)!

+\037 \037 (2n + 1)!.)

Thus)

x)

1 .y(x) = -(cocosx+CIsInx).x

We have thus found a general solutionexpressedasa linear combinationof the two

Frobeniusseriessolutions)

cosx sInxYI (x)= and Y2(X) =-.x x)

(34))

As indicated in Fig.3.3.1,one of theseFrobeniusseriessolutionsisboundedbut the

other is unboundednear the regular singular point x = O-acommonoccurrenceinthe caseof exponentsdiffering by an integer. .)

SummaryWhen confronted with a linear second-orderdifferentialequation)

\037) A (x)y// + B(x)y'+C(x)y= 0)

with analytic coefficientfunctions, in orderto investigatethe possibleexistenceofseriessolutions we first write the equation in the standard form)

\037) y// + P(x)y/+ Q(x)y = O.)

If P(x)and Q(x)are both analytic at x = 0,then x = 0 is an ordinary point, andthe equation has two linearly independentpowerseriessolutions.)))


Otherwise,x = 0 is a singular point, and we next write the differential equa-tion in the form)

\037)

II + p(x) / + q(x) _ 0y y \037y- .x x)

If p(x) and q(x)are both analytic at x = 0,then x = 0 is a regular singular point.In this casewe find the two exponentsrl and r2 (assumedreal, and with rl > r2) bysolving the indicial equation)

\037) r(r - 1)+por +qo = 0,)

where Po = p(O)and qo = q(O).There always existsa Frobeniusseriessolution

y = xrl L anx n associatedwith the larger exponent rl,and if rl - r2 is not an

integer, the existenceof a secondFrobeniusseriessolution Y2 = xr2 L bnxn isalso

guaranteed.)

IEI1Proble\"ms_)

In Problems1through 8, determine whether x = 0 is an ordi-nary point, a regular singular point, or an irregular singularpoint. If it is a regular singular point, find the exponents of the

differential equation at x = o.

1.xy\" + (x -x3)y'+ (sinx)y= 02.xy\" + x2y' + (eX - l)y = 03.x2

y\" + (cosx)y'+ xy = 04. 3x3

y\" + 2x2y'+ (1-x2)y = 0

5.x(1+ x)y\" + 2y' + 3xy = 06.x2(1-x2

)y\" + 2xy'- 2y = 07. x2

y\" + (6sinx)y'+ 6y = 08.(6x2 + 2x3

)y\" + 2lxy'+ 9(x2 - l)y = 0)

If x = a i= 0 is a singular point ofa second-orderlinear dif-ferential equation, then the substitution t = x -a transformsit into a differential equation having t = 0 asa singular point.We then attribute to the original equation at x = a the be-havior of the new equation at t = O. Classify(as regular orirregular) the singular points of the differential equations in

Problems 9 through 16.9.(1- x)y\" + xy' + x2

y = 010.(1-X)2y\" + (2x -2)y'+ y = 011.(1-x2

)y\"-2xy'+ 12y = 0

12.(x - 2)3y\" + 3(x-2)2y'+ x3y = 0

13.(x2 - 4)y\" + (x -2)y'+ (x + 2)y = 014.(x2 -

9)2y\" + (x2 + 9)y'+ (x2 + 4)y = 015.(x - 2)2y\" - (x2 -4)y'+ (x + 2)y = 016.x3(1- x)y\" + (3x+ 2)y'+ xy = 0)

Find two linearly independent Frobenius seriessolutions (forx > 0) of eachof the differential equations in Problems 17through 26.

17.4xy\" + 2y' + y = 018.2xy\" + 3y' - y = 0)

19.2xy\"- y' - y = 0

20.3xy\" + 2y' + 2y = 021.2x2y\" + xy'- (1+ 2x2

)y = 022.2x2y\" + xy' - (3 -2x2

)y = 023.6x2

y\" + 7xy' - (x2 + 2)y = 024. 3x2

y\" + 2xy'+ x2y = 0

25.2xy\" + (1+ x)y' + y = 026.2xy\" + (1- 2X2)y' -4xy = 0)

Usethe method ofExample 6 to find two linearly independentFrobenius seriessolutions ofthe differential equations in Prob-lems 27 through 31.Then construct a graph showing their

graphsfor x > O.

27.xy\" + 2y' + 9xy = 028.xy\" + 2y'-4xy = 029.4xy\" + 8y' + xy = 030.xy\"

- y' + 4x3y = 0

31.4x2y\"

-4xy' + (3 -4x2)y = 0)

In Problems 32 through 34,find the first three nonzero terms

ofeachoftwo linearly independent Frobenius seriessolutions.

32.2x2y\" + x(x+ l)y'- (2x + l)y = 033.(2x2 + 5x3

)y\" + (3x-x2)y'- (1+ x)y = 034. 2x2y\" + (sinx)y'- (cosx)y= 035.Note that x = 0 is an irregular point of the equation)

x2y\" + (3x - l)y' + y = o.)

(a) Show that y = xr L cnxn can satisfy this equation

only if r = O. (b) Substitute y = L cnxn to derive

the \"formal\" solution y = L n !xn .What is the radius ofconvergenceof this series?)))

36. (a) Supposethat A and B are nonzero constants. Showthat the equation x2

y\" + Ay' + By = 0 has at most onesolution of the form y = xr L cnx

n . (b) Repeatpart (a)with the equation x3

y\" + Axy' + By = O. (c) Showthat the equation x3

y\" + Ax2y' + By = 0 has no Frobe-nius seriessolution. (Suggestion: In eachcasesubstitute

y = xr L cnxn in the given equation to determine the pos-

siblevalues ofr.)37.(a) Usethe method of Frobenius to derive the solution

YI = x of the equation x3y\"

-xy' + y = O. (b) Verify

by substitution the secondsolution Y2 = xe-l/x .DoesY2

have a Frobenius seriesrepresentation?38.Apply the method of Frobenius to Bessel'sequation oforder 4,)

x2y\" +xy'+ (X2 -\037) Y = 0,

to derive its general solution for x > 0,)

cosx Slnxy(x) = Co

,JX+ CI

,JX.

Figure 3.3.2shows the graphs of the two indicated solu-tions.)

Y)

YI)

1)

x)

FIGURE3.3.2.Thesolutionscosx Slnx

Yl (x)=,JX

and Y2(X) =,JX

in

Problem 38.39.(a) Show that Bessel'sequation oforder1,

x2y\" + xy' + (x2 - l)y = 0,)

has exponents rl = 1 and r2 = -1at x = 0, and that theFrobenius seriescorrespondingto rl = 1 is

x 00 (_1)nx2n

J1(x)= 2 \037 n!(n + 1)!22n')

(b) Show that there is no Frobenius solution correspond-ing to the smaller exponent r2 = -1;that is,show that itis impossibleto determine the coefficientsin)

00

Y2(X) = X-I LCnxn.n=O)

40. Considerthe equation x2y\" + xy' + (1- x)y = O. (a)

Show that its exponentsare:l:i,so it has complex-valuedFrobenius seriessolutions)

00i\037 n

y+ =x \037Pnxn=O)

00and y_ = x-i

Lqnxnn=O)

with Po = qo = 1.(b) Show that the recursion formula

IS)

Cn-lCn = .

n 2 + 2rn

Apply this formula with r = i to obtain Pn = Cn , then with

r = -ito obtain qn = Cn. Concludethat Pn and qn arecomplexconjugates:Pn = an + ibn and qn = an - ibn ,where the numbers {an} and {bn } are real. (c) Deducefrom part (b) that the differential equation given in this

problem has real-valuedsolutions of the form)

YI (x) = A(x) cos(lnx)-B(x)sin(lnx),Y2(X) = A(x) sin(lnx) + R(x) cos(lnx),)

where A(x) = Lanxn and B(x)= Lbnxn .41.Considerthe differential equation)

x(x-1)(x+ 1)2y\" + 2x(x-3)(x+ l)y'-2(x- l)y = 0)

that appearedin an advertisement for a symbolic algebraprogram in the March 1984issueof the American Math-

ematical Monthly. (a) Show that x = 0 is a regular

singular point with exponents rl = 1 and r2 = O. (b) It

follows from Theorem1 that this differential equation has

a power seriessolution of the form)

YI (x) = x + C2X2+ C3x3 + ....)

Substitute this series(with CI = 1)in the differential equa-tion to show that C2= -2,C3 = 3, and)

Cn+2 =)

(n2 -n)cn-l+ (n2 -5n -2)cn- (n2 + 7n + 4)Cn+1

(n + 1)(n + 2))

for n > 2. (c) Usethe recurrencerelation in part (b)to prove by induction that Cn = (-I)n+1n for n ::::1 (!).Hencededuce(using the geometricseries)that)

XYI (x) =

(1+ X)2)

for 0 < x < 1.42. This problem is a brief introduction to Gauss'shypergeo-

metricequation)

x(1- x)y\" + [y - (a + fJ + l)x]y'- afJy = 0, (35))

where a,fJ, and yareconstants.This famous equation has

wide-ranging applications in mathematics and physics.(a) Show that x = 0 is a regular singular point ofEq.(35),with exponents0 and 1-y. (b) If y is not zeroor a neg-ative integer, it follows (why?) that Eq.(35)has a powerseriessolution)

00 00

y(x) = XO Lcnx

n =Lcnxn

n=O n=O)))

3.4Methodof Frobenius:The ExceptionalCases 233)

with Co i= O. Show that the recurrencerelation for thissenesIS)

00 a fJy(x) == 1+ L \037xn

n=On! Yn)

(36))

where an == a(a+ 1)(a+ 2) . ..(a + n - 1)for n\037 1,

and fJn and Yn aredefined similarly. (d) Theseriesin (36)is known as the hypergeometricseriesand is commonlydenotedby F(a,fJ, y, x).Show that

1(i) F(I,1,1,x) == (the geometric series);I-x

(ii) xF(I,1,2,-x)== In(1+x);(iii) xF (4,1,

\037,

-x2) == tan

-I x;(iv) F(-k,1,1,-x)== (l+x)k (the binomial series).)

(a + n)(fJ + n)C - Cn+1 -(y + n)(1+ n)

n

for n > O. (c) Concludethat with Co == 1 the seriesinpart (b) is)

_ MethodofFrobenius:TheExceptionalCases)

We continue our discussionof the equation)

\037)// + p(x) , + q(x) 0y y -y=

X x2)(1))

where p(x) and q (x)are analytic at x = 0,and x = 0 is a regular singular point. Ifthe roots rl and r2 of the indicial equation)

\037) cp(r) = r(r -1)+por +qo= 0) (2))

do not differ by an integer, then Theorem 1 of Section3.3guarantees that Eq.(1)has two linearly independentFrobeniusseriessolutions.We considernow the morecomplexsituation in which rl - r2 is an integer. If rl = r2, then there is only oneexponent available, and thus there can be only one Frobenius seriessolution. Butwe saw in Example6 of Section3.3that if rl = r2 + N, with N a positiveinteger,then it is possiblethat a secondFrobenius seriessolution exists.We will also seethat it is possiblethat such a solution doesnot exist.In fact, the secondsolutioninvolvesIn x when it is not a Frobeniusseries.As you will seein Examples3 and 4,theseexceptionalcasesoccurin the solutionof Bessel'sequation. For applications,this is the most important second-orderlinear differential equation with variablecoefficients.)

The NonlogarithmicCasewith rl = r2 + NIn Section3.3we derived the indicial equation by substituting the power seriesp (x)= L Pn xn and q (x)= L qnxn and the Frobeniusseries)

\037)

00 00

y(x) = xr Lcnxn = Lcnxn+r

n=O n=O)

(co i= 0)) (3))

in the differentialequation in the form)

\037) x2y// +xp(x)y'+q(x)y = O.) (4))

The result of this substitution, after collectionof the coefficientsof like powersofx, is an equation of the form)

00

LFn(r)xn+r = 0n=O)

(5))))

Example1)

in which the coefficientsdependon r.It turns out that the coefficientof xr is)

Fo(r)= [r(r- 1)+por +qo]co= cp(r)co,) (6))

which gives the indicial equation becauseCo i= 0 by assumption; also,for n > 1,the coefficientof xn+r has the form)

Fn(r)= cp(r+n)cn +Ln(r;Co,CI,...,Cn-I).) (7))

HereLn is a certain linear combination of Co, CI,...,Cn-I.Although the exactformula is not necessaryfor our purposes,it happensthat)

n-ILn = L[(r+k)Pn-k+qn-k]Ck.

k=O)

(8))

Becauseall the coefficientsin (5)must vanish for the Frobeniusseriesto beasolution of Eq.(4),it follows that the exponent r and the coefficientsCo,CI,..., Cn

must satisfy the equation)

cp(r +n)cn +Ln(r;Co,CI,...,Cn-I)= O.) (9))

This is a recurrencerelation for Cn in terms of Co,CI,..., Cn-I.Now supposethat rl = r2 + N with N a positive integer. If we usethe larger

exponent rl in Eq.(9),then the coefficientcp (ri +n) of Cn will benonzerofor everyn > 1 becausecp (r) = 0 only when r = rl and when r = r2 < ri.OnceCo,CI,..., Cn-I have beendetermined,we therefore can solve Eq.(9)for Cn and continueto computesuccessivecoefficientsin the Frobeniusseriessolutioncorrespondingtothe exponent rl .

But when we use the smallerexponent r2, there is a potential difficulty in

computing CN.For in this casecp (r2 + N) = 0,soEq.(9)becomes)

O.CN +LN(r2;Co,CI,...,CN-I)= o.) (10))

At this stageCo,CI,...,CN-Ihave already beendetermined.If it happensthat)

LN(r2;Co,CI,...,CN-I)= 0,)

then we can chooseCN arbitrarily and continue to determine the remaining coeffi-cientsin a secondFrobeniusseriessolution. But if it happensthat)

LN (r2; Co,CI,..., CN-I)i= 0,)

then Eq.(10)is not satisfiedwith any choiceof CN; in this casethere cannot existasecondFrobenius seriessolution correspondingto the smaller exponent r2. Exam-ples 1 and 2 illustrate thesetwo possibilities.)

....

Considerthe equation)

x2y// + (6x+x2)y' +xy = O.) (11))

HerePo = 6 and qo = 0,so the indicial equation is)

cp(r) = r(r - 1)+6r = r 2 +5r = 0) (12))))


with roots rl = 0 and r2 = -5;the roots differ by the integer N = 5.We substitute

the Frobeniusseriesy = L cnxn+r and get)

00 00

L(n+ r)(n+ r - l)cnxn+r +6 L(n+ r)cnxn+r

n=O n=O)

00 00

+L(n+ r)cnxn+r+1 +Lcnxn+r+1 = o.n=O n=O)

When we combinethe first two and alsothe last two sums,and in the latter shift the

index by -1,the result is)

00 00

L[(n+r)2+5(n+ r)]cnxn+r +L(n+ r)cn_lXn+r = o.n=O n=l)

The terms correspondingto n = 0 give the indicial equation in (12),whereasforn > 1 we get the equation)

[en+ r)2+5(n+r)]cn + (n + r)cn-l = 0,) (13))

which in this examplecorrespondsto the general solution in (9). Note that the

coefficientof Cn is c/J (n +r).We now follow the recommendationin Section3.3for the caserl = r2 +N:

We begin with the smaller root r2 = -5.With r2 = -5,Eq.(13)reducesto)

\037) n(n -5)cn + (n -5)Cn-l= o.) (14))

If n i= 5,we can solve this equation for Cn to obtain the recurrencerelation)

\037)

Cn-lCn =--

n)for n i= 5.) (15))

This yields)

Cl = -co,)Cl Co

C2 = -2 = 2')(16))

C2 CoC3 = -3= -6 ') and)

C3 CoC4------

4-

24.)

In the caserl = r2 + N, it is always the coefficientCN that requiresspecialconsid-eration. HereN = 5,and for n = 5 Eq.(14)takesthe form 0 .Cs+0 = O.HenceCs is a secondarbitrary constant, and we can compute additionalcoefficients,still

using the recursion formula in (15):)

CsC6= -6')

C6 CsC7 = -7=

6 .7')

C7 CsCg = -8 = -

6 .7 .8') (17))

and so on.)))


When we combinethe resultsin (16)and (17),we get)

00

Y = x-5Lcnxn

n=O)

(x2 x3 x4

)= cQx-5 1-x +2:-6+ 24

(x6 x7 X 8

)+C5X-5

X5--+-- +...

6 6.7 6.7.8)in terms of the two arbitrary constants Co and C5. Thus we have found the two

Frobeniusseriessolutions)

(x2 x3 x4

)YI(X) = x-5 1-x +---+-2 6 24)

and)00 (-I)nxn 00 (-I)nxn

Y2 (x)= 1 +L = 1 + 120Ln=l 6.7 ...(n +5) n=l (n +5)!)

of Eq.(11).) .)

Example2).,.

Determine whether or not the equation)

x2y//

-xy' + (x2- 8)y = 0) (18))

has two linearly independentFrobeniusseriessolutions.)

Solution HerePo = -1and qo = -8,so the indicial equation is)

cjJ(r) = r(r - 1)- r - 8 = r 2-2r- 8 = 0)

with roots rl = 4 and r2 = -2 differing by N = 6. On substitution of y

L cnxn+r in Eq.(18),we get)

00 00

L(n+ r)(n+ r - l)cnxn+r -L(n+r)cnxn+r

n=O n=O)

00 00

+Lcnxn+r+2- 8Lcnxn+r = o.n=O n=O)

If we shift the index by -2 in the third sum and combinethe other three sums,weget)

00 00

L[(n+ r)2-2(n+ r) -8]cnxn+r +LCn_2Xn+r = o.n=O n=2)

The coefficientof xr gives the indicial equation,and the coefficientof xr+l gives)

[(r+ 1)2-2(r+ 1)- 8] CI = o.)))


Becausethe coefficientof CI is nonzero both for r = 4 and for r = -2,it followsthat CI = 0 in eachcase.For n > 2 we get the equation)

\037) [(n+ r)2-2(n+ r) - 8] Cn +Cn-2= 0,) (19))

which correspondsin this exampleto the generalequation in (9);note that the coef-ficientof Cn is c/J (n +r).

We work first with the smaller root r = r2 = -2.Then Eg.(19)becomes)

\037) n(n -6)cn +Cn-2= 0) (20))

for n >2.For n i= 6 we can solve for the recurrencerelation)

\037)

Cn-2Cn =-

n(n -6)) (n > 2, n i= 6).) (21))

BecauseCI = 0,this formula gives)

CoC2 = 8'

C2 CoC - -4-8 -

64 ,)

C3 = 0,)

and) Cs= o.)

Now Eq.(20)with n = 6 reducesto)

Coo .C6+-= o.64)

But Co i= 0 by assumption, and hencethere is no way to chooseC6 so that this

equation holds. Thus there is no Frobenius seriessolution correspondingto the

smallerroot r2 = -2.To find the singleFrobenius seriessolution correspondingto the larger root

rl = 4, we substitute r = 4 in Eq.(19)to obtain the recurrencerelation)

\037)

Cn-2Cn =-

n(n + 6))(n >2).) (22))

This gives)Co

C2 = -2 .8

'The general pattern is)

C2 CoC4= - = .

4 .10 2 .4.8.10)

(-I)nco (-I)n6co\037= =.

2 .4 ...(2n).8 .10...(2n +6) 22n n!(n +3)!)

This yieldsthe Frobeniusseriessolution)

(00

(_ 1)nX 2n

)Yl (x)= x4 1 +6\037 22nn!(n +3)!)

of Eq.(18).) .)))


ReductionofOrder)When only a singleFrobeniusseriessolution exists,we needan additional tech-nique. We discusshere the method of reductionof order,which enablesus touseone known solution YI of a second-orderhomogeneouslinear differentialequa-tion to find a secondlinearly independent solution Y2. Considerthe second-orderequation)

\037) yl! + P(x)y'+ Q(x)y ==0) (23))

on an openinterval I on which P and Q are continuous. Supposethat we know onesolution YI of Eq.(23).By Theorem 2 of Section2.1,there existsa secondlinearlyindependent solution Y2; our problemis to find Y2. Equivalently, we would like tofind the quotient)

veX) = Y2(X) .YI (x))

(24))

Oncewe know vex), Y2 will then be given by)

Y2(X) == V(X)Yl (x).) (25))

We begin by substituting the expressionin (25) in Eq. (23), using the

derivatives)

Y\037

==vY\037 + v' Yl and

Y\037==

vY\037' +2v'Y\037

+ v// Yl.)

We get)

[vy\037'+ 2v'

Y\037+ vI!YI] + P

[vy\037+ v'YI] + QVYI ==0,

and rearrangementgives)

[I! P '

Q ]// 2 \" P ' 0V YI + YI + YI + v YI + v YI + v YI == .)

But the bracketedexpressionin this last equation vanishesbecauseYI is a solutionof Eq.(23).This leaves the equation)

vI!YI + (2y\037 + PYI)V' ==O.) (26))

The key to the successof this method is that Eq. (26)is linearin v'. Thus thesubstitution in (25)has reducedthe second-orderlinear equation in (23)to the first-

order(in v') linear equation in (26).If we write u == v' and assumethat YI (x)nevervanisheson I,then Eq.(26)yields)

u' + (2\037:

+P(X\302\273)

u = O.) (27))

An integrating factor for Eq.(27)is)

p=exp(1(2\037: +P(X\302\273)dX) =exP(21nIYll+1P(X)dX);)

thus)

p(X)= Y\037 exp(IP(x)dx) .)))

Example3)


We now integrate the equation in (27)to obtain)

uy;exp(IP(X)dX)= C, so Vi = U =

\037

exp(-IP(X)dX).)

Another integration now gives)

Y2 = V = Cfexp(-f P(x)dx)dx+K.

Yl Y;)

With the particular choicesC = 1 and K = 0 we get the reduction-of-orderformula)

\037) fexp(-f p(x)dx)

Y2 = Yl 2 dx.Yl)

(28))

This formula provides a secondsolution Y2(X) of Eq. (23)on any interval whereYl (x)isneverzero.Note that becausean exponentialfunction nevervanishes,Y2(x)is a nonconstantmultiple of YI (x),so Yl and Y2 are linearly independentsolutions.)

For an elementaryapplicationof the reduction-of-orderformula considerthe differ-ential equation

x2y\"-9xy'+25y = O.

In Section3.3we mentioned that the equidimensionalequation x2y\" + Poxy' +qoY = 0 has the powerfunction y(x) = xr as a solution if and only if r is a rootofthe quadraticequation r 2+ (Po- l)r +qo = O.HerePo = -9and qo = 25, soourquadraticequation is r 2- lOr+25 = (r-5)2 = 0 and has the single(repeated)rootr = 5.This gives the singlepowerfunction solution YI (x)= x5 of our differential

equation.Beforewe can apply the reduction-of-orderformula to find a secondsolution,

we must first divide the equation x2y\" -9xy' +25y = 0 by its leadingcoefficientx2 to get the standard form)

II 9 I 25y

- -y +-y= 0X x2)

in Eq. (23)with leading coefficient 1.Thus we have P(x)= -9/xand Q(x)==

25/x2, so the reduction-of-orderformula in (28)yieldsthe secondlinearly indepen-dent solution)

Y2 (x)= x5f (X\037)2

exp(-f -\037

dx)dx

= x5f x-IO exp(91nx)dx= x5

f x-lO x9dx= x51nx)

for x > O. Thus our particular equidimensionalequation has the two independentsolutions Yl (x)= x5 and Y2 (x)= x5 In x for x >O. .)

Similar applicationsof the reduction-of-orderformula can be found in Prob-lems37-44of Section2.2-wherewe introducedthe method of reductionof orderin Problem36 (though without deriving there the reduction-of-orderformula itself).)))


The LogarithmicCasesWe now investigatethe general form of the secondsolutionof the equation)

\037)

/!+ p(x) , + q(x) _ 0y y -z-Y- ,x x)(1))

under the assumption that its exponents'1and '2== '1- N differ by the integerN >O.We assumethat we have already found the Frobeniusseriessolution)

\037)

00

Yl (x)==xrl Lanxn

n=O)

(aO \037 0)) (29))

for x > 0 correspondingto the larger exponent '1.Let us write P(x)for p(x)/xand Q(x)for q(x)/x2.Thus we can rewriteEq.(1)in the form y\" +Py' + Qy == 0of Eq. (23).

Becausethe indicial equationhas roots '1and '2==,}-N, it can be factoredeasily:)

r2 + (Po- l)r+qo ==(,- '1)('- '1+N)==,2+ (N -2'1)'+ (,\037

- 'IN)==0,)

sowe seethat)

Po- 1 ==N -2'1;)that is,)

-Po-2'1==-1-N.) (30))

In preparation for useof the reduction of orderformula in (28),we write)

2Po+PIX +P2X +...PoP(x)== ==-+PI +P2X + ....X X)

Then)

exp(-f P(x)dx) = exp(-f [:0+PI +P2X +...]dx

))

==exp(-PoIn X - PIX -1P2x2- ...))

==x-poexp(-PIX -1P2x2- ...) ,)

so that)

exp(-f P(x)dx) = x-po(1+ Alx + A 2x2 + ...) . (31))

In the last step we have used the fact that a composition of analytic functions isanalytic and therefore has a powerseriesrepresentation; the initial coefficientofthat seriesin (31)is 1 becauseeO== 1.

We now substitute (29)and (31)in (28);with the choiceao == 1 in (29),this

yields)

fx-po(1+ A 1x + A 2x2 + ...)

Y2 == Yl 2 dx.-

X 2rl (1+a 1 X +a2x2 + ...))))


We expandthe denominatorand simplify:

fx-Po-2rl(1+ A1x + A2 X2 +...)

Y2 = Yl dx1 + B1x+ B2x2 + ...

= Yl f x-1-N (1+C1x+ C2X2 + ...) dx

(Herewe have substituted (30)and indicated the result of carrying out long divisionof seriesas illustrated in Fig.3.1.1,noting in particular that the constant term ofthe quotient seriesis 1.)We now considerseparately the casesN = 0 and N >O.We want to ascertain the general form of Y2 without keepingtrack of specificcoef-ficients.)

(32))

CASE1:EQUALEXPONENTS(rl = r2). With N = 0,Eq.(32)gives

Y2 = Yl f (\037

+ C1 +C2x2 + ...)dx

= y1lnx+ Yl (C1x+ 4C2X2 +...)= y1lnx+xrl (1+alX +...)(C1x+ 4C2X2+...)= Yl In x +xrt

(box+b 1x2 +b2x3 + ...) .Consequently,in the caseof equal exponents,the general form of Y2 is)

\037)

00

Y2(X) = Yl(x)lnx+xl+rlL:bnxn .

n=O)

(33))

Note the logarithmic term; it is alwayspresentwhen '1= '2.)

CASE2:POSITIVEINTEGRALDIFFERENCE(rl = r2 + N). With N > 0,Eq.(32)gives

Y2 = Yl f x-1-N (1+C1x+ C2x2 + \302\267 \302\267 \302\267 + CNXN + .\302\267 \302\267) dx)

f (CN 1 Cl

)= Yl-+ +-+ ... dx

X xN+l xN)

(-N C -N+l

)x IX

=CNy1lnx+Yl -+ +...-N -N + 1)

r +N

(L:OO

n

)N

(1 C1x

)=CN y 1 lnx+x2 ax x- --+ +...n=O

nN -N + 1

')

so that)

\037)

00

Y2(X) = CNYI (x)lnx +xr2 L:bnxn

,n=O)

(34))

where bo = -aolN i=- O.This gives the general form of Y2 in the caseof exponentsdiffering by a positive integer. Note the coefficientCN that appearsin (34)but not

in (33).If it happensthat CN = 0,then there is no logarithmic term; if so,Eq.(1)has a secondFrobeniusseriessolution (as in Example1).)))


In our derivationof Eqs.(33)and (34)-whichexhibit the general form of the

secondsolution in the cases'1= '2and '1- '2= N > 0,respectively-wehavesaidnothing about the radii of convergenceof the variouspowerseriesthat appear.Theorem 1 (next) is a summation of the precedingdiscussionand alsotells wherethe seriesin (33)and (34)converge.As in Theorem 1 of Section3.3,we restrictourattention to solutions for x > o.)

THEOREM1 TheExceptionalCasesSupposethat x = 0 is a regular singularpoint of the equation)

x2y\" +xp(x)y'+q(x)y = O.) (4))

Letp >0 denotethe minimum of the radii of convergenceof the powerseries)

00

p(x) = LPnxnn=O)

00

and q(x)= Lqnxn.n=O)

Let'land r2 be the roots,with rl > r2, of the indicial equation)

r(r - 1)+Por +qo = O.)

(a) If rl = r2, then Eq. (4)has two solutions Yl and Y2 of the forms)

\037)

00

Yl (x)= xr } Lanxn

n=O)

(aO\037 0)) (35a))

and)

\037)

00

Y2(X) = Yl (x)In x +xq+lLbnxn .

n=O)

(35b))

(b) If rl -r2 = N, a positive integer, then Eq.(4)has two solutionsYl and Y2 ofthe forms)

\037)

00

Yl (x)= xr } Lanxn

n=O)

(aO\037 0)) (36a))

and)

\037)

00

Y2(X) = CY1(X)In x +xr2 Lbnxn .n=O)

(36b))

In Eq. (36b),bo \037 0 but C may be either zero or nonzero, so the logarithmicterm mayor may not actually bepresentin this case.The radii of convergenceofthe powerseriesof this theorem are all at leastp.The coefficientsin theseseries(and the constant C in Eq.(36b))may bedeterminedby direct substitution of theseriesin the differentialequation in (4).)))

Example4)


. ....

We will illustrate the case'l= r2 by deriving the secondsolutionof Bessel'sequa-tion of orderzero,)

x2y

!I+xy' +x2y = 0,) (37))

for which '1= '2= O.In Example5 of Section3.3we found the first solution)

00 (_ 1)nX2n

Yl(X) = Jo(x)=\037 22n(n!)2)

(38))

According to Eq.(35b)the secondsolution will have the form)

00

Y2 = Yl lnx +Lbnxn .n=1)

(39))

The first two derivativesof Y2 are)

00, , I Yl '\"' b n-l

Y2=Yl nx+-+\037n n XX

n=1)

and)2 ' 00!! !!I Yl Yl '\"' ( l)b n-2

Y2=

Yl nx +--2\" + \037n

n -n X .

X Xn=2)

We substitute thesein Eq.(37)and usethe fact that Jo(x)alsosatisfiesthis equationto obtain)

O 2!/ , 2= X Y2 +XY2 +x Y2

=(x2y\037' + XY\037 +X

2Yl) In x + 2xy\037)

00 00 00

+Ln(n- l)bnxn +Lnbnxn+Lbnxn +2,

n=2 n=1 n=l)

and it follows that)

00( l)n 2 2n 00'\"' - nx 2 2 '\"' 2 no = 2

\037 22n n' 2 +b 1x +2 b2x + \037(n bn +bn-2)x .n=1 ( .) n=3)

(40))

The only term involving x in Eq. (40)is b 1x,so b 1 = O.But n 2bn +bn-2 = 0 if n

is odd,and it follows that all the coefficientsof oddsubscriptin Y2 vanish.Now we examine the coefficientswith even subscriptsin Eq. (40).First we

seethat)

(-1)(2)b2 = -2.22 .22 .(1!)2)

1

4)(41))

For n >2,we read the recurrencerelation)

2 (2)(-1)n (2n)(2n)b2n +b2n-2= -

22n (n!)2)(42))))


from (40).Note the \"nonhomogeneous\"term (not involving the unknown coeffi-cients)on the right-hand side in (42).Suchnonhomogeneousrecurrencerelationsare typical of the exceptionalcasesof the method of Frobenius,and their solutionoften requiresa bit of ingenuity. The usual strategy dependson detecting the mostconspicuousdependenceof b2n on n. We note the presenceof 22n (n !)2on the right-hand side in (42);in conjunction with the coefficient (2n)2 on the left-hand side,we are inducedto think of b2n as something divided by 22n (n !)2.Noting alsothealternationof sign,we make the substitution)

(_I)n+lCb - 2n

2n -22n (n !)2

') (43))

in the expectation that the recurrencerelation for C2n will be simplerthan the onefor b2n. We chose(_I)n+l rather than (-I)nbecauseb2 ==

\037

> 0;with n == 1 in

(43),we get C2 == 1.Substitution of (43)in (42)gives)

2 2 (_1)n+lC2n (-ltc2n-2( n) 22n(n!)2

+22n-2[(n _ 1)!]2)

(-2)(-2)n(2n)22n (n !)2)

which boilsdown to the extremelysimplerecurrencerelation)

1C2n ==C2n-2+-.

n)

Thus)

1 1C4==C2+ - == 1 +-,2 2

1 1 1C6 ==c4 +-==1+-+-

3 2 3'1 111

Cg ==C6+ 4== 1 +

2+

3+ 4 ')

and soon.Evidently,) 111C2 ==1+-+-+...+ -==Hn 2 3 n

n,) (44))

where we denoteby Hn the nth partial sum of the harmonicseriesL(I/n).Finally, keepingin mind that the coefficientsof oddsubscriptare all zero, we

substitute (43)and (44)in (39)to obtain the secondsolution)

00 (_I)n+lHnx2nY2(X) == lo(x)In x + '\"'

2 2\037 2 n (n!)n=l)

x2 3x4 l1x6== lo(x)In x +---+ -...

4 128 13824) (45))

of Bessel'sequationof orderzero.The powerseriesin (45)convergesfor all x.Themost commonlyusedlinearly independent [of lo(x)]secondsolution is)

2 2Yo (x)==-(y -ln2)Yl+ -Y2;J[ J[)))

Example5)


that is,)

2

[X

00 (-I)n+1Hnx2n

]Yo (x)==- (y +In-) lo(x)+\" 2 2 '

J[ 2 \037 2 n(n!))(46))

where y denotesEuler'sconstant:)

y == lim (Hn - Inn) \037 0.57722.n\037oo)

(47))

This particular combination Yo (x) is chosenbecauseof its nice behavioras x --++00;it is calledthe Besselfunction of orderzeroof the secondkind. .)As an alternative to the method of substitution, we illustrate the caserl - r2 == Nby employing the technique of reduction of order to derive a secondsolution ofBessel'sequationof order1,)

x2y\" +xy' + (x2- l)y ==0;) (48))

the associatedindicial equation has roots rl == 1 and r2Problem 39 of Section3.3,one solutionof Eq.(48)is)

-1.According to)

x 00(-1)n X

\037nX X

3X

5X

7

Yl(X) = J1 (X) =2 \037 22nn!(n + I)!

=2-

16+ 384-

18432+....(49))

With P(x)== l/xfrom (48),the reductionof orderformula in (28)yields)

Y2 == YI f \037 dxxYI)

f1

== Yl dxx(x/2-x3/16+x5/384-x7 /18432+ ...)2)

f4- dx-Yl x3(I-x2/8+x4/192-x6/9216+...)2)

f1

==4YI dxx3(1-x2/4 +5x4/192-7x6/4608+ ...))

f1

(x2 7x4 19x6

)==4Yl - 1 +-+-+ +...dxx3 4 192 4608) (by lOng

)division)

f (1 1 7x 19x3

)==4Yl -+-+-+ +...dx4x x3 192 4608)

(1 7x2 19x4

)==Yl lnx +4Yl --+-+ +....2x2 384 18432)

Thus)

1 X x3 l1x5Y2(X)==Yl(x)lnx--+-+-- +....x 8 32 4608) (50))))


Note that the techniqueof reductionof orderreadily yieldsthe first severalterms ofthe series,but doesnot provide a recurrencerelation that can be usedto detenninethe general term of the series.

With a computationsimilar to that shown in Example4 (but morecomplicated-seeProblem21),the method of substitution can beusedto derive the solution)

1 00 (-l)n(Hn + Hn_1)x2n-1Y3(X) = Yl(X) lox--+2: 22n '( _

1)\"(51)x n=l n. n .)

where Hn is defined in (44)for n > 1;Ho ==O.The readercan verify that the termsshown in Eq. (50)agreewith)

3Y2(X) ==-J1 (x)+Y3(X).

4)(52))

The most commonlyusedlinearly independent [of J1]solutionof Bessel'sequationof order1 is the combination)

2 2f1 (x) == -(y -ln2)Y1(x)+ -Y3(X)J[ J[)

2

[(X

)1 2:

00 (-1)n(Hn+Hn_1)x2n-l]

==- y + In - J1(x)--+ . (53)J[ 2 x n= 122nn!(n - I)!

.)

Examples4 and 5 illustrate two methodsof finding the solution in the logarith-mic cases-directsubstitution and reductionof order.A third alternative isoutlinedin Problem 19.)

BJ..rroblems. ..)

In Problems1through 8, either apply the method ofExample1to find two linearly independent Frobenius seriessolutions,orfind onesuch solution and show (as in Example 2) that asecondsuch solution doesnot exist.)

1.xy\" + (3 -x)y'-y = 02.xy\" + (5-x)y'-y = 03.xy\"+(5+3x)y'+3y=04. 5xy\" + (30+ 3x)y'+ 3y = 05. xy\"

- (4+ x)y'+ 3y = 06.2xy\"

- (6+ 2x)y'+ y = 07. x2

y\" + (2x+ 3x2)y'- 2y = 08.x(1-x)y\"

-3y' + 2y = 0)

In Problems9 through 14,first find the first four nonzero termsin a Frobenius seriessolution of the given differential equa-tion. Then usethe reduction ofordertechnique (asin Example4) tofind the logarithmic term and thefirst three nonzerotermsin a secondlinearly independent solution.

9.xy\" + y' -xy = 010.x2

y\"-xy'+ (x2 + l)y = 0

11.x2y\" + (x2 -3x)y'+ 4y = 0)

12.x2y\" + x2y' -2y = 0

13.x2y\" + (2x2 -3x)y'+ 3y = 0

14.x2y\" + x(1+ x)y'-4y = 0

15.Begin with)

x2 x4 x6

Jo(x)= 1--+ ---+ ....4 64 2304)

Using the method of reduction oforder,derive the secondlinearly independent solution)

x2 3x4 l1x6Y2(X) = Jo(x)In x + ---+ -...

4 128 13284)

ofBessel'sequation oforderzero.16.Find two linearly independent Frobenius seriessolutions

ofBessel'sequation oforder\037,)

x2y\" +xy' + (X2 -\037)Y = O.)

17.(a) Verify that YI (x) = xeX is onesolution of

x2y\"-x(1+ x)y'+ y = O.)))

(b) Note that rl = r2 = 1.Substitute)

00

Y2 = yIlnx+ Lbnxn+I

n=I)

in the differential equation to deducethat bI

that)

-1and)

1nb n

-bn-I =-- for n > 2.n!(c) Substitute bn = cn/n!in this recurrencerelation andconcludefrom the result that Cn = -Hn. Thus the secondsolution is)

00 H xn+IY2(X) = xex In x -L n

n=I n!)

18.Considerthe equation xy\"-

Y = 0, which has exponentsrl = 1 and r2 = 0 at x = O. (a) Derivethe Frobeniusseriessolution)

00 xn

Yl (x)= L '( _ 1)1'

n=I n. n .)

(b) Substitute)

00

Y2 = CyIlnx+ Lbnxn

n=O)

in the equation xy\"-

Yrelation)

o to derive the recurrence)

2n + 1n(n + l)bn+I

-bn = - C.(n + I)!n!)

Concludefrom this result that a secondsolution is)

Loo

Hn + Hn-IY2(X) = YI (x) In x + 1- xn .

n=I n!(n - I)!19.Supposethat the differential equation

L[y]= x2y\" + xp(x)y'+ q(x)y= 0) (54))

has equal exponents rl = r2 at the regular singular pointx = 0, so that its indicial equation is)

<p(r) = (r -rI)2 = O.)

Let Co = 1 and define Cn (r) for n > 1 by using Eq.(9);that is,)

Ln(r;Co,CI,...,Cn-I)cn(r) = - . (55)

<p(r + n))

Then define the function Y(x,r) ofx and r to be)

00

y(x,r) = Lcn(r)xn+r .n=O)

(56))

(a) Deducefrom the discussionprecedingEq.(9) that)

L[y(x,r)]= xr (r -rI)2.) (57))

Hencededucethat)

00

YI = y(x, rI) =LCn (rdxn+r\\

n=O)

(58))

is one solution of Eq.(54). (b) Differentiate Eq. (57)with respectto r to show that)

L[Yr(x,rl))= \037 [xr(r - rl)2] = O.8r

r=r\\)

Deducethat Y2 =Yr (x,rI) is a secondsolution of

Eq. (54). (c) Differentiate Eq. (58)with respectto r toshow that)

00

Y2 = ylin x + xr\\ Lc\037 (rI)xn . (59)

n=l)

20.Usethe method of Problem 19 to derive both the solu-tions in (38)and (45)of Bessel'sequation of order zero.The following stepsoutline this computation. (a) TakeCo= 1;show that Eq. (55)reducesin this caseto)

(r + 1)2CI(r) = 0 and)

(60))Cn-2(r)cn(r) = -

(n + r)2

(b) Next show that CI(0) = c;(0) = 0, and then deducefrom (60)that Cn (0) =

c\037 (0) = 0 for n odd. Henceyouneedto compute Cn (0)and

c\037 (0)only for n even. (c) De-ducefrom (60)that)

for n > 2.)

(-1)nC2n(r) = (61)

(r + 2)2(r + 4)2 ...(r + 2n)2

With r = rl = 0 in (58),this gives Jo(x).(d) Differenti-ate (61)to show that)

, (_I)n+IHnc2n (0)= 22n (n !)2.)

Substitution of this result in (59)gives the secondsolutionin (45).

21.Derivethe logarithmic solution in (51)of Bessel'sequa-tion oforder1 by the method of substitution. Thefollow-

ing stepsoutline this computation. (a) Substitute)

Y2=CJ1(x)lnx+x-1

(1+[\037 bnxn))

in Bessel'sequation to obtain)

00-bI + x + L[(n2 - l)bn+1 + bn_tJx

n

n=2)

[00 (-I)n(2n + 1)x2n+1

]+ C x + \" = O.\037 22n (n + I)!n!)

(62))))


(b) Deducefrom Eq. (62)that C = -1and that bn = 0for n odd. (c) Next deducethe recurrencerelation

[(2n+ 1)2-1]b2n+2 + b2n =(\037l)n(2n

+ 1)(63)2 n(n + I)!n!

for n > 1.Note that if b2 is chosenarbitrarily, then b2n isdetermined for all n > 1.(d) Takeb2 = i and substitute

b _ (-1)nC2n2n -

22n(n - 1)!n!)

in Eq.(63)to obtain)

1 1C2n+2

-C2n = + -.n+l n)

(e) Note that C2 = 1 = HI + Ho and deducethat)

C2n = Hn + Hn-I.)

lIDBessel'sEquation)We have already seenseveralcasesof Bessel'sequationof orderp >0,)

\037) x2y\" +xy' + (x2-p2)y ==O.) (1))

Its solutions are now calledBesselfunctions of orderp. Suchfunctions first ap-pearedin the 1730sin the work of Daniel Bernoulli and Euler on the oscillationsof a vertically suspendedchain. The equation itself appearsin a 1764article byEuleron the vibrations of a circular drumhead,and FourierusedBesselfunctions in

his classicaltreatise on heat (1822).But their general propertieswerefirst studiedsystematically in an 1824memoir by the German astronomer and mathematicianFriedrich W. Bessel(1784-1846),who was investigating the motion of planets.Thestandard sourceof information on Besselfunctions is G.N.Watson'sA Treatiseonthe Theoryof BesselFunctions, 2nd ed.(Cambridge:CambridgeUniversity Press,1944).Its 36 pagesof references,which cover only the periodup to 1922,givesomeideaof the vast literature of this subject.

Bessel'sequation in (1)has indicial equation r 2-p2 ==0,with roots r ==::l::p.If we substitute y ==L cmx

m+r in Eq.(1),we find in the usual manner that Cl ==0and that)

\037) [em+ r)2 -p2]Cm +Cm-2==0) (2))

for m >2.The verificationof Eq.(2)is left to the reader(Problem6).)

The Caser=p > 0)

If we use rformula)

p and write am in placeof Cm, then Eq. (2)yieldsthe recursion)

\037)

am-2am ==- .

m(2p+m))(3))

Becauseal ==0,it follows that am == 0 for all oddvalues of m.The first few evencoefficientsare)

ao aoa - -2--2(2p+2)--22(p+1)')a2 aoa -- -4 -

4(2p+4)-

24 .2(p+ l)(p +2)')a4 aoa6==- ==- .

6(2p+6) 26 .2.3(p+1)(p+2)(p+3))))

3.5Bessel'sEquation 249)

The general pattern is)

(-l)maoa2m == ,22m m!(p+1)(p+2)...(p+m))

so with the larger root r ==p we get the solution)

\037)

00 (-l)mx2m+pYI(X) ==ao\" .

f;;o22m m!(p + l)(p+2)...(p+m))(4))

If p == 0 this is the only Frobeniusseriessolution; with ao == 1 as well, it is the

function Jo(x)we have seenbefore.)

The Caser = -p< 0)

If we use r ==-p and write bm in placeof Cm,Eq.(2)takes the form)

\037) m(m -2p)bm +bm-2 ==0) (5))

for m >2,whereasb I == O. We seethat there is a potential difficulty if it happensthat 2p isa positiveinteger-thatis, if p iseithera positiveintegeror an oddpositiveintegral multiple of 1.For then when m ==2p,Eq.(5)is simply O.bm + bm-2 = o.Thus if bm -2 i= 0,then no value of bm can satisfy this equation.

But if p is an odd positive integral multiple of 1,we can circumvent this

difficulty. For supposethat p == k/2where k is an oddpositive integer. Then weneedonly choosebm ==0 for all oddvalues of m.The crucial stepis the kth step,)

k(k -k)bk +bk-2 ==0;)

and this equation will hold becausebk ==bk-2 ==O.Henceif p is not a positive integer,we take bm ==0 for m oddand definethe

coefficientsof even subscriptin terms of bo by means of the recursionformula)

\037)

bm-2bm ==- ,

m(m -2p))m >2.) (6))

In comparing (6)with (3),we seethat (6)will leadto the same result as that in (4),exceptwith p replacedwith -p.Thus in this casewe obtain the secondsolution)

\037)

00 (-l)mx2m -p

Y2(X) ==bo\" .f;;o22m m!(-p+ l)(-p+2)...(-p+m))

(7))

The seriesin (4)and (7)convergefor all x > 0 becausex ==0 is the only singular

point of Bessel'sequation. If p > 0,then the leading term in YI is aoxP, whereasthe leading term in Y2 is box

-p. HenceYI (0)==0,but Y2 (x) \037 ::1::00asx \037 0, soit is clearthat YI and Y2 are linearly independent solutions of Bessel'sequation oforderp > O.)))

TheGammaFunctionThe formulas in (4)and (7)can be simplifiedby useof the gamma function rex),which is definedfor x >0 by)

\037) r(x) = 100

e-ttx-1 dt.) (8))

It is not difficult to show that this improper integral convergesfor each x >O.Thegamma function is a generalization for x > 0 of the factorial function n!,which

is defined only if n is a nonnegative integer. To seethe way in which r (x) is ageneralizationof n!,we note first that)

100 b

reI)== e-t dt == lim [_e-t] == 1.o b\037oo 0)

(9))

Then we integrate by parts with u == t X and dv ==e-t dt:)

r(x + 1)= lim [be-ttX dt = lim ([_e-ttx]b+ [bxe-ttx-I dt)b\037oo 10 b\037oo 0 10)

==x ( lim [be\037ttx\037l dt) ;

b\037oo 10)

that is,)

\037) rex+ 1)==xr(x).) (10))

This is the most important property of the gamma function.If we combineEqs.(9)and (10),we seethat)

r(2)== 1 .reI)== I!, r(3)==2.r(2)==2!, r(4) ==3 .r(3)==3!,)

and in general that)

\037) r(n + 1)==n!) for n >0 an integer.) (11))

An important specialvalue of the gamma function is)

r G) = 100

e-tt-1/2dt = 2100

e-u2 du = ...(ii,) (12))

where we have substituted u 2 for t in the first integral; the fact that)

100

-u2d \037e u==-o 2)

is known, but is far from obvious.[See,for instance,Example5 in Section13.4ofEdwardsand Penney, Calculus:Early Transcendentals,7th edition (UpperSaddleRiver, NJ:Prentice Hall,2008).]

Although r (x) is defined in (8)only for x > 0,we can use the recursionformula in (10)to define r (x)wheneverx is neither zero nor a negativeinteger. If-1<x <0,then)

rex+ 1)rex)== ;x)))

FIGURE3.5.1.Thegraph of theextendedgamma function.)


the right-hand sideis definedbecause0 <x + 1 < 1.The same formula may then

be usedto extend the definition of f (x) to the openinterval (-2,-1),then to the

openinterval (-3,-2),and soon.The graph of the gamma function thus extendedis shown in Fig.3.5.1.The student who would liketo pursue this fascinatingtopicfurther shouldconsult Artin's TheGammaFunction (NewYork: Holt, Rinehart andWinston, 1964).In only 39 pages,this is one of the finest expositionsin the entireliterature of mathematics.)

BesselFunctionsof theFirstKindIf we chooseao == Ij[2P f(p+ 1)]in (4),where p >0, and note that)

f(p+m + 1)== (p +m)(p+m - 1)...(p +2)(p+ I)f(p+ 1))

by repeatedapplication of Eq. (10),we can write the Besselfunction of the firstkind of orderp very conciselywith the aid of the gamma function:)

\037)

00 (-I)m(

X

)2m+p

1p(x)=\037 m!r(p+m + 1) 2

.) (13))

Similarly, if p > 0 is not an integer, we chooseho == Ij[2-P f(-p + 1)]in (7) toobtain the linearly independentsecondsolution)

\037)

00 (-I)m(

X

)2m-p

Lp(X)=\037m!r(_p+m+l)2)(14))

of Bessel'sequation of orderp.If p is not an integer,we have the general solution)

y(x) ==C1Jp(X)+ C2J-p(X)) (15))

for x >0;x P must bereplacedwith Ix IP in Eqs.(13)through (15)to get the correctsolutionsfor x <o.

If p ==n, a nonnegativeinteger, then Eq.(13)gives)

\037)_ \037 (-l)m

(X

)2m+n

In(x)- \037-

m=om!(m+n)!2)(16))

for the Besselfunctions of the first kind of integral order.Thus)

00 (_I)mx 2m x2 x4 x6lo(x)=

\037 22m (m!)2= 1-

22 + 22.42-

22.42 .62 +... (17))

and)

00(-I)m22m+ 1 X 1 x 3 1 x 511 (x)=

\037 22m+lm!(m+1)!=

2-

2!(2 ) + 2!.3!(2)-....(18))

The graphs of Jo(x)and J1(x)are shown in Fig.3.5.2.In a general way they re-sembledampedcosineand sineoscillations,respectively(seeProblem27).Indeed,if you examine the seriesin (17),you can seepart of the reason why lo(x)andcosx might be similar-onlyminor changesin the denominatorsin (17)are neededto producethe Taylor seriesfor cosx. As suggestedby Fig.3.5.2,the zerosof the)))

y) .>,......,....';<.i n tll.<Zeoo.'.i.n

'<u:Df'l\"x.'.'.'.\\.>\"//\037O\037 J)

(nl}Jr)nth Zeroof JI(x)

3.83177.0156

10.173513.323716.4706)

(n +\037)

1r)

1 2.4048 2.35622 5.5201 5.4978

x 3 8.6537 8.63944 11.7915 11.78105 14.9309 14.9226)

3.92707.068610.210213.351816.4934)

FIGURE3.5.2.Thegraphs of the Besselfunctions Jo(x)and J1 (x).)

FIGURE3.5.3.Zerosof Jo(x)and J1 (x).)

functions Jo(x) and J1 (x) are interlaced-betweenany two consecutivezerosofJo(x) there is preciselyone zero of J1 (x) (Problem 26)and vice versa. The first

four zerosof Jo(x)are approximately2.4048,5.5201,8.6537,and 11.7915.Forn large,the nth zero of Jo(x)is approximately (n

-\037)

Jr; the nth zero of J1 (x)is approximately (n +

\037)Jr.Thus the interval between consecutivezerosof either

Jo(x) or J1 (x) is approximately Jr-anothersimilarity with cosx and sin x. You

can seethe way the accuracy of theseapproximationsincreaseswith increasingn

by rounding the entries in the table in Fig.3.5.3to two decimalplaces.It turns out that Jp (x) is an elementary function if the orderp is half an odd

integer. For instance, on substitution of p =\037

and p = -\037

in Eqs.(13)and (14),respectively,the resultscan berecognized(Problem2)as)

J1j2(X) = j 2sin x

JrX)and) Llj2(X)= j 2 cosx.

JrX)(19))

BesselFunctionsof theSecondKind)

The methods of Section3.4must be used to find linearly independent secondso-lutions of integral order. A very complicatedgeneralization of Example3 in that

sectiongives the formula)

2 x 1 n-l 2n-2m(n -m - I)!

Yn(x) = - (y + In -) In(x)--\"

Jr 2 Jr \037 m!x n-2mm=O)

- \037 f (-l)m(Hm + Hm+n) cr+2m

(20)Jr m=O m!(m +n)! 2 ')

with the notation usedthere.If n = 0 then the first sum in (20)is taken to bezero.Here,Yn (x) is calledthe Besselfunction of the secondkind of integralordern >o.

The general solution of Bessel'sequation of integral ordern is)

\037) y(x) = CIJn(X) +C2Yn(X).) (21))

It is important to note that Yn (x) --+ -00as x --+ 0 (Fig.3.5.4).HenceC2 = 0 in

Eq. (21)if y(x) is continuous at x = O. Thus if y(x) is a continuous solution ofBessel'sequation of ordern, it follows that)

y(x) = cJn(x))))


for someconstant c. BecauseJo(O) == 1,we seein addition that if n == 0, then

c ==y (0).In Section9.4we will seethat this singlefact regardingBesselfunctions

has numerousphysical applications.Figure 3.5.5illustrates the fact that for n > 1 the graphs of In (x)and Yn (x)

lookgenerallylikethoseof J1 (x)and Y 1 (x).In particular, In (0)==0 while Yn (x)--*-00asx --+0+, and both functions undergodampedoscillationas x --++00.)

y) y)

0.5) 12(x))0.5)

x)x)

-0.5) -0.5)

FIGURE3.5.4.Thegraphs of the BesselfunctionsYo(x) and Y1 (x).)

FIGURE3.5.5.Thegraphs of the BesselfunctionsJ2(x) and Y2(x).)

BesselFunctionIdentities)Besselfunctionsare analogousto trigonometricfunctions in that they satisfy a largenumber of standardidentitiesof frequentutility, especiallyin the evaluation of inte-gralsinvolving Besselfunctions. Differentiationof)

00 (-I)m(

X

)2m+p

Jp(x)= J;m!r(p+m + 1) 2)(13))

in the casethat p is a nonnegativeinteger gives)

d d 00 (-I)mx2m+2p-[xp J (x)] ==-'\"\"\"\"

dx p dxf:=o22m +Pm!(p +m)!)

00 (-I)mx2m+2p-l= J;22m +p-1m!(p +m - 1)!)

00 (-I)mx 2m +p-1-X P '\"\"\"\"-f:=o22m +p-lm !(p +m - I)!')

and thus we have shown that)

d[

p]_p (- x Jp(x) -X Jp-l x).

dx)(22))

Similarly,)

d-[x-p Jp(x)]==-x-p Jp+l(x).dx

If we carry out the differentiationsin Eqs.(22)and (23)and then dividethe resultingidentitiesby x P and x-P , respectively,we obtain (Problem8) the identities)

(23))

J;(x)= Jp-1 (x)- PJp(x)

x)(24))))


Example1)

Example2)

Example3)

and)

J;(x)= PJp(x)- Jp+I(X).

x)(25))

Thus we may expressthe derivativesof Besselfunctions in terms of Besselfunctions

themselves.Subtractionof Eq.(25)from Eq.(24)gives the recursion formula)

2pJp+ l (x) ==-Jp(x)- Jp-l (x),X)

(26))

which can be used to expressBesselfunctions of high order in terms of Besselfunctions of lower orders.In the form)

2pJp-l(x)== -Jp(x)- Jp+l(x),x)

(27))

it can be usedto expressBesselfunctions of large negativeorderin terms of Besselfunctions of numerically smaller negativeorders.

The identities in Eqs.(22)through (27)hold wherever they are meaningful-that is,wheneverno Besselfunctions of negativeintegral orderappear.In particular,

they hold for all nonintegral values of p.)

With p ==0,Eq.(22)gives)

f xJo(x)dx= xJI(X)+C.

Similarly, with p ==0,Eq.(23)gives)

f JI (x)dx= -Jo(x)+C.) .)

Usingfirst p ==2 and then p == 1 in Eq.(26),we get)

4 4

[2

]J3(X) == -J2(X)- Jl(x)==- -Jl(x)- Jo(x) - Jl(x),X X X)

so that)

J3(X)= -;Jo(x)+ (:2-

1)JI (x).

With similar manipulations every Besselfunction of positive integral order can beexpressedin terms of Jo(x)and Jl(x). .)

To antidifferentiatex J2(x),we first note that)

f x-If2(x)dx= -x-I JI (x)+ C

by Eq. (23)with p == 1.We therefore write)

f xf2(x)dx= f x2 [x-If2(x)]dx)))

and integrateby parts with)

u ==X2

,)

du ==2x dx,) and)

dv ==x-IJ2(X) dx,V ==_X-1 J1(x).)

This gives

f xh(x)dx= -xJ1(x)+2f J1(x)dx= -xJ1(x)-2Jo(x)+C,

with the aid of the secondresult of Example1.) .)The ParametricBesselEquationThe parametricBesselequationof ordern is

\037 x2y\" +xy' + (a2x2- n 2

)y ==0,) (28))

wherea isa positiveparameter.As we will seein Chapter9,this equationappearsin

the solutionof Laplace'sequation in polar coordinates.It is easyto see(Problem9)that the substitution t ==ax transforms Eq.(28)into the (standard)Besselequation

d2y dy

t 2_ +t-+ (t2- n 2

)y ==0dt2 dt)

(29))

with general solution yet) == C1Jn(t)+ C2Yn(t). Hencethe general solution ofEq.(28)is)

\037) y(x) ==C1In(ax)+C2Yn(ax).) (30))

Now considerthe eigenvalueproblem

x2y\" +xy' + (AX

2- n 2) ==0,

y(L) ==0)(31))

on the interval [0,L].We seekthe positive values of A for which there existsanontrivial solution of (31)that is continuous on [0,L].If we write A = a2, then

the differentialequation in (31)is that in Eq.(28),so its general solution isgiven in

Eq.(30).BecauseYn(x) --+-00asx --+0 but In(O)is finite, the continuity of y(x)requiresthat C2 == O. Thus y(x) == C1Jn(ax).The endpoint conditiony(L) = 0now impliesthat z ==aL must be a (positive)root of the equation)

In (z)==O.) (32))

For n > 1,In(x)oscillatesrather likeJ1(x) in Fig.3.5.2and hence has an infinite

sequenceof positive zerosYn1, Yn2, Yn3, ...(seeFig.3.5.6).It follows that the kth

positiveeigenvalueof the problem in (31)is

Ak = (ak)2 =(Y2\037)2

,) (33))

and that its associatedeigenfunctionis)

(Ynk

)Yk(X) ==In LX .) (34))

The roots Ynk of Eq. (32)for n < 8 and k < 20are tabulated in Table 9.5ofM.Abramowitz and I.A. Stegun,HandbookofMathematicalFunctions(New York:

Dover, 1965).)))


y)

y =Yn(X))

x)

FIGURE3.5.6.Thepositive zerosYnl, Yn2, Yn3, ...of theBesselfunction In (x).)

_.J>_r,obleltl\037_.)

1.Differentiate termwise the seriesfor Jo(x)to show directlythat J\037(x)

= -JI (x) (another analogy with the cosineandsine functions).

2.(a) Deducefrom Eqs.(10)and (12)thatII.3 .5 .(2n - 1)r (n + 2) =2n vn.

(b) Usethe result of part (a) to verify the formulas in

Eq.(19)for JI/2(X) and J-I/2(X), and construct a figureshowing the graphs of thesefunctions.

3.(a) Supposethat m is a positive integer. Show that

2 2.5.8...(3m-l)2r (m + 3) =3m

r (3).(b) Concludefrom part (a) and Eq.(13)that

J-1/3(X) =

(xj2)-1/3(

00(_I)m3mx2m

)r (D1+

\037 22m m! .2.5 ...(3m - 1).

4. Apply Eqs.(19),(26),and (27)to show that

J3/2(X) = J2

3 (sin x -x cosx)JrX)

and)

L3/2(X) = -J 23 (cosx+xsinx).JrX

Construct a figure showing the graphs of thesetwo func-tions.

5.ExpressJ4(x) in terms of Jo(x)and JI (x).6.Derive the recursion formula in Eq. (2) for Bessel's

equation.7. Verify the identity in (23)by termwise differentiation.8.Deducethe identities in Eqs.(24)and (25)from those in

Eqs.(22)and (23).9.Verify that the substitution t = ax transforms the para-

metric Besselequation in (28)into the equation in (29).10.Show that

4J;(x)= Jp-2(x) -2Jp(x)+ Jp+2(x).)

11.Use the relation rex+ 1) = xr(x) to deducefrom

Eqs.(13)and (14)that if p is not a negative integer, then

Jp(x)=

(xj2)P

[00 (-I)m(xj2)2m

]r(p+ 1)1+

\037 m! (p + 1)(p+ 2)...(p + m).)

This form is moreconvenient for the computation ofJp(x)becauseonly the singlevalue r(p+ 1)of the gamma func-

tion is required.12.Usethe seriesofProblem 11to find y (0)= lim y (x) ifx-+o)

y(x) = x2

[JS/2(X)+ J-S/2(X)

].JI/2(X) + J-I/2(X)

Usea computer algebrasystem to graph y (x) for x near O.Doesthe graph corroborateyour value of y (O)?

Any integral of the form f xm In (x) dx can be evaluated

in terms of Besselfunctions and the indefinite integral

f Jo(x)dx. The latter integral cannot be simplified further,but the function f; Jo(t) dt is tabulated in Table 11.1ofAbramowitz and Stegun. Usethe identities in Eqs.(22)and

(23)to evaluate the integrals in Problems13through 21.13.f x2Jo(x)dx 14.f x3Jo(x)dx

15.f x4Jo(x)dx 16.f xJ1(x)dx

17.f x2J1(x)dx 18.f x3J1(x)dx

19.f x4J1(x)dx 20.f ]z(x)dx

21.f i)(x)dx

22.Prove that)

1 17r

Jo(x)= - cos(xsin e)deJr 0

by showing that the right-hand sidesatisfiesBessel'sequa-tion of orderzeroand has the value Jo(0) when x = O.Explain why this constitutes a proof.)))

23.Prove that)

I 1n

JI (x) = - cos(e-x sin e)den

0)

by showing that the right-hand sidesatisfiesBessel'sequa-tion of order1 and that its derivative has the value J{(0)when x = O.Explain why this constitutes a proof.

24. It can be shown that)

I 1n

In (x)= - cos(ne-x sin e)de.n

0)

With n > 2, show that the right-hand side satisfiesBessel'sequation of ordern and alsoagreeswith the val-uesIn (0) and J

\037(0).Explain why this doesnot suffice to

prove the precedingassertion.25.Deducefrom Problem22 that)

1 12n

Jo(x)=- eix sine de.2n

0)

(Suggestion:Show first that)

{2n {n10

eixsinO dB =10 (eiXSinO + e-iXsinO) dB;)

then useEuler'sformula.)26.UseEqs.(22)and (23)and Rolle'stheorem to prove that

between any two consecutivezerosof In (x) there is pre-ciselyonezeroofIn+1(x).Usea computer algebrasystemto construct a figure illustrating this fact with n = 10(forinstance) .

27.(a) Show that the substitution y = X-1/2z in Bessel'sequation oforderp,)

x2y\" + xy'+ (x2 - p2)y = 0,)

lIDApplicationsof Bess\037lFlInctions)

3.6Applicationsof BesselFunctions 257)

yields)

Z\" +(1

-p2X\037

i)z = O.

(b) If x is so large that (p2 - i) /x2 is negligible, thenthe latter equation reducesto z\" + z \037 O.Explain why this

suggests (without proving it) that if y(x) is a solution ofBessel'sequation, then)

y(x) \037 X-1/2(A cosx+ B sin x)= CX-1/2cos(x- ex)) (35))

with C and ex constants, and x large.)

Asymptotic Approximations It is known that the choicesC = -J2/nand ex = (2n + l)n/4 in (35)yield the best ap-proximation to In (x) for x large:)

In(x) \037 j 2cas[x- i(2n+ l)n].nx)

(36))

Similarly,)

Yn (x)\037 j 2sin [x-i(2n + l)n] .

nx)(37))

In particular,)

Jo(x) \037 j 2cas(x - in)nx)

and)

Yo(x) \037 j:xsin (x - in)if x is large.Theseareasymptotic approximations in that theratio of the two sidesin eachapproximation approachesunityasx ----+ +00.)

The importanceof Besselfunctions stemsnot only from the frequent appearanceofBessel'sequation in applications,but alsofrom the fact that the solutionsof manyother second-orderlinear differentialequationscan beexpressedin terms of Besselfunctions. To seehow this comesabout, we begin with Bessel'sequationof orderpin the form)

and substitute)

d2w dwZ2-+z-+ (Z2 -p2)w ==0,dz2 dz)

(1))

w ==X-ex

y ,) z ==kx t3.) (2))

Then a routine but somewhattedioustransformation (Problem 14)of Eq.(1)yields)

x2y\" + (1-2a)xy'+ (a2- f32p2 + f3

2k2x 2t3 )y ==0;)))

that is,)

\037) x2y\" + Axy' + (B+ Cxq)y ==0,) (3))

where the constants A, B,C,and q are given by

A == 1 -2a, B ==a2 - fJ2 p2, C == fJ2k2

, and q ==2fJ. (4)

It is a simplematter to solve the equations in (4)for)

\037)

I-Aa==2

k = 2./C,

q)

and)

qfJ ==

2 '

J(1-A)2 -4Bp==

q)

(5))

Underthe assumption that the squareroots in (5)are real, it follows that the generalsolution of Eqo (3)is)

y(x) ==x<Xw(z) ==x<Xw(kxfJ ),)

where)

w(z) ==CIJp(z)+C2Y-p(z)

(assumingthat p is not an integer) is the general solution of the Besselequation in

(1).This establishesthe following result.)

THEOREM1 Solutionsin BesselFunctionsIf C > 0,q i= 0,and (1-A)2 >4B,then the general solution (for x > 0)ofEq.(3)is)

\037) y(x) ==xa[cIJp(kx,B)+c2J_p(kx,B)],) (6))

where a,fJ, k, and p are given by the equations in (5). If p is an integer, then

J-p is to bereplacedwith Yp .)

Example1)

- \"..\"

Solvethe equation)

4x2y\" +8xy'+ (x

4 -3)y ==o.Solution To compareEqo (7)with Eqo (3),we rewrite the former as

x2y\" +2xy' + (-\037 + ix4)Y ==0)

(7))

and seethat A == 2,B == -\037,

C ==\037,

and q ==4.Then the equations in (5)givea ==-

\037,fJ ==2,k ==

\037,and p ==

\037.

Thus the general solution in (6)of Eq.(7)is

y(x) ==x-I /2[CIJI/2(\037x2) +C2J-I/2(\037x2)].

Ifwerecallfrom Eqo (19)of Section3.5that

J1/2(Z) = J2

sinz and L1/2(Z) = J2 cosz,nz nz

we seethat a general solutionof Eq.(7)can be written in the elementaryform

(x2 X

2

)y(x) = x-3/2 Acos4+ B sin 4 \302\267).)))

Example2)


Solvethe Airy equation)

yll +9xy = O.) (8))

Solution First we rewrite the given equation in the form)

x=L)

FIGURE3.6.1.Thebucklingcolumn.)

x2yll +9x3y = O.)

This is the specialcaseof Eq. (3)with A = B = 0,C = 9,and q = 3.It followsfrom the equations in (5)that a = i,fJ =

\037,k = 2,and p =

\037.

Thus the generalsolution of Eq.(8)is)

y(x) = X 1/2[Cl11/3(2x3/2) +C21-1/3(2x3/2)] .) .)

Bucklingofa VerticalColumnFor a practical application, we now considerthe problemof determining when auniform vertical column will buckleunder its own weight (after, perhaps,beingnudged laterally just a bit by a passingbreeze).We take x = 0 at the free top endof the column and x = L > 0 at its bottom; we assumethat the bottom is rigidlyimbeddedin the ground, perhapsin concrete;seeFig.3.6.1.Denotethe angulardeflection of the column at the point x by ()(x). From the theory of elasticity it

follows that)

d2()EI\037 +gpx()= 0,

dx)(9))

where E is the Young'smodulus of the material of the column, I is its cross-sectionalmoment of inertia, p is the linear density of the column, and g is gravita-tional acceleration.For physicalreasons-nobending at the free top of the columnand no deflectionat its imbeddedbottom-theboundary conditionsare)

()'(0)= 0, ()(L)= O.) (10))

We will accept(9)and (10)as an appropriatestatementof the problemand attemptto solve it in this form. With)

A - 2 - gp-Y -EI') (11))

we have the eigenvalueproblem)

()II+ y2x() = 0; ()'(O)= 0, ()(L)= O.) (12))

The column can buckleonly if there is a nontrivial solution of (12);otherwisethecolumn will remain in its undeflectedverticalposition.

The differentialequation in (12)is an Airy equation similar to the one in Ex-ample 2. It has the form of Eq. (3)with A = B = 0,C = y2, and q = 3.Theequations in (5)givea = i,fJ =

\037,k =

\037y,and p =

\037.

Sothe general solution is)

()(x)= x1/2[Cl1l/3 (\037yx3/2) +C21-1/3(\037yx3/2)].) (13))))


In orderto apply the initial conditions,we substitute p = :i::\037

in)

00 (-l)m(

X

)2m+p

Jp(X)=\037m!r(p+m+l)2')

and find after somesimplificationsthat)

c1yl/3(

y2x4 y4x7

)B(x)=31/3r (1)

x-12 + 504

-...c231/3

(1 _ y2x3

y4x6 -...

)+y 1/3r (D 6

+ 180 .)

From this it is clearthat the endpoint condition()'(0)= 0 impliesthat Cl = 0, so)

()(x)= C2x1/2J-1/3 (\037yx3/2) .) (14))

The endpoint condition ()(L)= 0 now gives)

2) J-1/3 (\037yL3/2

) = o.) (15))

o)

Thus the column will buckleonly if z =\037y

L3/2 is a root of the equationJ-I /3(Z) = O.The graph of)

>-.)

(Z/2)-1/3

(00 (-1)m3m

z2m

)Ll/3(Z)= r n)1 +

\037 22m m!.2.5 .(3m- 1))(16))

-1o) 5) 10) 15)

z) (seeProblem 3 of Section3.5)is shown in Fig.3.6.2,where we seethat the smallestpositivezero z 1 is a bit lessthan 2.Mosttechnicalcomputing systemscan find rootslikethis one.For instance,eachof the computer system commands)

FIGURE3.6.2.Thegraph ofJ-1/3(Z).)

fsolve(BesselJ(-1/3,x)=O,x, 1..2)FindRoot[BesselJ[-1/3,x]==O,{x,2}]fzero(lbesselj(-1/3,x)l,2))

(Maple)(Mathematica)(MATLAB))

yield the value ZI = 1.86635(roundedaccurate to five decimalplaces).The shortest length L 1 for which the column will buckleunder its own weight)

IS)

Ll = ( \037\037 Y/3=

[3\0371 (;;Y/2r/

3

.

If we substitute ZI \037 1.86635and p = 8A, where 8 is the volumetric density of thematerial of the column and A is its cross-sectionalarea, we finally get)

(EI

)1/3

Ll \037 (1.986)-g8A)

(17))))


for the critical buckling length. For example,with a steelcolumn or rod for which

E = 2.8X 107 Ibjin.2 and g8 = 0.28Ibjin.3, the formula in (17)gives the resultsshown in the table in Fig.3.6.3.)

H ..,..,...-'__'_\",'_ '._.._. ,,-,._.

Cro.i.'.s.'.s...............Secti()u.....ot........R.....

.

. i>d,___ \"_.H_.H__H.,-, __ _ ____, '_' _',__.'_,)ShortestBucklingLength Ll)

Circular with r = 0.5in.

Circular with r = 1.5in.

Annular with nnner = 1.25in. and router = 1.5in.)

30ft 6 in.

63ft 5 in.

75ft 7 in.)

FIGURE3.6.3.)

We have used the familiar formulas A = Jr r 2 and Idisk.The data in the table show why flagpolesare hollow.)

iJrr 4 for a circular)

ED Problems)In Problems 1through 12,expressthe generalsolution of the

given differential equation in terms ofBesselfunctions.

1.x2y\"

-xy' + (1+ x2)y = 02.xy\" + 3y' + xy = 03.xy\"

-y' + 36x3y = 0

4. x2y\"

-5xy'+ (8+ x)y = 05. 36x2

y\" + 60xy'+ (9x3 -5)y = 06.16x2y\" + 24xy'+ (1+ 144x3)y = 07. x2

y\" + 3xy'+ (1+ x2)y = 08.4x2

y\"- 12xy'+ (15+ 16x)y= 0

9.16x2y\"- (5 - 144x3)y = 0

10.2x2y\" -3xy'-2(14-x5)y = 0

11.y\" + x4y = 0

12.y\" + 4x3y = 0

13.Apply Theorem1 to show that the general solution of)

\" 2 ' 0xy + y + xy =)

is y(x) = X-I (A cosx+ B sin x).14.Verify that the substitutions in (2) in Bessel'sequation

(Eq.(1))yield Eq.(3).15.(a) Show that the substitution)

1 duy =---

u dx)

transforms the Riccati equation dyjdx = x2 + y2 intou\" + x2u = O. (b) Show that the general solution ofdyjdx= x2 + y2 is)

J3/4 (\037X2)-CJ-3/4 (\037X2)y(x) = X (')(')'CJI/4 2X2 + J-I/4 2x2)

(Suggestion:Apply the identities in Eqs.(22)and (23)ofSection3.5.))

16.(a) Substitute the seriesof Problem 11of Section3.5in

the result of Problem 15here to show that the solution ofthe initial value problem

dy

dx= x2 + y2, y(O) = 0)

IS)

J3/4 (\037X2)

y(x) = x(1. 2

).J-I/4 2X

(b) Deducesimilarly that the solution of the initial value

problem)dy-= x2 + y2, y(O) = 1dx)

IS)

2r(*)J3/4 (\037X2)+ r

(\037)J-3/4 (\037X2)

y(x) = x(

3) (

1 2) (

1) (

1 2).2r 4 J-I/4 2X - r 4 J1/4 2X

Somesolution curvesof the equation dyjdx = x2+ y2 areshown in Fig. 3.6.4.Thelocation of the asymptotes where

y(x) \037 +00can be found by using Newton's method tofind the zerosof the denominators in the formulas for the

solutions as listed here.)

3)

2)

1)

\037 0)

-1)

-2)

-3-3 -2 -1 0) 2 3)

x)

dyFIGURE3.6.4.Solution curvesof-=x2 + y2.dx)))


17.Figure 3.6.5shows a linearly tapered rod with circularcrosssection,subject to an axial forceP of compression.As in Section2.8,its deflectioncurve y == y (x) satisfiesthe endpoint value problem)

Ely\" + Py == 0; yea) == y(b) == O. (18))

Note that if a == b, this result reducesto Eq.(28)of Sec-tion 2.8.

18.Considera variable-length pendulum as indicated in

Fig. 3.6.6.Assume that its length is increasing linearlywith time, L(t) == a + bt. It can be shown that the oscil-lations of this pendulum satisfy the differential equation)

y)

LO\" + 2L'O'+ gO == 0)

under the usual condition that 0 is so small that sin 0is very well approximated by 0:0 \037 sin O. SubstituteL == a + bt to derive the general solution)

x)

x=a) x=b)B(t) = .Jr[Al'(\037JgL)

+ BY,(\037JgL)]

.)

FIGURE3.6.5.Thetaperedrod of Problem 17.Here,however, the moment of inertia 1 == 1(x) of thecrosssectionat x is given by)

For the application of this solution to a discussionof the

steadily descendingpendulum (\"its nether extremity wasformed of a crescentof glittering steel,about a foot in

length from horn to horn; the horns upward, and the under

edgeas keen as that of a razor...and the whole hissedas it swung through the air ...down and still down it

came\") ofEdgar Allan Poe'smacabreclassic\"The Pit andthe Pendulum,\" seethe article by Borrelli, Coleman,andHobsonin the March 1985issueof Mathematics Maga-zine (Vol. 58,pp.78-83).)

lex)=\037n(kx)4

= 10'Gf '

where 10== I(b),the value of 1at x == b. Substitution of1(x) in the differential equation in (18)yields the eigen-value problem)

x4y\" + AY == 0, yea) == y(b) == 0,

where A == J12 == Pb4jE10. (a) Apply the theo-rem of this sectionto show that the general solution ofx4

y\" + J12y == 0 is)

n 2n 2

(a

)2

Pn ==

----v- bE 10.)

y(x) =x(ACOS\037

+ Bsin\037 ).

(b) Concludethat the nth eigenvalue is given by J1n

nnabjL, where L == b - a is the length of the rod, andhencethat the nth buckling forceis)

FIGURE3.6.6.A variable-length pendulum.)

3.6Application)...... \037ic\037ati_ Eq.\037ati\037_\037,\037,,_\037n\037..., \037,.\037.\037i\037ie\037 B\037ss\037__\" \037unc:.\037i\037\037s)

A Riccati equation is one of the form)

dy-= A(x)y2 + B(x)y+C(x).dx)

Many Riccati equations likethe oneslistednext can be solvedexplicitly in terms ofBesselfunctions.)

dydx

= x2 + l;) (1))

dy_ = x2 _ y2.dx)

(2))))


dydx

= y2 -X2; (3)

dydx

= x +y2; (4)

dy-= x- y2; (5)dxdy 2-=y -x. (6)dx)

For example,Problem 15in this sectionsaysthat the general solution of Eq.(1)isgiven by)

J3/4(\037x2)-cJ-3/4(\037x2)y(x) = x

(1

) (1

).

CJ1/4 2X2 +J-l/42X2)(7))

Seewhether the symbolicDEsolvercommand in your computeralgebrasys-tem, such as the Maplecommand

dsolve(diff(y(x),x)= x A 2 + y(x)A2, y(x\302\273

or the Mathematica command)

DSolve[y'[x]== x A 2 + y[x]A2, y[x],x ]

agreeswith Eq. (7). If Besselfunctions other than those appearing in Eq. (7)areinvolved, you may need to apply the identities in (26)and (27)of Section3.5totransform the computer's\"answer\" to (7).Then seewhether your systemcan takethe limit as x \037 0 in (7)to show that the arbitrary constant c is given in terms ofthe initial value y (0)by)

y(O)f (i)c= -2f (D

.) (8))

Now you should be able to use built-in Besselfunctions to plot typical solutioncurves likethoseshown in Fig.3.6.4.

Next, investigatesimilarly one of the other equations in (2)through (6).Eachhas a general solution of the samegeneral form in (7)-aquotient of linear com-binations of Besselfunctions. In addition to Jp(x)and Yp(x),thesesolutionsmayinvolve the modifiedBesselfunctions)

Ip(x)= i-P Jp(ix))

and)T(

Kp(x)= 2i-P [Jp(ix)+ Yp(ix)]

that satisfy the modifiedBesselequation

x2y\" +xy' - (x2 +p2)y = 0)

of orderp.For instance, the general solutionof Eq.(5)is given for x >0 by)

I (2 3/2

) I (2 3/2

)_ 1/2 2/3 3x -C -2/3 3xy(x) -X

(2

) (2

),

1-1/33x3/2 -C/l/3 3x3/ 2)(9))))


4)

2)

\037o)

-2)

-4)

-5) o) 5) 10)

x)

FIGURE3.6.7.Solution curvesdyof- == x -y2.dx)

where)

y(o)r (\037)c = - .31/3r

(\037))

(10))

Figure 3.6.7showssometypical solutioncurves, together with the parabola y2 = xthat appearsto bear an interesting relation to Eq. (6)-weseea funnel near y =+,JXand a spout near y = -,JX.

The Besselfunctions with imaginary argument that appear in the definitionsof Ip (x)and Kp (x)may lookexotic,but the powerseriesof the modified function

In (x)is simply that of the unmodified function In (x)exceptwithout the alternatingminus signs.For instance,)

x2 x4 x6Io(x)= 1 +-+-+ +...4 64 2304)

and)X x3 x5 X 7

II(x)=2

+ 16+ 384+ 18432+... .

Checkthesepowerseriesexpansionsusing your computeralgebra system-lookat

Bessellin either Mapleor Mathematica-andcomparethem with Eqs.(17)and(18)in Section3.5.

The second-orderdifferential equations of the form y\" = f (x,y) with thesameright-hand sidesas in Eqs. (1)through (6)have interesting solutions which,

however, cannot be expressedin terms of elementary functions and/or \"known\"

specialfunctions such as Besselfunctions. Nevertheless,they can be investigatedusing an ODEplotter. For instance, the interestingpattern in Fig.3.6.8showssolu-tion curves of the second-orderequation)

y\" = y2 _ X) (11))

with the sameinitial value y(O) = 0 but different slopesy/(O)= -3.3,-3.1,...,0.7.Equation(11)isa form ofthejirstPainlevetranscendant, an equation that arosehistorically in the classificationof nonlinear second-orderdifferentialequations in

terms of their critical points (seeChapter 14of E.L.Ince,Ordinary DifferentialEquations,New York: Dover Publications, 1956).Figure 3.6.8was suggestedbyan article by Anne Noonburgcontaininga similar figure in the Spring1993issueofthe C.ODE.E Newsletter.)

--------2 ---

y2 =x----,.\",.\";

0\037

-2

-4

0 2 4 6 8 10 12x)

FIGURE 3.6.8.Thefirst Painlevetranscendant y\" == y2 -x,y(O) == 0, y'(O)== -3.3,-3.1,...,0.7.)))


Finally, here'sa related examplethat was inspiredby a Mapledemonstration

package.The Mapledsolvecommandyieldsthe general solution)

y(x) = x-I(CIJIO(X)+C2Y IO(X))

+x-II(1857945600+51609600x2 +806400x4 +9600x6 + 100x8 +x IO) (12))

of the nonhomogeneoussecond-orderequation)

x2y\" +3xy'+ (x2-99)y = x.) (13))

Showthat Theorem 1 in this sectionexplainsthe \"Besselpart\" of the allegedsolu-tion in Eq.(12).Can you explain where the rational function part comesfrom, orat leastverify it? For further examplesof this sort, you can replacethe coefficient99 in Eq. (13)with r 2 - 1,where r is an even integer, and/or replacethe x on the

right-hand sidewith xS, where s is an odd integer. (With parities other than these,more exoticspecialfunctions are involved.))))

11II)

LaplaceTransforll1Methods)

LaplaceTransforD1sand InverseTransforD1s)

f(t),)

D{fU)} = f '(t))

f(t),)

\037{f(t)} = F(s)

FIGURE4.1.1.Transformationof a function: \302\243 in analogy withD.)

266)

In Chapter 2 we saw that linear differential equations with constant coefficientshave numerous applicationsand can be solved systematically.There are common

situations, however,in which the alternative methods of this chapter are preferable.For example,recallthe differentialequations)

mx\" +ex'+kx = F(t))1

and LI\"+RI'+-I= E'(t)C)

correspondingto a mass-spring-dashpotsystem and a seriesRLC circuit, respec-tively. It often happens in practice that the forcing term, F(t) or E'(t),hasdiscontinuities-forexample,when the voltage suppliedto an electricalcircuit isturned off and on periodically.In this casethe methods of Chapter 2 can be quiteawkward, and the Laplacetransform method is more convenient.

The differentiationoperator D can beviewedasa transformation which, when

appliedto the function f(t),yieldsthe new function D{f(t)}= f'(t).The Laplacetransformation \302\243 involves the operation of integration and yieldsthe new function

\302\243{f(t)}= F(s)of a new independent variables.The situation is diagrammed in

Fig.4.1.1.After learning in this sectionhow to compute the Laplacetransform F(s)of a function f(t),we will seein Section4.2that the Laplacetransform convertsa differential equation in the unknown function f(t) into an algebraicequation in

F(s). Becausealgebraicequations are generally easierto solve than differential

equations, this is onemethod that simplifies the problemof finding the solutionf (t).)))

Example1)

Example2)

4.1LaplaceTransforms and InverseTransforms 267)

DEFINITION TheLaplaceTransformGiven a function f(t) defined for all t > 0,the Laplacetransform of f is thefunction F definedas follows:)

>-) F(s)= \302\243{f(t)}=100

e-st f(t) dt) (1))

for all values of s for which the improper integral converges.)

Recall that an improperintegralover an infinite interval is definedas a limit

of integralsover boundedintervals; that is,)

100

g(t)dt = lim lb

g(t)dt.a b\037oo

a)

(2))

If the limit in (2)exists,then we say that the improperintegral converges;otherwise,it divergesor fails to exist.Note that the integrand of the improper integral in (1)contains the parameters in addition to the variableof integration t.Therefore,when

the integral in (1)converges,it convergesnot merely to a number, but to afunctionF of s.As in the following examples,it is typical for the improper integral in thedefinition of \302\243 {f(t)}to convergefor somevaluesof s and divergefor others.)

With f (t) = 1 for t >0,the definition of the Laplacetransform in (1)gives)

\302\243{l}= roo e-st dt =

[_\037e-st ]OO

= lim [_\037e-bS + \037

] ,Jo s 0 b\037oo S S)

and therefore)

1\302\243{1}=- for s>O.

s)(3))

As in (3),it'sgoodpracticeto specify the domain of the Laplacetransform-inproblemsas well as in examples.Also, in this computationwe have usedthe com-mon abbreviation)

[g(t)]OO

= lim [g(t)]b.a b\037oo a)

(4)

.)

Remark:The limit we computed in Example1 would not existif s < 0,for then (l/s)e-bS would becomeunboundedas b \037 +00.Hence

\302\243{1}is defined

only for s > O.This is typical of Laplacetransforms; the domain of a transform isnormally of the form s >a for somenumber a. .)

.......\037

With f(t) = eat for t >0,we obtain)

100

100

[-(s-a)t

]00

\302\243{e

at} = e-steat dt = e-(s-a)tdt = _ e .

o 0 s -a t =0)))

268 Chapter4 LaplaceTransform Methods)

Example3)

If s -a >0,then e-(s-a)t\037 0 as t \037 +00,so it follows that)

\037)\302\243{e

at} =)

1)

(5))for s > a.)s-a)Note here that the improper integral giving \302\243 {eat}diverges if s < a. It is worth

noting also that the formula in (5)holdsif a is a complexnumber. For then, with

a = a + if3,) -(s-a)t i{3t -(s-a)t 0e =e e \037)

as t \037 +00,provided that s >a = Re[a];recall that ei{3t = cosf3t + i sinf3t. .The Laplacetransform

\302\243{t

a} of a powerfunction is most convenientlyex-

pressedin terms of the gammafunction rex),which is defined for x > 0 by the

formula)

\037) r(x) = 100

e-ttx-1 dt.) (6))

For an elementarydiscussionof r (x),seethe subsectionon the gamma function in

Section3.5,where it is shown that)

r(l)= 1) (7))

and that)

rex+ 1)= xr(x)) (8))

for x >O.It then follows that if n is a positive integer, then)

r(n + 1)= nr(n)= n .(n - l)r(n- 1)= n .(n - 1).(n -2)r(n- 2))

= n(n - 1)(n-2)...2.r(2)= n(n - 1)(n-2)...2.1 .r(I);)

thus)

\037) r(n + 1)= n!) (9))

if n is a positive integer. Therefore, the function rex + 1),which is defined andcontinuous for all x > -1,agreeswith the factorial function for x = n, a positiveinteger.)

\037 ...... ..n .....

Supposethat f (t) = t a where a is real and a >-1.Then)

\302\243{t

a} = 1

00

e-stta dt.)

If we substitute u = st,t = ujs,and dt = dujs in this integral,we get)

\302\243{t

a} = \037 [00e-uua du = f(a+ 1)

sa+l Jo sa+l)(10))))

Example4)


for all s > 0 (sothat u = st > 0). Becauser(n + 1)= n! if n is a nonnegativeinteger,we seethat)

\037) {n

}n!Lt =-

sn+l)(11))for s > o.)

For instance,)

1L{t}= 2's)

2 2L{t } = 3's)

3 6and L{t}= 4.s)

As in Problems1 and 2,theseformulas can be derived immediately from the defi-nition, without the useof the gamma function. .)

LinearityofTransformsIt is not necessaryfor us to proceedmuch further in the computation of Laplacetransforms directly from the definition. Oncewe know the Laplacetransforms ofseveralfunctions,we can combine them to obtain transforms of otherfunctions.Thereasonis that the Laplacetransformation is a linearoperation.)

THEOREM1 LinearityoftheLaplaceTransformIf a and b are constants, then)

\037) L{af(t)+bg(t)}= aL{f(t)}+bL{g(t)}) (12))

for all s such that the Laplacetransforms of the functions f and g both exist.)

The proof of Theorem 1 follows immediatelyfrom the linearity of the opera-tions of taking limits and of integration:)

\302\243{af(t) +bg(t)}=100

e-st [af(t) +bg(t)]dt

= lirn t e-st [af(t) +bg(t)]dtc-+oo10)

= a ( lirn t e-st f(t) dt) +b ( lirn t e-st get)dt )c-+oo10 c-+oo10)

= aL{f(t)}+bL{g(t)}.)

The computationof L{tn/2} is basedon the known specialvalue)

r (\037 ) = 0T) (13))

of the gamma function. For instance, it follows that)

r (5

) = 3 r (3

) = 3 . \037r (\037

) = 30T2222224')))


Example5)

Example6)

using the formula rex+ 1)= xr(x) in (9),first with x =\037

and then with x = 4.Now the formulas in (10)through (12)yield)

2' 4r (\037

) 6

Hs\302\243{3t

2 +4t3/2} = 3 . \037 + 2 = -+3 -.s3 s5/2 s3 s5)

.)

\".., .., '\" .....n....;...'....\"N\" '\" ................\" VN,';,'N.V \037\"\"'....;....;..'N..'........ ,.,,\037 ... ....nnn \"\" \037_....\037 .V',','\",.\" ,....\".,....\".,....\". v'\" ,\037'\" \". ........ ... \037....\" \". \037,nnn.\" \037\037,....\". ....NU.N....\",....\".,....\". \".,,\037 N\037\"\"\"\" '\" ,..,,,'\037\037\"\" ,.., \037....'\" \037 \"',....\". __\037........ ...............,..............,...\"....;\037...............\037\037......,'-\"\"-\"\"'\"

Recall that coshkt = (ekt + e-kt )/2.If k > 0,then Theorem 1 and Example2together give)

1 1 1

(1 1

)\302\243{coshkt} =_\302\243{e

kt}+ _\302\243{e-

kt} = - + ;2 2 2 s-k s+k)

that is,)

s\302\243{coshkt} = 2 2 for s > k >O.

s -k) (14))

Similarly,)

. k\302\243{slnhkt}

= 2 2 for s > k >O.s -k) (15))

Becausecoskt = (eikt +e-ikt )/2,the formula in (5)(with a = ik) yields)

1

(1 1

)1 2s

\302\243 coskt = - = - .{ } 2 s- ik+

s + ik 2 S2- (ik)2')

and thus)

s\302\243{coskt}

= 2 2 fors>O.s +k)

(16))

(The domain follows from s >Re[ik]= 0.)Similarly,)

k\302\243{sinkt}

=2 2 for s >O.

s +k)(17).)

.h ..n.... ...,\037,.'N.\037\" \"'..d.........,... ,,\037 \"\"\"......, ..\"\" ..\"'.'N.\"\"...,....\".,....\".\"\" .n..........\037__,....\". .... ......\037__.\"...N..\"\"..n,....\".,....\".\"\"\"...,... ...,....\".'......'..........,.........\"......,....\".,....\". ...... .. .... \"\"'\037,....\".,........, \"\"...,.,...,...,...\"...,.,,\037...,.,...... ....,..................................\" _,....\".\037 ..........\",.......,,\"\".... \"'\" \"\"....\037.... \".................... ....

Applying linearity, the formula in (16),and a familiar trigonometricidentity, we get)

\302\243{3e

2t +2sin2 3t} =\302\243{3e

2t + 1-cos6t}3 1 s=

s -2+ s

-s2+36

3s3 + 144s-72s(s-2)(s2+ 36))

.)for s > O.)))

Example7)

E{s))

1)

t)

t n (n > 0))

t a (a > -1))

eat)

cosk t)

sin kt)

coshk t)

sinh k t)

u (t -a))

1)

s)

1)

S2)

n!sn+l)

(s > 0))

(s > 0))

(s > 0))

sa+l)

rea+ 1)(s > 0))

1)

s-a)S

S2+ k2)

k

S2+ k2)

SS2- k2)

k

S2- k2)

e-as)

s)

(s > 0))

(s > 0))

(s > 0))

(s > Ikl))

(s > Ikl))

(s > 0))

FIGURE4.1.2.A short table ofLaplacetransforms.)


InverseTransforms)

According to Theorem 3 of this section,no two different functions that are both

continuous for all t > 0 can have the sameLaplacetransform. Thus if F(s) is thetransform of somecontinuousfunction f(t),then f(t) is uniquely determined.Thisobservationallowsus to make the followingdefinition:If F(s) =

\302\243{f(t)}, then wecall f(t) the inverseLaplacetransform of F(s) and write)

\037) f(t) = \302\243-I{F(s)}.) (18))

Usingthe Laplacetransforms derived in Examples2,3, and 5 we seethat)

-I{

I

}1 2

\302\243- =-ts3 2') \302\243

-I{

1

}= e-2t

,s+2) \302\243-1

{ 22

}= 2

sin 3ts +9 3)

and so on.) .)

NOTATION:FUNCTIONSAND THEIRTRANSFORMS. Throughout this chapterwe denotefunctions of t by lowercaseletters.The transform of a function will al-ways bedenotedby that sameletter capitalized.Thus F(s)is the Laplacetransformof f (t) and x(t) is the inverseLaplacetransform of X (s).

A table of Laplacetransforms servesa purposesimilar to that of a table ofintegrals.The table in Fig.4.1.2lists the transforms derived in this section;manyadditional transforms can be derived from thesefew, using various generalproper-tiesof the Laplacetransformation (which we will discussin subsequentsections).)

PiecewiseContinuousFunctions)As we remarkedat the beginning of this section,we needto beableto handlecertaintypes of discontinuousfunctions. The function f (t) is saidto bepiecewisecontin-uous on the boundedinterval a < t <b provided that [a,b] can be subdividedinto

finitely many abutting subintervalsin such a way that)

1.f is continuous in the interior of eachof thesesubintervals;and2.f (t) has a finite limit as t approacheseach endpointof each subinterval from

its interior.)

We say that f is piecewisecontinuous for t > 0 if it is piecewisecontinuous onevery boundedsubinterval of [0,+(0).Thus a piecewisecontinuous function hasonly simplediscontinuities (if any) and only at isolatedpoints.At such points thevalue of the function experiencesa finite jump, as indicated in Fig.4.1.3.Thejumpin f (t) at the pointe is defined to be f (e+)- f(e-), where)

f(e+)= lim f(e+E) and f(e-)= lim f(e-E).E-+O+ E-+O+)

Perhaps the simplestpiecewisecontinuous (but discontinuous)function is theunit step function, whosegraph appearsin Fig.4.1.4.It is definedas follows:)

10 fort <0,

u(t) =1 for t >o.)

(19))))

y)

.........--).)

v) (0, 1))

u(t)) Ua(t) = u(t -a)(a, 1).)

a) b x) t = a)

FIGURE4.1.3.Thegraph ofapiecewisecontinuous function; thesoliddots indicate values of thefunction at discontinuities.)

FIGURE4.1.4.Thegraph of theunit step function.)

FIGURE4.1.5.Theunit stepfunction U a (t) has a jump at t = a.)

Becauseu(t) = 1 for t > 0 and becausethe Laplacetransform involvesonly thevalues of a function for t >0,we seeimmediately that)

1\302\243{u(t)}

= - (s >0).s)

(20))

The graph of the unit stepfunction ua(t) = u(t -a) appearsin Fig.4.1.5.Its jumpoccursat t = a rather than at t = 0;equivalently,)

10 for t <a,

ua(t) = u(t -a) =1 fort >a.)

(21))

Example8) Find\302\243{u a(t)}if a >O.)

Solution We beginwith the definition of the Laplacetransform. We obtain)

100

100

[-st

]b

\302\243{Ua(t)}= e-stua(t)dt= e-st dt = lim -=-- ;

o a b\037oo S t=a)

consequently,) e-as\302\243{Ua(t)}

=- (S>0, a > 0).S)

(22).)

GeneralPropertiesofTransformsIt is a familiar fact from calculusthat the integral)

lb

g(t)dt)

existsif g is piecewisecontinuous on the boundedinterval [a,b].Henceif J ispiecewisecontinuous for t >0,it follows that the integral)

lb

e-st !(t)dt)

existsfor all b < +00.But in orderfor F(s)-thelimit of this last integral asb \037 +oo-toexist,we needsomecondition to limit the rate of growth of J(t)as)))

t \037 +00.The function Iis saidto be of exponentialorderas t \037 +00if thereexistnonnegativeconstants M, c,and T such that)

I/(t)!<Mect for t > T.) (23))

Thus a function is of exponentialorderprovided that it grows no more rapidly (ast \037 +(0)than a constant multiple of someexponential function with a linearexponent. The particular values of M, c, and T are not so important. What isimportant is that somesuch valuesexistso that the condition in (23)is satisfied.

The condition in (23)merely saysthat I(t)/ect liesbetween -M and M andis therefore boundedin value for t sufficiently large.In particular, this is true (withc = 0) if I(t)itself is bounded.Thus every boundedfunction-suchas cosktorsin kt-isof exponentialorder.

If pet) is a polynomial, then the familiar fact that p(t)e-t \037 0 as t --++00impliesthat (23)holds (for T sufficiently large) with M = c = 1.Thus everypolynomial function is of exponentialorder.

For an exampleof an elementary function that is continuous and thereforeboundedon every (finite) interval, but neverthelessis not of exponentialorder,con-siderthe function I(t)= et2 = exp(t2). Whatever the valueof c,we seethat)

2. I(t) . et . 2_hm -= hm -= hm et ct = +00

t-+oo ect t-+oo ect t-+oo)

becauset 2 - ct \037 +00as t \037 +00.Hencethe condition in (23)cannot holdfor any (finite) value M, so we concludethat the function I(t) = et2 is not ofexponentialorder.

Similarly, becausee-st et2\037 +00as t \037 +00,we seethat the improperinte-

gral 1000 e-stet2 dt that would defineL{et2 }doesnot exist(for any s),and therefore

that the function et2 doesnot havea Laplacetransform. The followingtheoremguar-antees that piecewisefunctions of exponentialorderdohave Laplacetransforms.)

THEOREM2 ExistenceofLaplaceTransformsIf the function Iispiecewisecontinuousfor t >0 and is of exponentialorderast \037 +00,then its Laplacetransform F(s)= L{/(t)}exists.Moreprecisely,ifIispiecewisecontinuousand satisfiesthe condition in (23),then F(s)existsforall s > c.)

Proof:First we note that we can take T = 0 in (23). For by piecewisecontinuity, !I(t)!is boundedon [0,T].IncreasingM in (23)if necessary,we cantherefore assumethat

II(t) I< M if 0 < t < T.Becauseect > 1 for t > 0, it then

follows thatII(t) I

<Mect for all t >o.By a standardtheoremon convergenceof improperintegrals-thefact that ab-

solute convergenceimpliesconvergence-itsufficesfor us to prove that the integral)

100

le-stf (t)Idt)

existsfor s >c.To do this, it sufficesin turn to show that the value of the integral)

lb

le-stf(t)1dt)))

remains boundedasb \037 +00.But the fact thatI J (t) I

<Mect for all t >0 impliesthat)

lb

le-stf(t)1dt <lb

le-stMectIdt = Mlb

e-(s-c)tdt)

100 M<M e-(s-c)tdt =

o s-c)if s >c.This proves Theorem 2.) .)

We have shown, moreover,that)

100 M

IF(s)1< le-stJ(t)1dt <o s-c

if s >c.When we take limits ass \037 +00,we get the following result.)

(24))

COROLLARY F(s)fors LargeIf J(t) satisfiesthe hypothesesof Theorem2,then)

lim F(s)= O.s\037oo)

(25))

The condition in (25)severely limits the functions that can be Laplacetrans-forms. For instance,the function G(s)= s/ (s+1)cannotbe the Laplacetransformof any \"reasonable\"function becauseits limit as s \037 +00is 1,not O. Moregen-erally, a rational function-aquotient of two polynomials-canbe (and is, as weshall see)a Laplacetransform only if the degreeof its numerator is lessthan that ofits denominator.

On the other hand, the hypothesesof Theorem 2 are sufficient, but not neces-sary, conditions for existenceof the Laplacetransform of J(t).For example,thefunction J(t)= 1/,J!fails to be piecewisecontinuous (at t = 0),but nevertheless(Example3 with a = -!> -1)its Laplacetransform

\302\243{t-1/2} = r (1)= (if

s1/2 Y --;

both existsand violates the condition in (24),which would imply that sF(s)remainsboundedass \037 +00.

The remainder of this chapter is devoted largely to techniques for solving a

differenti\037l equation by first finding the Laplacetransform of its solution. It is then

vital for us to know that this uniquely determines the solution of the differential

equation; that is, that the function of s we have found has only one inverseLaplacetransform that couldbe the desiredsolution. The following theorem is proved in

Chapter 6 of Churchill'sOperationalMathematics, 3rd ed.(New York: McGraw-Hill,1972).)

THEOREM3 Uniquenessof InverseLaplaceTransforms

Supposethat the functions J(t)and get) satisfy the hypothesesof Theorem 2,so that their Laplacetransforms F(s)and G(s)both exist.If F(s)= G(s)forall s > c (for somec),then J(t)= get)whereveron [0,+(0)both J and g arecontinuous.)))


Thus two piecewisecontinuous functions of exponentialorder with the sameLaplacetransform can differ only at their isolatedpoints of discontinuity. This isof no importance in most practical applications,so we may regard inverseLaplacetransforms as beingessentiallyunique. In particular, two solutionsof a differential

equation must both becontinuous,and hencemust be the same solution if they havethe sameLaplacetransform.

HistoricalRemark: Laplacetransforms have an interesting history.The integral in the definition of the Laplacetransform probably appeared first in

the work of Euler.It is customary in mathematics to name a techniqueor theoremfor the next personafter Eulerto discoverit (elsethere would be severalhundred

different examplesof \"Euler'stheorem\.") In this case,the next personwas the

French mathematician Pierre Simonde Laplace(1749-1827),who employedsuch

integrals in his work on probability theory. The so-calledoperationalmethodsfor

solving differentialequations,which are basedon Laplacetransforms, werenot ex-ploitedby Laplace.Indeed,they werediscoveredand popularized by practicingengineers-notablythe EnglishelectricalengineerOliver Heaviside(1850-1925).Thesetechniques were successfullyand widely appliedbefore they had been rig-orously justified, and around the beginning of the twentieth century their validitywas the subjectof considerablecontroversy. Onereason is that Heavisideblithelyassumedthe existenceof functions whoseLaplacetransforms contradictthe condi-tion that F(s) \037 0 ass \037 0,therebyraisingquestionsas to the meaning and nature

of functions in mathematics. (This is reminiscentof the way Leibniz two centuriesearlierhad obtainedcorrectresultsin calculususing \"infinitely small\" real numbers,

thereby raising questionsas to the nature and roleof numbers in mathematics.))

11II\"\"Problems)Apply the definition in (1)to find directly the Laplacetrans-

Jorms oJthe Junctions described(by Jormula orgraph) in Prob-lems 1through 10.)

9.)(1,1))

1.J(t) = t 2.J(t) = t 2 t

3.J(t) = e3t+1 4. J(t) = cost FIGURE4.1.8.5. J(t) = sinht 6.J(t) = sin 2 t 10.

(0, 1))

7.)

(1,1))(1,0)

FIGURE4.1.9.)

t)

t) Usethe transJorms in Fig. 4.1.2to find the LaplacetransJorms

ojthe Junctions in Problems11through 22.A preliminary in-

tegration by parts may benecessary.)FIGURE4.1.6.)

t)

11.J(t) =0+ 3t

13.J(t) = t -2e3t

15.J(t) = 1+ cosh5t

17.J(t) = cos2 2t

19.J(t) = (1+ t)3

21.J(t) = t cos2t)

12.J(t) = 3t5/2 -4t3

14.J(t) = t 3/2 -e-lOt

16.J(t) = sin 2t + cos2t

18.J(t) = sin 3t cos3t

20. J(t) = tet

22.J(t) = sinh 23t)

8.)(1,1)

o)(2,1).)

c)

FIGURE4.1.7.)))

Usethe transforms in Fig. 4.1.2to find the inverse Laplacetransforms of the functions in Problems23 through 32.)

323.F(s)= 4s1 225.F(s)= --

5/2S S)

24. F(s)= S-3/2)

126. F(s)= s+5

3s+ 128. F(s)= 2s +4

9+s30.F(s)= 24-s)

327.F(s)= s-45 -3s29.F(s)= S2+ 910s- 331.F(s)= 225-s) 32.F(s)= 2s-1e-3s)

33.Derivethe transform of f (t) = sin kt by the method usedin the text to derive the formula in (16).

34. Derivethe transform off(t) = sinh kt by the method usedin the text to derive the formula in (14).

35.Usethe tabulated integral)

feaxeaxcosbxdx=2 2 (acosbx+bsinbx)+Ca +b)

to obtain \302\243 {coskt} directly from the definition of the

Laplacetransform.

36.Show that the function f(t) = sin(et2 ) is of exponentialorderas t \037 +00but that its derivative is not.)

37.Given a > 0, let f(t) = 1 if 0 < t < a, f(t) = 0 if

t > a. First, sketch the graph of the function f,makingclearits value at t = a. Then expressf in terms of unit

step functions to show that \302\243{f(t)}= s-l(1-e-as).

38.Given that 0 < a < b, let f (t) = 1 if a < t < b,f(t) = 0 if either t < a or t > b. First, sketch the graphof the function f, making clear its values at t = a andt = b. Then expressf in terms of unit step functions to

show that \302\243{f(t)}= s-l(e-as -e-bs

).)

39.Theunit staircasefunction is defined as follows:)

f(t) = n if n - 1 < t < n, n = 1,2,3,....)

(a)Sketch the graph off to seewhy its name is appropri-ate.(b) Show that)

00

f(t) = Lu(t -n)n=O)

for all t > o. (c) Assume that the Laplacetransform ofthe infinite seriesin part (b)can be taken termwise (it can).)

Apply the geometricseriesto obtain the result)

1\302\243{f(t)}

= .s(l-e-S)40. (a) Thegraph of the function f is shown in Fig. 4.1.10.

Show that f can be written in the form)

00

f(t) = L(-I)nu (t -n).n=O)

(b) Usethe method ofProblem39 to show that)

1\302\243{f(t)}

= s(l+ e-S).)

IL e---o e---o ...-.... 0 . 01 2 3 4 5 6 t)

FIGURE4.1.10.Thegraph of the function ofProblem40.)

41.The graph of the square-wavefunction get) is shown in

Fig.4.1.11.Expressg in terms of the function f ofProb-lem 40 and hencededucethat)

1-e-s 1 s\302\243{g(t)}

=s(1+ e-S)

=\037

tanh2.)

IL) e---o)e---o) ...-...)

1) 2 345 6) t)

-1 e---o e---o e---o)

FIGURE4.1.11.Thegraph of the function ofProblem41.)

42. Given constants a and b, defineh (t) for t > 0 by)

{a if n - 1 < t < nand n is odd;

h(t) = - .b if n - 1 < t < nand n IS even.)

Sketchthe graph ofh and apply oneof the precedingprob-lems to show that)

a + be-s\302\243{h(t)}

= .s(1+ e-S))))

4.2Transformationof Initial Value Problems 277)

III!JTransformationof InitialValueProblems)

y)

Continuous function)

a I b xI

II

II

I

II

I

I II

y' I II

\037 I I

I I

II

I I

I

I \037: I

I .I I

I

a I xI

I

II

\"-J)Piecewise continuous derivative)

FIGURE4.2.1.Thediscontinuities off' correspondto\"comers\" on the graph of f.)

We now discussthe applicationof Laplacetransforms to solve a linear differential

equation with constant coefficients,such as)

axil(t) +bx'(t)+cx(t) ==f (t),) (1))

with given initial conditions x(0) == Xo and x'(0) == xb. By the linearity of the

Laplacetransformation, we can transform Eq. (1)by separately taking the Laplacetransform of each term in the equation. The transformedequation is)

a\302\243{x\" (t)}+ b\302\243{x

'(t)}+ c\302\243{x(t)}==

\302\243{f(t)};) (2))

it involvesthe transforms of the derivativesx' and x\" of the unknown function x(t).The key to the method is Theorem 1,which tells us how to expressthe transform ofthe derivativeof a function in terms of the transform of the function itself.)

THEOREM1 Transformsof Derivatives)

Supposethat the function f (t) is continuousand piecewisesmoothfor t > 0 andispf exponentialorderas t --++00,so that there existnonnegativeconstantsM,c,and T suchthat)

If(t)1<Mect for t > T.) (3))

Then \302\243 {fl(t)}existsfor s >c,and)

...) L{f'(t)}==s\302\243{f(t)}

- f(O)==sF(s)- f(O).) (4))

The function f is calledpiecewisesmoothon the bounded interval [a,b] if it

is piecewisecontinuous on [a,b] and differentiableexceptat finitely many points,with f' (t) beingpiecewisecontinuous on [a,b].We may assignarbitrary valuesto f (t) at the isolatedpoints at which f is not differentiable. We say that f ispiecewisesmooth for t >0 if it is piecewisesmooth on everybounded subinterval

of [0,+(0).Figure 4.2.1indicates how \"comers\"on the graph of f correspondtodiscontinuities in its derivativef'.

The main ideaof the proof of Theorem 1 isexhibitedbestby the casein which

f'(t) is continuous (not merely piecewisecontinuous)for t > O. Then, beginningwith the definition of \302\243{f' (t)}and integrating by parts, we get)

\302\243 {f'(t)}= [00e-sfI'(t)dt = [e-st f (t )]

00+s [00e-sfI(t) dt .

10 t=O 10)

Becauseof (3),the integrated term e-st f(t) approacheszero (when s >c)as t --++00,and its value at the lower limit t == 0 contributes-f (0)to the evaluation ofthe precedingexpression.The integral that remains is simply \302\243{f(t)}; by Theorem2 of Section4.1,the integral convergeswhen s > c.Then \302\243{f'(t)} existswhen

s > c,and its value is that given in Eq. (4).We will defer the casein which f'(t)has isolateddiscontinuitiesto the end of this section.)))


SolutionofInitialValueProblems)In order to transform Eq. (1),we need the transform of the secondderivative aswell.If we assumethat get) = f'(t)satisfies the hypothesesof Theorem 1,then

that theorem impliesthat)

\302\243{f\"(t)}=

\302\243{g'(t)}=

s\302\243{g(t)}- g(O)

= S\302\243{f'(t)}- f'CO)

= s [s\302\243{f(t)}- f(O)]- f'CO),)

and thus)

\037)\302\243{f\" (t)}= s2F(s)-sf(O)- f'(0).) (5))

A repetition of this calculation gives)

\302\243{f\"l(t)}=

S\302\243{f\"(t)}- f\"(O)= s3F(s)-s2f(O)-Sf'(O)- f\"(O). (6))

After finitely many such stepswe obtain the followingextensionof Theorem 1.)

COROLLARY Transformsof HigherDerivatives

Supposethat the functions f, f', f\", ...,fCn-I) are continuous and piecewisesmooth for t >0,and that eachof thesefunctions satisfiesthe conditions in (3)with the samevalues of M and c.Then \302\243{f(n)(t)} existswhen s >c,and)

\037)

\302\243{fCn)(t)}= sn \302\243{f(t)}

-sn-If(O)-sn-2f'CO)-...-f Cn-l)(o)= snF(s)-sn-If(O)-...-sfCn-2)(0)-f Cn-l)(o). (7))

Example1)

....Solvethe initial value problem)

II I 6 0x-x-x=;)x(O)= 2, x'(O)= -1.)Solution With the given initial values,Eqs.(4)and (5)yield)

\302\243{X'(t)}=

s\302\243{x(t)}-x(O)= sX(s)- 2)

and)

\302\243{X\"(t)}= s2\302\243{x(t)}

-sx(O)-x'(O)= s2X(s)-2s + 1,where (accordingto our conventionabout notation) X (s)denotesthe Laplacetrans-form of the (unknown) function x(t).Hencethe transformedequation is)

[S2X(S)-2s + 1]- [sX(s)-2] -6 [Xes)]= 0,)

which we quickly simplify to)

(S2-S-6)X(s)-2s + 3 = O.)

Thus)2s- 3

Xes)= 2 6s -s-)2s-3

(s-3)(s+2))))

4.2Transformationof Initial Value Problems279)

By the method of partial fractions (of integral calculus),there existconstantsA and

B such that)

2s- 3 A B= + ,(s-3)(s+2)s-3 s+2and multiplication of both sidesof this equationby (s-3)(s+2) yieldsthe identity)

2s- 3 = A(s+2)+ B(s- 3).)

If we substitute sB =

\037.Hence)

3, we find that A =\037;

substitution of s) -2 shows that)

3 7- -X(s)=

\302\243{x(t)}= 5 + 5 .s-3 s+2

Because\302\243-l{l/(s-a)}= eat, it follows that)

x(t) =\037e3t

+\037e-2t)

is the solution of the original initial value problem.Note that we did not first find

the general solution of the differential equation. The Laplacetransform method

directly yieldsthe desiredparticular solution, automatically taking into account-via Theorem 1 and its corollary-thegiven initial conditions. .)

Remark:In Example1 we found the values of the partial-fractioncoeffi-cientsA and B by the \"trick\" of separately substituting the roots s = 3 and s = -2of the original denominators2- s -6 = (s-3)(s+2) into the equation)

2s- 3 = A(s+2)+ B(s- 3))

that resultedfrom clearing fractions. In lieu of any such shortcut, the \"sure-fire\"

method is to collectcoefficientsof powersof s on the right-hand side,)

2s- 3 = (A + B)s+ (2A - 3).)

Then upon equatingcoefficientsof terms of likedegree,we get the linearequations)

A + B = 2,2A - 3B = -3)

which are readily solved for the samevalues A =\037

and B =\037.)

.)

Example2)h \".'\037N\037'h... _\"'''\"'''''''\",,, \"'...,.... ..-,,,....'''' ...\" ..., '\" '.......'N..\"\"\"... ....,..._\"'.... ... ....... \"'... '\"

Solvethe initial valueproblem)

1/ 4 . 3x + x = SIn t;) x(O)= x'(0)= O.)

Such a problemarises in the motion of a mass-and-springsystem with externalforce, as shown in Fig.4.2.2.)))


Solution Becauseboth initial values are zero,Eq. (5)yields \302\243{x\" (t)}= s2X (s). We readthe transform of sin 3t from the table in Fig.4.1.2(Section4.1)and thereby get the

J(t) = sin 3t transformed equation)

FIGURE4.2.2.A mass-and-spring system satisfying the initialvalue problem in Example 2.Themassis initially at rest in its

equilibrium position.)

x)

12)

1-2)

FIGURE4.2.3.Thepositionfunction x(t) in Example 2.)

3s2X(s)+4X(s)= 2 .

s +9)

Therefore,)3

X(s)--(s2+4)(s2+9)

.)

The method of partial fractions callsfor)

3

(s2+4)(s2+ 9))

As + B Cs+ Ds2+4 + s2+9

.)

The sure-fireapproach would be to clearfractions by multiplying both sidesby thecommon denominator,and then collectcoefficientsof powersof s on the right-handside.Equating coefficientsof likepowerson the two sidesof the resulting equationwould then yield four linear equations that we couldsolve for A, B,C,and D.

However,here we can anticipate that A = C = 0,becauseneither the numer-ator nor the denominatoron the left involvesany oddpowersof s,whereasnonzerovalues for A or C would leadto odd-degreeterms on the right. Sowe replaceA andC with zero before clearing fractions. The result is the identity)

3 = B(s2+9)+ D(s2+4) = (B+ D)s2+ (9B+4D).)

When we equate coefficientsof likepowersof s we get the linear equations)

B + D = 0,9B+4D=3,)

which are readily solved for B =\037

and D =-\037.

Hence)

3 2 1 3X (s)=

\302\243{x(t)}= 10

\302\267

s2+4-

5.

S2+9.)

Because\302\243{sin 2t} = 2/(s2+4) and \302\243{sin 3t} = 3/(s2+9),it follows that)

x(t) =to sin 2t -

\037

sin 3t .)

Figure 4.2.3showsthe graph of this period2n position function of the mass.Notethat the Laplacetransform method again gives the solution directly, without the

necessityof first finding the complementaryfunction and a particularsolutionof the

original nonhomogeneousdifferential equation. Thus nonhomogeneousequationsare solved in exactly the samemanner as are homogeneousequations. .)

Examples1 and 2 illustrate the solutionprocedurethat isoutlined in Fig.4.2.4.)))


Differential

equationin x(t))

Solution x(t)of differential

equation)

Algebraic

equationin Xes))

Solution Xes)ofalgebraic

equation)

FIGURE4.2.4.Using the Laplacetransform to solvean initialvalue problem.)

LinearSystemsLaplacetransforms are frequently used in engineeringproblemsto solve a systemof two or more constant-coefficientlinear differential equations involving two ormore unknown functionsx(t),yet), ...of the independentvariablet.When initial

conditions are specified,the Laplacetransform reducessuch a linear systemof dif-ferential equations to a linear systemof algebraicequations in which the unknowns

are the transforms of the solutionfunctions. As Example3 illustrates,the techniquefor a system is essentiallythe sameas for a singlelinear differential equation with

constant coefficients.)

Example3) Solvethe system)

2x\" ==-6x+2y,y\" ==2x-2y +40sin 3t,)

(8))

subjectto the initial conditions)

x(0)==x'(0)== y (0)==y'(0)==O.) (9))

This initial valueproblemdeterminesthe indicateddisplacementfunctions x(t) and

yet) of the two massesshown in Fig.4.2.5,assuming that the forceJ(t)==40 sin 3tis suddenly appliedto the secondmassat the time t == 0 when both massesare at

rest in their equilibrium positions.)

J(t) =40sin 3t)

FIGURE4.2.5.A mass-and-springsystem satisfying the initialvalue problem in Example 3.Both massesare initially at rest in their

equilibrium positions.)

Solution We write Xes) ==\302\243{x(t)} and yes) ==

\302\243{y(t)}. Then the initial conditionsin (9)imply that)

\302\243 {X//

(t)}==S2X (s) and \302\243 {y//

(t )}==S2Y (s) .)))


y(t))

10)

-10)

FIGURE4.2.6.Thepositionfunctions x(t)and yet) in

Example 3.)

Because\302\243{sin3t}= 3/(s2+ 9), the transforms of the equations in (8)are the

equations)

2s2X(s)= -6X(s)+2Y(s),120

s2y(s)= 2X(s)-2Y(s)+ 2 .s +9)

Thus the transformedsystemis)

(s2+3)X(s)) -Y(s)= 0,)

120-2X(s)+ (s2+2)Y(s)= 2 .s +9)

(10))

The determinant of this pair of linear equations in X (s)and Y (s) is)

s2_i3

s2-12= (s2+3)(s2+2)-2 = (S2+ 1)(s2+4),)

and we readily solve-usingCramer'srule, for instance-thesystem in (10)for)

120 5 8 3X(s)=

(s2+1)(s2+4)(s2+9)= s2+1

-s2+4+ s2+9 (lla))

and)

120(s2+3) 10 8 18Y(s)=

(s2+1)(s2+4)(s2+9)= s2+1+ s2+4

-s2+9. (lIb))

The partial fraction decompositionsin Eqs.(1Ia)and (lIb)are readily found usingthe method of Example2. For instance, noting that the denominator factors arelinear in s2,we can write)

120 ABC= + + ,(s2+1)(s2+4)(s2+9)s2+1 s2+4 s2+9)and it follows that)

120= A(s2 +4)(s2+9) + B(s2+ 1)(s2+9)+C(s2+ 1)(s2+4). (12))

Substitution of s2 = -1(that is, s = i,a zero of the factor s2+ 1)in Eq.(12)gives120= A .3 .8, so A = 5. Similarly, substitution of s2 = -4in Eq. (12)yieldsB = -8,and substitution of s2 = -9yieldsC = 3.Thus we obtain the partialfraction decompositionshown in Eq.(1Ia).

At any rate, the inverse Laplacetransforms of the expressionsin Eqs.(11a)and (11b) give the solution)

x(t) = 5 sin t -4 sin 2t + sin 3t,

y (t) = 10sin t +4 sin 2t -6 sin 3t .)

Figure 4.2.6showsthe graphs of thesetwo period2n position functions of the two

masses. ..)))

J(t))

FIGURE4.2.7.A mass-spring-dashpot system with

external forceJ(t).)

Example4)


TheTransformPerspectiveLetus regard the general constant-coefficientsecond-orderequationas the equationof motion

mx\" +ex'+kx = f(t)of the familiar mass-spring-dashpotsystem (Fig.4.2.7).Then the transformed

equation is)

m [s2X (s)-sx(O)-x'(0)]+c[sX (s)-x(O)]+kX (s)= F(s). (13))

Note that Eq.(13)is an algebraicequation-indeed,a linear equation-inthe \"un-

known\" X (s).This is the sourceof the powerof the Laplacetransform method:)

Lineardifferential equationsare transformedinto readily solvedalgebraicequations.)

If we solve Eq.(13)for X(s),we get)

F(s) I(s)X(s)=

Z(s)+

Z(s)') (14))

where)

Z(s)= ms2 +cs+k and I(s)= mx(O)s+mx'(0)+cx(O).)

Note that Z (s)dependsonly on the physical system itself. Thus Eq.(14)presentsX (s) =

\302\243{x(t)} as the sum of a term dependingonly on the externalforce and onedependingonly on the initial conditions.In the caseof an underdampedsystem,thesetwo terms are the transforms)

F(s)\302\243{xsp(t)}

=Z(s))

I(s)and \302\243{Xtr(t)}

=-Z(s))

of the steadyperiodicsolutionand the transient solution,respectively.The only po-tential difficulty in finding thesesolutionsis in finding the inverseLaplacetransform

of the right-hand sidein Eq.(14).Much of the remainderof this chapteris devotedto finding Laplacetransforms and inverse transforms. In particular, we seekthosemethods that are sufficiently powerful to enableus to solve problemsthat-unlikethosein Examples1 and 2-cannotbe solvedreadily by the methodsof Chapter2.)

AdditionalTransformTechniques).....

Showthat)

1\302\243 {t ea

t} = .(s-a)2)

Solution If f (t) = teat, then f (0)= 0 and f'(t) = eat +at eat. HenceTheorem 1 gives)

\302\243{e

at +ateat} =

\302\243{f'(t)}= s\302\243{f(t)} =

s\302\243{te

at}.)

It follows from the linearity of the transform that)

\302\243{e

at}+ a\302\243{te

at} =

s\302\243{te

at}.)))

Hence)

\302\243 {eat} 1

\302\243 {teat}== ==s -a (s-a)2)

(15))

because\302\243{e

at } == 1/(s-a).) .)

Example5) Find\302\243{t

sin kt}.)

Solution Letf (t) == t sin kt.Then f (0)==0 and)

f'(t) ==sin kt +kt coskt.)

The derivative involves the new function t coskt, so we note that f'(0) == 0 and

differentiateagain. The result is)

f\" (t) ==2k coskt - k2t sin kt.)

But \302\243{f\"(t)}== s2\302\243{f(t)} by the formula in (5)for the transform of the second

derivative,and \302\243{coskt}==s/(s2+k2), sowe have)

2ks2 2

-k2\302\243{t sinkt} ==

s2\302\243{t sinkt}.s +k)

Finally, we solve this equation for)

2ks\302\243 {t sin k t} == 2 2 2.(s +k

))

(16))

This procedureis considerablymore pleasant than the alternative of evaluating the

integral)

\302\243{tsin kt} = 1

00

te-51sin kt dt.) .)

Examples4 and 5 exploitthe fact that if f (0) == 0,then differentiation of fcorrespondsto multiplication of its transform by s. It is reasonableto expecttheinverseoperation of integration (antidifferentiation) to correspondto divisionof the

transform by s.)

THEOREM2 Transformsof IntegralsIf f(t) is a piecewisecontinuousfunction for t >0 and satisfiesthe conditionofexponentialorderIf(t)I

<Mect for t > T, then)

{{t

}1 F(s)

\302\243

10f('e)d7: =\037\302\243{f(t)}

=S)

(17))

for s >c.Equivalently,)

\302\243-1

{F;S)}= 1

1

f(7:)d7:.) (18))))

Proof:Becausef is piecewisecontinuous,the fundamental theoremof cal-culus impliesthat)

g(t) = it f('c)dr:)

is continuousand that g'(t) = f(t) where f iscontinuous;thus g iscontinuous and

piecewisesmooth for t >O.Furthermore,)

it it M MIg(t)1< IfCr)1di<M eCTdi= _(ect - 1)< _ect

,o 0 C C)

so get)is of exponentialorderas t --+ +00.Hencewe can apply Theorem 1 to g;this gives)

\302\243{f(t)}=

\302\243{g'(t)}=

s\302\243{g(t)}-g(O).)

Now g(O)= 0,sodivisionby s yields)

\302\243 {itf(r:)dr:})

\302\243{g(t)}= \302\243{f(t)}

,s)

which completesthe proof.) \037)

Example6) Find the inverseLaplacetransform of)

G(s))1

s2(s-a).)

Solution In effect,Eq.(18)means that we can deletea factor of s from the denominator, find

the inverse transform of the resulting simplerexpression,and finally integrate fromo to t (to \"correct\"for the missingfactors).Thus)

\302\243-1

{1

}= t

\302\243-1

{1

}dr:= teaTdr:= !(eat -1).s(s-a) 10 s -a 10 a)

We now repeat the technique to obtain)

{I

} it

{I

} it 1\302\243-1 = \302\243-1 dT = _(eaT- 1)dT

s2(s-a) 0 s(s-a) 0 a)

[1

(1 aT

)]t 1 at=

a ae - r:0

=a2(e -at - 1).)

This technique is often a more convenientway than the methodof partial fractionsfor finding an inverse transform of a fractionof the form P(s)j[snQ(s)]. .)))

ProofofTheorem 1: We concludethis sectionwith the proof of Theorem 1in the general casein which f' is merely piecewisecontinuous. We needto provethat the limit)

lim {be-stI'(t)dtb\037oo 10)

existsand alsoneedto find its value. With b fixed, let tl, t2, ..., tk-I be the pointsinterior to the interval [0,b] at which f' is discontinuous.Let to == 0 and tk == b.Then we can integrate by parts on eachinterval (tn-I,tn ) where f' is continuous.This yields)

(bk

jtn

10e-stl'(t)dt =I: e-stI'(t)dto n=1 tn-l

k tnk

jtn

=\037[e-st

I(t)t-l+\037

Stn-l

e-st I(t)dt.) (19))

Now the first summation)

k t

I:[e-stI(t)r= [-I(to)+e-st1I(tl)]+ [-e-stlI(tl)+e-stzI(t2)]n=1 tn-l

+ ...+ [_estk-ZI(tk-2)+e-stk-lI(tk-l)]+ [_estk-lI(tk-l)+e-stk

I(tk)]in (19)telescopesdown to -f(to)+e-stkf(tk) ==-f(O)+e-sbf(b),and the sec-ond summation addsup to s times the integral from to ==0 to tk ==b.Therefore(19)reducesto)

(20))

lb

e-stI'(t)dt = -1(0)+e-sbI(b)+Slb

e-st I(t)dt.

But from Eq.(3)we get)

le-sbf(b)1<e-sb .Mecb ==Me-b(s-c)--+0)

if s >c.Therefore, finally taking limits (with s fixed)asb --++00in the precedingequation, we get the desiredresult)

\302\243{f'(t)}==

s\302\243{f(t)}- f(O).) \037)

ExtensionofTheorem1)

Now supposethat the function f is only piecewisecontinuous (instead of continu-ous),and let tl , t2, t3, ...be the points (for t >0)where either f or f' is discontin-uous.The fact that f is piecewisecontinuous includesthe assumptionthat-withineachinterval [tn-I,tn ] betweensuccessivepoints of discontinuity-f agreeswith afunction that is continuouson the whole closedinterval and has \"endpoint values\

f(t:_I ) == lim f(t) and f(t;)== lim f(t)t ---'>.. t+ t t--'n-l \037;;)

that may not agreewith the actual values f(tn-I) and f(tn ). The value of an in-

tegral on an interval is not affected by changing the values of the integrand at the)))

endpoints.However, if the fundamental theorem of calculusis applied to find the

value of the integral, then the anti derivative function must be continuous on the

closedinterval. We therefore use the \"continuous from within the interval\" end-point values above in evaluating (by parts) the integrals on the right in (19).Theresult is)

k t

L[e-st1(t)r = [-1(tt) +e-st ,1(t1)]+ [_e-st\\ I(t{)+e-st2l(t;)]

n=l tn-l

+ ...+ [_estk-21\302\253(L2)

+e-stk-Il(tk-I)]+ [_estk-Il(tL1)+e-stk

l(tk)]k-l

==-f(O+)-Ljj(tn ) +e-sbf(b-), (20/)n=l)

where)

jj(tn ) ==f(t:)- f(t;)) (21))

denotesthe (finite) jump in f(t) at t == tn. Assuming that \302\243{f'(t)} exists,wetherefore get the generalization)

00

\302\243{f'(t)}==sF(s)- f(O+)-Le-stn jj(tn )

n=l)

(22))

of \302\243{f'(t)}==sF(s)- f(O)when we now take the limit in (19)asb \037 +00.)

Example7)

J(t)6 .......5 .....-04 .....-03 .....-02 .....-01)

Let f (t) == 1 + [t]be the unit staircasefunction; its graph is shown in Fig.4.2.8.Then f(O) == 1,f'(t)= 0,and jj(n)== 1 for each integern == 1,2,3,....HenceEq.(22)yields)

00

0==sF(s)- 1-Le-ns,

n=l)

so the Laplacetransform of f(t) is)

1 00 1F(s)==-Le-ns ==

sn=O s(1-e-S

))123456t)

FIGURE4.2.8.Thegraph of theunit staircasefunction of Example7.)

In the last stepwe usedthe formula for the sum of a geometricseries,)

00

Lxn

n=O)

1)

,I-x)

with x ==e-s< 1.) .)

lIB Problems)UseLaplacetransforms to solvethe initial value problems in

Problems1through 16.1.x\" + 4x = 0;x(O)= 5,x'(O)= 02.x\" + 9x = 0;x(O)= 3,x'(O)= 4)

3.x\" -XI -2x = 0;x (0)= 0,x'(0)= 2

4. x\" + 8X' + 15x= 0;x(O)= 2,x'(O)= -35. x\" + x = sin 2t;x(O)= 0 = x'(O)6.x\" + 4x = cost; x(O)= 0 = x'(O)7. x\" + x = cos3t; x(O)= 1,x'(O)= 0)))

8.x\" + 9x = 1;x (0)= 0 = x'(0)9.x\" + 4X' + 3x = 1;x(O)= 0 = x'(O)10.x\" + 3X' + 2x = t; x(O)= 0,x'(O)= 211.x'= 2x + y, y' = 6x + 3y; x(O)= 1,y(O)= -212.x'=x +2y, y' = X + e-t ; x(O)= y(O)= 013.x'+ 2y' + x = 0,x'-y' + y = 0;x(O)= 0, y(O)= 114.x\" + 2x + 4y = 0, y\" + x + 2y = 0; x(O)= y(O) = 0,

x'(0)= y' (0)= -115.x\" + x'+ y' + 2x -y = 0, y\" + x'+ y' + 4x-2y = 0;

x(O)= y(O)= 1,x'(O)= y'(O)= 016.x'=x+z,y' = x+y, z' = -2x-z;x(O)= 1,y(O)= 0,

Z (0)= 0)

Apply Theorem 2 to find the inverse Laplacetransforms of the

functions in Problems17through 24.

1 317.F(s)= 18.F(s)=s(s- 3) s(s+ 5)1 2s+ 119.F(s)=

s(s2+ 4)20.F(s)=

S(S2+ 9)1 121.F (s) =

2 2 22.F (s) 2s (s + 1)=

s(s -9)1 123.F(s)= 24. F(s)=

S2(S2- 1) s(s+ l)(s+ 2)25.Apply Theorem1 to derive \302\243 {sin kt} from the formula for

\302\243{cos kt}.26.Apply Theorem1 to derive \302\243 {coshkt} from the formula

for \302\243{sinh kt}.27.(a) Apply Theorem1 to show that)

\302\243{tne

at} =)

n\302\243{t

n-1eat }.s-a)(b) Deducethat \302\243{tne

at} = n!j(s- a)n+1 for n = 1,2,

3, ....)

Apply Theorem 1as in Example 5 to derive the Laplacetrans-

forms in Problems28 through 30.

S2-k228.\302\243 {t coskt} =

2 2 2(s + k )2ks29.

\302\243{tsinh kt} =

(s2_ k2)2S2+ k2

30.\302\243{t

coshkt} =(S2_ k2)231.Apply the results in Example 5 and Problem 28 to show

that)

-1{

I

}

1 .\302\243 2 2 2

= _3 (sIn kt - kt coskt).

(s + k ) 2k)

Apply the extension of Theorem 1in Eq. (22) to derive the

Laplacetransforms given in Problems32 through 37.

32.\302\243{u(t-a)}= S-le-as for a > O.

33.If f(t) = 1 on the interval [a, b] (where 0 < a < b) and

f (t) = 0 otherwise, then)

-as -bse -e\302\243{f(t)}

= .s)

34. If f(t) = (-l)[t]is the square-wave function whose

graph is shown in Fig. 4.2.9,then)

1 s\302\243{f(t)}

= - tanh -.s 2)

(Suggestion:Usethe geometricseries.))

J(t)1 ........0........0.......)

123456t-1 ........0........0........0)

FIGURE4.2.9.Thegraph of the

square-wave function ofProblem 34.)

35.If f (t) is the unit on-offfunction whosegraph is shown in

Fig. 4.2.10,then)

1\302\243{f(t)}

= s(l+ e-S).)

J(t)1 ........0........0.......)

1 2 3 4 5)

FIGURE4.2.10.Thegraph of theon-offfunction ofProblem 35.

36.If get) is the triangular wave function whose graph isshown in Fig.4.2.11,then)

1 s\302\243{g(t)}

= 2 tanh -.s 2)

get))

2 3 456 t)


triangular wave function ofProblem 36.)

37.If f(t) is the sawtooth function whose graph is shown in

Fig. 4.2.12,then)

1 e-s\302\243{f(t)}

= 2-(1 -S)

.s s -

e)

(Suggestion:Note that f'(t) = 1 where it is defined.))

J(t))

1)

2 3 456 t)

FIGURE4.2.12.Thegraph of thesawtooth function of Problem 37.)))

4.3Translation and Partial Fractions 289)

11I1,TraJ?-sl\037ltio\037and

\037ar!\037\037I}:'\037\037\037\037.!\037ol1\037.w ...... ._ ..)

As illustratedby Examples1 and 2 of Section4.2,the solutionof a lineardifferential

equation with constant coefficientscan often bereducedto the matter of finding theinverseLaplacetransform of a rational function of the form

R(s)= P(s) (1)Q(s)

where the degreeof pes)is lessthan that of Q(s).The technique for finding

oC-I{R(s)}is basedon the samemethod of partial fractions that we use in ele-mentary calculusto integrate rational functions. The following two rules describethe partialfraction decompositionof R(s),in terms of the factorizationof the de-nominator Q(s) into linear factors and irreduciblequadratic factors correspondingto the real and complexzeros,respectively,of Q(s).)

RULE 1 LinearFactorPartialFractionsThe portion of the partial fraction decompositionof R(s)correspondingto thelinear factor s - a of multiplicity n is a sum of n partial fractions, having theform)

Al A 2 An+ +...+ ,s-a (s-a)2 (s-a)n) (2))

where A I,A 2, ..., and An are constants.)

RULE 2 QuadraticFactorPartialFractionsThe portionof the partial fractiondecompositioncorrespondingto the irreduciblequadratic factor (s- a)2+ b2 of multiplicity n is a sum of n partial fractions,having the form)

Als + BI A 2s + B2 An s + Bn

(s-a)2+b2 +[(s-a)2+b2]2

+ \302\267 \302\267 \302\267 +[(s_ a)2+b2

]n' (3))

where AI, A 2, ..., An, BI,B2,..., and Bn are constants.)

Finding oC-I{R(s)}involvestwo steps.First we must find the partial fractiondecompositionof R(s),and then we must find the inverseLaplacetransform of eachof the individual partial fractions of the types that appearin (2)and (3).The latter

stepis basedon the followingelementaryproperty of Laplacetransforms.)

THEOREM1 Translationonthes-AxisIf F(s)= oC{J(t)}existsfor s >c,then oC{eat J(t)}existsfor s >a +c,and)

>-) oC{eat J(t)}= F(s-a).) (4))

Equivalently,>) oC-I{F(s-a)}= eatJ(t).) (5))

Thus the translation s -+ s -a in the transform correspondsto multiplication ofthe original function of t by eat.)))


Example1)

Proof:If wesimply replaces with s-ain the definition of F(s)= \302\243{f(t)},

we obtain)

F(s-a) = 100

e-(s-a)tf(t) dt = 100

e-st[eatf(t)]dt =

\302\243{e

at f(t)}.)

This is Eq.(4),and it isclearthat Eq.(5)is the same.) .)If weapply the translation theorem to the formulasfor the Laplacetransforms

of tn, coskt, and sin kt that we already know-multiplyingeachof thesefunctions

by eat and replacings with s -a in the transforms-weget the followingadditionsto the table in Fig.4.1.2.)

eat t n n!(s > a) (6)

(s-a)n+l

s-a(7)eat coskt

(s-a)2+ k2 (s > a)

k(8)eat sin kt

(s-a)2+ k2 (s > a))

For ready reference,all the Laplacetransforms derived in this chapter arelistedin the table of transforms that appearsin the endpapers.)

--\"-Co\037\037id\037;-;'m\037_\037\037\037=\037\037-d=\037p;i\037g--\037y-\037t\037\037'O;ith\"';;;

\"_.\"

'''\037''\037'''k

= 17,and c = 3 in mks units

(Fig.4.3.1).As usual, let x(t) denote the displacementof the massm from its

equilibrium position.If the massis set in motion with x(0)= 3 and x/ (0)= 1,find

x(t) for the resulting dampedfree oscillations.)

Solution The differential equation is !x\"+ 3x'+ 17x= 0,so we needto solve the initial

value problem)k = 17)

FIGURE4.3.1.Themass-spring-dashpot system ofExample1.)

x)

3)

1t\"2)

2)

1)

FIGURE4.3.2.Thepositionfunction x(t)in Example 1.)

x\" +6x'+34x = 0; x(O)= 3, x/CO)= 1.)

We take the Laplacetransform of eachterm of the differential equation. Because(obviously) \302\243{O}

= 0,weget the equation)

[S2X(S)-3s-1]+6[sX(s)-3] +34X(s)= 0,)

which we solve for)

3s + 19 s +3 5X(s)= s2+6s+34=3.

(S+3)2+2S+2 .

(S+3)2+2S.)

t)Applying the formulas in (7)and (8)with a = -3and k = 5,we now seethat)

x(t) = e-3t (3cos5t +2 sin 5t) .)

Figure 4.3.2showsthe graph of this rapidly decayingdampedoscillation.) .)))

Example2)


Example2 illustratesa useful techniquefor finding the partial fraction coeffi-cientsin the caseof nonrepeatedlinear factors.)

Find the inverseLaplacetransform of)

S2+ 1R(s)= .

s3-2s2- 8s)

Solution Note that the denominatorof R(s)factors as Q(s)= s(s+2)(s-4).Hence)

Example3)

S2+ 1 ABC=-+ + .s3-2s2-8s s s +2 s - 4)

Multiplicationof eachterm of this equationby Q(s)yields)

S2+ 1 = A(s +2)(s-4) +Bs(s-4) +Cs(s+2).)

When we successivelysubstitute the three zeross = 0,s = -2,and s = 4 of the

denominator Q(s)in this equation,we get the results)

-8A= 1, 12B= 5, and 24C= 17.)

Thus A = -!,B =1

52' and C =

\037\037, so)

S2+ 1 ! \0372 \037\037= --+ +s3-2s2-8s s s +2 s -4)

and therefore)

-1{

S2 + 1

}1 5 _2t 17 4t

\302\243= --+-e +-e .

s3-2s2-8s 8 12 24).)

Example3 illustratesa differentiation techniquefor finding the partial fraction

coefficientsin the caseof repeatedlinear factors.)

................


y\" +4y' +4y = (2; y(O)= y/(O)= O.)

Solution The transformedequation is)

2s2y(s)+4sY(s)+4Y(s)= 3\".s)

Thus)

2 ABC D Eyes) = = -+-+-+ +. (9)s3(s +2)2 s3 s2 S (s +2)2 s +2)

To find A, B,and C,we multiply both sidesby s3 to obtain)

22 =A+Bs+Cs2+s3F(s),

(s+2))(10))))


where F(s)= D(s+ 2)-2+ E(s+ 2)-1is the sum of the two partial fractionscorrespondingto (s +2)2.Substitution of s = 0 in Eq. (10)yieldsA = 4. To find

BandC,wedifferentiateEq.(10)twice to obtain)

-43

= B+2Cs+3s2F(s)+s3F'(s)(s+ 2))

(11))

and)

124 = 2C+6sF(s)+6s2F'(s)+s3F\"(s).

(s + 2))(12))

Now substitution of s = 0 in Eq.(11)yieldsB = -4,and substitution of s = 0 in

Eq.(12)yieldsC =\037.

To find D and E,we multiply eachsidein Eq.(9)by (s +2)2 to get)

23\"

= D + E(s+2)+ (s +2)2G(s),s)

(13))

where G(s)= As-3 +Bs-2 +Cs-1 , and then differentiateto obtain)

6-4= E +2(s+2)G(s)+ (s +2)2G'(s).s)

(14))

Substitution of s = -2in Eqs. (13)and (14)now yieldsD =-\037

and E =-\037.

Thus)1 1 3 1- - - -

yes) = -1..._ -1...+ \037 _ 4s3 s2 S (s +2)2

so the solution of the given initial value problemis)

38

S +2')

yet) =\037t2

-4t +

\037

-\037te-2t

-\037e-2t.)

.)Examples4, 5,and 6 illustrate techniquesfor dealing with quadraticfactors in

partial fraction decompositions.)

Example4) Considerthe mass-spring-dashpotsystem as in Example1,but with initial condi-tions x(O)= x'(O)= 0 and with the imposedexternal force F(t) = 15sin2t.Findthe resulting transient motion and steadyperiodicmotion of the mass.)

Solution The initial value problemwe needto solve is)

x\" +6x'+34x= 30 sin 2t; x(O)= x'(0)= O.)

The transformedequation is)

60s2X(s)+6sX(s)+34X(s)= 2 .

s +4)

Hence) 60 As + B Cs + DX(s)=

(s2+4)[(s+3)2+2S]=

s2+4+ (S+3)2+2S\)

x1 Periodic

\\)0.5)

\0371Transient)

-0.5)

FIGURE4.3.3.Theperiodicforcedoscillationxsp(t), dampedtransient motion xtr(t), andsolution x(t)= xsp(t) + xtr(t) in

Example 4.)


When we multiply both sidesby the commondenominator,we get)

60== (As + B)[(s+3)2+25]+ (Cs+D)(s2+ 4). (15))

To find A and B,we substitute the zero s == 2i of the quadratic factors2+ 4in Eq.(15);the result is)

60== (2iA + B)[(2i+3)2+25],)

which we simplify to)

60== (-24A+30B)+ (60A + 12B)i.)

We now equate real parts and imaginary parts on each sideof this equationto obtainthe two linear equations)

-24A +30B==60 and 60A + 12B==0,)

which are readily solvedfor A ==-\037\037

and B ==\037\037

.To find C and D,we substitute the zero s == -3+ 5i of the quadraticfactor

(s + 3)2+25 in Eq.(15)and get)

60==[C(-3+5i)+D][(-3+ 5i)2+4],)


60== (186C- 12D)+ (30C-30D)i.)

Again we equate real parts and imaginary parts; this yieldsthe two linear equations)

186C-12D==60 and 30C-30D==0,)

and we readily find their solution to be C == D ==\037\037

.With thesevalues of the coefficients A, B,C,and D, our partial fractions

decompositionof X (s) is)

1

(-1Os+ 50 10s+ 10

)X (s)= 29 S2+4+

(s + 3)2+25

== \037

(-IOS+25 .2 + lO(s+ 3)-4 .5

) .29 s2+4 (s + 3)2+25)

After we compute the inverseLaplacetransforms,we get the position function)

x(t) == 19(-2cos2t +5 sin 2t) + {ge-3t(5cos5t -2 sin 5t).)

The terms of circular frequency2 constitute the steadyperiodicforcedoscillationofthe mass,whereasthe exponentiallydampedterms of circularfrequency5 constituteits transient motion, which disappearsvery rapidly (seeFig.4.3.3).Note that the

transient motion is nonzeroeven though both initial conditionsare zero. .)))


ResonanceandRepeatedQuadraticFactors)The following two inverse Laplacetransforms are useful in inverting partial frac-tions that correspondto the caseof repeatedquadratic factors:

-1{

S

}1 .

k\302\243 2 2 2 = -t sIn t,(s +k ) 2k)

(16))

-1{

I

}1 .

\302\243 2 22 ==_3 (sinkt -ktcoskt).

(s +k ) 2k

Thesefollow from Example5 and Problem31of Section4.2,respectively.Becauseof the presencein Eqs. (16)and (17)of the terms t sin kt and t coskt, a repeatedquadratic factor ordinarily signalsthe phenomenon of resonancein an undampedmechanicalor electricalsystem.)

(17))

Example5) UseLaplacetransforms to solve the initial value problem)II 2 D.X +WaX

= ra SInwt;) X(O) = 0 = x/CO))

that determinesthe undampedforced oscillationsof a masson a spring.)

Solution When we transform the differentialequation,we get the equation

2 2 Faw Faws Xes)+woX(s)= 2 2 ' so Xes)=

2 2 2 2 .S +w (s +w)(s +wa)

If w =1= Wa, we find without difficulty that

Faw

(1 1

)X (s)= 2 2- ,

w 2 -w a s2+w a s2+w 2)

so it follows that)

Faw

(1 1.

)x(t) =2 2

-sinwat --SInwt .w -wa Wa w

But if w ==Wa, wehave)

t)

FawoX (s)=2 2 2 '(s +wa)

soEq. (17)yieldsthe resonancesolution

Fa .x(t) = \037(sInwat -wat coswat).

2wa)

(18).)

4)

FIGURE4.3.4.Theresonancesolution in (18)with (Va =

\037

andFa = 1,together with its envelopecurves x = ::f:C(t).)

Remark:The solution curve defined in Eq. (18)bouncesbackand forth

(seeFig.4.3.4)betweenthe \"envelopecurves\" X = ::I::C(t)that are obtained by

writing (18)in the form)

x(t) == A(t) coswat +B(t)sin wat

and then defining the usual \"amplitude\" C = ,JA2 +B2.In this casewe find that

C(t) = Fa J{Jit2 + 1.2w2 a

a

This techniquefor constructingenvelopecurvesof resonancesolutionsis illustratedfurther in the applicationmaterial for this section. .)))

Example6)

4.3Translation and Partial Fractions 295)........... ...... ........................


y(4) + 2y\" + y = 4tet; y(O)= y/ (0)= y\" (0)= y(3)(0)= O.)

Solution First weobservethat)

oC{y\"(t)}= s2y(s),) t 1and \302\243{te}

=(s _ 1)2

')oC{y(4)(t)}= s4y(s),)

Hencethe transformedequation is)

4(s4+2s2+ l)y(s) = 2.(s- 1))

Thus our problemis to find the inverse transform of)

4yes) --

(s- 1)2(s2+ 1)2A B Cs+ D Es+ F= + + +

(s- 1)2 S- 1 (s2+ 1)2 s2+ 1.) (19))

If wemultiply by the common denominator (s - 1)2(s2+ 1)2,we get the equation)

A(S2 + 1)2+ B(s- 1)(s2+ 1)2+Cs(s- 1)2+ D(s- 1)2+ Es(s- 1)2(s2+ 1)+ F(s- 1)2(s2+ 1)= 4. (20))

Uponsubstituting s = 1 we find that A = 1.Equation (20)is an identity that holdsfor all valuesof s.To find the valuesof

the remaining coefficients,we substitute in successionthe values s = 0,s = -1,s = 2,s = -2,and s = 3 in Eq.(20).This yieldsthe system)

-B + D + F = 3,-8B- 4C + 4D - 8E+ 8F= 0,25B+ 2C+ D + 10E+ 5F = -21,

-75B- 18C+ 9D- 90E+ 45F= -21,200B+ 12C+ 4D + 120E+ 40F= -96)

(21))

of five linear equations in B,C,D, E, and F. With the aid of a calculatorpro-grammed to solve linear systems,we find that B = -2,C = 2,D = 0,E = 2,andF = 1.

We now substitute in Eq.(18)the coefficientswe have found, and thus obtain)

1 2 2s 2s+1yes) = - + + .

(s - 1)2 S- 1 (s2+ 1)2 s2+ 1)

RecallingEq.(16),the translation property,and the familiar transforms of cost andsin t, weseefinally that the solutionof the given initial value problem is)

y (t) = (t -2)et + (t + 1)sin t +2cost .) .)))


III)Problems)

Apply the translation theorem to find the Laplacetransformsofthe functions in Problems1through 4.

1.f(t) = t4eJrt 2. f(t) = t 3/2e-4t)

3.f (t) =e-2t sin 3n t) 4. f (t) = e-t12cos2 (t-

kn))

Apply the translation theorem to find the inverse Laplacetrans-

forms ofthe functions in Problems5 through 10.3 s - 15. F(s)= 6.F(s)= 32s-4 (s+ 1)

1 s+27. F(s)= s2+ 4s+ 48.F(s)= s2+ 4s+ 5

3s + 5 2s- 39.F (s) 10.F (s) ==s2-6s+ 25 9s2 - 12s+ 20)

Usepartial fractions to find the inverse Laplacetransforms ofthe functions in Problems11through 22.)

111.F(s)= 2s -45-2s13.F(s)= 2s+7s+10115.F (s)= 3 2S -5s)

5s-612.F(s)= s2_ 3s5s-414.F(s)= s3-s2-2s)

116.F(s)= 2 2(s + s-6))

117.F(s)= 4S - 16)

S318.F(s)= (s_ 4)4

120.F(s)= 4 2S -8s + 162s3 -S2

22.F(s)= (4s2 -4s+ 5)2)

S2-2s19.F(s)= 4 2S + 5s + 4

s2+ 321.F(s)=(s2+ 2s + 2)2)

Usethe factorization)

S4+ 4a4 = (s2-2as+ 2a2)(s2+ 2as+ 2a2))

to derive the inverse Laplacetransforms listed in Problems23through 26.

23.\302\243-1

{S4:34a4}= coshat cosat

24. \302\243-1 L4:4a4}= 2:2 sinh at sin at

25.\302\243-1

{ 4s2

4 }= \037 (coshat sin at + sinh at cosat)s +4a 2a

26.\302\243

-1{ 4

14 }

= \0373 (coshat sin at -sinh at cosat)s +4a 4a)

UseLaplacetransforms to solvethe initial value problems in

Problems27 through 38.

27.xl/ + 6X' + 25x= 0;x(O)= 2,x'(0)= 3

28.xl/ - 6X' + 8x = 2;x(O)= x'(O)= 0

29.xl/ -4x = 3t;x(O)=x'(0)= 0

30.xl/ + 4X' + 8x =e-t ; x(O)= x'(O)= 0

31.x(3) + xl/ - 6X' = 0;x(O)= 0,x'(O)= xl/(O) = 1

32.X(4) -x = 0;x(O)= 1,x'(O)= xl/(O) = X(3) (0)= 0

33.X(4) + x = 0;x (0)= x'(0)= xl/ (0)= 0,x(3)(0)= 1

34. X(4) + 13xl/+ 36x = 0; x (0) = xl/ (0) = 0, x'(0) = 2,x(3)(0)= -13

35.X(4) + 8x\" + 16x = 0; x(O) = x'(O) = xl/(O) = 0,x(3)(0)= 1

36.X(4) + 2xl/ + x = e2t ; x (0)=x'(0)= xl/ (0)= x(3)(0)= 0

37.xl/ + 4X' + 13x= te-t ; x(O)= 0,x'(O)= 2

38.xl/ + 6X' + 18x= cos2t;x(O)= 1,x'(0)= -1)

Problems 39 and 40 illustrate two types of resonancein amass-spring-dashpotsystem with given external force F(t)and with the initial conditions x (0)= x'(0)= o.

39.Supposethat m = 1,k = 9, c = 0, and F(t)= 6cos3t.Usethe inverse transform given in Eq.(16)to derive the

solution x(t) = t sin 3t. Construct a figure that illustrates

the resonancethat occurs.)

40. Supposethat m = 1,k = 9.04,c = 0.4,and F(t)6e-tIScos3t. Derivethe solution)

x(t)= te-tl5 sin 3t.)

Show that the maximum value of the amplitude function

A(t) = te-tj5 is A(5) = 5je.Thus (as indicated in

Fig. 4.3.5)the oscillations of the mass increasein am-

plitude during the first 5 s beforebeing damped out ast ---+ +00.)

2)

x =+ te -t15/)t

IOn

\037x =-te -t15)

-2FIGURE4.3.5.Thegraph of the dampedoscillationin Problem 40.)))

4.4Derivatives,Integrals,and Productsof Transforms 297)

11IIDe\037i\0373:ti\037.c::\037.! I'.1\037c::.gE\037I.\037! a.\037d Pr_\037\037uct\037__\037fTransforms)

The Laplacetransform of the (initially unknown) solutionof a differential equationis sometimesrecognizableas the product of the transforms of two known functions.For example,when we transform the initial valueproblem

x\" +x ==cost; x(O)==x'(0)==0,)

weget)s s 1 .

Xes)=(S2+ 1)2

=s2+ 1

\302\267

s2+ 1= \302\243{cost}. \302\243{smt}.

This strongly suggeststhat there ought to be a way of combiningthe two functions

sint and costto obtain a function x(t) whosetransform is the product of theirtransforms. But obviouslyx(t) is not simply the product of cost and sin t, because

\302\243 {cost sin t} =\302\243 H sin 2t}=

21

=1= 2S

2.S +4 (s + 1)Thus \302\243{cos t sin t} =1= \302\243{cos t} .

\302\243{sin t}.Theorem 1 of this sectionwill tell us that the function

h(t) =11

f('r)g(t- i)di) (1))

has the desiredproperty that)

\302\243{h(t)}==R(s) == F(s) .G(s).) (2))

The new function of t definedas the integral in (1)dependsonly on I and g and iscalledthe convolutionof Iand g.It is denotedby I* g, the ideabeing that it is anew type of product of Iand g, so tailored that its transform is the productof thetransforms of I and g.)

DEFINITION TheConvolutionof Two Functions)The convolution I* g of the piecewisecontinuousfunctions I and g is definedfor t >0 as follows:)

\037) (f * g)(t) =11f(i)g(t- i)di.) (3))

We will alsowrite I(t)* get)when convenient. In terms of the convolutionproduct, Theorem 1 of this sectionsaysthat)

\037)\302\243{I

* g}==\302\243{/}

.\302\243{g}.)

If we make the substitution u == t -T in the integral in (3),we seethat)

f(t) * get)= 11

f(i)g(t-i)di= fOf(t -u)g(u)(-du)

= 11

g(u)f(t - u) du = g(t) * f(t).)

Thus the convolutionis commutative:I* g ==g * I.)))


Example1)

Example2)

The convolutionof cost and sin t is

(cost) * (sint) = itCOST sin(t -T)dT.

We apply the trigonometric identity

cosA sin B == 4 [sin(A+ B)- sin(A -B)])

to obtain)

(cost) * (sint) = it \037[sin t - sin(2T- t)] dT

== 1[Tsint + 1COS(2T_ t)]t

;T=O)

that is,)

(cost) * (sint) == 4t sin t.And we recallfrom Example5 of Section4.2that the Laplacetransform of 1tsin t

is indeeds/(s2+ 1)2. .Theorem 1 is proved at the end of this section.)

THEOREM1 TheConvolutionPropertySupposethat f(t) and get)are piecewisecontinuous for t > 0 and that If(t)!and !g(t)!are boundedby Mect as t -+ +00.Then the Laplacetransform of theconvolutionf(t) * get)existsfor s >c;moreover,)

\037)\302\243{f(t)

* get)}== \302\243{f(t)}. \302\243{g(t)}) (4))

and)

\037) \302\243-l{F(s).G(s)}==f(t) * get).) (5))

Thus we can find the inverse transform of the product F(s).G(s),providedthat wecan evaluate the integral)

\037) aC-1{F(s).G(s\302\273)

= it f(T)g(t-T)dT.) (5'))

Example2 illustrates the fact that convolution often provides a convenientalternative to the useof partial fractions for finding inverse transforms.)

cO ,.,With f(t) ==sin 2t and get)==et , convolutionyields

\302\243-1

{2

2 }= (sin2t)*et = t et-r sin2TdT

(s-1)(s+4) 10{t

[-T

]t

= et

10e-r sin2T dT = et e

5 (-sin2T - 2 COS2T)0')

so)

-1{

2

}2 t 1. 2

\302\243 ==-e --SIn 2t - -cos2t.(s-I)(s2+4) 5 5 5).)))

Example3)


DifferentiationofTransformsAccording to Theorem 1 of Section4.2,if J(O) == 0 then differentiation of f(t)correspondsto multiplication of its transform by s.Theorem 2, proved at the endof this section,tells us that differentiation of the transform F(s) correspondstomultiplication of the original function f (t) by -t.)

THEOREM2 DifferentiationofTransformsIf J(t)is piecewisecontinuousfor t >0and IJ(t)1<Mect as t -+ +00,then)

L{-tJ(t)}== F'(s)) (6))


\037)

1J(t)==L-1{F(s)}==--L-1{F'(s)}.

t)

(7))

Repeatedapplicationof Eq.(6)gives)

oC{tn J(t)}== (_I)np(n) (s)) (8))

for n == 1,2,3, ....)

Find oC{t2sin kt}.)

Solution Equation (8)gives)

Example4)

oC{t2sinkt}= (-1)2:S22C2:k2)

d

[-2ks

]6ks2-2k3

==ds (s2+k2)2

==(s2+k2)3

.) (9))

.)The form of the differentiation property in Eq. (7)is often helpful in finding

an inversetransform when the derivativeof the transform iseasierto work with than

the transform itself.)

Find oC-1{tan-1(1Is)}.)

Solution The derivativeof tan- 1 (IIs) is a simplerational function, sowe apply Eq.(7):)

-1{

_II}

1 -1{

d -1I

}oC tan - ==--oC -tan -

s t ds s)

1 -1{

-I/s2}

=--roC 1 + (l/S)2)

1

{-I

}1= --roC

-t

S2+ 1= --r(-sint).)))


Example5)

Therefore,)

-1{

-I 1

}sin t

\302\243 tan - ==---.s t)

.)

Equation (8)can be appliedto transform a linear differentialequation having

polynomial, rather than constant, coefficients.The result will bea differential equa-tion involving the transform; whether this procedureleadsto successdepends,ofcourse,on whether we can solve the new equationmore readily than the old one.)

,.\" ...,,'\" \"'\"\"Nunn,.. \"\" n\037.....,_,\", '\" '\" NY'''' \"\" ,...\"\",.\037....,....,,... \"'\".... .-.v .\"...., .\" ...\".n...'..o,...,.....\" ...... '\" ...., ,.\"

Letx(t) be the solution of Bessel'sequation of orderzero,)

tx\" +x' + tx ==0,)

such that x(0)== 1 and x'(0)==O.This solutionof Bessel'sequation is customarilydenotedby Jo(t).Because)

\302\243{x'(t)}==sX(s)- 1 and \302\243{X\"(t)}

==s2X(s)- s,)

and becausex and x\" are each multiplied by t, application of Eq. (6)yieldsthetransformed equation)

d d--[s2X(s)-s]+ [sX(s)-1]--[Xes)]==o.ds ds)

The result of differentiationand simplificationis the differentialequation)

(S2+ I)X'(s)+sX(s)==o.)

This equation is separable-)

X'es) sXes) s2+1')

its general solution is)

cX (s) == .

v's2+ 1

In Problem 39 we outline the argument that C == 1.BecauseXes) ==\302\243{Jo(t)}, it

follows that)

1\302\243{Jo(t)}

== .v's2+ 1)

(10))

.)))

Example6)


IntegrationofTransformsDifferentiation of F(s) correspondsto multiplication of f (t) by t (togetherwith

a change of sign). It is therefore natural to expectthat integration of F(s) will

correspondto division of J(t)by t. Theorem 3, proved at the end of this section,confirms this, provided that the resulting quotient f (t )/t remains well behavedast \037 0 from the right; that is,provided that)

1. J(t)Im-

t--+O+ t)

existsand is finite.) (11))

THEOREM3 Integrationof Transforms

Supposethat J(t)is piecewisecontinuous for t >0,that f(t) satisfiesthe con-ditionin(11),and that IJ(i)1<Mect as t \037 +00.Then)

\302\243

{J;t)}=

[00F(a)da) (12))


J(t)= \302\243-I{F(s)} = t\302\243-I

{[(X)F(a)da}.) (13))

.... .... .\037...

Find \302\243, {(sinht )/t}.)

Solution We first verify that the condition in (11)holds:)

, . ...Example7)

sinh t et -e-t et +e-tlim = lim = lim = 1,t--+O t t--+O 2t t--+O 2)

with the aid of l'Hopital'srule.Then Eq.(12),with J(t)= sinh t, yields)

{sinh t

} [00 [<X> dO'\302\243,

=\302\243'{sinht} dO'= 2_t s s a 1

l 100

(1 1

)1

[0'-1

]00-- - dO'-- In-2 s 0'-10'+1-2a+ls

.)

Therefore,)

{sinh t

}_ 1 s + 1

\302\243,-- In ,

t 2 s - 1)

becauseIn 1 = O.) .)The form of the integration property in Eq. (13)is often helpful in finding an

inverse transform when the indefinite integral of the transform is easierto handlethan the transform itself.)

Fi\037d\037:/:'

-f{i;i(\037i\037

=\037\037{)Ii:)))

Solution We couldusepartial fractions, but it is much simplerto apply Eq.(13).This gives)

{2S

} {100 2a

}\302\243-1

-t\302\243-1 da

(s2- 1)2-

s (a2- 1)2)

-1{[

-1]

oo

}-1

{I

}=

t\302\243

cr2 _ 1 s=

t\302\243

S2- 1')

and therefore)

1

{2s

}\302\243-

(s2- 1)2)t sinh t .) .)

* ProofsofTheoremsProofofTheorem 1:The transforms F(s) and G(s) existwhen s > c by

Theorem 2 of Section4.1.For any i > 0 the definition of the Laplacetransform

gIves)

G(s)= 100

e-SU g(u)du = 100

e-S(t-T)get -T)dt (u = t - T),)

and therefore)

G(s)= eST100

e-stg(t-T)dt,

becausewe may defineJ(t)and get)to be zero for t < O.Then)

F(s)G(s)= G(s)100

e-STJ(T)dT =100

e-STf(T)G(S)dT

= 100

e-STJ(T)(eST1

00

e-stget -T)dt)

dT

= 100

(100

e-st f(T)g(t-T)dt)

dT.)

Now our hypotheseson f and g imply that the orderof integration may bereversed.(The proof of this requiresa discussionof uniform convergenceof improper inte-grals,and can be found in Chapter 2 of Churchill'sOperationalMathematics,3rded.(New York: McGraw-Hill,1972).)Hence

F(s)G(s)= 100

(100

e-st J(T)g(t-T)dT)dt

= 100

e-st

(itf(T)g(t-T)dT) dt

= 100

e-st [f(t)* get)]dt,)

and therefore,)F(s)G(s)= \302\243{f(t)

* get)}.We replacethe upper limit of the inner integral with t becauseget-i)= 0 wheneveri > t.This completesthe proof of Theorem 1. \037)))


ProofofTheorem2:Because)

F(s)= 100

e-st f(t) dt,)

differentiationunder the integral sign yields)

d 100

F(s)= - e-st f(t) dtds 0)

100 d 1

00= -[e-st f(t)]dt = e-st [-tf(t)]dt;

o ds 0)

thus)

F'(s)=\302\243{ -tf(t)},

which is Eq.(6).We obtain Eq.(7)by applying \302\243-1 and then dividing by -t.Thevalidity of differentiationunder the integral sign dependson uniform convergenceof the resulting integral; this is discussedin Chapter2 of the bookby Churchill justmentioned. \037)

ProofofTheorem3:By definition,)

F(eT)= 100

e-at f(t)dt.)

Sointegration of F(a)from s to +00gives)

[00F(eT)deT =[00(1

00

e-at f(t) dt)deT.)

Underthe hypothesesof the theorem, the orderof integration may be reversed(seeChurchill'sbookonceagain); it follows that)

[00F(eT)deT= 100

([00e-at f(t) deT) dt

100

[-at

]00

= e f(t)dto t a=s)

= 100

e-st

f;t)dt = 4-{f;t)}

.)

This verifiesEq.(12),and Eq.(13)followsupon first applying \302\243-1 and then multi-

plying by t. \037)

11IIProblems)Find the convolution f(t) * g(t)in Problems1through 6.) 6.f(t) = eat, g(t) = ebt (a i= b))

1.f(t) = t, g(t)= 13.f(t) = g(t)= sin t

5. f(t) = g(t)= eat)

2. f(t) = t, g(t)= eat

4. f(t) = t 2, g(t)= cost) Apply the convolution theorem to find the inverse Laplacetransforms of the functions in Problems 7 through 14.)))

17. F(s)= s(s-3)19.F(s)= 2 2(s + 9)

S211.F (s) -- (s2+4)2s13.F(s)=

(s-3)(S2+ 1))

18. F (s) --S(S2+ 4)

110.F (s) --S2(S2+ k2)

112.F(s)=S(S2+ 4s+ 5)

s14.F (s) -- S4+ 5S2+ 4)

In Problems15through 22, apply either Theorem2 or Theo-rem 3 to find the Laplacetransform off(t).15.f(t) = t sin 3t 16.f(t) = t 2 cos2t17.f (t) = te2t cos3t 18.f (t) = te-t sin 2 t

sin t f )1-cos2t19.f(t) =- 20. (t =

t te3t - 1 et -e-t21.f(t) = 22.f(t) =

t t

Find the inverse transforms of the functions in Problems23through 28.)

s-223.F(s)= In s+2S2+ 1

25. F(s)= In(s+ 2)(s-3)

27.F (s)= In (1+s12))

S2+ 124. F(s)= In 2s +4

326. F(s)= tan-1

s+2)s28.F (s) --

(S2+ 1)3)

In Problems 29 through 34, transform the given differentialequation to find a nontrivial solution such that x (0)= o.

29.tx\" + (t -2)x'+ x = 030. tx\" + (3t - I)x'+ 3x = 031.tx\" - (4t + I)x'+ 2(2t+ I)x= 032.tx\" + 2(t - I)x'-2x = 033.tx\" -2x'+ tx = 034. tx\" + (4t -2)x'+ (I3t-4)x = 035.Apply the convolution theorem to show that)

\302\243-1

{I

vis }=

2\037[./Ie-u2 du = e/erf.Jt.

(s- 1) s V JT 10

(Suggestion: Substitute u = ,Jt.))

In Problems 36 through 38, apply the convolution theoremto derive the indicated solution x(t) of the given differentialequation with initial conditions x (0) =x'(0)= o.)

l it

36.x\" + 4x = f(t);x(t)= - f(t -T)sin 2Td-r2 0

37.x\" + 2x'+ x = f(t);x(t)= 1/-re-r f(t - -r)dr

38.x\" + 4x'+ 13x= f(t);l it

x(t)= - f(t -T)e-2Tsin 3TdT3

0)

TermwiseInverseTransformation of SeriesIn Chapter2 of Churchill's OperationalMathematics, the fol-lowing theorem isproved. Supposethat f (t) is continuous fort > 0, that f(t) is ofexponential orderas t --++00,and that)

00anF(s)=L n+k+lS

n=O)

where 0 < k < 1 and the seriesconvergesabsolutely fors > c.Then)

00ant

n+k

f(t)=\037 rcn+k+l)'

Apply this result in Problems39 through 41.39.In Example 5 it was shown that)

C C(

1

)

-1/2\302\243{Jo(t)}

= = - 1+ -y'S2+ 1 s S2)

Expand with the aid of the binomial seriesand then com-pute the inverse transformation term by term to obtain)

00 (_I)nt2n

Jo(t) = C\037 22n(n!)2

.)

Finally, note that Jo(O)= 1 implies that C= 1.40. Expand the function F(s)= S-I/2e-l/s in powersofS-1

to show that)

\302\243-1

{\037e-l/S

}= \037cos2vt.

vis 5i)41.Show that)

\302\243-1

{\037e-I/s}= Jo(2.Jt).)

lIED Per\037odicand.\037ie\037\037\037_\037\037._\037.\037\037t.\037\037.':!\037 \"\037 \037.!.\037p!:l\037.E\037\037 \037t \037\037_....._____________.___.)

Mathematical modelsof mechanical or electricalsystemsoften involve functionswith discontinuitiescorrespondingto external forcesthat are turned abruptly on oroff. Onesuch simpleon-off function is the unit stepfunction that we introduced in)))

x)

.)x =

Ua(t))

a)

FIGURE4.5.1.Thegraph of theunit step function at t = a.)

a)

FIGURE4.5.2.Translation ofJ(t)a units to the right.)

4.5Periodicand PiecewiseContinuousInput Functions 305)

Section4.1.Recall that the unit stepfunction at t = a is definedby)

10 if t < a,ua(t)=u(t-a)= . >11ft_ a.)

(1))

The notation Ua(t) indicatessuccinctly where the unit upward step in value takesplace(Fig.4.5.1),whereasu(t -a) connotesthe sometimesuseful idea of a \"time

delay\" a before the stepis made.In Example8 of Section4.1we saw that if a >0,then)

\037)

e-asoC{u(t-a)}=-.

s)(2))

BecauseoC{u(t)}= lis,Eq.(2)impliesthat multiplication of the transform of u(t)bye-ascorrespondsto the translation t \037 t -ain the originalindependentvariable.Theorem 1 tellsus that this fact, when properly interpreted,is a generalproperty ofthe Laplacetransformation.)

THEOREM1 Translationonthe'-AxisIf oC{f(t)}existsfor s >c,then)

oC{u(t-a)f(t-a)}= e-asF(s)) (3a))

and)

oC-1 {e-asF(s)}= u(t -a)f(t- a)) (3b))

for s >c+a.)

Note that)

u(t -a)f(t-a) =1

0

f(t - a))

if t < a,.f >1 t _ a.)

(4))

Thus Theorem 1 impliesthat oC-1{e-asF(s)}is the function whosegraph for t > ais the translation by a units to the right of the graph of f (t) for t > O.Note that the

part (if any) of the graph of f (t) to the left of t = 0 is \"cut off\" and isnot translated(Fig.4.5.2).In someapplications the function f(t) describesan incoming signalthat starts arriving at time t = O.Then u(t -a)f(t-a)denotesa signalof the same\"shape\" but with a time delay of a, so it doesnot start arriving until time t = a.

ProofofTheorem 1:From the definition of oC{f(t)},we get)

e-asF(s)= e-as100

e-STJC-C)di= 100

e-S(T+a)J(i)di.)

The substitution t = i+a then yields)

e-asF(s)=100

e-st J(t-a) dt.)))

From Eq.(4)we seethat this is the sameas)

e-asF(s)= 100e-S1u(t-a)f(t-a)dt= \302\243(u(t-a)f(t-a)},)

becauseu(t - a)f(t- a) == 0 for t < a.Theorem 1.)

This completesthe proof of\037)

Example1) With J(t)== 4t2, Theorem 1 gives)

{e-as

}1

10

\302\243-1_ ==u(t -a)-(t-a)2==s3 2 4(t-a)2)

if t <a,if t > a)

(Fig.4.5.3)..)

Example2)\037\037 \"_'\037'_' \037.... ;,\";,'\" \037,...... ..........\"........,.......,. ........ ...............\037'n.'n.'n_\"\"\"\",,,,,,,,, \"'.,;,..... \037\"'\" ,...-....,,,.. \"..ou,\", .... ..... ........\" ..._..,............. ... nn..... \"\"',............. ..... \",Mo. \"\"'..... \"\"',......... .. --..._... ...........'\"\"\"',...,. ......,,................. N \"'.....,,\"\"'... '\" \"'\" \"',...... ... \"\"\"'--............ __.. ,___ ........... .........................,...... ....\" ........ \037..............., .., \037 ... ........... \037 '\"'

Find \302\243{g(t)} if)

get)=1\0372)

if t < 3,if t > 3)

(Fig.4.5.4).)

Solution Beforeapplying Theorem 1,we must first write get)in the form u(t -3)f(t-3).The function f(t) whosetranslation 3 units to the right agrees(for t > 3) with

get)== t 2 is f(t) == (t +3)2becauseJ(t-3)= t 2.But then

2 6 9F(s)==

\302\243{t

2 +6t +9}==3\" + 2\" +-,s s s)

sonow Theorem 1 yields)

(2 6 9

)\302\243{g(t)}==

\302\243{u(t-3)J(t-3)}==e-3sF(s)==e-3s

3\" + 2\" +- .s s s)

.)

Example3) Find \302\243{f(t)} if

f(t) =I \037os

2t) if 0 < t <2n,if t >2n)

(Fig.4.5.5).)

x)

x)

5)2 \",/ \\

x=t // x=g(t)

-----;/)

\037x =f(t))x) 20)

15)

Tt) 2Tt) 3Tt t)

10)

1 2x ='2ua(t)(t

-a))

a) 1) 2) 3) 4)t)

FIGURE4.5.3.Thegraph ofthe inverse transform ofExample 1.)

FIGURE4.5.4.Thegraph of thefunction g(t) ofExample 2.)

FIGURE4.5.5.Thefunction

J(t) ofExamples 3 and 4)))

Solution We note first that)

...Example4)

f(t) ==[1- u(t -2n)]cos2t ==cos2t - u(t -2n)cos2(t -2JT))

becauseof the periodicityof the cosinefunction. HenceTheorem 1 gives)

s(l-e-2ns )oC{f (t)}== oC{cos2t}-e-2nsoC{cos2t} == 2 .

s +4).)

.....\037 \"\"\" .....\"\"\"',___ N ,___,___ ,.., ..... \037\"\" ,-..0,..,,___ '-' \"\" -.. ,___ ,.-<'\" '\"'..... '\"' '-' \037 '-' '\"

A massthat weighs32 lb (massm == 1 slug)is attached to the free end of a long,light spring that is stretched 1 ft by a force of 4 lb (k ==4 lb1ft).The massis initiallyat rest in its equilibrium position.Beginning at time t == 0 (seconds),an externalforce f(t) ==cos2t is appliedto the mass,but at time t ==2JT this force is turned off

(abruptly discontinued)and the massis allowed to continue its motion unimpeded.Find the resulting position function x(t) of the mass.)

Solution We needto solve the initial value problem)

n)

nl2)x =n 12)/

/)

-n 12)'--

x =-n12

2n 4n 6nt)

-no)

FIGURE4.5.6.Thegraph of thefunction x(t)of Example 4.)

xl!+4x ==f(t); x(O)==x/CO)==0,)

where f (t) is the function of Example3.The transformedequation is)

s(l-e-2ns )(s2+4)X(s)==F(s)== 2 's +4)

so)s s

X (s)=(s2+4)2

-e-2ns

(s2+4)2.)

Because)-1

{S

}1 . 2oC 2 2

== 4t Sin t(s +4)

by Eq.(16)of Section4.3,it follows from Theorem 1 that)

x(t) ==\037

t sin 2t - u (t -2n) .\037

(t -2n) sin 2(t -2JT)

==\037

[t - u (t -2n) .(t -2n)]sin 2t .)

If we separatethe casest < 2n and t >2n,we find that the position function maybe written in the form)

I\037

t sin 2tx(t) ==

!nsin 2t)

if t < 2n,)

if t >2n.)

As indicated by the graph of x(t) shown in Fig.4.5.6,the massoscillateswith

circular frequency w == 2 and with linearly increasing amplitude until the forceis removed at time t == 2n.Thereafter, the masscontinues to oscillatewith thesamefrequency but with constant amplitude n12.The force F(t) == cos2t would

producepure resonanceif continued indefinitely, but we seethat its effect ceasesimmediatelyat the moment it is turned off. .)))

\037)EO)

T)

Example5)

If we were to attack Example4 with the methods of Chapter 2, we would

need to solve one problemfor the interval 0 < t < 2n and then solve a new

problemwith different initial conditionsfor the interval t >2n.In such a situation

the Laplacetransform method enjoys the distinct advantage of not requiring the

solution of differentproblemson different intervals.)

......

Considerthe RLCcircuit shown in Fig.4.5.7,with R = 110Q,L = 1H,C = 0.001F, and a battery supplying Eo = 90V. Initially there is no current in the circuitandno charge on the capacitor.At time t = 0 the switch is closedand left closedfor1 second.At time t = 1 it is openedand left openthereafter. Find the resultingcurrent in the circuit.)

Solution We recallfrom Section2.7the basicseriescircuit equation)L)

'VV'vR)

T

C)

FIGURE4.5.7.TheseriesRLCcircuit ofExample 5.)

di 1L-+ Ri + -q= e(t);dt C)

(5))

we use lowercaseletters for current, charge, and voltage and reserve uppercaselet-ters for their transforms. With the given circuit elements,Eq.(5)is)

didt

+ 110i+ 1000q= e(t),

where e(t) = 90[1- u(t - 1)],correspondingto the opening and closingof the

switch.In Section2.7our strategy was to differentiateboth sidesof Eq.(5),then apply

the relation)

(6))

. dq1 =-

dt)(7))

to obtain the second-orderequation)

d2i di 1. ,L-+R- +-1= e (t).dt2 dt C)

Herewe do not use that method, becausee'(t)= 0 exceptat t = 1,whereas the

jump from e(t) = 90when t < 1 to e(t) = 0 when t > 1 would seemto requirethat e'(1)= -00.Thus e'(t) appearsto have an infinite discontinuity at t = 1.Thisphenomenon will be discussedin Section4.6.For now, we will simply note that it

is an odd situation and circumvent it rather than attempt to dealwith it here.To avoid the possibleproblemat t = 1,we observethat the initial value

q (0)= 0 and Eq.(7)yield,upon integration,

q(t) = 11

i(i)di.) (8))

We substitute Eq.(8)in Eq.(5)to obtain)

di 1 itL-+Ri+- i(-r)d-r=e(t).dt C 0)

(9))

This is the integrodifferentialequationof a seriesRLC circuit; it involves both

the integral and the derivativeof the unknown function i (t).The Laplacetransformmethod workswell with such an equation.)))

In the presentexample,Eq.(9)is)

di 1t-+ 110i+ 1000 i(-r)d-r==90[1- u(t -1)].dt 0)

(10))

Because)

{[t

}/ (s)

\302\243.

10i(-r)d-r =---;-)

by Theorem 2 of Section4.2on transforms of integrals, the transformedequation is)

/ (s) 90s/(s)+ 110/(s)+1000-==-(1-e-S).s s)

We solve this equation for / (s) to obtain)

90(1-e-S)I(s)= s2+ lIOs+ 1000.)

But)

90s2+ 11Os+ 1000)

1

s + 10)

1

s + 100')

sowehave)

1 1 -s(

1 1

)I(s)= s+lO-

s+l00-e s+10- s+100.)

We now apply Theorem 1 with f(t) ==e-10t -e-100t; thus the inversetransform is)

i(t) ==e-10t -e-100t - u(t - 1)[e-10(t-l)_ e-100(t-I)].)

After we separatethe casest < 1 and t > 1,we find that the current in the circuit isgiven by)

I

-l Ot -lOOte -ei (t) == e-10t _ e-10(t-l)_ e-100t +e-100(t-l))

if t < 1,if t > 1.)

The portion e-10t -e-100t of the solution would describethe current if the switchwereleft closedfor all t rather than beingopen for t > 1. .)))

I

I

I

I I

\037P-J I)

FIGURE4.5.8.Thegraph of afunction with periodp.)

TransformsofPeriodicFunctions)Periodicforcing functions in practical mechanical or electricalsystemsoften aremore complicatedthan pure sinesor cosines.The nonconstantfunction f (t) definedfor t >0is saidto beperiodicif there is a number p >0such that)

f(t +p) == f(t)) (11))

for all t >O.The leastpositivevalueof p (if any) for which Eq.(11)holdsiscalledthe periodof f. Sucha function is shown in Fig.4.5.8.Theorem 2 simplifiesthe

computationof the Laplacetransform of a periodicfunction.)

THEOREM2 Transformsof PeriodicFunctions)Let f(t) beperiodicwith periodp and piecewisecontinuousfor t >O.Then thetransform F(s)== oC{f(t)}existsfor s >0and is given by)

1 1p

F(s)== e-st f(t) dt.1-e-PS0)

(12))

Proof:The definition of the Laplacetransform gives)

100 00 lcn+l)p

F(s)== e-st f(t) dt ==L e-st f(t) dt.o n=O np)

The substitution t ==i+np in the nth integral followingthe summation sign yields)

lcn+l)p

1p

1p

e-st f(t) dt == e-sc-r+np)f(i +np)di==e-nps e-STf(i)dinp 0 0)

becausef(i+np) ==f (i) by periodicity.Thus)

F(s)=\037 (e-

nps lPe-srI(r:)dr:)

= (1+e-Ps+e-2ps+ ...)lPe-ST:I(r:)dr:.)

Consequently,)

F(s)=1 (Pe-ST:I(r:)dr:.

1 -e-PS 10)

We use the geometricseries)

1 23==1+x+x+x +...,I-x)with x == e-Ps < 1 (for s > 0) to sum the seriesin the final step. Thus we havederivedEq.(12). .)

The principal advantageof Theorem 2 is that it enablesus to find the Laplacetransform of a periodicfunction without the necessityof an explicitevaluationof an

improperintegral.)))

Example6)

/(t)1 --.0 --.0 .-...)

a 2a 3a 4a 5a 6a-1 --.0 --.0 --.0)


square-wavefunction of Example6.)

Example7)

g(t))

a)

a 2a 3a 4a 5a 6a t)

FIGURE4.5.10.Thegraph ofthe triangular-wave function ofExample 7.)

Example8)

4.5Periodicand PiecewiseContinuousInput Functions 311)..,.. ,..,,, .

Figure 4.5.9showsthe graph of the square-wave function f(t) == (_I)[t/aD ofperiodp == 2a;[x]denotesthe greatest integer not exceedingx. By Theorem2 the Laplacetransform of f (t) is)

t) F(s)=1

2(2ae-slf(t) dt

1-e-as Jo)

1

(r e-S1dt+ 1

2a(-l)e-S1dt )1-e-2as Jo a)

1)

([-;e-SIJ:

- [-;e-slIa))1 -e-2as)

(1-e-as)2

s(1-e-2as))

1-e-as

s(1+e-as))

Therefore,)

I -as-eF(s) - s(1+e-as))

(13a))

eas/2-e-as/2 1 as==- tanh -

s(eas/2 +e-as/2) s 2)(13b)

.)

Figure4.5.10showsthe graph of a triangular-wavefunction g(t) of periodp == 2a.Becausethe derivativeg/ (t) is the squarewave function of Example6,it followsfrom the formula in (13b)and Theorem 2 of Section4.2that the transform of this

triangular-wavefunction is)

F(s) 1 asG(s)== ==2tanh-.

s s 2)(14)

.)

\" v '.on.... v \"'.... \037 \037,....\" \037 \037 __ ... v\"\"'\037\037 \" ,... ,..,. \"\"\037.'N\"\" \" ,... \" '\" V\" \" \"'.... \037 '\" \"\" n.';\037\" \037n\037 v \"\"\037

Considera mass-spring-dashpotsystem with m == 1,c == 4, and k == 20 in ap-propriate units. Supposethat the system is initially at rest at equilibrium (x (0) ==

x/ (0) == 0) and that the massis acted on beginning at time t == 0 by the externalforce f (t) whosegraph is shown in Fig.4.5.11:the squarewavewith amplitude 20and period2n.Find the position function f (t).)

Solution The initial value problemis)

x\" +4x'+20x==f(t); x(O)==x/(O)== O.)

The transformedequation is)

S2X(s)+4sX(s)+20X(s)== F(s).) (15))))


f(t)20) .---.0

I I

I I

I I

I I

I I

I I)

.---.0I I

I I

I I

I I

I I

I I)

\037I

I

I

I

I

I)

1t 21t 31t 41t 51t 61t tI I I I I I

I I I I I I

I I I I I I

I I I I I I

-20 \037 \037 \037)

FIGURE4.5.11.Thegraph ofthe external-forcefunction ofExample 8.)

From Example6 with a ==n we seethat the transform of f(t) is)

20 1-e-ns

F(s)==-.s 1 +e-ns)

20==-(1-e-ns

) (1-e-ns +e-2ns _ e-3ns + ...)s)

20==-(1-2e-ns +2e-2ns _ 2e-3ns + ...) ,

s)

so that)

20 40 00

F(s)==-+-L(-I)ne-nns.s S

n=1)

(16))

Substitution ofEq.(16)in Eq. (15)yields)

F(s)X (s) -- s2+4s +20)

20 00 20e-nns=

s[(s+2)2+ 16]+28(-l)ns[(s+2)2+ 16]\"

(17))

From the transform in Eq.(8)of Section4.3,we get

L-1

{20

}==5e-2t sin 4t

(s+2)2+ 16 ')

soby Theorem 2 of Section4.2we have)

g(t) ==L-1

{ 2\302\2602 }= r 5e-2,sin4-rdi.s[(s+2) + 16] 10)

Usinga tabulated formula for f eat sin bt dt, we get)

g(t) == 1 -e-2t(cos4t +!sin 4t) == 1 - h (t ) ,) (18))

where)

h(t) ==e-2t(cos4t + 4 sin 4t) .) (19))

Now weapply Theorem 1 to find the inverse transform of the right-hand termin Eq.(17).The result is)

00x(t) ==g(t)+2 L(-I)nu (t -nn)g(t-nn),

n=1)

(20))

and we note that for any fixedvalue of t the sum in Eq.(20)is finite. Moreover,

g(t-nn) == 1 -e-2(t-nn) [cos4(t-nn) + 4 sin4(t-nn)])

== 1 -e2nn e-2t(cos4t + 4 sin 4t) .)))

Therefore,)

g(t-nn) == 1-e2n:rr:h(t).) (21))

Henceif 0 < t <n,then)

X(t) == 1-h(t).)

If n < t <2n,then)

x(t) ==[1- h (t)]-2 [1-e2:rr: h (t )] ==-1 + h (t)-2h (t) [1-e2:rr:]

.)

If 2n < t <3n,then)

x(t) ==[1- h (t)]-2 [1-e2:rr: h (t)]+2 [1-e4:rr: h (t )]== 1 +h(t)-2h(t)[1-e2:rr: +e4

:rr:].)

The general expressionfor nn < t < (n + l)n is)

x(t) ==h(t)+ (-l)n-2h(t)[1-e2:rr: +...+ (_1)ne2n:rr:])

1 + (-l)ne2(n+ l):rr:

==h (t)+ (-1)n -2h(t ) 2 'l+e:rr:)(22))

which we obtained with the aid of the familiar formula for the sum of a finite ge-ometric progression.A rearrangement of Eq. (22)finally gives, with the aid ofEq.(19),)

e2:rr: - 1x(t) ==

2 e-2t(cos4t + _

21 sin 4t) + (_l)ne:rr:+1

2 .(-1)ne2:rr:- e-2(t-n:rr:)(cos4t +!sin 4t)e2:rr: + 1 2) (23))

for nn < t < (n + l)n.The first term in Eq.(23)is the transient solution)

Xtr(t) \037 (O.9963)e-2t(cos4t+ 4 sin4t) \037 (1.1139)e-2t cos(4t-0.4636).(24))

The last two terms in Eq. (23)give the steady periodicsolutionxsp.To investigateit, we write i == t -nn for t in the interval nn < t < (n + l)n.Then)

Xsp(t) ==(-l)n[1 - 22e2:n:

e-2,(cos4r + 4 sin 4r)]e :rr: + 1 (25)\037 (-l)n[1- (2.2319)e-2TCOS(4i-0.4636)].)

Figure 4.5.12showsthe graph of xsp(t).Its most interesting feature is the appear-anceof periodicallydampedoscillationswith a frequencyfour times that of the

imposedforce f(t).In Chapter 8 (Fourier Series)we will seewhy a periodicex-ternal force sometimesexcitesoscillationsat a higher frequency than the imposedfrequency. .)))

Xsp)

\\

\\

\\)

\\

\\

\\ 1 + (2.23)e-2t

\037/\"'- -)

1) /1/\\

/ 1 -(2.23)e-2t)

-1)

-1+ (2.23)e-2(t-n)

II)\\

\\

\\

\\

\" \ -1 v\\I -1-(2.23)e-2(t -n)

I)

FIGURE4.5.12.Thegraph of the steady periodicsolution for

Example 8;note the \"periodically damped\" oscillationswith

frequency four times that of the imposedforce.)

lID.Problems.)

Find the inverse Laplacetransform f(t) ofeachfunction givenin Problems 1through 10.Then sketch the graph of f.)

e-3s1.F(s)= \037s

e-S3.F(s)= s+2)

e-S _ e-3s2.F(s)=)

S2-S 2-2se -e)4. F(s)=)s- 1)

e-ns5. F(s)=

S2 + 11-e-271:S

7. F(s)=S2 + 1

s(1+ e-3s)9.F(s)= 2 2S + n)

se-S6.F (s)=2 2S + n

s(1-e-2s)8.F(s)= 2 2S + n2s(e-71: S _ e-271:s)10.F(s)= S2+ 4)

Find the Laplacetransforms ofthe functions given in Problems11through 22.

11.!(t) = 2 if 0 < t < 3; f (t) = 0 if t > 312.!(t)= 1 if 1 < t < 4; f(t) = 0 if t < 1or if t > 413.!(t) = sin t if 0 < t < 2n;f (t) = 0 if t > 2n14.f(t) = cosntif 0 < t < 2; f(t) = Oift > 215.!(t)= sint if 0 < t < 3n;f(t) = 0 ift > 3n16.f (t) = sin 2t if n < t < 2n;f (t) = 0 if t < n or if

t > 2n17.f (t) = sin n t if 2 < t < 3;f (t) = 0 if t < 2 or if t > 318.f(t) = cos4nt if3 < t < 5; f(t) = Oift < 3 orift > 519.f (t) = 0 if t < 1;f (t) = t if t > 120.f (t) = t if t < 1;f (t) = 1if t > 121.f(t) = t if t < 1;f(t) = 2- t if 1 < t < 2; f(t) = 0 if

t > 222.f(t) = t 3 if 1 < t < 2; f(t) = 0 if t < 1or if t > 2)

23.24.25.)

Apply Theorem2 with p = 1 to verify that oC{ I}= 1/s.Apply Theorem2 to verify that oC{coskt} = S/(S2+ k2).Apply Theorem2 to show that the Laplacetransform ofthe square-wavefunction ofFig.4.5.13is)

1oC{f(t)}= s(1+ e-as))

r-I r-I r-I I I I I)

t)a 2a 3a 4a 5a 6a)


square-wavefunction ofProblem 25.)

26.Apply Theorem2 to show that the Laplacetransform ofthe sawtooth function f (t) ofFig.4.5.14is)

1 e-as

F(s)=-- .as2 s(l-e-as

))

J(t))

a 2a 3a 4a 5a 6a) t)

FIGURE4.5.14.Thegraph of thesawtooth function ofProblem 26.)))

27.Letget) bethe staircasefunction ofFig.4.5.15.Show that

get) = (tja)- f(t), where f is the sawtooth function ofFig. 4.5.14,and hencededucethat)

e-as

\302\243{g(t)}= s(1-e-as)

')

get))

4) ,...)3)

2)

a) 2a) 4a t)3a)

FIGURE4.5.15.Thegraph of thestaircasefunction ofProblem 27.)

28.) Supposethat f (t) is a periodicfunction ofperiod2a with

f(t) = t if 0 < t < a and f(t) = 0 if a < t < 2a.Find

oC{f(t)}.Supposethat f (t) is the half-wave rectification of sin kt,shown in Fig. 4.5.16.Show that)

29.)

k\302\243{f (t)} =

(S2+ k2)(1 _ e-1Csfk).)

J(t))

7r

k)

27r

k)

t)37r

k)

FIGURE4.5.16.Thehalf-wave rectification ofsin kt .)

30.Let g(t) = u (t - n jk) f (t - njk), where f (t) is thefunction of Problem 29 and k > O. Note that h(t) =f(t) + get) is the full-wave rectification of sinkt shownin Fig.4.5.17.Hencededucefrom Problem29 that)

k nsoC{h(t)}= 2 2 coth -.s +k 2k)

h(t))

7r

k)

37r

k)

FIGURE4.5.17.Thefull-wave rectification ofsin kt .)

In Problems31through 35, the values ofmass m, spring con-stant k, dashpot resistancec, and force f (t) are given for a)

mass-spring-dashpotsystem with external forcing Junction.Solvethe initial value problem)

mx// + cx/+ kx = f(t), x(O)= x/CO)= 0)

and construct the graph of the position function x(t).

31.m = 1,k = 4, c = 0; f(t) = 1 if 0 < t < n, f(t) = 0 if

t > n32.m = 1,k = 4, c = 5; f(t) = 1 if 0 < t < 2, J(t) = 0 if

t > 233.m = 1,k = 9, c = 0; f(t) = sin t if 0

\037t < 2n,

f (t) = 0 if t > 2n34. m = 1,k = 1,c = 0; f (t) = t if 0 < t < 1,J(t) = 0 if

t > 135.m = 1,k = 4, c = 4; f(t) = t if 0 < t < 2, f(t) = 0 if

t > 2)

In Problems 36 through 40, the values of the elements ojan

RLCcircuit aregiven. Solvethe initial value problem)

di 1 itL-+ Ri + - i(-r) di= e(t);dt C 0

with the given impressedvoltage e(t).

36.L = 0, R = 100,C = 10-3;e(t) = 100if 0 ::::t < 1;e(t) = 0 if t > 1

37.L = 1,R = 0, C = 10-4; e(t) = 100if 0 < t < 2n;e(t) = 0 if t > 2n

38.L = 1,R = 0, C = 10-4;e(t) = 100sin lOt if

o < t < n; e(t) = 0 if t > n39.L = 1,R = 150,C = 2 X 10-4;e(t) = lOOt if 0 ::::t < 1;

e(t) = Oift > 140. L = 1,R = 100,C = 4 X 10-4;e(t) = 50t if 0

\037t < 1;

e(t) = 0 if t >1)

i (0) = 0)

In Problems 41and 42, a mass-spring-dashpotsystem with

external force f(t) is described.Under the assumption that

x (0) = x/(0) = 0, use the method of Example 8 to find thetransient and steadyperiodicmotions of the mass. Then con-struct the graph of the position function x(t).Ifyou would like

to checkyour graph using a numerical DE solver, it may beuseful to note that the function)

f(t) = A[2u((t -n)(t-2n)(t -3n) .

(t -4n)(t-5n)(t -6n))- 1])

has the value +A if 0 < t < n, the value -A ifn < t < 2n,and soforth, and henceagreeson the interval [0,6n] with

the square-wavefunction that has amplitude A andperiod2n.(Seealso the definition ofa square-wavefunction in terms ojsawtooth and triangular- wave functions in the application ma-terialfor this section.)

41.m = 1,k = 4, c = 0; f (t) is a square-wave function with

amplitude 4 and period2n.42. m = 1,k = 10,c = 2; f(t) is a square-wave function

with amplitude 10and period 2n.)))

lID !\037pulsesand DeltaFunctions)Considera force f(t) that acts only during a very short time interval a < t < b,with f(t) = 0outsidethis interval. A typical examplewould be the impulsiveforceof a bat striking a ball-theimpact is almost instantaneous. A quicksurgeof voltage (resulting from a lightning bolt, for instance) is an analogous electricalphenomenon.In such a situation it often happensthat the principal effect of theforce dependsonly on the value of the integral)

p =lb

f(t)dt) (1))

and doesnot dependotherwiseon preciselyhow f(t) varies with time t.The num-berp in Eq.(1)is calledthe impulseof the force f(t) over the interval [a,b].

In the caseof a force f(t) that acts on a particle of massm in linear motion,integration of Newton'slaw)

df(t) = mv/(t) = -[mv(t)]

dt)

yields)

lbdp = -[mv(t)] dt = mv(b) -mv(a).

a dt)(2))

Thus the impulseof the force is equal to the change in momentum of the particle.Soif change in momentum is the only effect with which weare concerned,we needknow only the impulseof the force;weneedknow neither the precisefunction f (t)nor even the precisetime interval during which it acts.This is fortunate, becausein a situation such as that of a batted ball, we are unlikely to have such detailedinformation about the impulsiveforce that acts on the ball.

Our strategy for handling such a situation is to setup a reasonablemathemat-ical modelin which the unknown force f(t) is replacedwith a simpleand explicitforce that has the sameimpulse.Supposefor simplicity that f (t)has impulse 1 andacts during somebrief time interval beginning at time t = a > O. Then we canselecta fixed number E > 0that approximates the length of this time interval and

replacef (t) with the specificfunction)

x)1)

if a < t <a +E,)I--\342\202\254---1r

ITI

I I

I I

I I

I Area = 1 I!I I \342\202\254

I I

I I

I I

I I)

da,E (t) =) E) (3))

o otherwise.)

This is a function of t, with a and E beingparameters that specify the time interval

[a,a +E].If b >a +E, then we see(Fig.4.6.1)that the impulseof da,E over [a,b]IS)

a a+\342\202\254)

t) lb la+E 1p= da,E(t)dt= -dt=l.a a E

Thus da,E has a unit impulse,whatever the number E may be.Essentiallythe samecomputationgives)


impulse function da,E (t).)

100

da,E(t)dt= 1.) (4))))

Function

get),)

FIGURE4.6.2.A diagramillustrating how the delta function\"sifts out\" the value g(a).)

4.6Impulsesand DeltaFunctions 317)

Becausethe precisetime interval during which the force acts seemsunimpor-tant, it is tempting to think of an instantaneous impulse that occurspreciselyat the

instant t = a.We might try to model such an instantaneousunit impulseby takingthe limit as E \037 0,thereby defining)

8a(t) = limda,E(t),E\037O)

(5))

where a >O.If we couldalsotake the limit under the integral sign in Eq.(4),then

it would follow that)

100

8a(t)dt= 1.) (6))

But the limit in Eq.(5)gives)

1+00

8a(t) =0)

if t = a,if t i= a.)

(7))

Obviously, no actual function can satisfy both (6)and (7)-ifa function is zeroexceptat a singlepoint, then its integral is not 1 but zero.Nevertheless,the symbol8a(t) is very useful. Howeverinterpreted, it is calledthe Diracdelta function at aafter the British theoreticalphysicistP. A. M.Dirac (1902-1984),who in the early1930sintroduceda \"function\" allegedlyenjoying the propertiesin Eqs.(6)and (7).)

DeltaFunctionsasOperatorsThe following computation motivates the meaning that we will attach here to the

symbol 8a(t).If g(t) is continuous function, then the mean value theoremfor inte-

gralsimpliesthat)

la+\342\202\254

g(t)dt = Eg (1)

for somepoint t in [a,a +E].It follows that)

100 la+E 1

lim g(t)da,E(t) dt = lim g(t) .-dt = lim g (t) = g(a)E\037O 0 E\037O a E E\037O)

(8))

by continuity of g at t = a. If 8a(t) were a function in the strict senseof the

definition, and if we could interchange the limit and the integral in Eq. (8),wetherefore couldconcludethat)

100

g(t)8a(t)dt = g(a).) (9))

We take Eq. (9)as the definition (!) of the symbol 8a(t).Although we call it

the delta function, it is not a genuine function; instead, it specifiesthe operation)

100...8a(t) dt)

which-whenappliedto a continuous function g(t)-siftsout or selectsthe value

g(a) of this function at the point a > O. This idea is shown schematicallyin

Fig.4.6.2.Note that we will use the symbol 8a(t) only in the contextof integralssuch as that in Eq.(9),or when it will appear subsequentlyin such an integral.)))


For instance, if we take g(t) = e-st in Eq.(9),the result is)

100

e-S18a(t)dt = e-as.) (10))

We therefore definethe Laplacetransform of the delta function to be)

oC{8a(t)}= e-as (a >0).) (11))

If wewrite)

8(t)= 8o(t) and 8(t -a) = 8a(t),) (12))

then (11)with a = 0 gives)

oC{8(t)}= 1.) (13))

Note that if 8(t)werean actual function satisfying the usual conditionsfor existenceof its Laplacetransform, then Eq.(13)would contradict the corollary to Theorem2of Section4.1.But there is no problemhere;8(t) is not a function, and Eq.(13)isour definition of oC{8(t)}.

DeltaFunctionInputsNow, finally, supposethat we are given a mechanical system whoseresponsex(t)to the external force f(t) is determinedby the differentialequation)

Ax\" +Bx'+Cx= f(t).) (14))

To investigate the responseof this system to a unit impulseat the instant t = a, it

seemsreasonableto replacef(t) with 8a(t) and begin with the equation)

Ax\" +Bx'+Cx= 8a(t).) (15))

But what is meant by the solution of such an equation? We will call x(t) a solutionof Eq.(15)provided that)

x(t) = lim XE (t),E\037O)

(16))

where XE (t) is a solution of)

Ax\" +Bx'+Cx= da,E(t).) (17))

Because)

1da E(t) = - [ua(t)-ua+E(t)],

E)(18))

is an ordinary function, Eq. (17)makessense.For simplicity supposethe initial

conditions to be x(O) = x/(0) = O. When we transform Eq. (17),writing XE =oC{xE },we get the equation)

1

(e-as e-(a+E)S

)1-e-SE

(As2 +Bs+C)XE(s)= - -- = (e-as

) .E S S SE)))

o , ,

Example1)


If we take the limit in the last equationas E \037 0,and note that)

1 -SE-e) = 1)limE\037O SE)

by I'Hopital'srule, we get the equation)

(As2 + Bs+C)X(s)= e-as

,) (19))

if)

Xes)= lim XE(x).E\037O)

Note that this is preciselythe sameresult that we would obtain if we transformed

Eq.(15)directly,using the fact that oC{oa(t)}= e-as.On this basisit is reasonableto solve a differentialequation involving a delta

function by employing the Laplacetransform method exactly as if oa(t) were an

ordinary function. It is important to verify that the solution so obtained agreeswith the one defined in Eq.(16),but this dependson a highly technicalanalysisofthe limiting proceduresinvolved; we considerit beyond the scopeof the presentdiscussion.The formal method is valid in all the examplesof this sectionand will

producecorrectresultsin the subsequentproblemset.)

A massm = 1 is attached to a spring with constant k = 4; there is no dashpot.Themassis releasedfrom restwith x(O)= 3.At the instant t = 2Jr the massis struck

with a hammer, providing an impulsep = 8.Determine the motion of the mass.)

Solution According to Problem 15,we needto solve the initial valueproblem)

x\" +4x = 802n (t); x(O)= 3, x'(0)= O.)

We apply the Laplacetransform to get)

S2X (s)-3s +4X(s)= 8e-2ns,)

so)3s 8e-2ns

X (s)=2 + 2 .

s +4 s +4Recalling the transforms of sineand cosine,as well as the theoremon translations

on the t-axis(Theorem 1 of Section4.5),we seethat the inversetransform is)

x(t) = 3cos2t +4u(t - 2Jr) sin 2(t -2Jr)= 3cos2t +4U2n (t)sin 2t.)

Because3cos2t+4sin2t= 5cos(2t-a) with a = tan- 1 (4/3) \037 0.9273,sepa-ration of the casest <2Jr and t >2Jr gives)

( )1

3cos2tx t \037

5 cos(2t-0.9273))if t <2Jr,if t >2Jr.)

The resulting motion is shown in Fig.4.6.3.Note that the impulseat t = 2n resultsin a visible discontinuity in the velocity at t = 2Jr, as it instantaneously increasesthe amplitude of the oscillationsof the massfrom 3 to 5. .)))

t= 2n)

6)

x=3)

\037 0)

x= -3)-- - x = -5)

-6)

o) 2Jt) 4Jtt)

6Jt)

FIGURE4.6.3.Themotion of the mass ofExample 1.)

DeltaFunctionsandStepFunctionsIt isuseful to regard the delta function oa(t)asthe derivativeof the unit stepfunction

Ua(t). To seewhy this is reasonable,considerthe continuousapproximationUa,E (t)to Ua(t) shown in Fig.4.6.4.We readily verify that)

x)

d-Ua E (t) = da E (t).dt' ,)

Because)a a+\302\243)

t) Ua(t) = lim Ua,E (t) and oa(t) = lim da,E (t),E\037O E\037O

an interchangeof limits and derivativesyields)FIGURE4.6.4.Approximationofua(t) by Ua,E(t).)

d . d .-ua(t)= hm -uaE(t)= hmdaE(t),dt E\037O dt' E\037O')

and therefore)

d-ua(t)= oa(t) = o(t -a).dt)

(20))

We may regard this as the formal definition of the derivativeof the unit stepfunction,

although Ua(t) is not differentiable in the ordinary senseat t = a.)

Example2).-<.d., ...... _

We return to the RLCcircuit of Example5 of Section4.5,with R = 110Q, L =1 H, C = 0.001F, and a battery supplying eo = 90V. Supposethat the circuit isinitially passive-nocurrent and no charge.At time t = 0 the switch is closedandat time t = 1 it is openedand left open.Find the resulting current i (t) in the circuit.)

Solution In Section4.5we circumventedthe discontinuity in the voltageby employingthe in-

tegrodifferentialform of the circuit equation. Now that delta functions are available,we may beginwith the ordinary circuit equation)

1Li\" + Ri'+-i= e'(t).

C)

In this examplewe have)

e(t) = 90-90u(t- 1)= 90-90Ul(t),)))

Example3)


soe'(t) = -900(t-1)by Eq.(20).Hencewe want to solve the initial value problem

i\" + 11Oi'+ IOOOi= -900(t -1); i (0)= 0, i'(0)= 90. (21)The fact that i'(0)= 90comesfrom substitution of t = 0 in the equation

1Li'(t)+Ri(t)+ Cq(t) = e(t)

with the numericalvalues i (0)= q (0)= 0 and e(O)= 90.When we transform the problem in (21),we get the equation

s2/(s)-90+ 110s/(s)+ 1000/(s)==-90e-s .)

Hence)90(1-e-S)I(s)= S2+ lIOs+ 1000.

This is preciselythe sametransform I(s)we found in Example5 of Section4.5,soinversionof I(s)yieldsthe samesolution i (t) recordedthere. .)

\037 \"\"_\"f\",\037,'''''.?'..N''N..N.'..''.........,.....,,,,,,,,,,,,,..\037.....,\037\037,,, ..........,,,v\"\"\".'.'..N..u..u..nnun\037..u.. '.... ''''_\037\037''''\037\037'\"' '\"' \037\037\037\037\"\"\"\" \" \" \" \"\" \"\037\",-\"..\037\".....,,....,.,, \"\" ... __ ,... \"\" '-\"'\" \"\" \".....,\037.....,..,..,,... c\"'''

Considera masson a springwith m = k = 1 and x(0)= x'(0)= O.At each of the

instants t = 0,Jr, 2Jr, 3Jr, ..., nJr, ..., the massis struck a hammer blow with aunit impulse.Determine the resulting motion.)

Solution We needto solve the initial value problem)

x(t))

n 2n 3n 4n)

FIGURE4.6.5.Thehalf-waverectification of sin t.)

00

x\" +x = Lonn(t); x(O)= 0 = x'(O).n=O)

BecauseoC{onn (t)}= e-nn s, the transformedequation is)

00

s2X(s)+ Xes)= Le-nns,n=O)

so)00 e-nn s

Xes)= L 2 .n=O

S + 1

We compute the inverseLaplacetransform term by term; the result is)

00

x(t) = Lu(t -nJr) sin(t -nJr).n=O)

Becausesin(t- nJr) = (-I)nsin t and u(t - nJr) = 0 for t < nJr, we seethat ifnJr < t < (n + I)Jr,then

x(t) = sin t - sin t + sin t - ...+ (-1)n sin t;)

that is,)

x(t) ==

losin t if n is even,

if n is odd.

Hencex(t) is the half-waverectificationof sin t shown in Fig.4.6.5.The physicalexplanation is that the first hammer blow (at time t = 0)starts the massmoving to

tthe right; just as it returns to the origin, the secondhammer blow stopsit dead;it

remains motionlessuntil the third hammer blow starts it moving again, and soon.Of course,if the hammer blowsare not perfectly synchronized then the motion ofthe masswill be quite different. .)))


SystemsAnalysisandDuhamel'sPrincipleConsidera physicalsystem in which the output or responsex(t) to the input func-tion f (t) is describedby the differentialequation)

axil +bx'+ex= f(t),) (22))

where the constantcoefficientsa,b, and eare determinedby the physicalparametersof the systemand are independentof f(t).The mass-spring-dashpotsystemand the

seriesRLCcircuit are familiar examplesof this general situation.For simplicity weassumethat the system is initially passive:x(0)= x'(0) =

O.Then the transform of Eq.(22)is)

as2X(s)+bsX(s)+eXes)= F(s),)

so)

F(s)Xes)=

2= W(s)F(s).as +bs + e)

(23))

The function)

1W(s) =

as2 +bs + e)(24))

is calledthe transferfunction of the system.Thus the transform of the responsetothe input f(t) is the product of W(s) and the transform F(s).

The function)

wet) = oC-1{W(s)}) (25))

is calledthe weightfunction of the system.From Eq. (24)we seeby convolutionthat)

x(t) = itw('r)f(t -T)dT.) (26))

This formula is Duhamel'sprinciplefor the system. What is important is that

the weight function w (t) is determinedcompletelyby the parametersof the system.Oncewet) has beendetermined,the integral in (26)gives the responseof the systemto an arbitrary input function f (t).

In principle-thatis, via the convolution integral-Duhamel'sprinciplere-ducesthe problemof finding a system'soutputs for all possibleinputs to calculationof the singleinverseLaplacetransform in (25)that isneededto find its weight func-tion. Hence,a computationalanalogue for a physical mass-spring-dashpotsystemdescribedby (22)can beconstructed in the form of a \"black box\" that ishard-wiredto calculate (and then tabulate or graph, for instance) the responsex(t)given by (26)automaticallywhenevera desiredforce function f(t) is input. In engineeringprac-tice,all manner of physicalsystemsare \"modeled\"in this manner, sotheir behaviorscan be studied without needfor expensiveor time-consumingexperimentation.)))

Example4)


Considera mass-spring-dashpotsystem (initially passive)that respondsto the ex-ternal force J(t)in accordwith the equation x\" +6x'+ lOx ==J(t).Then)

1 1W(s) - --

s2+6s + 10- (s+3)2+ 1')

so the weight function is wet) ==e-3t sint.Then Duhamel'sprincipleimplies that

the responsex(t) to the force J (t) is)

x(t) = ite-3T(sinT)f(t -T) dT.) .)

Note that)

1 oC{o(t)}W(s) == ==

as2 +bs +c as2 +bs +c)

Consequently, it follows from Eq. (23)that the weight function is simply the re-sponseof the system to the delta function input oCt).For this reason wet) is some-timescalledthe unit impulseresponse.A responsethat isusually easierto measurein practiceis the responseh(t) to the unit stepfunction u(t);h(t) is the unit stepresponse.BecauseoC{u (t)}== 1/s,we seefrom Eq. (23)that the transform of h (t )IS)

H(s) = W(s) .s

It follows from the formula for transforms of integrals that)

h(t) = it W(T) dT, so that w(t) = h'(t).) (27))

Thus the weight function, or unit impulseresponse,is the derivative of the unit stepresponse.Substitution of (27)in Duhamel'sprinciplegives)

x(t) = ith'(t)f(t-T)dT) (28))

for the responseof the system to the input J (t).)

ApPLICATIONS:To describea typical application of Eq. (28),supposethat weare given a complexseriescircuit containing many inductors, resistors,and capac-itors. Assume that its circuit equation is a linear equation of the form in (22),but

with i in placeof x. What if the coefficientsa, b, and c are unknown, perhapsonly becausethey are too difficult to compute? We would still want to know the

current i(t) correspondingto any input J(t) == e'(t).We connect the circuit to alinearly increasingvoltagee(t) == t, so that J(t)==e'(t) == 1 ==u(t),and measurethe responseh(t) with an ammeter. We then compute the derivative h'(t), eithernumericallyor graphically.Then accordingto Eq.(28),the output current i (t) cor-respondingto the input voltagee(t) will be given by)

i (t) = 1t

h'(T)e'(t -T)dT

(usingthe fact that J(t) ==e'(t)).)))

HISTORICALREMARK: In conclusion,we remark that around 1950,after engi-neersand physicistshad beenusing delta functions widely and fruitfully for about20yearswithout rigorousjustification, the French mathematician LaurentSchwartzdevelopeda rigorousmathematicaltheory of generalizedfunctionsthat suppliedthe

missinglogicalfoundation for delta function techniques.Every piecewisecontinu-ousordinary function is a generalizedfunction, but the delta function is an exampleof a generalizedfunction that is not an ordinary function.)

_\"Problems)Solvethe initial value problems in Problems1through 8, andgraph eachsolution function x (t).)

1.x\" + 4x = 8(t);x(O)= x'(O)= 02.x\" + 4x = 8(t) + 8(t -n);x(O)= x'(0)= 03.x\" + 4x'+ 4x = 1+ 8(t -2);x (0)= x'(0)= 04. x\" + 2x'+ x = t + 8(t);x(O)= 0,x'(O)= 15. x\" + 2x'+ 2x = 28(t-n);x(O)= x'(O)= 06.x\" + 9x = 8(t -3n) + cos3t; x(O)= x'(O)= 07. x\" +4x'+5x = 8(t -n)+8(t-2n);x(O)= 0, x'(0)= 28.x\" + 2x'+x = 8(t)-8(t -2);x(O)= x'(O)= 2)

Apply Duhamel'sprinciple to write an integral formula for thesolution ofeach initial value problem in Problems9 through12.9.x\" + 4x = f(t); x(O)= x'(O)= 0

10.x\" + 6x'+ 9x = f(t); x(O)= x'(O)= 011.x\" + 6x'+ 8x = f(t); x(O)= x'(0) = 012.x\" + 4x'+ 8x = f(t); x(O)= x'(0) = 013.This problem dealswith a mass m, initially at rest at the

origin, that receivesan impulse p at time t = O. (a) Findthe solution XE (t) of the problem

mx\" = pdO,E (t); x (0) = x'(0)= O.

(b) Show that limxE(t) agreeswith the solution of theE--+O

problem)

mx\" = p8(t); x(O)= x'(0)= O.

(c)Show that mv = p for t > 0 (v = dx/dt).14.Verify that u' (t -a) = 8(t -a) by solving the problem

x'= 8(t -a); x(O)= 0)

to obtain x(t) = u(t -a).15.This problem dealswith a mass m on a spring (with con-

stant k) that receivesan impulse Po = mvo at time t = O.Show that the initial value problems)

mx\" + kx = 0; x(O)= 0, x'(0) = vo

and

mx\" + kx = po8(t); x(O)= 0, x'(O)= 0)

have the samesolution. Thus the effectof po8(t) is, in-

deed,to impart to the particle an initial momentum po.)

16.This is a generalization of Problem 15.Show that the

problems)

ax\" + bx'+ cx = f (t) ; x (0)= 0, x'(0)= vo)

and)

ax\" + bx'+ cx = f(t) + avo8(t); x(O)= x'(O)= 0)

have the samesolution for t > O.Thus the effectof the

term avo8(t) is to supply the initial condition x'(O)= vo.

17.Consideran initially passiveRC circuit (no inductance)with a battery supplying eovolts. (a) If the switch to

the battery is closedat time t = a and openedat timet = b > a (and left openthereafter), show that the current

in the circuit satisfiesthe initial value problem)

1Ri' + -i= eo8(t -a) -eo8(t -b); i(O)= O.

C)

(b) Solvethis problem if R = 100Q, C = 10-4 F,eo= 100V, a = 1 (s),and b = 2 (s).Show that i (t) > 0if 1 < t < 2 and that i (t) < 0 if t > 2.

18.Consideran initially passiveLC circuit (no resistance)with a battery supplying eo volts. (a) If the switch isclosedat time t = 0 and openedat time t = a > 0, show

that the current in the circuit satisfiesthe initial value prob-lem)

1Li\" + -i= eo8(t)-eo8(t -a);C

i (0)= i'(0)= O.)

(b) If L = 1 H, C = 10-2 F, eo= 10V, and a = n (s),show that)

.(

{sin lOt

1 t) =o)

if t < n,if t > n.)

Thus the current oscillatesthrough five cyclesand then

stops abruptly when the switch is opened(Fig. 4.6.6).)))

i(t)

1)

-1)

t)

FIGURE4.6.6.Thecurrent function ofProblem 18.

19.Considerthe LCcircuit ofProblem 18(b),exceptsupposethat the switch is alternately closedand openedat timest = 0, n/10,2n/10,....(a) Show that i (t) satisfiestheinitial value problem)

00

i\"+lOOi= 10L(-lt.s(t- nJT

); i(O)= i'(O)\302\267 O.

n=O10

(b) Solvethis initial value problem to show that

nn (n + l)n-<t< .10 10)i(t) = (n + 1)sin lOt) if)

Thus a resonancephenomenon occurs(seeFig. 4.6.7).i(t)

10)

t)

-10FIGURE4.6.7.Thecurrent function ofProblem 19.)

20.RepeatProblem 19,exceptsupposethat the switch is al-ternately closedand openedat times t = 0,n/5, 2n/5, ...,nn /5, ....Now show that if

nn (n + l)n-<t <5 5)

then)

i (t) ={\037in

lOt) if n is even;if n is odd.)

Thus the current in alternate cyclesof length n/5 first ex-ecutesa sine oscillationduring onecycle,then is dormant

during the next cycle,and soon (seeFig. 4.6.8).)


i(t))

t)

FIGURE4.6.8.Thecurrent function ofProblem 20.)

21.Consideran RLCcircuit in serieswith a battery, with

L = 1 H, R = 60 Q, C = 10-3 F, and eo = 10V. (a)Supposethat the switch is alternately closedand openedat times t = 0, n/10,2n/10,....Show that i (t) satisfies

the initial value problem)

00

(nn

)iff + 60i'+ 1000i= 10L(-1)n8t -- ;n=O

10)

i (0)= i'(0)= O.)

(b) Solvethis problem to show that if)

nn (n + l)n-<t <10 10)

then)

i (t) =)e3mr+3n - 1e-30t sin lOt.

e3n -1)

Construct a figure showing the graph of this current func-

tio n.22.Considera mass m = 1 on a spring with constant k = 1,

initially at rest, but struck with a hammer at eachofthe in-

stants t = 0,2n, 4n, ....Supposethat eachhammer blow

imparts an impulse of+1.Show that the position function

x(t)of the mass satisfiesthe initial value problem)

00

x\" + x =L8(t -2nn); x(O)= x'(O)= O.n=O)

Solvethis problem to show that if 2nn < t < 2(n + l)n,then x (t) = (n + 1)sin t. Thus resonanceoccursbecausethe mass is struck eachtime it passesthrough the origin

moving to the right-incontrast with Example 3, in which

the masswas struck eachtime it returned to the origin. Fi-nally, construct a figure showing the graph of this positionfunction.)))

LinearSystelllsof DifferentialEquations)

11IIFirst-Ordel'\037ystemsandApplicatio\037s)

In the precedingchapterswe have discussedmethods for solving an ordinary dif-ferential equation that involvesonly one dependentvariable. Many applications,

however, require the use of two or more dependentvariables,eacha function of asingleindependent variable (typically time).Sucha problemleadsnaturally to asystemof simultaneousordinary differentialequations.We will usually denote the

independentvariableby t and the dependentvariables (the unknown functions of t)by Xl, X2, X3, ...or by x,y, z, ....Primeswill indicate derivativeswith respecttot.

We will restrict our attention to systemsin which the number of equations isthe sameas the number of dependentvariables (unknown functions). For instance,a systemof two first-order equations in the dependentvariables X and y has the

generalform)

f (t,x,y, x',y') = 0,g(t,x,y, x',y') = 0,)

(1))

where the functions f and g are given. A solutionof this system is a pair x(t),yet)of functions of t that satisfy both equations identicallyover someinterval of valuesof t.

For an exampleof a second-ordersystem,considera particle of massm that

moves in spaceunder the influenceof a force fieldF that dependson time t, the po-sition (x(t),yet),z(t))of the particle, and its velocity (x/(t),y'(t),Z'(t)).ApplyingNewton'slaw ma = F componentwise,we get the system)

II F (I I ,

)mx = It,x,y, z, x , y , z ,II F (

I I ,)my = 2t,x,y,z,x,y,z,

II F (I I ,

)mz = 3 t,x,y,z,x,y,z)(2))

326)))

Example1)

!(t))

Equilibrium positions)

FIGURE5.1.1.Themass-and-spring system of Example 1.)

k1x......) m...........,)

k2(y -x)\037)

k2(y -x)......)

!(t))

FIGURE5.1.2.The\"free

body diagrams\" for the

system of Example 1.)

Example2)

20gal/minFresh water)

Tank 2)

20gal/min)

FIGURE5.1.3.Thetwo brinetanks of Example 2.)

5.1First-OrderSystemsand Applications 327)

of three second-orderequationswith independentvariable t and dependentvariablesx, y, z; the three right-hand-side functions FI , F2, F3 are the componentsof the

vector-valuedfunction F.)

InitialApplications)Examples1 through 3 further illustrate how systemsof differential equationsarisenaturally in scientificproblems.)

..\037.. \" ..\"\" ..

Considerthe system of two massesand two springsshown in Fig.5.1.1,with agiven external force f(t) acting on the right-hand massm2. We denote by x(t)the displacement(to the right) of the massmI from its static equilibrium position(when the system is motionlessand in equilibrium and f(t) = 0)and by yet) the

displacementof the massm2 from its static position. Thus the two springs areneither stretched nor compressedwhen x and y are zero.

In the configuration in Fig.5.1.1,the first spring is stretchedx units and the

secondby y-x units. We apply Newton'slaw of motion to the two \"free body

diagrams\" shown in Fig.5.1.2;we thereby obtain the system)

mIx\" = -kIx+k2(y -x),m2Y\" = -k2(y -x) + f(t))

(3))

of differential equations that the position functions x(t) and yet) must satisfy. Forinstance, if mI = 2,m2 = 1,k I = 4, k2 = 2,and f(t) = 40sin3tin appropriatephysicalunits, then the system in (3)reducesto)

2x\" = -6x+2y,y\" = 2x-2y +40sin 3t.)

(4)

.)

J)

......

Considertwo brine tanks connectedas shown in Fig.5.1.3.Tank 1 containsx(t)poundsof salt in 100gal of brine and tank 2 contains y (t) poundsof salt in 200galof brine. The brine in each tank is keptuniform by stirring, and brine is pumpedfrom eachtank to the other at the rates indicated in Fig.5.1.3.In addition, fresh wa-ter flows into tank 1 at 20gal/min,and the brine in tank 2 flows out at 20 gal/min(sothe total volume of brine in the two tanks remains constant). The salt concen-trations in the two tanks are x/I00poundspergallon and y /200poundsper gallon,respectively.When we compute the rates of change of the amount of salt in the two

tanks, we therefore get the system of differentialequations that x(t) and yet) must

satisfy:)

I X Y 3 1x = -30.-+ 10.-= --x+-y100 200 10 20'I X Y Y 3 3

y = 30.--10.--20.-= -x--y.100 200 200 10 20')

that is,)

20x'= -6x+y,

20y' = 6x - 3y.)(5))

.)))

328 Chapter5 LinearSystemsof Differential Equations)

Example3)

'CircuitElement)

Voltage...

.Dr&pp:)

Inductor)dIL-dt

RI

1-QC)

Resistor)

Capacitor)

FIGURE5.1.5.Voltage dropsacrosscommon circuit elements.)

L:2henries) c:0.008farads)

+8Eo: -=- I} R}:100volts

_50ohms) t\037)

R2:25 ohms)

FIGURE5.1.4.Theelectricalnetwork ofExample 3.)

....\"\"\037\037 ..\037\037\037\037\037\037_\037\037....,....,.\" '\"\"' ....., ,,\037\037......... \037...\037,....,.,....,.\037\037 \".,....,....;;..,.;,.,.0..\"\"..\"'.,.;..,...,..,.. '\" \037\037\037\037....'-\".,\"v\037\037'\"' .....- '-\"

Considerthe electricalnetwork shown in Fig.5.1.4,where 11(t) denotesthe cur-rent in the indicated direction through the inductor Land 12(t) denotesthe current

through the resistorR2.The current through the resistorR 1 is I = 11-12 in the di-rection indicated.We recallKirchhoff's voltage law to the effect that the (algebraic)sum of the voltage dropsaround any closedloopof such a network is zero.As in

Section2.7,the voltage dropsacrossthe three types of circuit elementsare thoseshown in Fig.5.1.5.We apply Kirchhoff's law to the left-hand loopof the networkto obtain)

dl12-+50(/1 -12) - 100= 0,dt)

(6))

becausethe voltage drop from the negative to the positive pole of the battery is-100.The right-hand loopyieldsthe equation)

125Q2+25/2 +50(/2- 11)= 0,) (7))

where Q2(t)is the charge on the capacitor.BecausedQ2/dt= 12, differentiation

of eachsideof Eq.(7)yields)

dl2 dl1125/2 +75--50-= O.dt dt)

(8))

After dividing Eqs. (6)and (8)by the factors 2 and -25,respectively,we get the

system)

dl1-+25/1 -25/2 = 50,dt

dl1 dl22--3--5/2 = 0dt dt)

(9))

of differentialequations that the currents 11(t) and 12(t)must satisfy.) .)))

Example4)


First-OrderSystemsConsidera system of differentialequations that can be solved for the highest-orderderivatives of the dependentvariables that appear, as explicitfunctions of t andlower-orderderivativesof the dependentvariables. For instance, in the caseof asystemof two second-orderequations,our assumptionis that it can bewritten in the

form)

II J (\"

)xl == I t,XI,X2,XI ,X2 ,)

(10))II 1'. (\"

)X2 == J 2 t, Xl , X2, X I 'X2 .)

It is of both practical and theoretical importance that any such higher-ordersystemcan be transformed into an equivalentsystem offirst-orderequations.

To describehow such a transformation is accomplished,we considerfirst the

\"system\" consistingof the singlenth-order equation)

\037)(n) -J( , (n-I))X - t,x,X,...,X .) (11))

We introduce the dependentvariablesXl, X2, ..., Xn definedas follows:)

Xl ==X,),

X2 ==X ,) Xn ==x(n-I).) (12))II

X3 ==X ,) ...,)

Note thatx\037

==x' ==X2,x\037

==x\" ==X3, and soon.Hencethe substitution of (12)\037in

Eq.(11)yieldsthe system)

,xl ==X2,)

,X2 ==X3,)

\037) (13)),

Xn-l ==Xn ,)

X\037

==J(t,Xl, X2,'..,Xn ))

of n first-orderequations.Evidently, this system is equivalent to the original nth-

orderequation in (11),in the sensethat x(t) is a solution of Eq.(11)if and only ifthe functions Xl (t),X2(t), ...,xn(t) defined in (12)satisfy the systemof equationsin (13).)

The third-orderequation)

X (3) + 3X\" +2x'-5x==sin 2t)

is of the form in (11)with

J (t, X , x',x\") ==5x -2x'-3xII + sin 2t .Hencethe substitutions)

Xl ==X,)\" \"'

X2 ==X ==X I ' X3 ==X ==X2)

yield the system)

,xl ==X2,

,X2 ==X3,

X\037

==5XI - 2X2 - 3X3 + sin 2t

of three first-orderequations.) .)))


It may appearthat the first-order system obtained in Example4 offers little

advantagebecausewe coulduse the methodsof Chapter 2 to solve the original (lin-ear) third-order equation. But supposethat we wereconfronted with the nonlinear

equation) I! 3 + (')3

X =X X,to which none of our earlier methods can be applied.The correspondingfirst-ordersystem is)

Ixl = X2,)

(14))X\037

= (XI)3 + (X2)3,)

and we will seein Section6.4that there existeffective numerical techniques for

approximating the solution of essentiallyany first-ordersystem.Soin this casethe

transformation to a first-ordersystem is advantageous.From a practical viewpoint,large systemsof higher-orderdifferentialequations typically are solvednumericallywith the aid of the computer, and the first stepis to transform such a system into afirst-ordersystem for which a standardcomputerprogram is available.)

Example5) The system)

2xI!= -6x+ 2y,

yl! = 2x- 2y + 40sin 3t)(4))

of second-orderequations was derived in Example1.Transform this system into an

equivalentfirst-ordersystem.)

Solution Motivatedby the equations in (12),we define)

I I I IXI=X, X2=X=XI , YI=y, Y2=Y=YI.)Then the systemin (4)yieldsthe system)

Ixl = X2,)

2x\037= -6XI+ 2YI,)

IY I

= Y2,

Y\037

= 2XI - 2YI +40sin 3t)

(15))

of four first-orderequations in the dependentvariablesXl, X2, YI, and Y2.) .)

SimpleTwo-DimensionalSystemsThe linear second-orderdifferentialequation)

Xl! +px'+qx = 0) (16))

(with constant coefficientsand independentvariable t) transforms via the substitu-

tions x'= Y, xl!= y' into the two-dimensionallinear system)

IX = y,)

IY = -qX - py.)

(17))))

Example6)

54321

\037 0-1-2-3-4-5

-5-4-3-2-10 1 2 3 4 5x)

FIGURE5.1.6.Directionfieldand solution curves for the systemx'= -2y,y' =

\037xofExample 6.)


Conversely,wecan solve this system in (17)by solving the familiar singleequationin (16).)

To solve the two-dimensionalsystem)

X' = -2y,)I 1

Y =2\"x,)

(18))

we beginwith the observation that)

x\" = -2y' = -2(!x)= -x.)

This gives the singlesecond-orderequation x\" +x = 0 with general solution)

x(t) = Acost+ Bsint = Ccos(t-a))

where A = Ccosaand B = Csina.Then)

yet) = -4x/(t)= -!(-Asint+Bcost)= !csin(t -a).)

The identity cos2 () +sin2 () = 1 therefore impliesthat, for each value of t, the point

(x(t),yet)) lieson the ellipse)

x2 y2C2

+(Cj2)2

= 1)

with semiaxesC and C/2.Figure5.1.6showsseveralsuchellipsesin the xy-plane..)A solution (x(t),yet)) of a two-dimensionalsystem)

x' = J(t,x,y),

y' = g(t,x,y))

may be regardedas a parametrizationof a solutioncurve or trajectory of the sys-tem in the xy-plane.Thus the trajectories of the system in (18)are the ellipsesofFig.5.1.6.The choiceof an initial point (x(O),y(O))determineswhich one of thesetrajectoriesa particular solutionparametrizes.

The picture showing a system'strajectories in the xy-plane-itsso-calledphaseplaneportrait-failsto reveal preciselyhow the point (x(t),yet)) movesalong its trajectory.If the functions J and g donot involve the independentvariablet, then a directionfield-showingtypical arrows representing vectors with com-ponents (proportional to) the derivativesx' = J(x,y) and y' = g(x,y)-canbeplotted. Becausethe moving point (x(t),yet)) has velocity vector (x'(t),y'(t)),this direction field indicatesthe point's direction of motion along its trajectory.For instance, the direction field plotted in Fig.5.1.6indicates that each such pointmovescounterclockwisearound its ellipticaltrajectory.Additional information canbe shown in the separategraphs of x(t) and yet) as functions of t.)))


Example6)

Continued)

Example7)

4321

\037 0-1-2-3-4

\037 \037

-4-3-2-10 1 2 3 4x)

FIGURE5.1.8.Direction fieldand solution curvesfor the systemx'= y, y' = 2x + y ofExample 7.)

..\037.. \",,,,..,\037. ,.. ... .................., ,,'\" ,'\" ,.. \"..\"

With initial valuesx(0)= 2,y (0)= 0,the general solution in Example6 yields

x(0)=A=2, y(O)=-!B=O.)The resulting particular solution is given by)

x(t) = 2 cost, Y (t) = sin t.The graphs of the two functions are shown in Fig.5.1.7.We seethat x(t) initially

decreaseswhile yet) increases.It follows that, as t increases,the solution point

(x(t),yet)) traverses the trajectory \037x2+y2 = 1 in the counterclockwisedirection,

as indicated by the direction field vectors in Fig.5.1.6. .)4321

\037 0\037-1-2-3-4 5 10 150

t)

FIGURE5.1.7.x-and y-solution curves for the initial

value problem x'= -2y,y' =\037x, x(O)= 2,y(O)= O.)

...... ,...n........\" \" ,...........,..

To find a general solution of the system)I

X = y,)

(19))y' = 2x+ y,)


x\" = y' = 2x+ y = x'+2x.

This gives the singlelinear second-orderequation)

1/ I 2 0x -x - x=)

with characteristicequation

r2 - r -2 = (r + l)(r-2)= 0)

and general solution)

x(t) = Ae-t + Be2t.) (20))

Then)

yet) = x'(t) = -Ae-t +2Be2t.) (21))

Typicalphaseplanetrajectoriesof the system in (19)parametrizedby Eqs.(20)and(21)are shown in Fig.5.1.8.Thesetrajectories may resemblehyperbolas sharingcommonasymptotes,but Problem23 showsthat their actual form is somewhatmorecomplicated. .)))

Example8)

2)

\037 0)

-1)

-2)

-2 -1) ox)

1) 2)

5.1First-OrderSystemsand Applications 333)......\037 ..................... \037 .. \037..\" .,.... .......... .........\"............

To solve the initial value problem)

x' == -y,)

y' ==(1.01)x- (0.2)y,x(O)==0, y(O) ==-1,)

(22))


x\" ==-y'==-[(1.01)x- (0.2)y]==(-1.01)x- (0.2)x'.)

This gives the singlelinear second-orderequation)

x\" + (O.2)x'+ (1.01)x==0)

with characteristicequation)

r 2 +(0.2)r+ 1.01== (r +0.1)2+ 1 ==0,)

characteristicroots -0.1::l::i,and general solution)

x(t) ==e-tllO(Acost + B sin t).)

Then x(0)== A ==0,so)

x(t) ==Be-tllO sin t,)

FIGURE5.1.9.Direction field yet) ==-x'(t) == /0Be-tllO sin t -Be-tllO cost.and solution curve for the systemx'= -y,y' = (l.Ol)x- (O.2)yof Finally, y(O) ==-B==-1,so the desiredsolutionof the system in (22)isExample 8.)

1.2)

0.8) x =x(t))

0.4)

\037 0.0\037)

-0.4)

-0.8)

-1.2o 5 10 15 20 25 30

t)

FIGURE5.1.10.x-and

y-solution curves for the initialvalue problem of Example 8.)

x(t) ==e-tllO sin t,)

yet) == /oe-tllO(sint - 10cost).)(23))

Theseequations parametrize the spiral trajectory in Fig.5.1.9;the trajectory ap-proachesthe origin as t \037 +00.Figure 5.1.10showsthe x-and y-solutioncurvesgiven in (23). .

When we study linear systemsusing the eigenvaluemethod of Section5.4,wewillieamwhy the superficiallysimilar systemsin Examples6 through 8 have the

markedly different trajectories shown in Figs.5.1.6,5.1.8,and 5.1.9.)

LinearSystemsIn addition to practical advantagesfor numericalcomputation,the general theory ofsystemsand systematicsolution techniquesare more easily and moreconciselyde-scribedfor first-ordersystemsthan for higher-ordersystems.For instance,considera linearfirst-ordersystem of the form)

x\037

== P11(t )Xl + P12(t )X2 + ...+ PInXn + 11(t),X\037

== P2l(t)Xl + P22(t)X2 + ...+ P2n Xn + I2(t),)(24))

x\037== Pnl (t)Xl + Pn2 (t)X2 + ...+ PnnXn + In (t).)))


We say that this systemis homogeneousif the functions fI, f2, ...,in are all

identicallyzero;otherwise,it is nonhomogeneous.Thus the linear system in (5)ishomogeneous,whereasthe linear system in (15)is nonhomogeneous.The systemin (14)isnonlinearbecausethe right-hand sideof the secondequationisnot a linearfunction of the dependentvariablesXl and X2.

A solutionof the system in (24)is an n-tuple of functions Xl (t),X2(t), ...,Xn (t) that (on someinterval) identicallysatisfy eachof the equationsin (24).We will

seethat the general theory of a system of n linear first-orderequationssharesmanysimilaritieswith the general theory of a singlenth-order linear differential equation.Theorem 1 (proved in the Appendix) is analogous to Theorem 2 of Section2.2.Ittellsus that if the coefficient functions Pij and fj in (24)are continuous, then the

systemhas a unique solution satisfying given initial conditions.)

THEOREM1 ExistenceandUniquenessfor LinearSystemsSupposethat the functions PII,PI2,...,Pnn and the functions fI, f2, ...,inare continuous on the openinterval I containing the point a.Then, given the n

numbers bI , b2, ...,bn , the systemin (24)has a unique solution on the entireinterval I that satisfiesthe n initial conditions)

Xl (a) = b I , x2(a)= b2,) ... ,) Xn(a) = bn .) (25))

Thus n initial conditions are neededto determine a solution of a system ofn linear first-orderequations, and we therefore expecta general solutionof such a

systemto involve n arbitrary constants.For instance, we saw in Example5 that the

second-orderlinear system)

2xI/ = -6x + 2y,yl/= 2x-2y+40sin3t,)

which describesthe positionfunctions x(t) and yet) of Example1,is equivalent tothe system offour first-orderlinear equations in (15).Hencefour initial conditionswouldbeneededto determine the subsequentmotionsof the two massesin Example1.Typical initial values would be the initial positionsX (0)and y (0)and the initial

velocitiesx'(0)and y'(0). On the other hand, we found that the amounts X (t) and

yet) of salt in the two tanks of Example2 are describedby the system)

20x'= -6x + y,

20y' = 6x -3y)

of two first-orderlinearequations.Hencethe two initial valuesx(O)and y(O)shouldsufficeto determine the solution. Givena higher-ordersystem,we often must trans-form it into an equivalentfirst-ordersystem to discoverhow many initial conditionsare neededto determine a unique solution. Theorem 1 tellsus that the number ofsuch conditions is preciselythe sameas the number of equations in the equivalentfirst-order system.)))

11IIProblems)

In Problems 1 through 10,transform the given differentialequation or system into an equivalent system offirst-orderdif-ferential equations.

1.x\" + 3X' + 7x = t 2

2.X(4) + 6X\" - 3X' + x = cos3t3.t 2x\" + tx '+ (t 2 - l)x= 04. t 3x (3) -2t2x\" + 3tx' + 5x = In t

5.x (3) = (X')2 + cosx6.x\" - 5x + 4y = 0, y\" + 4x- 5y = 0

7 IIkx II ky

\302\267 x = -(x2 + y 2)3/2 'Y = -

(x2 + y 2)3/2

8.x\" + 3X' + 4x- 2y = 0, y\" + 2y'- 3x + y = cost9.x\" = 3x - y + 2z,y\" = x + y

-4z, z\" = 5x - y- z

10.x\" = (1- y)x, y\" = (1-x)y)

Usethe method ofExamples6, 7, and 8 to find generalsolu-tions of the systems in Problems11through 20. If initial con-ditions are given, find the correspondingparticular solution.For eachproblem, usea computer system or graphing calcu-lator to construct a direction fieldand typical solution curvesfor the given system.

11.x'= y, y' = -x12.x'= y, y' = x13.x'= -2y,y' = 2x;x(O)= 1,y(O)= 014.x'= 10y,y' = -lOx;x(O)= 3, y(O)= 415.x'=

\037y, y' = -8x16.x'= 8y, y' = -2x17.x I = y, y' = 6x - y; x (0) = 1,y (0) = 218.x'= -y,y' = lOx- 7y; x(O)= 2, y(O)= -719.x'= -y,y' = 13x+ 4y;x(O)= 0, y(O)= 320.x'= y, y' = -9x+ 6y21.(a) Calculate[x(t)]2+ [y(t)]2 to show that the trajecto-

riesof the system x' = y, y' = -xof Problem 11arecircles.(b) Calculate[X(t)]2 - [y(t)]2to show that the

trajectoriesof the system x' = y, y' = x of Problem 12arehyperbolas.

22.(a) Beginning with the general solution of the systemx' = -2y,y' = 2x of Problem 13,calculatex2 + y2 toshow that the trajectoriesarecircles.(b) Show similarlythat the trajectoriesof the system x' =

\037y, y' = -8x

of Problem 15 are ellipseswith equations of the form

16x2 + y2 = C2.23. First solveEqs.(20)and (21)for e-t and e2t in terms of

x(t),yet), and the constants A and B.Then substitute theresults in (e2t )(e-t )2 = 1 to show that the trajectoriesofthe system x' = y, y' = 2x + y in Example 7 satisfy an

equation of the form)

4x3 - 3xy2 + y3 = C (constant).)

Then show that C = 0 yields the straight lines y = -xand y = 2x that arevisible in Fig. 5.1.8.)


24. Derivethe equations

mlx\037'= -(kl + k2)xl + k2X2,

m2x{ = k2X I- (k2 + k3)X2

for the displacements (from equilibrium) of the two

massesshown in Fig. 5.1.11.)

FIGURE5.1.11.Thesystem ofProblem 24.)

25.Twoparticleseachofmassm are attached to a string under

(constant) tension T, as indicated in Fig. 5.1.12.Assumethat the particlesoscillatevertically (that is, parallel to the

y-axis)with amplitudes so small that the sinesof the an-

glesshown areaccurately approximated by their tangents.Show that the displacementsYl and Y2 satisfy the equa-ti 0n s

ky\037'= -2YI + Y2, ky{ = YI

-2Y2

where k = mL/T.)y)

m)

x)L) L)

FIGURE5.1.12.Themechanical system ofProblem 25.)

26.)Three 100-galfermentation vats are connectedas indi-

cated in Fig. 5.1.13,and the mixtures in each tank arekept uniform by stirring. Denoteby Xi (t) the amount (in

pounds) of alcoholin tank Ii at time t (i = 1,2,3).Sup-posethat the mixture circulates between the tanks at the

rate of 10gal/min. Derive the equations)

10x\037= -XI + X3)

10x\037= XI

-X2)

10x\037= X2

-X3.)

FIGURE5.1.13.Thefermentation tanksofProblem 26.)))


27.Set up a system of first-order differential equations forthe indicated currents I} and 12 in the electricalcircuit ofFig. 5.1.14,which shows an inductor, two resistors,anda generator which supplies an alternating voltage drop ofE (t) = 100sin 60t V in the direction of the current I}.)

E(t) = 100sin 60t)

\037\037enri8\037Lhms)

\037)R}:50ohms)

FIGURE5.1.14.Theelectricalcircuit ofProblem27.28.Repeat Problem 27, exceptwith the generator replaced

with a battery supplying an emf of 100V and with theinductor replacedwith a I-millifarad (mF) capacitor.29.A particle ofmass m moves in the plane with coordinates(x(t),y(t\302\273

under the influence of a forcethat is directedtoward the origin and has magnitude k/(x2 + y2)-aninverse-squarecentral forcefield. Show that

H kx H kymx = -r3 and my = ---;3'

where r = Jx2 + y2.30.Supposethat a projectileof mass m moves in a vertical

plane in the atmosphere near the surfaceof the earth un-der the influence of two forces:a downward gravitational)

5.1Application)

forceof magnitude mg, and a resistive forceFR that isdirectedoppositeto the velocity vector v and has magni-tude k v 2 (where v =

I v Iis the speedof the projectile;

seeFig. 5.1.15).Show that the equations ofmotion of the

projectileare)

Hk

/mx = - vx,)H k

/my = -

vy-mg,)

where v = J(X/)2 + (y/)2.)

y) v)

,,)

x)

FIGURE5.1.15.Thetrajectory of the

projectileof Problem 30.

31.Supposethat a particle with mass m and electricalchargeq moves in the xy-plane under the influence of the mag-netic field B = Bk (thus a uniform field parallel to the

z-axis),so the forceon the particle is F = qv x B if its

velocity is v. Show that the equations of motion of the

particle are)

mx H = +qBy', myH = -qBx'.)

Gravitationand K\037pler'sLawsof PlanetaryMotion)

Around the turn of the 17th century, Johannes Kepleranalyzed a lifetime of plane-tary observationsby the astronomerTychoBrahe.Keplerconcludedthat the motionof the planets around the sun is describedby the followingthree propositions,now

known asKepler'slawsof planetary motion:)

1.The orbit of each planet is an ellipsewith the sun at one focus.2.The radius vector from the sun to each planet sweepsout area at a constant

rate.3.The squareof the planet'speriodof revolution is proportional to the cube of

the major semiaxisof its elliptical orbit.

In his PrincipiaMathematica (1687)IsaacNewton deducedthe inverse-squarelaw of gravitation from Kepler'slaws.In this applicationwe leadyou (in the oppo-sitedirection) through a derivationof Kepler'sfirst two laws from Newton'slaw ofgravitation.

Assumethat the sun is locatedat the origin in the plane of motion of a planet,and write the positionvector of the planet in the form)

r(t) = (x(t),y(t)) = xi+ yj,) (1))

where i = (1,0)and j = (0,1)denote the unit vectors in the positive x- and y-directions.Then the inverse-squarelaw of gravitation implies(Problem29)that the)))

y)

,/,/,/,/,/,/,/,/,/,/,/,/,/,/)

FIGURE5.1.16.Theradial andtransverse unit vectorsOr and 00.)

FIGURE5.1.17.Area swept out

by the radius vector.)


accelerationvector r\" (t) of the planet is given by)

\" krr =--3 '

r)(2))

where r = Jx2 + y2 is the distancefrom the sun to the planet. If the polar co-ordinates of the planet at time tare (r(t),O(t)),then the radial and transverseunit

vectors shown in Fig.5.1.16are given by)

Ur = i cos0 +j sin 0 and De = -i sin 0 +jcosO.) (3))

x)

The radial unit vector Ur (when locatedat the planet'sposition) always points di-rectly away from the origin, so Ur = rjr, and the transverse unit vector Ue is ob-tained from Ur by a 90\302\260 counterclockwiserotation.)

STEP1:Differentiatethe equations in (3)componentwiseto show that)

dUr dOdt

= Uedt)

due dO-= -Ur-.dt dt)(4))and)

STEP2: Usethe equations in (4) to differentiatethe planet'spositionvectorr ==

rUr and thereby show that its velocity vector is given by)

dr dr dOv = -= Ur-+ r -Ue.dt dt dt)

(5))

STEP3: Differentiate again to show that the planet'sacceleration vector advjdt is given by)

a =[

d2r _ r (dO

)2

]Ur + [

\037\037

(r2dO

)]ue.dt2 dt r dt dt)(6))

STEP4: The radial and transversecomponentson the right-hand sidesin Eqs.(2)and (6)must agree.Equating the transverse components-thatis, the coefficientsof Ue-weget)

\037\037

(r 2dO

) = 0,r dt dt)

(7))

so it follows that)

r 2dO = hdt ') (8))

where h is a constant. Becausethe polar-coordinatearea element-forcomputationof the area A(t) in Fig.5.1.17-isgiven by dA =

\037r2dO, Eq. (8)implies that thederivativeA' (t) is constant, which is a statementof Kepler'ssecondlaw.)

STEP5: Equate radial components in (2)and (6)and then use the result in (8)to show that the planet'sradial coordinate function r(t) satisfies the second-orderdifferentialequation)

d2r h 2---dt2 r 3)

k2.

r)(9))))

y) ////8=a)

FIGURE5.1.18.Theellipticalorbit)

Lr= l+ecos(B-a))with perihelion distancerl = Lj(l+ e) and apheliondistance r2 = Lj(l- e).)

c) \037)FIGURE5.1.19.Theshapeofthe orbit ofHalley'scomet.)

STEP6: Although the differential equation in (9) is nonlinear, it can be trans-

formed to a linear equation by means of a simplesubstitution. For this purpose,assumethat the orbit can be written in the polar-coordinateform r = reO),and first

use the chain rule and Eq.(8)to show that if r = l/z, then)

dr dz- - -h-dt

-dO.)

Differentiate again to deducefrom Eq.(9)that the function z(O) = l/r(O)satisfiesthe second-orderequation)

x)

d2z k

d02 +zh 2 .) (10))

STEP7:) Showthat the general solutionof Eq.(10)is)

. kz(0)= A SIn 0 + Bcos0 +

h 2 .) (11))

STEP8: Finally, deducefrom Eq.(11)that reO) = l/z(O)is given by)

LreO) =

1 +ecos(0-a))(12))

with e = Ch2/k, Ccosa= A, C sin a = B,and L = h 2/k. The polar-coordinategraph of Eq. (12)is a conicsectionof eccentricitye-anellipseif 0 < e < 1,aparabola if e = 1,and a hyperbola if e > I-withfocus at the origin. Planetaryorbits are boundedand therefore are ellipseswith eccentricitye < 1.As indicatedin Fig.5.1.18,the major axisof the ellipseliesalong the radial line 0 = a.)

STEP9: Plot sometypical elliptical orbits as describedby (12)with different

eccentricities,sizes,and orientations. In rectangularcoordinatesyou can write)

x(t) = r(t) cost, yet) = r(t) sin t, 0 < t <2n)

to plot an elliptical orbit with eccentricitye, semilatus rectum L (Fig.5.1.18),and

rotation angle a. The eccentricity of the earth'sorbit is e \037 0.0167,so closetozero that the orbit looksnearly circular (though with the sun off center),and the

eccentricitiesof the other planetary orbits range from 0.0068for Venus and 0.0933for Marsto 0.2056for Mercury and 0.2486for Pluto. But many cometshavehighlyeccentricorbits,likeHalley'scomet with e \037 0.97(Fig.5.1.19).)

_ TheMethodof Elimination)The most elementary approach to linear systemsof differential equations involvesthe elimination of dependentvariables by appropriatelycombining pairs of equa-tions.The objectof this procedureis to eliminatedependentvariables in successionuntil there remains only a singleequation containing only one dependentvariable.This remaining equation will usually be a linear equationof high orderand can fre-quently besolvedby the methodsof Chapter2.After its solutionhas beenfound, the

other dependentvariablescan be found in turn, using either the original differentialequations or thosethat have appearedin the elimination process.)))

Example1)

5.2The Methodof Elimination 339)

The method ofelimination for linear differential systemsis similar to thesolution of a linear system of algebraicequations by a processof eliminating the

unknowns one at a time until only a singleequationwith a singleunknown remains.It is most convenient in the caseof manageably small systems:those containingno more than two or three equations. For such systemsthe method of elimina-tion provides a simpleand concreteapproach that requireslittle preliminary theoryor formal machinery. But for larger systemsof differential equations, as well asfor theoretical discussion,the matrix methods of later sectionsin this chapterarepreferable.)

Find the particular solution of the system)

x'=4x-3y, y'=6x-7y) (1))

that satisfies the initial conditionsx(0)= 2,y (0)= -1.)

Solution If we solve the secondequation in (1)for x,we get)l' 7X = 6Y + 6Y ') (2))

so that), 1\" 7'

X = 6Y + 6Y .) (3))

We then substitute theseexpressionsfor x and x' in the first equationof the systemin (1);this yields) 1\" 7' 4(1' 7

) 36Y + 6Y = 6Y + 6Y-

y,)


y\" +3y' - 10y= O.

This second-orderlinear equation has characteristicequation)

r2 +3r-10= (r -2)(r+5)= 0,)

so its general solution is)

y(t) = Cle2t +C2e-5t.) (4))

Next, substitution of (4) in (2)gives)

x(t) =\037 (2Cl e2t -5C2e-5t)+ \037 (Cle2t +C2e-5t);)

that is,)

x(t) =\037Cl

e2t + \037C2e-5t.) (5))

Thus Eqs.(4)and (5)constitute the general solutionof the system in (1).The given initial conditions imply that)

x(O)= \037Cl+

\037C2= 2)

and that)

y(O)= C}+C2= -1;)))

340 Chapter5 LinearSystems of Differential Equations)

54321

>-. 0-1-2-3-4-5 \"

-5-4-3-2-1)x)

FIGURE5.2.1.Direction fieldand solution curves for the systemx'= 4x-3y, y' = 6x -7y ofExample 1.)

theseequations are readily solved for Cl == 2 and C2 == -3.Hencethe desiredsolution is

x(t) ==3e2t -e-St, yet) ==2e2t -3e-St.

Figure 5.2.1showsthis and other typical solutioncurves parametrizedby the equa-tionsx(t)==

\037Cle2t + \037C2e-St,yet) == Cle2t +C2e-Stwith different values of the

arbitrary constants Cl and C2.We seetwo families of curves resemblinghyperbolassharing the samepair of (oblique)asymptotes. .)

Remark:The general solutiondefinedby Eqs.(4)and (5)may beregardedas the pair or vector (x(t),yet)).Recalling the componentwiseaddition of vectors(and multiplication of vectors by scalars),we can write the general solution in (4)and (5)in the form)

(x(t),yet)) ==(\037Cle2t + \037C2e-St,Cle2t +C2e-St

)

==Cl (\037e2t,e2t

) +C2 (\037e-St,e-St) .)

This expressionpresentsthe general solutionof the system in (1)as a linearcombi-nation of the two particular solutions)

(Xl, Yl) ==(\037e2t,

e2t) and (X2, Y2) ==

(\037e-St,e-St) .) .)

PolynomialDifferentialOperatorsIn Example1 we used an ad hoc procedureto eliminate one of the independentvariables by expressingit in terms of the other. We now describea systematicelimination procedure.Operator notation is most convenient for thesepurposes.Recall from Section2.3that a polynomialdifferential operatoris one of the form)

L ==anDn +an_lDn-l +...+aID+ao,) (6))

where D denotesdifferentiationwith respectto the independentvariablet.If L 1 and L2 are two such operators,then their product L 1 L2 is defined this

way:)

\037) LlL2[X]==Ll[L2x].) (7))

For instance, if Ll == D +a and L2 == D +b, then)

LlL2[X]== (D+a)[(D+b)x]== D(Dx+bx)+a(Dx+bx)== [D2 + (a+b)D+ab]x.)

This illustrates the fact that two polynomialoperators with constantcoefficientscanbe multiplied as if they wereordinary polynomials in the \"variable\" D.Becausethe

multiplication of such polynomials is commutative, it follows that)

LlL2[X]==L2L l[X]) (8))

if the necessaryderivativesof X (t)exist.By contrast, this propertyof commutativitygenerallyfails for polynomialoperatorswith variablecoefficients-seeProblems21and 22.)))


Any system of two linear differentialequations with constantcoefficientscanbe written in the form)

L 1x +L2y = 11(t),L3X + L4Y = I2(t),)

(9))

where L 1 , L2, L3, and L4 are polynomialdifferentialoperators(perhapsof different

orders)as in Eq. (6),and II(t) and I2(t) are given functions. For instance, the

system in (1)(Example1)can be written in the form)

(D-4)x+ 3y = 0,-6x+ (D+7)y = 0,)

(10))

with L 1 = D -4, L2 = 3, L3 = -6,and L4 = D +7.To eliminate the dependentvariablex from the system in (9),we operatewith

L3 on the first equation and with LIonthe second.Thus we obtain the system)

L3L IX +L3L2y = L3Il(t),LIL3X +L 1 L4y = L 1 I2(t).)

(11))

Subtractionof the first from the secondof theseequationsyieldsthe singleequation)

(L1 L4 -L2L3)Y = L 1 I2(t)-L3Il(t)) (12))

in the singledependentvariabley. After solving for y = yet) we can substitute the

result into either of the original equations in (9)and then solve for x = x(t).Alternatively, we couldeliminate in likemanner the dependentvariabley from

the original system in (9).If so,we would get the equation)

(L1 L4 -L2L3)X = L4Il(t)-L2I2(t),) (13))

which can now be solvedfor x = x(t).Note that the sameoperator L 1 L4-L2L3appearson the left-handsidein both

Eq.(12)and Eq.(13).This is the operationaldeterminant)

L 1 L2L3 L4

= LIL4 -L2L3) (14))

of the system in (9).In determinant notation Eqs.(12)and (13)can be rewritten as)

L 1 L2 II(t) L2L3 L4

x= 12(t ) L4 ,

(15)Ll L2 L 1 II(t )L3 L4 y= L3 12(t ))

It is important to note that the determinants on the right-hand sidein (15)are eval-uated by means of the operators operating on the functions. The equations in (15)are strongly reminiscentof Cramer'srule for the solutionof two linearequationsin

two (algebraic)variables and are thereby easyto remember.Indeed,you can solvea system of two linear differential equations either by carrying out the systematiceliminationproceduredescribedhere or by directly employing the determinant no-tation in (15).Either processis especiallysimpleif the system is homogeneous(11(t) = 0 and I2(t)= 0),becausein this casethe right-hand sidesof the equationsin (12),(13),and (15)are zero.)))

342 Chapter5 LinearSystems of Differential Equations),.. .....Nn\037..N \037.-. \037NN\037 ;o;N /....\" NNN IV oN UN

Example2) Find a general solution of the system)

(D-4)x+ 3y ==0,-6x+ (D+7)y ==O.)

(10))

Solution The operationaldeterminant of this system is)

(D-4)(D+7)-3 .(-6)== D2 +3D- 10.) (16))

HenceEqs.(13)and (12)are)

x\" + 3X' - lOx ==0,y\" +3y' - 10y==O.)

The characteristicequation of eachis)

r2 +3r-10== (r -2)(r+5) ==0,)

so their (separate)general solutions are)

x(t) ==ale2t +a2e-5t,yet) ==b l e2t +b2e-5t

.)

(17))

At this point we appearto havefour arbitrary constantsaI,a2, b l , and b2.Butit follows from Theorem 1 in Section5.1that the general solutionof a systemof two

first-orderequations involvesonly two arbitrary constants.This apparent difficulty

demandsa resolution.The explanation is simple:There must be somehidden relations among our

four constants.We can discoverthem by substituting the solutions in (17)into eitherof the original equations in (10).On substitution in the first equation,we get)

o ==x'-4x +3y== (2ale2t -5a2e-5t)-4(ale2t +a2e-5t

) +3(b1e2t +b2e-5t);)

that is,)o ==(-2al+3bl )e2t +(-9a2 +3b2)e-5t.

But e2t and e-5t are linearly independentfunctions, so it follows that al ==\037bl

and

a2 ==\037

b2.Therefore, the desiredgeneral solution is given by)

x(t) ==\037ble2t + \037b2e-5t, yet) ==b 1e2t +b2e-5t

.)

Note that this result is in accordwith the general solution (Eqs.(4)and (5))that weobtained by a differentmethod in Example1. .)

As illustrated by Example2,the elimination procedureusedto solve a linearsystemfrequently will introduce a number of interdependent constants that mayappearto be arbitrary, but actually are not independent.The \"extra\" constantsmustthen beeliminatedby substitution of the proposedgeneral solution into one or moreof the originaldifferentialequations.The appropriatenumber of arbitrary constantsin a general solution of a linear system is determinedby the followingproposition:)))


If the operational determinant in (15)is not identically zero, then thenumber of independent arbitrary constants in a general solution of the

system in (9)is equal to the orderof its operational determinant-thatis, its degreeas a polynomial in D.)

(For a proof of this fact, seepages 144-150of E.L.Ince'sOrdinary DifferentialEquations(New York: Dover, 1956).)Thus the general solution of the system in

(10)of Example2 involvestwo arbitrary constants,becauseits operationaldetermi-nant D2 +3D- 10is of order2.

If the operationaldeterminant is identicallyzero,then the systemis said to bedegenerate.A degeneratesystem may have either no solution or infinitely many

independentsolutions.For instance, the equations)

Dx- Dy = 0,2Dx- 2Dy = 1)

with operationaldeterminant zero are obviouslyinconsistentand thus have no solu-tions.On the other hand, the equations)

Dx+ Dy = t,2Dx+ 2Dy = 2t)

with operational determinant zero are obviously redundant; we can substitute any

(continuously differentiable) function for x(t) and then integrate to obtain yet).Roughly speaking,every degeneratesystem is equivalent to either an inconsistentsystem or a redundant system.

Although the aforementionedproceduresand resultsare describedfor the caseof a systemof two equations, they can be generalizedreadily to systemsof threeormore equations.For the system)

LIIX + L12y + L13Z = 11(t),L21X + L22y + L23Z = 12(t),L31X + L32y + L33Z = 13(t))

(18))

of three linearequations, the dependentvariablex(t) satisfiesthe singlelinearequa-tion)

L11 L12 L13L21 L22 L23 X =L31 L32 L33)

11(t) L12 L13

12(t) L22 L23

13(t) L32 L33)

(19))

with analogousequationsfor y = yet) and z = z(t).For most systemsof more than

three equations, however,the method of operationaldeterminants is too tedious tobe practical.)

MechanicalVibrations)

A mechanicalsystem typically vibratesor oscillatesperiodicallyin oneor morespe-cific ways.The methodsof this sectionoften can beappliedto analyzethe \"natural

modesof oscillation\"of a given system.Example3 illustrates this approach.)))


Example3) In Example1 of Section5.1,wederived the equations)

(D2 +3)x+ (-l)y= 0,-2x+ (D2 +2)y = 0)

(20))

for the displacementsof the two massesin Fig.5.2.2.HereJ(t)= 0 becauseweassumethat there is no external force.Find the general solution of the system in

(20).)

Solution The operationaldeterminant of the system in (20)is)

Equilibrium positions)

FIGURE5.2.2.Themass-and-spring system ofExample 3.)

(D2 +3)(D2 +2)- (-1)(-2)= D4 +5D2 +4= (D2 + 1)(D2 +4).)

Hencethe equations for x(t) and yet) are)

(D2 + 1)(D2 +4)x= 0,(D2 + 1)(D2 +4)y = o.)

(21))

The characteristic equation (r2 + 1)(r2 +4) = 0 has roots i,-i,2i,and -2i.Sothe general solutions of the equations in (21)are)

x(t) = aI cost +a2 sin t +b I cos2t +b2sin 2t ,

yet) = CICOSt +C2sin t +dl cos2t +d2 sin 2t.)(22))

Becausethe operationaldeterminant is of order4, the general solution should

contain four (rather than eight) arbitrary constants.When we substitute x(t) and

yet) from (22)in the first equation in (20),weget)

o = x// +3x - y

= (-alcost -a2 sin t -4bI cos2t -4b2sin 2t)+3(alcost +a2 sin t +b I cos2t +b2 sin 2t)- (CIsin t +C2sin t +dl cos2t +d2sin 2t);)

thus)

o = (2aI-CI)cost + (2a2-C2)sin t

+ (-bI-dI ) cos2t + (-b2-d2) sin 2t.)

Becausecost, cos2t, sin t, and sin 2t are linearly independent, it follows that their

coefficientsin the last equation are zero.Thus)

CI = 2aI, C2= 2a2, dl = -bI , and d2 = -b2.)

Therefore)

x(t) = alcost+a2sint+blcos2t+b2sin2t,yet) = 2aI cost + 2a2sin t - b I cos2t - b2sin 2t)

(23))

is the desiredgeneral solution of the system in (20).) .)))

3)

2)

1)

::0\037)

-1)

-2)

-3o) 21t) 41t)

t)

FIGURE5.2.3.Thetwo massesmove in the samedirection, eachwith frequency WI = 1.)

3

2

1

::0\037

-1-2-3

0 1t 21t 31t 41tt)

FIGURE5.2.4.Thetwo massesmove in oppositedirectionswith

frequency W2 = 2.)

lIB Pro\302\273.\037em\037)


The equations in (23)describefreeoscillationsof the mass-and-springsys-tem of Fig.5.2.2-motionsubjectto no external forces.Four initial conditions(typically, initial displacementsand velocities)would be required to determinethevalues of aI,a2, b I , and b2.The expression)

(X(t),yet)) = al(cost, 2cost) +a2(sint, 2 sin t)

+b I (cos2t,-cos2t) +b2(sin2t, -sin 2t))(24))

then presentsthe general solution of the system in (20)as a linear combinationof four particular solutions.Moreover, the first two of theseparticular solutions

representphysically similar oscillationsof the masses,as do the latter two.Indeed,we can (by the usual trigonometricmachinations)write)

alcost+ a2sint = A cos(t-a),2aIcost + 2a2sin t = 2A cos(t-a))

and)

b I cos2t + b2 sin 2t = Bcos(2t- fJ),-bI cos2t - b2sin 2t = -Bcos(2t - fJ)

with A = Jar + a\037,tan a = a2/aI,B = JbI + b\037,

and tanfJ = b2/bl . ThenEq.(24)takes the form)

(X, y) = A (xI,YI)+ B(X2, Y2),) (25))

where the particular solutions)

(Xl (t),YI (t))= (cos(t-a),2cos(t-a))) (26))

and)

(X2(t), Y2(t)) = (cos(2t- fJ), -cos(2t- fJ))) (27))

describethe two naturalmodesof oscillationof the mass-and-springsystem.More-over, they exhibit its two (circular)naturalfrequenciesWI = 1 and W2 = 2.

The linear combination in Eq. (25)representsan arbitrary free oscillationofthe mass-and-springsystem as a superpositionof its two natural modesof oscilla-tion, with the constants A, a,B,and fJ determinedby the initial conditions.Figure5.2.3(wherea = 0) illustrates the natural mode(Xl, YI) of Eq. (26),in which the

two massesmove in synchrony in the samedirection with the samefrequencyof os-cillation WI = 1,but with the amplitude of m2 twice that of mI (becauseYI = 2XI).Figure 5.2.4(wherefJ = 0) illustrates the natural mode (X2, Y2) of Eq. (27),inwhich the massesmove in synchrony in oppositedirections,with the same fre-quency W2 = 2 and with equal amplitudesof oscillation(becauseY2 = -X2).)

Find general solutions of the linear systems in Problems1through 20. If initial conditions are given, find the particu-lar solution that satisfiesthem. In Problems1through 6, useacomputer system or graphing calculatorto construct a direc-)

tion fieldand typical solution curvesfor the given system.

1.x'= -x+ 3y, y' = 2y

2.x'= x -2y, y' = 2x - 3y)))


3.x'= -3x+ 2y, y' = -3x+ 4y;x(O)= 0, y(O)= 24. x'= 3x - y, y' = 5x-3y; x(O)= 1,y(O)= -15.x'= -3x-4y, y' = 2x + y

6.x'=x + 9y, y' = -2x-5y;x(O)= 3, y(O)= 27. x'= 4x+ y + 2t, y' = -2x+ y

8.x'= 2x + y, y' = x + 2y -e2t

9.x'= 2x -3y + 2 sin 2t, y' = x -2y -cos2t10.x'+2y'=4x+5y,2x'-y' = 3x;x(0)= 1,y(0)=-111.2y'-x'= x + 3y + et , 3x'-4y' = x - 15y+ e-t12.x II = 6x + 2y, y\" = 3x + 7Y

13.x\" = -5x+ 2y, y\" = 2x- 8y14.x\" = -4x+ sin t, y\" = 4x- 8y15.x\" -3y'-2x = 0, y\" + 3x'-2y = 016.x\" + 13y'-4x = 6sint,y\" -2x'-9y = 017.x\" + y\"

-3x'-y' -2x + 2y = 0,2X\" + 3y\" -9x'-2y'-4x + 6y = 0

18.x'= x + 2y + z, y' = 6x-y, z' = -x-2y - z19.x'= 4x-2y, y' = -4x+ 4y -2z,z' = -4y + 4z20.x' = y + z + e-t , y' = x + z, z' = x + y (Suggestion:

Solvethe characteristicequation by inspection.)21.Supposethat LI = alD2 + b l D + CI and L2 = a2D2+

b2D + C2, where the coefficientsare all constants, and

that x(t) is a twice differentiable function. Verify thatLI L2x = L2Llx.

22.Supposethat LIX = t Dx + x and that L2x = Dx + tx.Show that LI L2x f:= L2LIx. Thus linear operatorswith

variable coefficientsgenerally donot commute.)

Show that the systems in Problems23 through 25 are degen-erate. In eachproblem determine-by attempting to solvethe system-whetherit has infinitely many solutions or nosolutions.)

23.(D+ 2)x+ (D+ 2)y = e-3t

(D+ 3)x+ (D+ 3)y = e-2t

24. (D+ 2)x+ (D+ 2)y = t

(D+ 3)x+ (D+ 3)y = t 2

25. (D2 + 5D+ 6)x + D(D+ 2)y = 0(D+ 3)x+ Dy = 0)

In Problems26 through 29,first calculatethe operationalde-terminant of the given system in orderto determine how many

arbitrary constants should appear in a generalsolution. Then

attempt to solve the system explicitly soas to find such a gen-eralsolution.)

26.(D2 + l)x+ D2y = 2e-t(D2 - l)x+ D2y = 0

27. (D2 + l)x+ (D2 + 2)y = 2e-t(D2 - l)x+ D2y = 0

28.(D2 + D)x+ D2y = 2e-t(D2 - l)x+ (D2 -D)y = 0

29.(D2 + l)x-D2y = 2e-t(D2 - l)x+ D2y = 0)

30.) Supposethat the salt concentration in eachof the two brine

tanks of Example 2 of Section5.1initially (t = 0) is 0.5lb/gal.Then solvethe system in Eq.(5) there to find the

amounts x(t)and yet) of salt in the two tanks at time t.

Supposethat the electricalnetwork of Example 3 ofSec-tion 5.1is initially open-nocurrents are flowing. As-sume that it is closedat time t = 0; solvethe system in

Eq.(9)there to find II (t) and I2(t).RepeatProblem 31,exceptuse the electricalnetwork ofProblem27 of Section5.1.RepeatProblem 31,exceptuse the electricalnetwork ofProblem 28 of Section5.1.Assume that 11 (0) = 2 and

Q(O)= 0, so that at time t = 0 there is no chargeon the

capacitor.

Three 100-galbrine tanks are connectedas indicated in

Fig. 5.1.13of Section5.1.Assume that the first tank ini-

tially contains 100lb of salt, whereas the other two arefilled with fresh water. Find the amounts of salt in eachofthe three tanks at time t. (Suggestion:Examine the equa-tions to bederived in Problem26of Section5.1.)From Problem 31 of Section5.1,recallthe equations ofmotion)

31.)

32.)

33.)

34.)

35.)

II B 'mx = q y,)II B '

my = -q x)

36.)

for a particle of mass m and electricalchargeq under the

influence of the uniform magnetic field B = Bk. Sup-posethat the initial conditions arex (0) = ro, y(O) = 0,x'(O)= 0, and y'(O)= -wro where w = qBlm. Show

that the trajectory of the particle is a circleof radius roo

If, in addition to the magnetic field B = Bk, the charged

particleofProblem35moves with velocity v under the in-

fluenceof a uniform electricfield E = Ei, then the forceacting on it is F = q(E+ v x B).Assume that the particlestarts from rest at the origin. Show that its trajectory is the

cycloid)

x = a(l-coswt), y = -a(wt- sinwt))

where a = EI(wB)and w = qBlm. Thegraph of such acycloidis shown in Fig. 5.2.5.)

FIGURE5.2.5.Thecycloidalpath ofthe particleofProblem36.)

37.) In the mass-and-spring system of Example 3, supposein-

steadthat ml = 2, m2 = 0.5,k l = 75,and k2 = 25. (a)Find the general solution of the equations ofmotion of the

system. In particular, show that its natural frequenciesareWI = 5 and W2 = 5,J3.(b) Describethe natural modesofoscillationof the system.)))

5.3Matricesand LinearSystems 347)

38.Considerthe system of two massesand three springsshown in Fig. 5.2.6.Derivethe equations of motion

mIx\" == -(kl + k2)x + k2y,

m2Y\" == k2x - (k2 + k3)y.) FIGURE5.2.7.Themechanical system ofProblem47.)

48. Supposethat the trajectory (x(t),y(t\302\273of a particle mov-

ing in the plane satisfiesthe initial value problem)

Equilibrium positions) XII - 2yl + 3x == 0,yll + 2Xl + 3y == 0;x(O)== 4, y(O) == x/(O)== yl(O) == O.)

FIGURE5.2.6.Themechanicalsystem ofProblem38.

In Problems 39 through 46, find the generalsolution of the

system in Problem 38 with the given massesand spring con-stants. Find the natural frequenciesof the mass-and-springsystem and describeits natural modesof oscillation.Useacomputer system or graphing calculatorto illustrate the twonatural modesgraphically (asin Figs. 5.2.3and 5.2.4).39.m I == 4, m2 == 2,k l == 8,k2 == 4, k3 == 040. m I == 2,m2 == 1,k I == 100,k2 == 50,k3 == 041.m) == l,m2 == l,kl == l,k2 ==4,k3 == 142. m I == 1,m2 == 2,k I == 1,k2 == 2, k3 == 243. m I == 1,m2 == 1,k l == 1,k2 == 2,k3 == 144. m) == 1,m2 == 1,k l == 2,k2 == 1,k3 == 245. m I == 1,m2 == 2,k I == 2,k2 == 4, k3 == 446. m I == 1,m2 == 1,k I == 4, k2 == 6, k3 == 447. (a) For the system shown in Fig. 5.2.7,derive the equa-

tions of motion)

Solvethis problem. You should obtain)

x(t) == 3cost + cos3t,

y(t) == 3 sin t - sin 3t.)

Verify that theseequations describethe hypocycloid traced

by a point P(x,y) fixed on the circumferenceof a circleof radius b == 1 that rolls around inside a circleof radius

a == 4. If P begins at A(a, 0) when t == 0, then the pa-rameter t representsthe angle A OCshown in Fig. 5.2.8.)

y)

mx ll== -2kx+ ky,

myll == kx - 2ky + kz,mz ll

== ky - 2kz.)

A(a,0))

x)

(b) Assume that m == k == 1.Show that the natural fre-quenciesofoscillation of the system are

WI =../2,W2 =j2 -../2,and W3 =j2 + ../2.) FIGURE5.2.8.Thehypocycloid ofProblem 48.)

_ }\\1atricesandLinea\037._\037Y\037\037\037I1?:s)

A lthough the simple elimination techniques of Section5.2suffice for the

solution of small linear systemscontaining only two or three equationswith

constant coefficients,the general propertiesof linear systems-aswell as solutionmethods suitable for larger systems-aremost easily and conciselydescribedusingthe language and notation of vectors and matrices.For ready referenceand review,this sectionbeginswith a completeand self-containedaccountof the matrix notation

and terminology that is needed.Specialtechniquesof linear algebra-specifically,thoseassociatedwith eigenvalues and eigenvectors-areintroducedas neededin

subsequentsectionsof this chapter.)))

NUNn \037nnNnNNNNUNNnNNn....n \"'\" N N \037N.'U.','u..n..\".'U,'U,' \037....\"\",\037,'

Example1)

ReviewofMatrix NotationandTerminologyAn m x n matrixA is a rectangulararray of mn numbers (or elements)arrangedin

m (horizontal)rowsand n (vertical)columns:)

all al2 al3 alj alna21 a22 a23 a2j a2n

a31 a32 a33 a3j a3n

A= (1)ail ai2 ai3 a.. ainlJ)

amI am2 am 3)amn)amj)

We will ordinarily denotematrices by boldfacecapital letters.Sometimeswe usethe abbreviationA = [aij] for the matrix with the element aij in the ith row and

jth column, as in Eq.(1).We denote the zeromatrix,each entry of which is zero,by)

o 0o 0)

oo)

0=) (2))

o 0) o)

Actually for each pair of positive integers m and n there is an m x n zero matrix,but the singlesymbol 0 will sufficefor all thesezero matrices.

Two m x n matrices A = [aij]and B = [bij]are said to be equal if

correspondingelementsare equal;that is, if aij = bij for 1 < i <m and 1 <j < n.We add A and B by adding correspondingentries:)

\037) A +B = [ aij ] + [ bij ] = [ aij +bij ] .) (3))

Thus the element in row i and columnj of C = A +Biscij = aij +bij .Tomultiplythe matrix A by the number c,we simply multiply each of its elementsby c:)

\037) cA = Ac = [caij].) (4))

If)

[2 -3

]A= 4 7') and C = [; -

\037 l)[-13B=

7)\037\037l)

then)

[2 -3

] [-13A+B= 4 7 +

7)\037])

10]

_[-11

-5 -11)

and)

6C= 6 \302\267

[;)0

] [18 0

]-7-

30 -42 .) .)))


We denote(-l)Aby -Aand definesubtractionof matricesas follows:)

A -B = A + (-B).) (5))

The matrix operations just defined have the following properties,each ofwhich is analogous to a familiar algebraicpropertyof the real number system:)

A +0 = 0 +A = A, A -A = 0;A+B=B+AA + (B+C)= (A +B)+C

c(A +B) = cA +cB,(c+d)A = cA +dA.)

(commutativity) ;

(associativity);)

(6)

(7)

(8))

(distributivity)) (9))

Each of thesepropertiesis readily verified by elementwiseapplicationof a corre-spondingproperty of the real numbers.For example,ai}+ hi} = hi} +au for all iand j becauseadditionof real numbers is commutative.Consequently,)

A +B = [ ai}+ hi} ] = [ hi} +au ] = B+A.)

The transposeAT of the m x n matrix A = [ai}]is the n x m (note!)matrix

whosejth column is the jth row of A (and consequently,whoseith row is the ith

column of A). Thus AT = [aji ],although this is not notationally perfect; youmust rememberthat AT will not have the same shapeas A unlessA is a squarematrix-thatis,unlessm = n.

An m x 1 matrix-onehaving only a singlecolumn-iscalleda columnvector, or simply a vector.We often denote column vectorsby boldfacelowercaseletters, as in)

b =[-\037])

Xl

X2)

or x =)

Xm)

Similarly, a row vector is a 1 x n matrix-onehaving only a singlerow, suchas c = [5 17 0 -3].For aesthetic and typographicalreasons,we will

frequently write a column vector as the transposeof a row vector; for example,the

two precedingcolumn vectors may be written in the forms)

b = [3 -7) o ]T

and x= [Xl X2) Xm ]T .)

Sometimesit is convenientto describean m x n matrix in terms of either itsm row vectors or its n column vectors.Thus if we write)

a}a2)

and B = [ b}) b2)...b ]n ,)A=)

am)

it is understood that a},a2, ..., and am are the row vectors of the matrix A and that

b},b2, ..., and bn are the column vectorsof the matrix B.)))


Example2)

Matrix MultiplicationThe propertieslistedin Eqs.(6)through (9)are quite natural and expected.The first

surprisesin the realm of matrix arithmetic comewith multiplication.We definefirst

the scalarproductof a row vector a and a column vector b, each having the samenumber p of elements.If)

a == [al a2) ... b ]T

p ,)ap ])and b == [bl b2)

then a \302\267 b is definedas follows:)

\037)

p

a.b ==Lakbk==a}bl +a2b2+...+apbp ,k=l)

(10))

exactly as in the scalaror dot product of two vectors-afamiliar topic from elemen-tary calculus.

The product AB of two matrices is definedonly if the number of columnsofA is equal to the number of rows of B.If A is an m x p matrix and B is a p x n

matrix, then their productAB is the m x n matrix C == [ci}],where ci} is the

scalarproduct of the ith row vectorai of A and the jth columnvectorbj of B.Thus)

\037) C ==AB == [ ai \302\267 bj ] .) (11))

In terms of the individual entries of A == [ai}] and B == [bi}],Eq.(11)can berecastin the form)

\037)

p

ci} ==Laikbkj.k=l)

(12))

For purposesof hand computation, the definition in Eqs.(11)and (12)is easy to

rememberby visualizing the picture)

all al2 alp bll b l2 PI} bIn

a21 a22 a2p b21 b22 \0372jb2n)

ai ----+)

am I am 2) bpI bp2)

bpn)tJpj

tb.})

amp)

which showsthat one forms the dot product of the row vector ai with the columnvector b} to obtain the element ci} in the ith row and the jth column of AB.It mayhelp to think of \"pouring the rows of A down the columns of B.\" This alsoremindsus that the number of columns of A must beequal to the number of rows of B.)

Checkyour understanding of the definition of matrix multiplication by verifyingthat if)

[2 -3

]A== -1 5)

\037l)[13

and B ==4)))

5.3Matricesand LinearSystems 351then

AD =[-i

-3] [13 9]

_[

14\037\037l

5 4 0-

7

Similarly, verify that

[:-3 1

] [X] [2X

- 3y + z

]5 -2 y = 4x+5y-2z-7 o z 6x -7y

and that

1 2 0 7 -13 4

[-i1 -;]= 2 15 1 .5 6 3 4 23 3

7 8 6 31 5)

It can be shown by direct (though lengthy) computationbasedon its definition

that matrix multiplication isassociativeand isalsodistributive with respectto matrix

addition; that is,)

A(BC)= (AB)C) (13))

and)

A(B+C)= AB +AC,) (14))

provided that the matrices are of such sizes that the indicated multiplications and

additions are possible.But matrix multiplication is not commutative. That is, if A and B are both

n x n matrices (sothat both the productsAB and BA are definedand have the samedimensions-nx n),then, in general,)

AB i= BA.) (15))

Moreover,it can happen that)

AB = 0 even though A i= 0 and B i= O.) (16))

Examplesillustrating the phenomenain (15)and (16)may befound in the problems,although you can easilyconstructyour own examplesusing 2x2 matriceswith small

integral elements.)

InverseMatrices)A squaren x n matrix is saidto haveordern. The identity matrix of ordern is the

squarematrix)

1 000o 100o 0 1 0000 1)

oooo) (17))1=)

o 0 0 0 1)))


for which each entry on the principaldiagonalis 1 and all off-diagonalentriesarezero.It is quite easyto verify that)

\037) AI = A = IA) (18))

for every squarematrix A of the sameorderasI.If A is a squarematrix, then an inverseof A is a square matrix B of the same

orderasA such that both)

AB = I and BA = I.)

It is not difficult to show that if the matrix A has an inverse, then this inverse is

unique. Consequently,we may speakof the inverseof A, and we will denote it byA-I.Thus)

\037) AA- I = I = A-IA ,) (19))

given the existenceof A-I. It is clearthat some square matrices do not haveinverses-considerany squarezero matrix. It is alsoeasyto show that if A -Iexists,then (A

-I)-1existsand (A-I)-1= A.

In linear algebra it isproved that A -Iexistsif and only the determinant det(A)of the squarematrix A is nonzero. If so,the matrix A is saidto be nonsingular;ifdet(A) = 0,then A is calleda singularmatrix.)

Determinants)We assumethat the student has computed 2 x 2 and 3 x 3 determinants in earliercourses.If A = [ai}] is a 2 x 2 matrix, then its determinantdet(A) = IAI isdefined as)

\037) IAI-_ all al2 = alla22 -a12a21.a21 a22)

Determinantsof higherordermay bedefinedby induction, asfollows.If A = [ aij ]is an n x n matrix, let Ai} denote the (n - 1)x (n - 1)matrix obtained from A by

deleting its ith row and its jth column. The expansionof the determinant IAI alongits i th row is given by)

n

IAI = L(-I)i+j ai}IAi}1j=1)

(i fixed),) (20a))

and its expansionalong its jth column is given by)

n

IAI = L(_1)i+jai}IAi} I

i=1)(j fixed).) (20b))

It is shown in linear algebra that whicheverrow we use in Eq. (20a)and whichevercolumn we use in Eq. (20b),the resultsare the samein all 2n cases.HenceIAI iswell definedby theseformulas.)))

Example3)

Example4)


If)

A=[

\037

-2)

1 -2

]2 1 ,3 5)

then the expansionof IAI along its secondrow is)

1 -2 +2 . 3 -2 _ 1 . 3 1IAI = -4.

33 5 -2 5 -2= -4.11+2 .11- 1 .11= -33.)

And the expansionof IAI along its third column is)

4IAI = -2. -2)

2 3- 1 .3 -2)

1 +5. 3 13 4 2)

= -2.16- 1 .11+5 .2=-33.) .)

Calculatorsand computersare convenientfor the calculationof higher-dimen-sional determinants and inverse matrices, but determinants and inversesof 2 x 2matrices are easyto compute by hand. For instance, if the 2 x 2 matrix)

A=[\037 \037])

has nonzero determinant IAI ==ad-be i= 0,then its inversematrix is)

\037) A-I _ \037

[d -b

]-

IAI -e a.) (21))

Note that the matrix on the right-hand sideof Eq.(21)is obtained from A by inter-

changing the diagonalelementsand changingthe signsof the off-diagonalelements.)

If)

A=[\037 \037l

then IAI = 6 .7-5 .8 = 2.HenceEq.(21)gives)

A-I=![

7 -8]

=[

\037

-4]

.2 -5 6 -\037 3

2)

You shouldpauseto verify that)

A-IA =[_ j -;][:\037]

=[\037

\037 l) .)))


Example5)

Matrix-ValuedFunctions)A matrix-valuedfunction, or simply matrixfunction, is a matrix such as)

XI (t)

X2 (t)x(t) = (22a)

Xn (t)

or

all(t) aI2(t) aln(t)a21(t) a22(t) a2n (t )

(22b)A(t) =

ami (t) am2(t) amn (t))

in which each entry is a function of t. We say that the matrix function A(t) iscontinuous(or differentiable)at a point (oron an interval) if eachof its elementshas the sameproperty. The derivative of a differentiablematrix function is definedby elementwisedifferentiation;that is,)

\037) A'(t)= dA =[

daij] .

dt dt)(23))

If)

x(t) =[\037t ])

and A(t) =[

sint 1

]t cost ')

then)

dx

[1

]-- 2tdt -e-t)

I

[cost 0

]and A (t) =1

. .-SIn t)

.)

The differentiationrulesd dA dBdt (A +B) = dt + dt) (24))

and)

d dB dA-(AB)= A-+-Bdt dt dt)

(25))

follow readily by elementwiseapplicationof the analogousdifferentiation rules ofelementary calculusfor real-valuedfunctions. If c is a (constant) real number and

C is a constant matrix, then)

d dA-(cA)= c-,dt dt)

d dA-(AC)= -C.dt dt)

(26))d dA-(CA)= C-,dt dt)

and)

Becauseof the noncommutativity of matrix multiplication, it is important not toreverse the orderof the factors in Eqs.(25)and (26).)))

Example6)


First-OrderLinearSystemsThe notation and terminology of matrices and vectors may seemrather elaboratewhen first encountered,but it is readily assimilatedwith practice.Our main useformatrix notation will be the simplificationof computations with systemsof differ-ential equations,especiallythose computationsthat would beburdensomein scalarnotation.

We discusshere the general systemof n first-orderlinear equations)

\037)

x\037

= P11(t) XI + P12(t )X2 + ...+ Pin (t )xn + 11(t) ,

X\037

= P21(t)Xl+ P22(t)X2 +...+ P2n(t)X n + 12(t),X\037

= P31(t)Xl+ P32(t)X2 +...+ P3n(t)X n + 13(t),) (27))

X\037

= Pnl(t)Xl + Pn2(t)X2 +...+ Pnn(t)X n + In(t).)

If we introduce the coefficientmatrix)

pet)= [Pij (t) ])

and the column vectors)

x= [ Xi] and f(t) = [ Ii(t) ] ,)

then the system in (27)takes the form of a singlematrix equation)

\037)

dx-= P(t)x+f(t).dt)

(28))

We will seethat the general theory of the linear system in (27)closelyparallelsthat of a singlenth-order equation. The matrix notation used in Eq. (28)not only

emphasizesthis analogy,but alsosavesa great dealof space.A solutionof Eq. (28)on the open interval I is a column vector function

x(t) = [Xi (t) ] such that the component functions of x satisfy the system in (27)identically on I. If the functions Pij (t) and Ii(t) are all continuous on I, then

Theorem 1 of Section5.1guarantees the existenceon I of a unique solution x(t)satisfying preassignedinitial conditionsx(a)= b.)

The first-ordersystem)

X\037

= 4Xl - 3X2,

x\037= 6Xl - 7X2)

can be written as the singlematrix equation)

\037;

= [:=\037 ]x= Px.)

To verify that the vector functions)

[3e2t

]Xl (t) = 2e2t) [-5t

]and X2(t) = 3:-51)))

are both solutions of the matrix differentialequation with coefficientmatrix P, weneedonly calculate)

PxI = [:=;][ ;:\037: ]=

[\037:\037: ]=

X/I)

and)

[4 3

] [-5t

] [5 -5t

]Px2 =6 =7 3:-51 =

_\0375:-51 =x;.) .)

To investigatethe general nature of the solutionsof Eq.(28),we considerfirst

the associatedhomogeneousequation)

\037)

dx-= P(t)x,dt)

(29))

which has the form shown in Eq. (28),but with f(t) = O. We expectit to have n

solutions Xl, X2, ..., Xn that are independent in someappropriate sense,and suchthat everysolutionof Eq.(29)isa linearcombinationof thesen particularsolutions.Given n solutions Xl, X2, ..., Xn of Eq.(29),let us write)

Xlj(t))

Xj(t) =xij(t)) (30))

Xnj (t))

Thus Xij (t) denotesthe ith component of the vector Xj (t), so the secondsubscriptrefers to the vector function Xj (t),whereasthe first subscriptrefers to a componentof this function. Theorem 1 is analogous to Theorem 1 of Section2.2.)

THEOREM1 PrincipleofSuperpositionLetXl, X2, ..., Xn be n solutions of the homogeneouslinear equation in (29)onthe openinterval I.If CI,C2,..., Cn are constants, then the linear combination)

X(t) = ClXI (t)+C2X2(t) + ...+ cnxn(t)) (31))


Proof:We know thatx\037

immediately that)

P(t)Xi for each i (1< I < n), so it follows)

I I I IX = CIX I +C2x2 + ...+cnxn

= CIP(t)XI+C2P(t)X2 + ...+cnP(t)xn= pet)(CIXI +C2X2 + ...+ CnX n ).)

That is,x'= pet)x,asdesired.The remarkablesimplicity of this proofdemonstratesclearly one advantageof matrix notation. .)))

Example6)

Continued)


If Xl and X2 are the two solutionsof)

dx _[

4 -3]dt

-6 -7 X)

discussedin Example6,then the linear combination)

[3e2t

] [e-5t

]x(t)= CIXI (t) +C2X2(t) = CI 2e2t +C2 3e-5t)

is alsoa solution. In scalarform with X = [Xl X2]T,this gives the solution)

Xl (t) = 3CIe2t + C2e-5t,

X2(t) = 2Cle2t+3C2e-5t,)

which is equivalent to the general solution we found by the method of eliminationin Example2 of Section5.2. .)

IndependenceandGeneralSolutions)Linear independenceis defined in the sameway for vector-valuedfunctions as forreal-valued functions (Section2.2).The vector-valuedfunctions Xl, X2, ..., Xn arelinearlydependenton the interval I provided that there existconstantsCI,C2,...,Cn not allzero such that)

\037) CIXI (t) +C2X2(t)+ ...+CnX n (t) = 0) (32))

for all t in I. Otherwise,they are linearly independent.Equivalently, they arelinearly independent provided that no one of them is a linear combinationof the

others.For instance, the two solutionsXl and X2 of Example6 are linearly indepen-dent because,clearly,neither is a scalarmultiple of the other.

Just as in the caseof a singlenth-order equation, there is a Wronskian deter-minant that tellsus whether or not n given solutionsof the homogeneousequationin (29)are liJ).early dependent.If Xl, X2, ..., Xn are such solutions,then their Wron-skian is the n x n determinant)

XII (t) XI2(t) Xln(t)

X21 (t) X22(t) X2n (t )\037 Wet) = (33)

Xn I (t) Xn2(t) Xnn (t))

using the notation in (30)for the componentsof the solutions.We may write eitherWet) or W(XI, X2, ...,xn ). Note that W is the determinant of the matrix that has asits column vectors the solutionsXl, X2, ..., Xn. Theorem2 is analogousto Theorem3 of Section2.2.Moreover,its proof is essentiallythe same,with the definition ofW (Xl, X2, ..., xn ) in Eq. (33)substituted for the definition of the Wronskian of n

solutionsof a singlenth-order equation (seeProblems42 through 44).)))


Example7)

THEOREM2 Wronsklansof Solutions)

Supposethat Xl, X2, ..., Xn are n solutions of the homogeneouslinear equationx' = P(t)xon an openinterval I.Supposealsothat P(t) is continuouson I.Let)

w = w (xI,X2, ..., Xn ).)

Then:. If Xl, X2, ..., Xn are linearly dependenton I, then W = 0 at every pointof I.. If Xl, X2, ..., Xn are linearly independenton I, then W 1=0 at each pointof I.

Thus there are only two possibilitiesfor solutionsof homogeneoussystems:Ei-ther W = 0at everypoint of I,or W = 0 at no point of I.)

,, \" '\" .. ......... \037.,...,...,....\"..,..,..\"......... . .. \037 _...'....n....,... ........... .\"..\"\".......

It is readily verified(as in Example6) that)

[2et

]Xl (t) =

2;1,)

[2e3t

]X2 (t) = 0 ,

_e3t) [2e5t

]and X3(t) = -2e5t

e5t)

are solutionsof the equation)

dx =[-i -; -\037

3]x.

dt 0-1) (34))

The Wronskian of thesesolutions is)

W=)2et

2et

et)

2e3t

o_e3t)

= e9t)

2222 0 -2 = -16e9t

,1 -1 1)

2e5t-2e5t

e5t)

which is never zero.HenceTheorem 2 impliesthat the solutions Xl, x2, and X3 arelinearly independent(on any openinterval). .

Theorem 3 is analogous to Theorem 4 of Section2.2.It says that a generalsolutionof the homogeneousn x n systemx'= P(t)xis a linear combination)

\037) X = CIXI +C2X2 + ...+ CnX n) (35))

of any n given linearly independentsolutionsXl, X2, ..., Xn.)

THEOREM3 GeneralSolutionsof HomogeneousSystemsLetXl, X2, ..., Xn ben linearly independentsolutionsof the homogeneouslinearequation x' = P(t)x on an open interval I,where pet) is continuous. If x(t)is any solution whatsoever of the equation x' = P(t)x on I, then there existnumbers CI,C2,..., Cn such that)

X(t) = CIXl (t)+C2X2(t) + .. \302\267 + cnxn(t)) (35))

for all t in I.)))


Proof:Leta bea fixedpoint of I.We show first that thereexistnumbers CI,C2,..., Cn such that the solution)

y (t) = CI X I (t)+C2X2 (t)+ ...+Cn Xn (t )) (36))

has the sameinitial values at t = a asdoesthe given solutionx(t);that is,such that)

CI X I (a)+C2X2(a)+ ...+Cn Xn (a) = x(a).) (37))

LetX(t) be the n x n matrix with column vectors XI, X2, ...,Xn , and let cbe the

column vector with componentsCI,C2,..., Cn. Then Eq.(37)may bewritten in the

form)

X(a)c= x(a).) (38))

The Wronskian determinant W(a) = IX(a)1is nonzero becausethe solutionsXI,

X2, ..., Xn are linearly independent.Hencethe matrix X(a)has an inversematrix

X(a)-I.Therefore the vectorc = X(a)-Ix(a)satisfiesEq.(38),as desired.Finally, note that the given solution x(t) and the solution yet) of Eg.(36)-

with the values of Ci determined by the equation c = X(a)-Ix(a)-havethe sameinitial values (at t = a).It follows from the existence-uniquenesstheoremof Sec-tion 5.1that x(t) = yet) for all t in I.This establishesEq.(35). .

Remark:Every n x n systemx' = P(t)xwith continuouscoefficientma-trix does have a set of n linearly independent solutions XI, X2, ...,Xn as in the

hypothesesof Theorem 3.It suffices to choosefor Xj (t) the unique solution suchthat)

ooo)

Xj(a)= 01o)

+--position j)

o)

-thatis, the column vector with all elementszero exceptfor a 1 in row j. (In otherwords,xj(a)is merely the jth column of the identity matrix.) Then)

W (xI,X2, ..., Xn )I t=a =

II

I= 1 i= 0,)

so the solutionsXI, X2, ..., Xn are linearly independentby Theorem2.Howactuallyto find thesesolutions explicitly is another matter-onethat we addressin Section5.4(for the caseof constant coefficientmatrices). .)

InitialValue ProblemsandElementaryRow OperationsThe general solution in Eq. (35)of the homogeneouslinear systemx' ==P(t)xcanbe written in the form)

X(t) = X(t)c,) (39))))


where)

X(t) = [XI(t) X2(t)) Xn (t) ]) (40))

is the n x n matrix whosecolumn vectorsare the linearly independentsolutionsXl,

X2, ..., Xn , and c = [ CI C2 Cn ]T is the vector of coefficientsin the linear

combination)

x(t) = CI X I (t) +C2x2(t) + ...+cn Xn (t ) .) (35))

Supposenow that we wish to solve the initial valueproblem

dxdt

= Px, x(a)= b,) (41))\037)

where the initial vector b = [bl b2 bn ]T is given. Then, according to

Eq.(39),it sufficesto solve the system)

X(a)c= b (42))

to find the coefficientsCI,C2,..., Cn in Eq.(35).We thereforereviewbriefly the elementarytechniqueof row reductionto solve

an n x n algebraiclinear system)

allXI+ al2x2+...+ alnX n = b l ,

a2lX I + a22X2 + ...+ a2n Xn = b2,)

(43))

anI Xl + an2X2 + ...+ annX n = bn)

with nonsingular coefficient matrix A = [aij],constant vector b = [bi],and un-

knowns xI,X2, ..., Xn. The basicideais to transform the system in (43) into the

simpleruppertriangular form)

a llXI + a l2X2 +...+ a lnX n = b l ,

a22X2 + ...+ a2nXn = b2,)

(44))

annX n = bn)

in which only the unknowns xj,Xj+l,..., Xn appear explicitly in the jth equation(j = 1,2, ..., n). The transformed system is then easilysolved by the processofbacksubstitution. First the last equation in (44)is solvedfor Xn , then the next-to-Iastis solvedfor Xn-l,and so forth, until the first equation is finally solvedfor Xl.

The transformation of the systemin (43) to upper triangular form is most

easilydescribedin termsof elementaryrow operationson the augmentedcoefficientmatrix)

all al2 al n b l

[A:b]=a21 a22 a2n b2

(45)

anI an2 ann bn)

that is obtained by adjoining the vector b to the matrix A as an additional column.The admissibleelementaryrow operationsare of the followingthree types:)))

Example8)


1.Multiply any (single)row of the matrix by a nonzero constant.2. Interchangeany two rows of the matrix.

3.Subtract a constant multiple of one row from any other row.

The goal is to usea sequenceof such operations(oneby one, in turn) to trans-form [A : b ] into an upper triangular matrix, one that has only zerosbeneath its

principal diagonal. This upper triangular augmented coefficientmatrix then cor-respondsto an upper triangular system as in (44). The processof transforming

[A : b ] is carriedout one column at a time, from left to right, as in the next exam-ple.)

\037 \"\037\037\037\"\"''''''''W'-'V\037.'N\037\037''' \"\"',....,,,...,\"\"........,....,,......,,, .nuNN.......... \"\"\"\"\"\"''''' __,..\"\"\"\"U.'N..U...,....u....,.....\"\"'__\037\"..............V\"\"''''..N....NU....U... ... \"\"'''''''''''\"...\".,''''''''''\"...\".,........... \"'\" \"\"\"

Usethe solution vectors given in Example7 to solve the initial value problem)

dx =[

_i -\037 -\037

]x, X(0)==

[\037

6].

dt 0 -1 3)

(46))

Solution It follows from Theorem 3 that the linear combination)

x(t) = CIXI(t) +C2X2(t) +C3X3(t))

[2et

] [2e3t

] [2e5t

]= CI 2et +C2 0 +C3 -2e5t

et _e3t e5t)

is a general solutionof the 3 x 3 linear system in (46).In scalarform, this givesthe

general solution)

Xl (t) = 2CIet + 2C2e3t+ 2C3e5t,X2 (t) = 2CIet - 2C3e5t,X3 (t) = CIet - C2e3t + C3e5t.

We seekthe particular solution satisfying the initial conditions)

Xl (0)= 0, X2(0) = 2, X3 (0)= 6.)

When we substitute thesevalues in the three precedingscalarequations,we get the

algebraiclinear system)

2CI+ 2C2+ 2C3= 0,2CI - 2C3==2,

CI - C2+ C3 ==6

with augmentedcoefficientmatrix)

[2 2 2:0

]2 0 -2

i

2 .1 -1 1 I 6

I)

Multiplicationof each of the first two rows by\037

gives)

[:)

1 1o -1-1 1)

: 0

]i1 ,

: 6I)))


then subtraction of the first row both from the secondrow and from the third row

gives the matrix)

[1 1 1:0

]o -1 -2

i

1 .o -2 0 I 6

I

The first column of this matrix now has the desiredform.Now we multiply the secondrow by -1,then add twice the result to the third

row. Thereby we get the upper triangular augmentedcoefficientmatrix)

[\037)

11o)

1 : 0

]2 !-14: 4

I)

that correspondsto the transformedsystem

Cl + C2 + C3 == 0,C2 + 2C3==-1,

4C3== 4.We finally solve in turn for C3 == 1,C2 ==-3,and Cl ==2.Thus the desiredparticularsolution is given by)

[4et - 6e3t + 2e5t

]x(t) ==2Xl (t) - 3X2(t) +X3(t) == 4et - 2e5t .

2et + 3e3t + e5t)

.)

NonhomogeneousSolutionsWe finally turn our attention to a nonhomogeneouslinear systemof the form

dxdt

==P(t)x+ f(t).)\037) (47))

The following theorem is analogous to Theorem 5 of Section2.2and is proved in

preciselythe sameway, substituting the precedingtheorems in this sectionfor the

analogous theorems of Section2.2.In brief, Theorem 4 means that the generalsolutionofEq.(47)has the form)

\037) X(t) ==xc(t)+xp(t),) (48))

wherexp(t)isa singleparticularsolutionofEq.(47)and the complementaryfunc-tion Xc (t) is a general solutionof the associatedhomogeneousequationx'==P(t )x.)

THEOREM4 SolutionsofNonhomogeneousSystemsLet xp be a particular solution of the nonhomogeneouslinear equation in (47)on an openinterval I on which the functions P(t) and f(t) are continuous. LetXl, X2, ..., Xn be linearly independent solutionsof the associatedhomogeneousequationon I.Ifx(t) is any solutionwhatsoeverofEq.(47)on I, then thereexistnumbers Cl,C2,..., Cn suchthat)

\037) X(t) =C1Xl(t)+C2X2(t) + ...+CnX n (t)+xp(t)) (49))

for all t in I.)))

Example9)


Thus finding a general solution of a nonhomogeneouslinear system involvestwo separatesteps:)

1.Finding the general solution Xc (t) of the associatedhomogeneoussystem;2.Finding a singleparticular solutionxp (t) of the nonhomogeneoussystem.

The sum x(t) ==xc(t)+xp(t)will then bea generalsolutionof the nonhomogeneoussystem.)

I,\"'.\".-f. ,'u.,\", \037'\"\"' \037-''''';'''NN'''N.'.-''.'.-N.-N \" '\" .... '\"

The nonhomogeneouslinear system)

X\037

== 3XI - 2X2 - 9t + 13,

x\037== -Xl + 3X2 - 2X3 + 7t - 15,

x\037

== -X2 + 3X3 - 6t + 7)

is of the form in (47)with)

P(t) =[-!=i -\037],) [

-9t+ 13

]f(t) == 7t - 15 .-6t+ 7)

In Example7 we saw that a general solution of the associatedhomogeneouslinear

system)

dxdt) [-!=i -\037 ]

x)

is given by)

[2CI et + 2C2e3t+ 2C3e5t

]xc(t)== 2clet - 2c3e5t ,CIet - C2e3t+ C2e5t)

and we can verify by substitution that the function)

Xp(t) =[:\037 ])

(found using a computer algebra system,or perhapsby a human being using amethod discussedin Section5.8)is a particular solution of the original nonhomo-geneoussystem. Consequently,Theorem 4 impliesthat a general solution of the

nonhomogeneoussystem is given by)

x(t) ==xc(t)+xp(t);)

that is, by)

Xl (t) ==2CIet + 2C2e3t+ 2C3e5t+ 3t,X2(t) ==2clet - 2C3e5t+ 5,X3(t) == clet - C2e3t+ C3e5t+ 2t.)

.)))

364 Chapter5 LinearSystemsof Differential Equations)_ Problems)

1.Let)

A = [; -\037]and B =

U -\037 J.Find (a) 2A + 3B; (b) 3A -2B; (c) AB; (d) BA.

2.Verify that (a) A(BC)== (AB)Cand that (b) A(B+C)==

AB + AC, where A and B are the matrices given in Prob-lem 1 and)

c=[\037-;J.3.Find AB and BA given)

A= [;) and B =[-

i)\037

].

-2)o

-4) -;])4. LetA and B be the matricesgiven in Problem 3 and let)

x = [;!t]) and y ==

[s:\037 t

].

cost)

Find Ay and Bx. Are the products Ax and By defined?Explain your answer.

5.Let

A =[_\037 \037 -\037]

and B =[r -: =\037 l

Find (a) 7A + 4B; (b) 3A -5B; (c) AB; (d) BA;(e) A - tI.

6.Let)

Al =[ _\037)

\037J.

B= [;)

A 2 =[_

\037

\037J.)

-\037 J.)

(a) Show that AIB == A 2B and note that Al i= A 2. Thusthe cancellationlaw doesnot hold for matrices; that is, ifAIB == A 2B and B i= 0, it doesnot follow that Al == A 2.(b) LetA == Al -A2 and usepart (a) to show that AB == O.Thus the product of two nonzeromatrices may be the zeromatrix.

7. Compute the determinants of the matrices A and B inProblem 6.Are your results consistent with the theoremto the effect that)

det(AB) == det(A) .det(B))

for any two squarematrices A and B of the sameorder?8.Supposethat A and B arethe matrices ofProblem5.Ver-

ify that det(AB) == det(BA).)

In Problems 9 and 10,verify the product law for differentia-tion, (AB)' == A'B + AB'.)

9.A(t) =[;3

[et

10.A(t) == -t8t)

2t -1

] [1- t 1+ t

]\037

and B(t)= 3t2 4t3

t t 2

] [3

]o 2 and B(t)== 2e-t-1 t 3 3t)

In Problems11through 20,write the given system in the formx'== P(t)x + f(t).)

11.x' == -3y,y' == 3x12.x' == 3x -2y, y' == 2x + Y13.x' == 2x + 4y + 3et , y' == 5x- y

-t 2

14.x' == tx -ety + cost,y' == e-tx+ t 2y - sint

15.x' == y + z, y' == z + x,z' == x + Y

16.x' == 2x-3y, y' == x + y + 2z,z' == 5y -7z17.x' == 3x -4y + z + t, y' == x -3z+ t 2, Z' == 6y -7z+ t3

18.x' == tx -y +etz, y' == 2x+t 2y-Z, z' == e-tx +3ty+t 3z

19.x\037

== X2, x\037

== 2X3, x\037

== 3X4,x\037

== 4XI

20.x\037

== X2 + X3 + 1,x\037

== X3 + X4 + t,x\037

== Xl + X4 + t 2,x\037

== Xl + X2 + t 3)

In Problems21through 30,first verify that the given vectorsare solutions of the given system. Then use the Wronskian toshow that they are linearly independent. Finally, write the gen-eral solution of the system.)

21.x'= [ _\037)

2] [

2et

] [e2t

]-1 x; Xl == -3et , X2 ==_e2t)

[-322.x'== _ 3)

2] [

e3t

] [2e-2t

]4 x; Xl ==3e3t , X2 ==

e-2t)

23.x'=U =j]x; Xl = e2t

[ \037 lX2 = e-b [;]24. x'= [ _\037 \037 ]x; Xl = e3t

[ _\037 J X2 = e2t

[ _\037])

[4 -3

] [3e2t

] [e-5t

]25.x'==6 _ 7 x; Xl ==

2e2t , X2 ==3e-5t)

26.x'==[

-\037 -\037 -\037

]X;XI ==et

[;]

,o -1 3 1

X2 = e3t

[ -\037 JX3 = eSt

[ -\037]

27.x'=[r \037 1]

x; Xl = eb

[ i JX2 = e-t

[_\037 JX3 = e-t

[_ r])))

28.x'==

[\037

-i \037

]x; XI ==

[\037

],-1 -2 -1 -13

X2 = e3t

[ _\037 1X3 = e-4t

[ -\037 ]

29.X/=[=\037-\037\037 -\037]x;xl=e-2t[-ilX2 = et

[-

i 1X3 = e3t

[-l]

\037 ]x;Xl = e-t

[ \037 1X2 = e-t [!1X3 = et

[ _\037 1\037= et

[\037])

30.X/=[\037)

-41

-12-4)

oo-1o)

In Problems31through 40,find a particular solution ofthe in-dicatedlinear system that satisfiesthe given initial conditions.

31.Thesystem ofProblem22:Xl (0) == 0, X2 (0) == 532.Thesystem of Problem23:Xl (0) == 5,X2 (0) == -333.Thesystem of Problem24:Xl (0) == 11,X2 (0) == -734. Thesystem of Problem25:Xl (0) == 8,X2(0) == 035.The system of Problem 26:Xl (0) == 0, X2(0) == 0,

X3(0) == 436.The system of Problem 27: Xl (0) == 10,X2(0) == 12,

X3(0) == -137.The system of Problem 29:Xl (0) == 1,X2(0) == 2,

X3(0) == 338.The system of Problem 29:Xl (0) == 5, X2 (0) == -7,

X3(0) == 11)

5.3Application)


39.The system of Problem 30:Xl (0) == X2 (0) = X3(0)X4(0) == 1

40. The system of Problem 30: Xl (0) == 1,X2 (0) = 3,X3(0) == 4, X4 (0) == 7

41.(a) Show that the vector functions

XI (I)= [t;] and X2 =[ \037\037 ]

arelinearly independent on the real line. (b) Why doesit

follow from Theorem2 that there is no continuous matrix

P(t) such that Xl and X2 areboth solutions ofx'=P(t)x?42. Supposethat oneof the vector functions)

Xl (t) ==

[Xll (t)

]X21 (t))and X2(t) == [

X12(t)

]X22 (t))

is a constant multiple of the other on the open interval I.Show that their Wronskian Wet) == l[xij(t)]1must vanish

identically on I.This provespart (a) ofTheorem2 in the

casen == 2.43. Supposethat the vectorsXl (t) and X2(t) ofProblem 42are

solutions of the equation x'== P(t)x,where the 2 x 2 ma-trix pet) is continuous on the open interval I.Show that if

there existsa point a of I at which their Wronskian W(a)is zero,then there exist numbers CI and C2not both zerosuch that CIXI (a) + c2x2(a)== O.Then concludefrom the

uniqueness of solutions of the equation x'== P(t)x that)

CIXI (t) + C2X2(t) == 0)

for all t in I;that is, that Xl and X2 arelinearly dependent.This provespart (b) ofTheorem2 in the casen == 2.

44. GeneralizeProblems42 and 43 to prove Theorem2 for n

an arbitrary positive integer.45. LetXl (t), X2 (t), ..., Xn (t) be vector functions whose ith

components (for somefixed i) Xii (t), Xi2(t), ..., Xin (t) arelinearly independent real-valuedfunctions. Concludethat

the vector functions arethemselves linearly independent.)

AutomaticSolutionof LinearSystemsu.-N.... \037 \037\"'\037..__ ....,,, w..'h.....d ... \"'\" w ,...., w_ \" ........\"... .... \" ___ v \"\"'\" \"\"'.... '\037\".\"h.-U\" \"\"'\"',...., ... \"\",,,,,,,..,,,,, \"\" Y<''' h....\"'\" \"U.-,N,... \"\" \".. \" '\" .... ...., .....-.- \037... \"\",,au.. .\" '\" ..,..)

[ [2\037 2 \037 2][2\037 0\037 -2][1\037 -1\0371]]\037A) [[22 2]

[2 0 -2][1 -1 1 ]][ [0][2][6]] \037B)

Linear systemswith more than two or three equations are most frequently solvedwith the aid of calculators or computers.For instance, recall that in Example8 weneededto solve the linear system)

that can be written in the form AC = B with 3 x 3 coefficientmatrix A, right-

hand sidethe 3 x 1 column vectorB = [0 2 6]T,and unknown column vector

C = [CI C2 C3]T.Figure 5.3.1showsa TI calculator solution for C = A-IB,with the result that CI = 2, C2 = -3,and C3 = 1.Oncethe matricesA and B havebeenentered,the sameresult can be found using the Maplecommand)

A-'*B\037C)

[ [0][2][6]][[2 ]

[ -3][1 ]])

FIGURE5.3.1.TI-86solutionof the system AC == B in (1).)

2CI+ 2C2+ 2C3= 0,2CI - 2C3= 2,

CI - C2+ C3 = 6)

(1))))


C :=multiply(inverse(A),B);the Mathematica command)

C = Inverse[A].Bor the MATLAB commands)

C = inv(A)*BUseyour own calculator or availablecomputer algebra system to solve \"automati-

cally\" Problems31through 40 in this section.)

__:!:E:\302\253?\037\037!g\302\253?\037\037\037lue

l\\!\302\253?!\037 \037\037._!\037!:._!!.\037 \037\037.g\037\037\037_\037\037_\037 _\037I\037\037\037_\037\037____.____. ___ _.___.__._.___)

We now introducea powerfulalternative to the methodof elimination for construct-ing the general solution of a homogeneousfirst-order linear system with constantcoefficients,)

X\037

= allXI+ al2X2 + ...+ alnX n ,

x\037

= a2lX I + a22X2 + ...+ a2nXn,)(1))

X\037

= anlXI + an2X2 + ...+ annXn.)

By Theorem 3 of Section5.3,we know that it sufficesto find n linearly independentsolution vectors Xl, X2, ..., Xn ; the linear combination)

\037) X(t) ==CIXI +C2X2 + ...+ CnX n) (2))

with arbitrary coefficientswill then be a general solutionof the system in (1).To searchfor the n neededlinearly independentsolution vectors, we proceed

by analogy with the characteristic root method for solving a singlehomogeneousequation with constant coefficients(Section2.3).It is reasonableto anticipatesolu-tion vectors of the form)

Xl VI eAt VI

X2 V2 eAt V2

x(t) = X3 V3eAt V3 eAt = veAt (3)

Xn vneAt Vn)

where A, VI, V2, ..., Vn are appropriatescalarconstants.For if we substitute)

AtXi = vie ,)

, '\\ AtX. = A Vl'el ') (i = 1,2,...,n))in (1),then eachterm in the resulting equations will have the factor eAt, so wecan cancelit throughout. This will leave us with n linear equations which-forappropriate values of A-wecan hopeto solve for values of the coefficientsVI, V2,..., Vn in Eq.(3)so that x(t) = veAt is, indeed,a solutionof the system in (1).

To investigate this possibility,it is more efficientto write the system in (1)in

the matrix form)

\037) x'=Ax) (4))))

5.4The EigenvalueMethodfor HomogeneousSystems 367)

where A = [aij].When we substitute the trial solutionx = veAl with derivative

x'= Ave At in Eq.(4),the result is)

AveAt = Ave At.)

We cancelthe nonzero scalarfactor eAt to get)

\037) Av = AV.) (5))

This means that x = veAt will bea nontrivial solutionof Eq.(4)providedthat v is anonzerovectorand A is a constant such that Eq.(5)holds;that is, the matrix productAv is a scalarmultiple of the vector v. The question now is this:Howdo we find vand A ?

To answer this question,we rewriteEq.(5)in the form)

\037) (A -AI)v = O.) (6))

GivenA, this is a systemof n homogeneouslinearequations in the unknowns VI, V2,..., Vn . By a standard theorem of linear algebra, it has a nontrivial solution if and

only if the determinant of its coefficientmatrix vanishes; that is, if and only if)

IA -All = det(A -AI) = O.) (7))

In its simplestformulation, the eigenvalue methodfor solving the systemx' == Axconsistsof finding A so that Eq.(7)holdsand next solvingEq.(6)with this value ofA to obtain VI, V2, ..., Vn .Then x= veAt will bea solutionvector.The name of the

methodcomesfrom the followingdefinition.)

DEFINITION EigenvaluesandEigenvectors)The number A (eitherzero or nonzero)iscalledan eigenvalue of the n xn matrixA provided that)

IA -All = O.) (7))

An eigenvectorassociatedwith the eigenvalueA is a nonzerovector v such that

Av = AV, so that)

(A -AI)v = O.) (6))

Note that if v is an eigenvectorassociatedwith the eigenvalueA, then sois any

nonzero constant scalarmultiple cv of v-thisfollows upon multiplication of eachsidein Eq.(6)by c i= O.

The prefix eigenis a German word wit\037 the approximate translation charac-teristicin this context; the terms characteristicvalue and characteristicvector arein commonuse.For this reason,the equation)

all- A al2 alna21 a22- A a2n

\037 IA -All = =0 (8)

anI an2 ann-

A)))


is calledthe characteristicequationof the matrix A; its roots are the eigenval-ues of A. Uponexpanding the determinant in (8),we evidently get an nth-degreepolynomialof the form)

(-1)n An +bn-1An-1+ ...+b I A +bo ==O.) (9))

By the fundamental theorem of algebra, this equation has n roots-possiblysomeare complex,possiblysomeare repeated-andthus an n x n matrix has n eigenval-ues (counting repetitions,if any). Although we assumethat the elements of A arereal numbers, we allow the possibilityof complexeigenvaluesand complex-valuedeigenvectors.

Our discussionof Eqs.(4) through (7)provides a proof of the followingthe-orem, which is the basisfor the eigenvalue method of solving a first-order linearsystemwith constant coefficients.)

THEOREM1 EigenvalueSolutionsofx'= Ax

Let A be an eigenvalue of the [constant] coefficient matrix A of the first-orderlinear system)

\037)

dx-==Ax.dt)

If v is an eigenvectora\037sociated with A, then)

\037) x(t) ==veAl)

is a nontrivial solution of the system.)

TheEigenvalueMethodIn outline, this methodfor solving the n x n homogeneousconstant-coefficientsys-tem x' ==Ax proceedsas follows:)

1.We first solve the characteristicequation in (8)for the eigenvaluesAI, A2, ...,An of the matrix A.

2.Next we attempt to find n linearly independent eigenvectorsVI, V2, ...,Vn

associatedwith theseeigenvalues.3.Step2 is not always possible,but when it is, we get n linearly independent

solutions)

Xl (t) ==VleA}t,) X2 (t) ==V2eA2t,) ...,) Xn (t) ==vneAnt.) (10))

In this casethe general solutionof x' ==Ax is a linear combination)

x(t) ==CIXI (t) +C2X2(t) +...+ cnxn(t))

of thesen solutions.)

We will discussseparately the various casesthat can occur, dependingon whether

the eigenvalues are distinct or repeated,real or complex.The caseof repeatedeigenvalues-multipleroots of the characteristicequation-willbedeferred to Sec-tion 5.6.)))

I

Example1)


DistinctRealEigenvaluesIf the eigenvaluesAI, A2, ..., An are real and distinct, then we substitute each ofthem in turn in Eq. (6)and solve for the associatedeigenvectorsVI, V2, ...,Vn .In this caseit can be proved that the particular solution vectors given in (10)arealways linearly independent.(For instance, seeSection6.2of Edwards and Pen-ney, Elementary LinearAlgebra (EnglewoodCliffs, NJ: Prentice Hall, 1988).)Inany particular examplesuch linear independencecan always be verified by usingthe Wronskian determinant of Section5.3.The following example illustrates the

procedure.)

Find a general solutionof the system)

x\037

= 4xI + 2X2,

x\037

= 3xI - X2.)(11))

Solution The matrix form of the system in (11)is)

X' =[\037

_i] x.) (12))

The characteristicequation of the coefficientmatrix is)

4-A3)

2 =(4-A)(-I-A)-6-1-A)

= A2- 3A - 10= (A +2)(A -5) = 0,)

so we have the distinct real eigenvaluesAl = -2and A2 = 5.For the coefficientmatrix A in Eq.(12)the eigenvectorequation(A-AI)v == 0

takesthe form)

[4;A -12_ A ][ \037 ]

=[\037 ])

(13))

for the associatedeigenvectorV = [a b]T.)

CASE1:Al = -2. Substitution of the first eigenvalueAl = -2in Eq.(13)yieldsthe system)

[\037 i][\037]=

[\037l)

that is, the two scalarequations)

6a + 2b = 0,3a + b = O.)

(14))

In contrast with the nonsingular (algebraic)linear systemswhosesolutionswe dis-cussed in Section5.3,the homogeneouslinear system in (14)is singular-thetwo scalarequations obviously are equivalent(eachbeinga multiple of the other).Therefore, Eq. (14)has infinitely many nonzero solutions-wecan choosea arbi-trarily (but nonzero) and then solve for b.)))


Substitution of an eigenvalue A in the eigenvectorequation (A - AI)v = 0alwaysyieldsa singular homogeneouslinear system,and among its infinity of solu-tions we generally seeka \"simple\" solution with small integer values (if possible).Lookingat the secondequation in (14),the choicea = 1 yieldsb = -3,and thus)

VI = [_j ])

is an eigenvectorassociatedwith Al = -2(as is any nonzero constant multiple ofVI ).)

Remark:If instead of the \"simplest\" choicea = 1,b = -3,we had madeanother choicea = C i= 0,b = -3c,we would have obtained the eigenvector)

VI =[ -3\037]

= c[_j ] ,)

Becausethis is a constant multiple of our previousresult, any choicewe make leadsto (a constant multiple of) the samesolution)

XI(t) = [_j ]e-2t,)

CASE2:A2 = 5. Substitution of the secondeigenvalueA = 5 in (13)yieldsthe

paIr)

-a+ 2b = 0,3a - 6b = 0)

(15))

of equivalentscalarequations.With b = 1 in the first equation we get a = 2, so)

V2 =[i])

is an eigenvectorassociatedwith A2 = 5. A different choicea = 2c,b = c :I0would merely give a [constant]multiple of V2.

Thesetwo eigenvaluesand associatedeigenvectorsyield the two solutions)

XI(t) = [_j ]e-2t and X2(t) = [i]eSt,)

They are linearly independentbecausetheir Wronskian)

-2te3 -2t- e)

2 5te7 3t

5t = ee)

is nonzero.Hencea general solutionof the system in (11)is)

) [1

]-2t

[2

]5tx(t = CIXI (t)+C2X2(t) = CI -3 e +C2 1

e ;)

in scalarform,)

Xl (t) = CIe-2t + 2c2e5t,X2 (t) = -3CI e-2t + C2e5t.)))

4321

N 0\037

JC'.-1v-2-3-4

-4-3-2-10 1 2 3 4xl)

FIGURE5.4.1.Direction fieldand solution curves for the linear

system x\037

= 4Xl + 2X2,x\037

= 3XI -X2 of Example 1.)

\037r

FIGURE5.4.2.Thethree brinetanks of Example 2.)

Example2)


Figure 5.4.1showssome typical solution curves of the system in (11).We seetwo families of hyperbolas sharing the same pair of asymptotes: the line Xl == 2X2obtained from the general solution with CI = 0,and the line X2 = -3XIobtainedwith C2 = O. Given initial values Xl (0) = b l , X2 (0)= b2, it is apparentfrom the

figure that). If (bl ,b2) liesto the right of the line X2 = -3XI,then XI(t) and X2(t) both

tend to +00as t ---+ +00;. If (bl , b2) liesto the left of the line X2 = -3XI,then Xl (t) and X2 (t) both tendto -00as t ---+ +00. .)

Remark:As in Example1,it isconvenientwhen discussinga linearsystemx' = Ax to usevectors Xl, X2, ..., Xn to denote different vector-valuedsolutions ofthe system,whereasthe scalarsXl, X2, ..., Xn denote the componentsof a singlevector-valuedsolutionx. .)

CompartmentalAnalysisFrequently a complexprocessor system can be broken down into simplersubsys-tems or \"compartments\" that can be analyzed separately. The whole system canthen be modeledby describingthe interactionsbetween the various compartments.Thus a chemicalplant may consistof a successionof separatestages(or evenphys-ical compartments) in which variousreactants and products combineor are mixed.It may happen that a singledifferentialequationdescribeseach compartment of the

system,and then the whole physical system is modeledby a systemof differential

equations.As a simpleexampleof a three-stagesystem,Fig.5.4.2shows three brine

tanks containing VI, V2, and V3 gallons of brine, respectively. Fresh water flowsinto tank 1,while mixed brine flows from tank 1 into tank 2,from tank 2 into tank

3, and out of tank 3.Let Xi (t) denote the amount (in pounds)of salt in tank i at

time t for i = 1,2,and 3.If each flow rate is r gallons perminute, then a simpleaccounting of salt concentrations,as in Example2 of Section5.1,yields the first-

ordersystem)

X\037

= -kIXI,x\037

= klXI - k2X2,

x\037

= k2X2 - k3X3,)

(16))

where)

rkl

, --, i = 1, 2'3.-Vi)

(17))

_ ....... .w,.\".......-.... .... .........N .....d \"',...,,,,, ....,yU\".,..... \" \"\"\"\",'''''' w\"\",,.....,,,,, ,-.-,...-.,.....'.... ...N....U.\"'.-\".-n.. \037 \"...... \",....., ..... ...-.,.. .......\" \037

If VI = 20,V2 = 40, V3 = 50,r = 10(gal/min),and the initial amounts of salt in

the three brine tanks, in pounds,are)

Xl (0)= 15, X2 (0)= X3 (0)= 0,)

find the amount of salt in each tank at time t >O.)))


Solution Substituting the given numerical values in (16)and (17),we get the initial value

problem)

[-0.5 0.0 0.0

]x'(t) = 0.5 -0.25 0.0 x,

0.0 0.25-0.2)x(O)=

[lg])(18))

for the vector x(t) = [ Xl (t) X2(t) X3(t)]T.The simpleform of the matrix)

[-0.5-A 0.0

A -AI = 0.5 -0.25- A

0.0 0.25)

0.0

]0.0

-0.2- A)

(19))

leadsreadily to the characteristicequation)

IA -All = (-0.5-A)( -0.25-A)(-0.2-A) = O.)

Thus the coefficientmatrix A in (18)has the distinct eigenvaluesAl = -0.5,A2 =-0.25,and A3 = -0.2.)CASE1:Al = -0.5.Substituting A = -0.5in (19),we get the equation)

[0.0 0.0 0.0

] [a

] [0

][ A + (0.5).I ] v = 0.5 0.250.0 b = 00.0 0.250.3 c 0)

for the associatedeigenvectorv = [a be]T. The last two rows,after division

by 0.25and 0.05,respectively,yield the scalarequations)

2a + b = 0,5b + 6c= O.)

The secondequation is satisfiedby b = -6and c = 5,and then the first equationgives a = 3.Thus the eigenvector)

VI = [ 3 -6 5 ]T)

is associatedwith the eigenvalueAl = -0.5.)CASE2:A2 = -0.25.Substituting A = -0.25in (19),we get the equation)

[-0.25

[ A + (0.25).I ] v =\037.5)

oo0.25) g.05][\037]

=[g])

for the associatedeigenvectorv = [a be]T. Each of the first two rows impliesthat a = 0,and divisionof the third row by 0.05gives the equation)

5b+c = 0,)

which is satisfiedby b = 1,c = -5.Thus the eigenvector)

V2 = [ 0) 1 -5]T)

is associatedwith the eigenvalueA2 = -0.25.)))

15)

x =Xl (1))

10)

:..:)

5)

5 10 15 20 25 30t)

FIGURE5.4.3.Thesalt contentfunctions ofExample 2.)


CASE3:A3 = -0.2. Substituting A = -0.2in (19),we get the equation)

[-0.3 0.0

[ A + (0.2).I ] v = 0.5 -0.050.0 0.25)

0.0

] [a

] [

0

]0.0 b = 00.0 C 0)

for the eigenvector v. The first and third rows imply that a = 0,and b 0,respectively,but the all-zerothird column leavesCarbitrary (but nonzero).Thus)

V3 = [0 0 1]T)

is an eigenvectorassociatedwith A3 = -0.2.The general solution)

( ) All A2 1 A3 1xt =CIVle +C2V2e +C3V3e)

therefore takesthe form)

x(t)= Cl

[-\037]

e<-O.5)t+C2

[_\037]

e<-O.25)t+C3

[\037]

e(-O.2)t.)

The resulting scalarequations are)

Xl (t) = 3CIe(-0.5)1,X2 (t) = -6CI e(-0.5)1+ C2e(-0.25)1,X3 (t) = 5CIe(-0.5)1- 5C2e(-0.25)1+ C3e(-0.2)1

.)

When we imposethe initial conditionsXl (0)= 15,X2(0) = X3(0) = 0, we get the

equations)

3CI = 15,-6Cl+ C2 0,

5CI- 5C2+ C3 = 0)

that are readily solved (in turn) for CI = 5,C2 = 30,and C3 = 125.Thus, finally,the amounts of salt at time t in the three brine tanks are given by)

XI(t) = 15e(-0.5)1,X2(t) = -30e(-0.5)1+ 30e(-0.25)1,X3 (t) = 25e(-0.5)1- 150e(-0.25)1+ 125e(-0.2)1

.)

Figure 5.4.3showsthe graphs of Xl (t),X2(t), and X3(t).As we would expect,tank

1 is rapidly \"flushed\" by the incomingfresh water, and Xl (t) --+0 as t --++00.Theamounts X2(t) and X3(t) of salt in tanks 2 and 3 peakin turn and then approachzeroas the whole three-tanksystem is purgedof salt as t --++00. .)))


ComplexEigenvaluesEven if someof the eigenvaluesare complex,so long as they are distinct the methoddescribedpreviously still yieldsn linearly independentsolutions.The only compli-cation is that the eigenvectorsassociatedwith complexeigenvaluesare ordinarily

complexvalued, sowe will have complex-valuedsolutions.To obtain real-valued solutions, we note that-becausewe are assuming that

the matrix A has only real entries-thecoefficientsin the characteristicequation in

(8)will all bereal.Consequentlyany complexeigenvaluesmust appear in complexconjugate pairs. Supposethen that 'A = P +qi and 'A = P -qi are such a pair ofeigenvalues.If v is an eigenvectorassociatedwith 'A, so that)

(A - 'AI)v = 0,)

then taking complexconjugates in this equation yields)

(A - 'AI) v = 0)

sinceA = A and I = I (thesematrices beingreal) and the conjugateof a complexproduct is the product of the conjugates of the factors. Thus the conjugate v of vis an eigenvectorassociatedwith 'A . Of coursethe conjugate of a vector is definedcomponentwise;if)

al +b l i al b l

a2 +b2i a2 b2v= + i = a +bi, (20)

an + bni an bn)

then v = a -bi.The complex-valuedsolutionassociatedwith 'A and v is then)

x(t) = veAt = ve(p+qi)t = (a+bi)ePt (cosqt + i sin qt);)

that is,)

x(t) = ePt (acosqt -b sin qt) + iePt (bcosqt +a sin qt). (21))

Becausethe real and imaginary parts of a complex-valuedsolution are alsosolu-tions, we thus get the two real-valued solutions)

XI(t) = Re[x(t)]= ePt(acosqt-bsinqt),X2 (t) = Im[x(t)]= ePt (bcosqt +a sin qt))

(22))

associatedwith the complexconjugate eigenvaluesp :!:qi. It is easyto checkthat

the sametwo real-valued solutions result from taking real and imaginary parts ofveAt . Rather than memorizing the formulas in (22),it is preferable in a specificexampleto proceedas follows:). First find explicitly a singlecomplex-valuedsolutionx(t)associatedwith the

complexeigenvalue 'A;. Then find the real and imaginary parts Xl (t) and X2(t) to get two independentreal-valuedsolutionscorrespondingto the two complexconjugateeigenvalues'A and 'A.)))

Example3)



dXI-= 4XI - 3X2,dt

dX2-= 3XI +4X2.dt)

(23))

Solution The coefficientmatrix)

108642

\0370

-2-4-6-8-10

-10-8-6-4-20 2 4 6 8 10xl)


system x\037

= 4Xl - 3X2,x\037

= 3Xl + 4X2 ofExample 3.)

[4 -3

]A= 3 4)

has characteristicequation)

4-AIA -All =

3)

-3 2=(4-A) +9=0,4-A)

and hencehas the complexconjugateeigenvaluesA = 4 -3i and A = 4 +3i.Substituting A = 4-3i in the eigenvectorequation (A -AI)v = 0, we get the

equation

[A - (4-3i) .I]v =[3\037 -;\037] [ \037 ] =

[ \037 ]for an associatedeigenvaluev = [a b]T. Division of each row by 3 yields the

two scalarequations)ia - b = 0,a + ib = 0,

each of which is satisfiedby a = 1 and b = i. Thus v = [1 i]T is a complexeigenvectorassociatedwith the complexeigenvalueA = 4 -3i.

The correspondingcomplex-valuedsolutionx(t)= veAt of x'= Ax is then)

( ) _[

1

](4-3i)t_

[1

]4t

( 3 ..3 ) _ 4t

[cos3t - i sin 3t

]x t - . e -.e cos t - 1SIn t -e . 3 + .3 .

1 1 1COS t SIn t)

The real and imaginary parts of x(t)are the real-valuedsolutions)

( ) _ 4t

[cos3t

]Xl t -e .3SIn t)

4t

[-sin3t

]and X2 (t) = e 3 .cos t)

A real-valuedgeneral solutionof x'= Ax is then given by)

4t

[CICOS3t - C2sin 3t

]x(t)= CIXI (t) +C2X2(t) = e .3 + 3t

.CISIn t C2COS)

Finally, a general solutionof the system in (23)in scalarform is)

Xl (t) = e4t(CIcos3t -C2sin 3t),

X2(t) = e4t(CIsin 3t +C2COS3t).)

Figure 5.4.4showssometypical solution curves of the system in (23).Eachappearsto spiral counterclockwiseas it emanatesfrom the origin in the Xlx2-plane.Actually, becauseof the factor e4t in the general solution, we seethat)))


Flow rate: r1--)

FIGURE5.4.5.Thethree brinetanks ofExample 4.)

Example4)

. Along eachsolution curve, the point (Xl (t),X2 (t)) approachesthe origin ast \037 -00,whereas. The absolutevalues of Xl (t) and X2 (t) both increasewithout bound as t --++00. .)

Figure 5.4.5showsa \"closed\"system of three brine tanks with volumes VI,V2, and V3.The differencebetweenthis system and the \"open\" systemof Fig.5.4.2is that now the inflow to tank 1 is the outflow from tank 3.With the samenotation

as in Example2,the appropriatemodificationof Eq.(16)is)

dXI-== -kIXIdt

dX2

dt)

+ k3X3,)

(24))klXI - k2X2,)

dX3

dt)k2X2 - k3X3,)

where ki ==r / Vi as in (17).)

Find the amounts Xl (t),X2(t), and X3(t) of salt at time t in the three brine tanks ofFig.5.4.5if VI ==50gal, V2 ==25 gal, V3 ==50 gal, and r == 10gal/min.)

Solution With the given numericalvalues, (24)takesthe form)

dx

[-0.2 0 0.2

]-== 0.2 -0.4 0 xdt 0 0.4 -0.2)

(25))

with x == [ Xl X2 X3]Tas usual.When we expand the determinant of the matrix)

[-0.2- 'A 0.0 0.2

]A - 'A .I == 0.2 -0.4- 'A 0.0

0.0 0.4 -0.2- 'A)

(26))

along its first row, we find that the characteristicequationof A is)

(-0.2- 'A)(-0.4- 'A)(-0.2- 'A) + (0.2)(0.2)(0.4)== -'A3 - (0.8).'A

2- (0.2).'A

== -'A [('A+0.4)2+ (0.2)2]==O.)

Thus A has the zero eigenvalue 'A Q== 0 and the complexconjugate eigenvaluesA,

'A ==-0.4:!:(0.2)i.)

CASE1: 'A Q==O. Substitution of 'A ==0in Eq.(26)gives the eigenvectorequation)

[-0.2 0.0 0.2

] [a

] [0

](A -0.I)v == 0.2 -0.4 0.0 b == 00.0 0.4 -0.2 c 0)))


for v == [a b C]T.The first row gives a ==c and the secondrow givesa == 2b,so vo == [ 2 1 2]T is an eigenvectorassociatedwith the eigenvalueAO == O.The

correspondingsolutionxo(t)==voeAo1 of Eg.(25)is the constant solution)

xo(t)= [!].) (27))

CASE2:A ==-0.4-(0.2)i.Substitution of A ==-0.4- (0.2)iin Eg.(26)givesthe eigenvectorequation

[0.2+(0.2)i 0.0

[A - (-0.4- (0.2)i)I]v == 0.2 (0.2)i0.0 0.4)

0.2

] [

a

]0.0 b

0.2+ (0.2)i c)

=[\037].)The secondequation (0.2)a+ (0.2)ib == 0 is satisfiedby a == 1 and b == i.Thenthe first equation)

[0.2+(0.2)i]a+(0.2)c==0

gives c ==-1- i.Thus v == [1 i (-1- i) ]T is a complexeigenvectorassoci-

ated with the complexeigenvalueA ==-0.4- (0.2)i.The correspondingcomplex-valuedsolutionx(t) ==veAl of (25)is)

x(t) ==[1 1 -1- i ]Te(-0.4-0.2i)l

==[1 1 -1- i]Te(-0.4)t(cosO.2t- i sinO.2t)

[

cos0.2t- i sin 0.2t

]==e(-0.4)1 sin 0.2t+ i cos0.2t .-cos0.2t- sin 0.2t- i cos0.2t+ i sin 0.2t)

The real and imaginary parts of x(t)are the real-valuedsolutions)

[

cos0.2t

]Xl (t) ==e(-0.4)1 sin 0.2t ,-cos0.2t- sin 0.2t)

(28))

[

-sin0.2t

]X2 (t) ==e(-O.4)t cos0.2t .-cos0.2t+ sin 0.2t)

The general solution)

x(t) ==coxo(t)+c]x](t) +C2X2(t))

has scalarcomponents)

x] (t) ==2co+e(-0.4)1(CICOS0.2t-C2sin 0.2t),X2 (t) == Co+e(-O.4)l(c)sin 0.2t+C2COS0.2t), (29)X3(t) ==2co+e(-O.4)t[(-c]-C2)COS0.2t+ (-c)+C2)sin 0.2t])))


giving the amounts of salt in the three tanks at time t .Observethat)

5045403530

\03725

2015105

00)

Xl (t) +X2(t) +X3(t) = 5co.) (30))

Of coursethe total amount of salt in the closedsystem is constant; the constantCoin (30)is one-fifth the total amount of salt.Becauseof the factors of e(-0.4)1in (29),we seethat)

lim XI(t) = 2co, lim X2(t) = Co, and lim X3(t) = 2co.1 ---+ 00 1 ---+ 00 1---+ 00)

Thus as t \037 +00the salt in the system approachesa steady-statedistribution with

40% of the salt in each of the two 50-gallontanks and 20% in the 25-gallon tank.

Sowhateverthe initial distribution of salt among the three tanks, the limiting distri-bution is one of uniform concentration throughout the system.Figure 5.4.6showsthe graphs of the three solution functions with Co= 10,CI = 30,and C2 = -10,inwhich case)

x =x2(1))

10 15t)

5) 20)

FIGURE5.4.6.Thesalt contentfunctions ofExample 4.)

_\037E\037blems)

Xl (0)= 50 and X2(0) = X3(0) = O.) .)

In Problems1through 16,apply the eigenvalue method of thissectionto find a generalsolution of the given system. If initial

values are given, find also the correspondingparticular solu-tion. For eachproblem, use a computer system or graphingcalculator to construct a direction field and typical solutioncurves for the given system.

1.x\037

== Xl + 2X2,X\037

== 2XI + X2

2.x\037

== 2XI + 3X2,x\037

== 2XI + X2

3.x\037

== 3XI + 4X2, x\037

== 3XI + 2X2; Xl (0) == X2 (0) == 14.

x\037

== 4XI + X2, x\037

== 6XI -X2

5.x\037

== 6XI - 7X2,x\037

== Xl - 2X26.

x\037

== 9XI + 5X2,x\037

== -6XI-2X2; Xl (0) == 1,X2(0) == 07.

x\037

== -3XI+ 4X2,x\037

== 6XI - 5X28.

x\037

== Xl - 5X2,X\037

== Xl -X2

9.x\037

== 2XI - 5X2,x\037

== 4XI - 2X2; Xl (0) == 2, X2(0) == 310.

x\037

== -3XI- 2X2, x\037

== 9XI + 3X211.x\037

== Xl - 2X2,X\037

== 2XI + X2; Xl (0) == 0, X2 (0) == 412.

x\037

== Xl - 5X2, X\037

== Xl + 3X213.

x\037

== 5XI - 9X2, x\037

== 2XI -X2

14.x\037

== 3XI - 4X2,x\037

== 4XI + 3X215.

x\037

== 7Xl - 5X2,x\037

== 4XI + 3X216.

x\037

== -50XI+ 20X2,x\037

== 100XI - 60X2)

In Problems 17through 25, the eigenvaluesof the coefficientmatrix can befound by inspection and factoring. Apply the

eigenvalue method to find a generalsolution ofeachsystem.

17.x\037

== 4XI + X2 + 4X3, x\037

== Xl + 7X2 + X3,

X\037

== 4XI + X2 + 4X318.

x\037

== Xl + 2X2 + 2X3, X\037

== 2XI + 7X2 + X3,

X\037

== 2xI + X2 + 7X319.

x\037

== 4XI +X2 +X3, x\037

== Xl +4X2+X3,X\037

== Xl +X2 +4X3)

20.)X\037

== 5XI + X2 + 3X3, X\037

== Xl + 7X2 + X3,

X\037

== 3XI + X2 + 5X3

X\037

== 5XI-6x3,X\037

== 2XI -X2-2X3,X\037

== 4XI-2x2-4x3X\037

== 3XI + 2X2 + 2X3, X\037

== -5XI- 4X2 - 2X3,

X\037

== 5XI + 5X2 + 3X3

X\037

== 3Xl + X2 + X3, X\037

== -5XI- 3X2 -X3,

X\037

== 5XI + 5X2 + 3X3

X\037

== 2XI + X2-X3, X\037

== -4XI- 3X2 -X3,

X\037

== 4XI + 4X2 + 2X3

X\037

== 5XI + 5X2 + 2X3, X\037

== -6XI- 6X2 - 5X3,

X\037

== 6XI + 6X2 + 5X3Find the particular solution of the system

dXI

dt)

21.22.)

23.)

24.)

25.)

26.)

3XI) + X3,)

dX2

dt

dX3-== -9X I + 4X2 - X3dt)

9XI -X2 + 2X3,)

that satisfies the initial conditions Xl (0) == 0, X2 (0) == 0,X3(0) == 17.)

The amounts Xl (t) and X2(t) of salt in the two brine tanks ofFig. 5.4.7satisfy the differential equations)

dXI

dt== -kIXI,)

dX2-== klXI - k2X2,dt)

where ki == r/V i for i == 1, 2. In Problems27and 28 the vol-

umes VI and V2 are given. First solvefor Xl (t) and X2(t), as-suming that r == 10(gal/min), X I (0) == 15(lb), and X2 (0) == o.)))


Then find the maximum amount ofsalt everin tank 2.Finally,construct afigure showing the graphs ofXl (t) and X2(t).)

Fresh waterFlow rate r)

Tank 1

Volume VISalt x I (t))

Tank2Volume V2Salt x2(t))

\037

r)

FIGURE5.4.7.Thetwo brine tanks ofProblems27 and 28.)

27. VI = 50 (gal), V2 = 25 (gal)28.VI = 25 (gal), V2 = 40 (gal))

The amounts Xl (t) and X2(t) ofsalt in the two brine tanks ofFig. 5.4.8satisfy the differential equations)

dXI

dt= -klxI+ k2x2,)

dX2-= klxI -k2x2,dt)

where ki = r/V i as usual. In Problems29 and 30, solveforXl (t) and X2(t), assuming that r = 10(gal/min), Xl (0) = 15(lb), and X2 (0) = O. Then construct a figure showing the

graphs ofXl (t) and X2 (t).)

r\037)

\037r)

FIGURE5.4.8.Thetwo brine tanks ofProblems29and 30.)

29.VI = 50 (gal), V2 = 25 (gal)30.VI = 25 (gal), V2 = 40 (gal))

Problems31through 34deal with the open three-tank systemofFig. 5.4.2.Fresh water flows into tank 1;mixed brine flowsfrom tank 1into tank 2,from tank 2 into tank 3, and out oftank

3; all at the given flow rate r gallonsperminute. The initial

amounts Xl (0) = Xo (lb), X2(0) = 0, and X3(0) = 0 of saltin the three tanks are given, asare their volumes VI, V2, andV3 (in gallons).First solvefor the amounts ofsalt in the threetanks at time t, then determine the maximal amount ofsalt thattank 3 evercontains. Finally, construct a figure showing the

graphs ofXl (t), X2(t), and X3(t).

31.r = 30,Xo = 27, VI = 30, V2 = 15,V3 = 1032. r = 60,Xo = 45, VI = 20, V2 = 30, V3 = 6033. r = 60,Xo = 45, VI = 15,V2 = 10,V3 = 30)

34. r = 60,Xo = 40, VI = 20, V2 = 12,V3 = 60)

Problems 35 through 37 deal with the closedthree-tank sys-tem ofFig. 5.4.5,which is describedby the equations in (24).Mixed brine flows from tank 1into tank 2, from tank 2 intotank 3, andfrom tank 3 into tank 1,all at the given flow rate rgallonsperminute. The initial amounts Xl (0) = Xo (pounds),X2 (0) = 0, and X3(0) = 0 ofsalt in the three tanks aregiven,as are their volumes VI, V2, and V3 (in gallons). First solvefor the amounts ofsalt in the three tanks at time t, then deter-mine the limiting amount (ast \037 +(0)ofsalt in eachtank.

Finally, construct a figure showing the graphs ofXl (t), X2(t),and X3(t).35.r = 120,Xo = 33, VI = 20, V2 = 6, V3 = 4036.r = 10,Xo = 18,VI = 20, V2 = 50, V3 = 2037.r = 60,Xo = 55, VI = 60, V2 = 20, V3 = 30)

For eachmatrix A given in Problems38 through 40,the zerosin the matrix make its characteristicpolynomial easyto calcu-late.Find the generalsolution ofx'= Ax.)

38.A=[\037

0 0

\037]

2 03 30 4

[-\037

0 0

-1\037]

39.A = 2 00 -10 0

[

2 0 0

-\037]

-21 -5 -2740. A =\037

0 50 -21 -2

41.Thecoefficientmatrix A of the 4 x 4 system

x\037

= 4XI + X2 + X3 + 7X4,

x\037

= Xl + 4X2 + 10x3+ X4,

x\037

= Xl + 10x2+ 4X3 + X4,

x\037

= 7XI + X2 + X3 + 4X4)

has eigenvalues A I = -3, A2 = -6,A3 = 10, and

A4 = 15.Find the particular solution of this system that

satisfiesthe initial conditions)

Xl (0)= 3, X2(0) = X3(0) = 1, X4(0)= 3.)

In Problems42 through 50,use a calculatoror computer sys-tem to calculatethe eigenvaluesand eigenvectors(as illus-trated in the 5.4Application below) in orderto find a generalsolution ofthe linear system x'= Ax with the given coefficientmatrix A.)

[-40

42. A = 35-25

[-20

43. A = 12-48)

-1213-711-121)

54

]-46

3413

]-7

31)))

380 Chapter5 LinearSystemsof Differential Equations

[147 23 -202

] [

47 -8 5

-5]44. A = -90 -9 129 -10 32 18 -2

90 15 -123 48. A =139 -40 -167 -121

[

9 -7 -5

-\037]

-232 64 360 248-12 7 11 139 -14 -52 -14 2845. A == 24 -17 -19 -22 5 7 8 -7-18 13 17 49. A == 370 -38 -139 -38 76

[13 -42 106 139

]

152 -16 -59 -13 35

46.A= i -16 52 70 95 -10 -38 -7 236 -20 -31 9 13 0 0 0 -13-1 -6 22 33 -14 19 -10 -20 10 4

[

23 -18 -16

-\037]

50.A ==-30 12 -7 -30 12 18

-8 6 7 -12 10 -10 -9 10 247. A ==34 -27 -26 6 9 0 6 5 -15

-26 21 25 12 -14 23 -10 -20 10 0)

5.4Application)

. \" .,.[0,.25,-.2]] \037A[[-.500.000.00]

[.50 -.25 0.00][0.00 .25 -.20]]ei9UlA)

ei9UcA[[0.000.00 1.00][0.001.00 -2.00][1.00 -5.00 1.67 ]])

FIGURE5.4.9.TI-86calculation of the eigenvalues and

eigenvectors of the matrix A.)

AutomaticCalculationof Eigenvaluesand Eigenvectors...... .. \",..,'...0..\"._ .. c<\" ....)

Mostcomputational systemsoffer the capability to find eigenvaluesand eigenvec-tors readily. For instance,Fig.5.4.9showsa graphing calculatorcomputationof the

eigenvaluesand eigenvectorsof the matrix)

[-0.5 0.0 0.0

]A = 0.5 -0.25 0.0

0.0 0.25-0.2)of Example2.We seeeacheigenvectordisplayedas a column vector beneath its

eigenvalue. Note that with resultspresentedin decimal form, it is up to us to guess(and verify by matrix multiplication) that the exacteigenvectorassociatedwith the

third eigenvalue A = -\037

is v = [1 -2\037 ]T. Oncethe matrix A has been

entered,the Maplecommand

eigenvec'ts(A) ;the Mathematica command)

Eigensys'tem[A]or the MAT LAB command)

[V,D]= eig(A)(whereD will be a diagonal matrix displaying the eigenvaluesof A and the columnvectors of v are the correspondingeigenvectors)producesimilar results. You canusethesecommands to find the eigenvaluesand eigenvectorsneededfor any of the

problemsin this section.For a more substantial investigation,choosea positive integer n < 10(n = 5,

for instance) and let ql,q2, ..., qn denote the first n nonzero digits in your student

ID number. Now consideran opensystem of brine tanks as in Fig.5.4.2,exceptwith n rather than three successivetanks having volumes Vi = 10qi(i = 1,2, ...,n) in gallons.If each flow rate is r = 10gallonsperminute, then the salt amounts

Xl (t),X2(t), ...,xn(t) satisfy the linear system)

X\037

= -kIXI,x;= ki-IXi-l- kiXi (i = 2,3,..., n),)))

5.5Second-OrderSystemsand MechanicalApplications 381)

where ki r/Vi . Apply the eigenvaluemethod to solve this system with initial

conditions)

Xl (0)= 10, X2(0) = X3(0) = ...= xn(O)= O.)

Graph the solution functions and estimate graphically the maximum amount of saltthat eachtank ever contains.

For an alternative investigation,supposethat the system of n tanks is closedas in Fig.5.4.5,so that tank 1 receivesas inflow the outflow from tank n (ratherthan fresh water). Then the first equationshouldbereplacedwith

x\037

= knxn -kIXI.Now show that, in this closedsystem, as t --+ +00the salt originally in tank 1distributes itself with constant density throughout the various tanks. A plot likeFig.5.4.6should make this fairly obvious.)

Hi \037.\037.c\037\037\037.\037Q\037\037\037:r...\037Ee\037\037\037...\037\037d .\037!\037\037\037\037\037\037\037!. \037pp!\037\037.\037!!ons

*)

FIGURE5.5.1.Threespring-coupledmasses.)

In this sectionwe apply the matrix methods of Sections5.3and 5.4to investigatethe oscillationsof typical mass-and-springsystemshaving two or more degreesoffreedom.Our examplesare chosento illustrate phenomena that are generallychar-acteristicof complexmechanicalsystems.

Figure 5.5.1showsthree massesconnectedto eachother and to two walls bythe four indicated springs. We assumethat the massesslidewithout friction and

that each spring obeysHooke'slaw-itsextensionor compressionX and forceF ofreaction are related by the formula F = -kx.If the rightward displacementsXl,

X2, and X3 of the three masses(from their respectiveequilibrium positions)are allpositive, then)

\302\267 The first spring is stretched the distanceXl ;\302\267 The secondspring is stretched the distanceX2 -Xl ;\302\267 The third spring is stretched the distanceX3 -X2;

\302\267 The fourth spring is compressedthe distanceX3.)

Therefore, applicationof Newton'slaw F = ma to the three masses(asin Example1 of Section5.1)yieldstheir equationsof motion:)

II kmixi= -

IXI) + k2(X2 -Xl),)

m2X\037= -k2(X2 -Xl) + k3(X3 -X2),

m3x\037= -k3(X3 -X2) - k4X3.)

(1))

Although we assumedin writing theseequationsthat the displacementsof the massesare all positive, they actually follow similarly from Hooke'sand Newton's laws,whatever the signsof thesedisplacements.

In terms of the displacementvector x= [Xl X2 X3]T, the massmatrix)

[m l

M=\037)

o)

13])(2))m2

o)

*This optional section may be omitted without lossof continuity. It provides a sample of the moretechnical applications ofeigenvalues to physics and engineering problems.)))


FIGURE5.5.2.A system ofn

spring-coupled masses.)

and the stiffnessmatrix)

[-(kl+k2)

K == k2o)

o

]k3 ,

-(k3 +k4))

(3))

k2-(k2 +k3)

k3)

the system in (1)takesthe matrix form)

\037) Mx\" == Kx.) (4))

The notation in Eqs.(1)through (4)generalizesin a natural way to the systemof n spring-coupledmassesshown in Fig.5.5.2.We needonly write)

ml 00 m2

M==

0 0

and

-(k1 + k2) k2 0

k2 -(k2 +k3) k3

0 k3 -(k3 +k4)

K== 0 0 k4)

oo)

(5))

m n)

o

o

o

o (6))

o

o)

o

o)

-(kn-1 +kn ) kn

kn -(kn +kn+1 ))

for the massand stiffnessmatrices in Eq.(4).The diagonal matrix M is obviously nonsingular; to get its inverseM-1 we

needonly replaceeachdiagonal element with its reciprocal.Hencemultiplicationof eachsidein Eq.(4)by M-1 yieldsthe homogeneoussecond-ordersystem)

\037) x\" ==Ax,) (7))

where A == M-1K. There is a widevariety offrictionlessmechanical systemsforwhich a displacementor position vector x, a nonsingular mass matrix M, and astiffnessmatrix K satisfying Eq.(4)can be defined.)

SolutionofSecond-OrderSystemsToseeka solutionof Eq.(7),we substitute (asin Section5.4for a first-ordersystem)a trial solution of the form)

x(t) ==veat,) (8))

where v is a constant vector. Then x\" == a2veat , so substitution of Eq. (8)in (7)gIves)

a2veat ==Ave at,)))

which impliesthat)

Av ==a2v.) (9))

Therefore X(t) ==Veal is a solutionof x\" ==Ax if and only if a2 ==A, an eigenvalueof the matrix A, and v is an associatedeigenvector.

If x\" ==Ax modelsa mechanicalsystem, then it is typical that the eigenvaluesof A are negativereal numbers. If)

a2 == A ==-w2 < 0 ,)

then a ==::!::wi.In this casethe solutiongiven by Eg.(8)is)

x(t) ==ve[(1J!==v(coswt + i sin wt).)

The real and imaginary parts)

Xl (t) ==V coswt and X2(t) ==v sin wt) (10))

of x(t)are then linearly independentreal-valued solutionsof the system.Thisanal-ysisleadsto the following theorem.)

THEOREM1 Second-OrderHomogeneousLinearSystemsIf the nxn matrix A has distinctnegativeeigenvalues-wi,-wi,..., -w\037 with

associated[real] eigenvectorsVl, V2, ..., vn , then a general solutionof)

\037) XII ==Ax)

is given by)

\037)

n

x(t) ==L(aiCOSWit + hi sinwit)vii=l)

(11))

with ai and hi arbitrary constants. In the specialcaseof a nonrepeatedzeroeigenvalueAo with associatedeigenvectorYo,)

xo(t)== (ao+hot)vo) (12))

is the correspondingpart of the general solution.)

Remark:The nonzerovector Vo is an eigenvectorcorrespondingto AO == aprovided that Avo ==O.If x(t) == (ao + bot)vo, then)

XII ==0 .va == (ao+hot) .0==(ao+hot) . (Avo) ==Ax,)

thus verifying the form in Eg.(12).) .)))


Example1)

\037Equilibrium positions)

X2 (t))

FIGURE5.5.3.Themass-and-spring system of Example 1.)

........... - \"., \037.....

Considerthe mass-and-springsystemwith n ==2 shown in Fig.5.5.3.Becausethereis no third spring connectedto a right-hand wall, we setk3 ==O.If ml ==2, m2 ==1,k l == 100,and k2 ==50,then the equation MX\" ==Kx is)

[\037 \037]

x\" =[-1;\037_;\037]

x,) (13))

which reducesto XII ==Ax with)

[-75 25

]A == 50 -50 .)

The characteristicequation of A is)

(-75-A)(-50-A) -50.25==A2 + 125A+2500

== (A +25)(A+ 100)==0,)

so A has the negative eigenvalues Al == -25and A2 == -100.By Theorem 1,the systemin (13)therefore has solutions with [circular] frequenciesWI == 5 andW2 == 10.)

CASE1:Al ==-25. The eigenvectorequation (A -AI)v ==0 is)

[-50 25

] [a

] [0

]50 -25 b-

0 ')

so an eigenvectorassociatedwith A I ==-25 is v I ==[1 2]T.)

CASE2:A2 ==-100.The eigenvectorequation (A -AI)v ==0 is)

[;\037 ;\037] [ \037 ] = [ \037 l)soan eigenvectorassociatedwith A2 ==-100is V2 == [1 -1]T.)

By Eq.(11)it follows that a general solutionof the system in (13)is given by)

x(t) == (alcos5t +b l sin 5t)VI + (a2coslOt +b2sin 10t)v2. (14))

As in the discussionof Example3 of Section5.2,the two terms on the right in

Eq. (14)representfreeoscillationsof the mass-and-springsystem.They describethe physicalsystem'stwo naturalmodesof oscillationat its two [circular]naturalfrequenciesWI ==5 and W2 == 10.The natural mode)

Xl (t) = (alcas5t +h l sin 5t)Vl = Clcas(5t-ad[ \037 ])

(with CI ==Jar +br, cosal ==al/CI,and sin al ==b l /c})has the scalarcomponentequations)

Xl (t) == CIcos(5t-al),X2(t) ==2CIcos(5t-al),)

(15))))

Example2)

FIGURE5.5.6.Thethree

railway carsof Example 2.)


and therefore describesa free oscillation in which the two massesmove in syn-chrony in the samedirection and with the samefrequency WI = 5, but with the

amplitude of motion of m2 twice that of mI (seeFig.5.5.4).The natural mode)

X2(t) = (a2coslOt +b2 sin IOt)v2= C2cos(10t- (2)[_\037 ])

has the scalarcomponentequations)

Xl (t) = C2cos(IOt- (X2),

X2(t) = -C2cos(IOt- (X2),)(16))

and therefore describesa free oscillation in which the two massesmove in syn-chrony in oppositedirectionswith the samefrequency W2 10and with equalamplitudesof oscillation(seeFig.5.5.5). .)

\037)

x =X2(t))

x=x\\(t))

\037)

X=Xl(t))

o) 2n) o) nl2 1tt)

31t/2)nt)

FIGURE5.5.4.Oscillationsinthe samedirection with frequencyWI = 5; the amplitude of motionofmass 2 is twice that ofmass 1.)

FIGURE5.5.5.Oscillationsinoppositedirections with frequencyW2 = 10;the amplitudes ofmotionof the two massesarethe same.)

.... ..... \"............\037'\" \"..,...,\" \"U,'.\037',,_'n.\" ...... \037\037 \" ......\"'\" .-.\" ... \"\",,,,,-, \"\"'''''''\"\"'\" \"'... \"-' ,... \",,,\"\"',,... A.' \"\"'\037 ...\037 ..... ....\"....\"Figure 5.5.6showsthree railway carsconnectedby buffer springsthat react when

compressed,but disengageinstead of stretching. With n = 3, k2 = k3 = k, andk I = k4 = 0 in Eqs.(2)through (4),we get the system)

[\037]

0 o

] [-k k

\037]

x,m2 o x\" = k -2k0 m3 0 k -k

which is equivalentto

[-c] CI

c\037]

Xx\" = C2 -2C20 C3 -C3

withk

Ci=- (i=I,2,3).mi)

(17))

(18))

(19))

If we assumefurther that mI = m3, so that CI = C3,then a brief computation gives)

-A(A +CI)(A +CI+2C2)= 0) (20))))


for the characteristic equation of the coefficientmatrix A in Eq. (18).Hencethe

matrix A has eigenvalues)

A1=0, A2=-C1,A3=-Cl-2c2) (21a))

correspondingto the natural frequencies)

WI = 0, W2 =,JC1, W3 = JC1 + 2C2) (21b))

of the physicalsystem.For a numericalexample,supposethat the first and third railway carsweigh 12

tons each,that the middlecar weighs8 tons, and that the spring constant isk = 1.5tons/ft; i.e.,k = 3000Ib/ft.Then, using fps units with massmeasuredin slugs(aweight of 32 poundshas a massof 1 slug),we have)

m1 = m3 = 750, m2 = 500,)

and)3000

CI = = 4,750Hencethe coefficientmatrix A is)

3000C2=

500= 6.)

A =[-\037 -1\037 _\037],)

(22))

and the eigenvalue-frequencypairsgiven by (21a)and (21b)are Al = 0,wI == 0;A2 = -4,W2 = 2;and A3 = -16,W3 = 4.)

CASE1:Al = 0,WI = O. The eigenvectorequation (A -AI)v = 0 is)

Av =[-\037 -1\037

_\037] [\037 ]=

[g],)

so it is clearthat VI = [1 1 1]T is an eigenvector associatedwith Al == o.According to Theorem 1,the correspondingpart of a general solutionof x\" == Ax is)

Xl (t) = (aI +blt)VI.)

CASE2:A2 = -4,W2 = 2. The eigenvectorequation (A -AI)v = 0 is)

[0 4

(A +4I)v = 6 -8o 4) \037][\037]=[g])

so it is clearthat V2 = [ 1 0 -1]T is an eigenvectorassociatedwith A2 ==-4.According to Theorem 1,the correspondingpart of a general solutionof x\" = Ax is)

X2(t) = (a2cos2t +b2sin 2t)V2.)))


CASE3:A3 = -16,LV3 == 4. The eigenvectorequation (A -AI)v ==0 is)

[12

(A + 16I)v=\037) :

1\037] [\037]

=[g]

,)

so it isclearthat V3 = [1 -3 1 ]T is an eigenvectorassociatedwith A3 ==-16.According to Theorem 1,the correspondingpart of a general solutionof x\" ==Ax is)

X3(t) = (a3cos4t+b3 sin4t)v3.)

The general solutionx = Xl +X2 +X3 of x\" = Ax is thereforegiven by)

x(t) = al[i ]

+blt[:]+a2

[_\037 ]

cos2t

+b2

[_\037]

sin2t+a3 [-\037]

cos4t+b3 [-\037]

sin4t. (23))

To determine a particular solution, let us supposethat the leftmostcar is moving tothe right with velocity Va and at time t = 0 strikesthe other two cars,which aretogether but at rest.The correspondinginitial conditionsare)

Xl (0)= X2(0) = X3(0) ==0,x\037(O)

= Va, x\037(O)=

x\037(O)==O.)

(24a)(24b))

Then substitution of (24a)in (23)gives the scalarequations)

al + a2 + a3 = 0,al - 3a3 = 0,al - a2 + a3 ==0,)

which readily yield al == a2 = a3 = O. Hencethe position functions of the threecarsare)

xI(t)=bI t+b2sin2t+b3 sin4t,x2(t)==bl t -3b3 sin4t,X3(t) = bIt - b2sin 2t + b3 sin 4t,)

(25))

and their velocity functions are)

X\037(t) = b i + 2b2cos2t + 4b3 cos4t,

x\037 (t) = b i- 12b3 cos4t,

x\037 (t) = b i- 2b2cos2t + 4b3 cos4t.)

(26))

Substitution of (24b)in (26)gives the equations)

b l + 2b2 + 4b3 = Va,

b i- 12b3 == 0,

b i - 2b2 + 4b3 == 0)))

that readily yield b l ==\037

Va, b2 ==\037

Va, and b3 ==3\037

Va. Finally, the positionfunctions

in (25)are)

Xl (t) ==3

1

2va(12t + 8 sin 2t + sin 4t),

X2(t) ==3

1

2va(12t - 3 sin4t),

X3(t) ==3

1

2va(12t - 8 sin 2t + sin 4t).)

(27))

But theseequations hold only so long as the two buffer springsremain com-pressed;that is, while both)

X2 -Xl <0 and X3-

X2 <o.)

To discoverwhat this impliesabout t, we compute)

X2(t) -Xl (t) ==3

1

2va(-8sin 2t -4 sin4t)

==-3\037

va(8sin 2t + 8 sin 2t cos2t)

==-\037va(sin 2t)(1+cos2t))

and, similarly,)

X3(t) -X2(t) ==-\037va(sin 2t)(1-cos2t).

It follows that X2-

Xl < 0 and X3- X2 < 0 until t == Jr/2 \037 1.57(seconds),at

which time the equations in (26)and (27)give the values)

(Jr

) (Jr

) (Jr

)3Jrva

Xl2\"

==X22\"

==X32\"

== 16 ')

x; ( \037 )=

x\037 ( \037 )= 0, x\037 (\037 )

= Vo.)

We concludethat the three railway cars remain engagedand moving to the rightuntil disengagementoccursat time t == Jr/2.Thereafter, cars 1 and 2 remain at

rest(!),while car 3 continues to the right with speedVa. If, for instance, Va ==48feet per second(about 33 milesper hour), then the three cars travel a distanceof9Jr \037 28.27(ft) during their 1.57secondsof engagement,and)

Xl (t) ==X2(t) ==9Jr, X3(t) ==48t - 15Jr) (27'))

for t > Jr/2.Figure5.5.7illustratesthe \"before\"and\"after\"situations, and Fig.5.5.8showsthe graphs of the functions Xl (t),X2(t), and X3(t) in Eqs.(27)and (27'). .)

ForcedOscillationsandResonance)Supposenow that the ith massof the mass-and-springsystem in Fig.5.5.2is subjectto an external force Fi (i == 1,2,..., n) in addition to the forcesexertedby the

springsattached to it. Then the homogeneousequation MX\" ==Kx is replacedwith

the nonhomogeneousequation)

MX\" ==Kx + F) (28))))


---..) 75)

0.5 1.0 1.5 2.0 2.5t)

(a))

50

\037

---.. 25)

(b)

FIGURE5.5.7.(a) Before; (b) after.) FIGURE5.5.8.Positionfunctions of the three railway carsof Example 2.)

where F == [FI F2

tiplication by M-Iyields)

Fn JT is the externalforcevector for the system.Mul-)

xl! ==Ax + f) (29))

where f is the external force vectorperunit mass. We are especiallyinterestedin

the caseof a periodicexternalfa rce)

f(t) ==Fo coswt) (30))

(where F0 is a constant vector). We then anticipatea periodicparticular solution)

xp(t)==Ccoswt) (31))

with the known external frequencywand with a coefficientvectorcyet to bedeter-mined.Because

x\037== -w2ccoswt, substitution of (30)and (31)in (29),followed

by cancellationof the common factorcoswt, gives the linear system)

(A +w 2I)c== -Fa) (32))

to be solved for c.Observethat the matrix A + w 2I is nonsingular-inwhich caseEq.(32)can

be solved for c-unless-w2 == A, an eigenvalueof A. Thus a periodicparticularsolution of the form in Eq. (31)existsprovided that the external forcing frequencydoesnot equal one of the natural frequenciesWI, W2, ..., Wn of the system. Thecasein which w isa natural frequencycorrespondsto the phenomenonof resonancediscussedin Section2.6.)

Example3) Supposethat the secondmass in Example 1 is subjectedto the external periodicforce 50coswt. Then with m I == 2, m2 == 1,k I

== 100,k2 == 50,and Fo == 50 in

Fig.5.5.9,Eq.(29)takes the form)

I!

[-75

x ==50) _;\037 ]

x + [ 5\037 ]caswt ,) (33))

FIGURE5.5.9.Theforcedmass-and-spring system ofExample 3.)

and the substitution x ==ccoswt leadsto the equation)

[w 2

5075

w 225

50]c= [ -5\037])(34))))

for the coefficientvector c == [CI C2]T. This system is readily solved for)

1250C -

1 -(w2 -25)(w2- 100)')

50(w2-75)C2==-

(w2 _ 25)(w2- 100).) (35))

For instance, if the external squaredfrequency is w 2 == 50,then (35)yieldsCI ==-1,C2==-1.The resulting forced periodicoscillation is describedby)

Xl (t) ==-coswt, X2(t) ==-coswt.)

15)

Thus the two massesoscillatein synchrony with equal amplitudes and in the samedirection.

If the external squaredfrequency is w 2 == 125,then (35)yieldsCI == 1,C2 ==-1.The resulting forcedperiodicoscillation is describedby)

0)10\"'0

\037.-;::0..S<

5)

XI(t) == 1coswt,X2(t) ==-coswt,)

oo 5 10

Forced frequency)

15)

and now the two massesoscillatein synchrony in oppositedirections,but with the

amplitudeof motion of m2 twice that of mI.It is evident from the denominators in (35)that CI and C2approach +00as

w approacheseither of the two natural frequenciesWI == 5 and W2 == 10(foundin Example1).Figure 5.5.10showsa plot of the amplitude Jcy +

c\037of the forced

periodicsolutionx(t) ==ccoswt asa function of the forced frequencyw. The peaksat W2 ==5 and W2 == 10exhibit visually the phenomenonof resonance. .)

FIGURE5.5.10.Frequency-amplitude plot for Example 3.)

PeriodicandTransientSolutions)It follows from Theorem 4 of Section5.3that a particular solution of the forcedsystem)

x\" ==Ax +F0coswt) (36))

will be of the form)

\037) x(t) ==Xc (t) +xp (t ) ,) (37))

where xp(t)is a particular solution of the nonhomogeneoussystem and xc(t)is asolution of the correspondinghomogeneoussystem.It is typical for the effects offrictionalresistancein mechanicalsystemsto damp out the complementaryfunction

solution xc(t),so that)

xc(t)\037 0 as t \037 +00.) (38))

Hencexc(t)is a transientsolution that dependsonly on the initial conditions; it

dies out with time, leaving the steady periodicsolutionxp(t)resulting from theexternal driving force:)

x(t) \037 xp(t) as t \037 +00.) (39))

As a practical matter, every physical systemincludesfrictional resistance(howeversmall) that dampsout transient solutions in this manner.)))


E.\037roblems_)

Problems 1through 7 deal with the mass-and-springsystemshown in Fig. 5.5.11with stiffness matrix)

K=[-(\\7k2 )) k2

]-(k2 + k3))

and with the given mks values for the massesand spring con-stants. Find the two natural frequenciesof the system and de-scribeits two natural modesofoscillation.)

FIGURE5.5.11.Themass-and-springsystem for Problems 1 through 6.

1.m I = m2 = 1; k I = 0,k2 = 2,k3 = 0 (no walls)2.m I = m2 = 1; k I = 1,k2 = 4, k3 = 13.ml = 1,m2 = 2; k l = 1,k2 = k3 = 24. m I = m2 = 1; k I = 1,k2 = 2,k3 = 15.m I = m2 = 1; k l = 2, k2 = 1,k3 = 26.ml = 1,m2 = 2; k l = 2,k2 = k3 = 47. ml = m2 = 1; k l = 4, k2 = 6, k3 = 4)

In Problems 8 through 10the indicated mass-and-springsys-tem is set in motion from rest

(x\037 (0) =x\037 (0) = 0) in its equi-

librium position (XI (0) = X2(0) = 0) with the given external

forcesFI (t) and F2(t) acting on the massesm I and m2, respec-tively. Find the resulting motion of the system and describeitasa superposition ofoscillationsat three different frequencies.8.Themass-and-springsystem of Problem 2, with FI (t) =

96cos5t, F2(t) = 09.Themass-and-spring system of Problem 3, with FI (t) =

0, F2(t) = 120cos3t10.Themass-and-springsystem of Problem 7, with FI (t) =

30cost, F2(t) = 60cost11.Consider a mass-and-spring system containing two

massesml = 1 and m2 = 1 whose displacement func-tions x(t) and y (t) satisfy the differential equations)

x\" = -40x+ 8y,y\" = 12x- 60y.)

(a) Describethe two fundamental modesof freeoscilla-tion of the system. (b) Assume that the two massesstartin motion with the initial conditions)

x(O)= 19, x/CO)= 12and

y(O)= 3, y/(O) = 6 17.)

and are actedon by the sameforce, FI (t) = F2(t) =-195cos7t. Describethe resulting motion as a superpo-sition ofoscillationsat three different frequencies.)

In Problems 12and 13,find the natural frequenciesof the

three-masssystem of Fig. 5.5.1,using the given massesand

spring constants. For eachnatural frequency w, give the ra-tio al:a2:a3of amplitudes for a correspondingnatural mode

XI = al coswt, X2 = a2coswt, X3 = a3coswt.)

12.13.)

ml=m2=m3=1;kl=k2=k3=k4=1ml = m2 = m3 = 1; kl = k2 = k3 = k4 = 2(Hint: Oneeigenvalue is A = -4.)In the system ofFig.5.5.12,assumethatml = 1,k} = 50,k2 = 10,and Fa = 5 in mks units, and that w = 10.Then

find m2 sothat in the resulting steady periodicoscillations,the mass m I will remain at rest(!).Thus the effectof the

secondmass-and-spring pair will be to neutralize the ef-fectof the forceon the first mass.This is an example ofa dynamic damper. It has an electricalanalogy that somecablecompaniesuse to prevent your receptionof certain

cablechannels.)

14.)

F(t) =Fa coswt)

15.)

FIGURE5.5.12.Themechanicalsystem ofProblem 14.

Supposethat ml = 2, m2 = k, k} = 75, k2 = 25,Fa = 100,and w = 10(all in mks units) in the forcedmass-and-spring system ofFig.5.5.9.Find the solution ofthe system Mx\" = Kx + F that satisfiesthe initial condi-tions x(O)= x/ (0)= O.Figure 5.5.13shows two railway carswith a buffer spring.We want to investigate the transfer ofmomentum that oc-curs after car1with initial velocity Va impacts car2at rest.Theanalog ofEq.(18)in the text is)

16.)

x\" =[-c}

C2)

CI

]-C2X)

with Cj = k/mj for i = 1,2.Show that the eigenvalues ofthe coefficientmatrix A areA I = 0 and A2 = -c}-Cz,with associatedeigenvectorsv I = [1 1]T and V2 ==

[ CI -C2]T

.)

x,'(O)=0) X2'(O) =0)

X2(t))

FIGURE5.5.13.Thetwo railwaycarsofProblems 16through 19.

If the two carsof Problem 16both weigh 16tons (sothat

ml = m2 = 1000(slugs\302\273

and k = 1 ton/ft (that is, 2000lb /ft), show that the carsseparateafter 1(/2seconds,and

thatx\037 (t) = 0 and

x\037 (t) = Va thereafter. Thus the originalmomentum ofcar 1 is completely transferred to car2.)))


18.If cars 1 and 2 weigh 8 and 16tons, respectively, andk = 3000lb/ft, show that the two carsseparateafter n /3seconds,and that

x\037(t)= -1vo and

x\037(t)=

+\037vo

thereafter. Thus the two cars rebound in oppositedirec-tions.

19.If cars 1 and 2 weigh 24 and 8 tons, respectively, andk = 1500lb/ft, show that the carsseparateafter Jr/2 sec-onds, and that

x\037 (t) =+\037vo and

x\037(t)=

+\037vo

thereafter. Thus both carscontinue in the original direc-tion of motion, but with different velocities.)

Problems 20 through 23 deal with the samesystem of three

railway cars (samemasses)and two buffer springs (samespring constants) as shown in Fig. 5.5.6and discussedin Ex-ample 2. The cars engageat time t = 0 with x I (0) =Xz(0)= X3(0)= 0 and with the given initial velocities(wherevo = 48ftls). Show that the railway cars remain engageduntil

t = n /2 (s),after which time they proceedin their respectiveways with constant velocities.Determine the values of theseconstant final velocities

x\037 (t), x\037 (t), andx\037 (t) ofthe three cars

for t > n/2.In eachproblem you should find (as in Example2) that the first and third railway carsexchangebehaviors in

someappropriate sense.)

20.x\037 (0)= vo, x\037 (0)= 0,

x\037 (0) = -vo21.

x\037 (0)= 2vo, x\037(O)= 0, x\037(O)

= -Vo22.

x\037 (0)= Vo, x\037 (0)= Vo, x\037 (0) = -2vo23.

x\037 (0)= 3vo, x\037 (0)= 2vo,x\037 (0) = 2vo

24. In the three-railway-car system ofFig. 5.5.6,supposethatcars1 and 3 eachweigh 32 tons, that car2 weighs 8 tons,and that eachspring constant is 4 tons/ft. If

x\037 (0) = Vo

andx\037 (0) =

x\037 (0) = 0, show that the two springs arecompresseduntil t = n /2 and that

x\037 (t) =-\037vo

andx\037(t)

=x\037(t)

=+\037 Vo

thereafter. Thus car 1rebounds,but cars2 and 3 continuewith the same velocity.)

The Two-AxleAutomobileIn Example 4 ofSection2.6we investigated the vertical oscil-lations of a one-axlecar-actually a unicycle. Now we cananalyze a more realisticmodel:a car with two axlesand with

separatefront and rearsuspensionsystems. Figure 5.5.14rep-resents the suspension system of such a car. We assume thatthe car body actsas would a solidbar ofmass m and lengthL = LI + Lz. It has moment of inertia I about its centerofmass C, which is at distanceLI from the front of the car. Thecar has front and back suspensionsprings with Hooke'scon-stants k l and k2, respectively. When the car is in motion, letx(t) denote the vertical displacementof the centerofmass ofthe carfrom equilibrium; let e(t)denoteits angular displace-ment (in radians) from the horizontal. Then Newton's laws of)

motion for linear and angular accelerationcanbeusedto de-rive the equations)

mx\" = -(kl + kz)x + (kiLl -kzLz)e,)(40))

Ie\" = (k l LI-kzLz)x- (k l Li + kzL\037)e.)

Equilibrium

position)

FIGURE5.5.14.Modelof thetwo-axleautomobile.)

25.Supposethat m = 75 slugs (the carweighs 2400lb), LI =7 ft, Lz = 3 ft (it'sa rear-enginecar),k l = kz = 2000lb/ft, and I = 1000ft.lb.sz. Then the equations in (40)take the form)

75x\" + 4000x-) 8000e= 0,)

1000e\"- 8000x+ 116,000e= O.)

(a) Find the two natural frequenciesWI and Wz of the car.(b) Now supposethat the caris driven at a speedof v feetper secondalong a washboard surface shaped like a sinecurve with a wavelength of40 ft. Theresult is a periodicforceon the car with frequency W = 2nv/40 = nv/20.Resonanceoccurswhen with W = WI or W = Wz. Find the

corresponding two critical speedsof the car (in feet persecondand in miles perhour).

26.Supposethat k l = kz = k and LI = Lz =\037L

in

Fig. 5.5.14(the symmetric situation). Then show that ev-ery freeoscillation is a combination of a vertical oscilla-tion with frequency)

WI = J2k/m)

and an angular oscillationwith frequency)

Wz = JkLz/(2I).)

In Problems27 through 29, the system ofFig. 5.5.14is taken

asa modelfor an undamped car with the given parametersinfps units. (a) Find the two natural frequenciesofoscillation(in hertz). (b) Assume that this car is driven along a sinu-

soidalwashboardsurfacewith a wavelength of40ft. Find the

two criticalspeeds.)27.m = 100,I = 800,LI = Lz = 5, k l = kz = 200028.m = 100,I = 1000,LI = 6, Lz = 4, k l = kz = 200029.m = 100,I = 800,LI = Lz = 5,k l = 1000,kz = 2000)))

5.6MultipleEigenvalueSolutions 393)

___Mu1tie!eEi_g\037\037value SoIu\037ions)

In Section5.4we saw that if the n x n matrix A has n distinct (real or complex)eigenvaluesAI,A2, ..., An with respectiveassociatedeigenvectorsVI, V2, ..., Vn,then a general solutionof the system)

dx-==Axdt)

(1))

is given by)

x(t) ==cIvIe)qt + c2v2eA2t +...+ CnVneAnt) (2))

with arbitrary constantsCI,C2,..., Cn . In this sectionwe discussthe situation when

the characteristicequation)

IA -All ==0) (3))

doesnot have n distinct roots,and thus has at leastone repeatedroot.An eigenvalue is of multiplicity k if it is a k-fold root of Eq. (3).For each

eigenvalueA, the eigenvectorequation)

(A -AI)v ==0) (4))

has at leastone nonzero solution v, so there is at leastone eigenvectorassociatedwith A. But an eigenvalueof multiplicity k > 1 may havefewer than k linearly

independentassociatedeigenvectors.In this casewe are unable to find a \"completeset\" of n linearly independent eigenvectors of A, as neededto form the generalsolution in (2).

Let us call an eigenvalue of multiplicity k completeif it has k linearly in-

dependentassociatedeigenvectors.If every eigenvalue of the matrix A is com-plete,then-becauseeigenvectorsassociatedwith different eigenvaluesare linearly

independent-itfollows that A doeshave a completesetof n linearly independenteigenvectors VI, V2, ..., Vn associatedwith the eigenvaluesAI, A2, ...,An (eachrepeatedwith its multiplicity). In this casea general solution of x' == Ax is still

given by the usual combination in (2).)

Example1) Find a general solutionof the system)

x =[-\037)

4-14)

\037]

x.) (5))

Solution The characteristicequationof the coefficientmatrix A in Eq.(5)is)

IA -All ==)

9-A-66)

4-1-A4)

oo

3-A)

== (3 -A)[(9-A)(-1-A) +24]== (3 -A)(15- 8A +A2)

== (5 -A)(3 -A)2 ==o.)

Thus A has the distinct eigenvalueAl == 5 and the repeatedeigenvalueA2 == 3 ofmultiplicity k == 2.)))


CASE1:)q ==5. The eigenvectorequation (A-AI)v ==0,wherev == [a b C]T,)IS)

(A -5I)v=[-i -\037 _g] [\037] [g]

.

Each of the first two equations, 4a +4b == 0 and -6a- 6b == 0,yieldsb == -a.Then the third equation reducesto 2a -2c==0,so that c ==a.The choicea == 1then yieldsthe eigenvector)

VI ==[1 -1 1 JT)

associatedwith the eigenvalueA I ==5.)

CASE2:A2 ==3. Now the eigenvectorequation is)

[64

(A -3I)v== -6 -46 4) g][\037] [g],)

so the nonzero vector V == [a b c JT is an eigenvectorif and only if)

6a +4b ==0;) (6))

that is, b ==-\037a.

The fact that Eq. (6)doesnot involve c means that c is arbitrary,

subjectto the condition V -:FO.If c == 1,then we may choosea ==b ==0;this givesthe eigenvector)

V2 == [0 0 1 JT

associatedwith A2 == 3. If c == 0,then we must choosea to be nonzero. Forinstance, if a ==2 (to avoidfractions), then b ==-3,so)

V3 == [2 -3 OJT)

isa secondlinearlyindependenteigenvectorassociatedwith the multiplicity 2 eigen-value A2 == 3.)

Thus we have found a completesetv I,V2, V3 of three eigenvectorsassociatedwith the eigenvalues5,3, 3.The correspondinggeneral solutionof Eq.(5)is)

( )5t 3t 3txt ==CIVle +C2v2e +C3v3e) (7))

= Cl

[-!]eSt +C2

[\037 ]e3t +C3 [

-\037 ]

e3t,)

with scalarcomponentfunctions given by)

Xl (t) == Cle5t + 2c3e3t,

X2(t) ==-cle5t - 3c3e3t,

X3 (t) == CIe5t +C2e3t .) .)))

Example2)


Remark:Our choicein Example1 of the two eigenvectors)

V2 == [0 0 1 ]T

and v3 == [2 -3 0 ]T)

associatedwith the repeatedeigenvalueA2 == 3 bearscomment. The fact that b ==-\037a

for any eigenvectorassociatedwith A2 == 3 means that any such eigenvectorcan be written as)

a 0 23 0 + l.a -3 Iv== --a ==c ==CV2 + 2av3 ,2 2

C 0)

and thus is a linear combination of V2 and V3. Therefore, given a and c not both

zero,we couldchoosev rather than V3 as our third eigenvector,and the new generalsolution)

x(t) ==C}VI eSt+C2v2e3t +C3ve3t

would beequivalentto the one in Eq.(7).Thus we neednot worry about making the

\"right\" choiceof independent eigenvectorsassociatedwith a multiple eigenvalue.Any choicewill do;we generally make the simplestone we can. .)DefectiveEigenvaluesThe followingexampleshowsthat-unfortunately-notall multiple eigenvaluesarecomplete.)

The matrix)

A = [; -\037])(8))


IA -All ==)I-A

3)

-37-A)

== (1-A)(7 -A) +9

== A2- 8A + 16== (A -4)2 ==O.)

Thus A has the singleeigenvalueA I ==4 of multiplicity 2.The eigenvectorequation)

[-3

(A -4I)v ==3) -\037 ] [ \037 ]

= [ \037 ])

then amounts to the equivalentscalarequations)

-3a-3b==O,3a+3b==O.)

Henceb ==-a if v == [a b ]T is to be an eigenvectorof A. Thereforeany eigen-

vector associatedwith A I== 4 is a nonzero multiple of v == [1 -1]T. Thus the

multiplicity 2 eigenvalueA) == 4 has only one independenteigenvector,and henceis incomplete. .)))

An eigenvalueA of multiplicity k > 1 is calleddefective if it is not complete.If A has only p <k linearly independenteigenvectors,then the number)

d==k-p) (9))

of \"missing\" eigenvectors is calledthe defectof the defective eigenvalueA. Thusthe defective eigenvalue Al == 4 in Example2 has multiplicity k == 2 and defectd == 1,becausewe saw that it has only p == 1 associatedeigenvector.

If the eigenvaluesof the n x n matrix A are not all complete,then the eigen-value method as yet describedwill producefewer than the neededn linearly inde-pendent solutionsof the systemx' ==Ax. We thereforeneedto discoverhow to find

the \"missing solutions\" correspondingto a defective eigenvalue A of multiplicityk > 1.)

TheCaseofMultiplicityk = 2Let us beginwith the casek == 2,and supposethat we have found (as in Example2)that there is only a singleeigenvectorv I associatedwith the defectiveeigenvalueA. Then at this point we have found only the singlesolution)

\037) Xl (t) ==VieAt) (10))

of x' == Ax. By analogy with the caseof a repeatedcharacteristicroot for a singlelinear differential equation (Section2.3),we might hopeto find a secondsolutionof the form)

X2(t) == (v2t)eAt ==V2 teAt .) (11))When we substitute x ==V2teAt in x' ==Ax, we get the equation)

V2 eAt +AV2 teAt ==AV2teAt.)

But becausethe coefficientsof both eAt and teAt must balance,it follows that V2 ==0,and hencethat X2(t) = O. This means that-contrary to our hope-thesystemx' ==Ax doesnot have a nontrivial solutionof the form assumedin (11).

Insteadof simply giving up on the ideabehindEq.(11),let us extend it slightlyand replaceV2t with Vlt + V2. Thus we explorethe possibilityof a secondsolutionof the form)

\037) ( ) ( )At At At

X2 t == Vlt+V2 e ==Vlte +V2e) (12))

whereVI and V2 are nonzeroconstantvectors.When we substitute x ==vIte At+V2eAt

in x' ==Ax, we get the equation)

VieAt +AVIte

At +AV2eAt ==AVIteAt +AV2eAt.) (13))

We equate coefficientsof eAt and teAt here,and thereby obtain the two equations)

(A -AI)VI ==0 (14)

and

(A -AI)v2 ==VI (15))

that the vectorsVI and V2 must satisfy in orderfor (12)to givea solutionof x'==Ax.)))


Note that Eq. (14)merely confirms that VI is an eigenvectorof A associatedwith the eigenvalueA. Then Eq.(15)saysthat the vector V2 satisfiesthe equation)

(A -AI)2 v2 == (A -AI)[(A -AI)v2J == (A -AI)VI ==O.)

It follows that, in orderto solve simultaneouslythe two equations in (14)and (15),it sufficesto find a solution V2 of the singleequation (A -AI)2v2 ==0 such that the

resulting vector VI == (A -AI)v2 is nonzero. It turns out that this is always possibleif the defectiveeigenvalueA of A is of multiplicity 2.Consequently,the proceduredescribedin the following algorithm always succeedsin finding two independentsolutionsassociatedwith such an eigenvalue.)

ALGORITHM DefectiveMultiplicity 2Eigenvalues1.First find a nonzero solution V2 of the equation)

(A -AI)2v2 ==0) (16))

such that)

(A -AI)v2 ==VI) (17))

is nonzero, and therefore is an eigenvectorV I associatedwith A.

2.Then form the two independentsolutions)

Xl (t) ==VI eAt) (18))

and)

X2(t) == (Vlt +V2)eAt) (19))

of x' ==Ax correspondingto A.)

Example3)\"\037_.N..... ........ '''' ,_.\"..\037_.\".... ... \037\"\"'\" JOn\"'.... \037.... A A\" '\" J' \"'..., ..._... '\" ..., \"'\" ... \", \",... .... , n\"'''''''''''' .... \"\" ...,....


I

[1

X ==3)

-3]7

x.) (20))

Solution In Example2 we found that the coefficientmatrix A in Eq. (20)has the defectiveeigenvalueA ==4 of multiplicity 2.We thereforebegin by calculating)

(A -41)2=[-\037 -\037] [-\037 -\037]

=[g gJ.)

HenceEq.(16)is)

[g g]V2 = 0,

and therefore is satisfied by any choiceof V2. In principle,it could happen that

(A -4I)v2 is nonzero (asdesired)for somechoicesof V2 though not for others. If)))

4321)

\0370

-1-2-3)

Xl)


system x\037

= Xl - 3X2,X\037

= 3x]+ 7X2 ofExample 3.)

we try V2 == [1 0 JT we find that

[-3

(A -4I)v2 ==3) -\037 ] [ \037 ] = [ -\037 ] =

VI)

is nonzero, and therefore is an eigenvectorassociatedwith A == 4. (It is -3 timesthe eigenvectorfound in Example2.)Therefore the two solutionsof Eq.(20)givenby Eqs.(18)and (19)are)

( )4t

[-3

]4t

Xl t ==VI e ==3 e ,)

4t

[-3t + 1

]4t

X2(t) == (Vlt +v2)e ==3t

e.)

The resulting general solution)

X(t) ==CIXI (t)+C2X2(t))

has scalarcomponentfunctions

Xl (t) ==(-3C2t+C2-3CI)e4t,

X2 (t) == (3C2t+3CI)e4t.)

With C2 == 0 these solution equations reduceto the equations X I (t) == -3CIe4t ,X2 (t) == 3CIe4t , which parametrize the line Xl ==-X2 in the xI X2 -plane.The point

(Xl (t),X2(t)) then recedesalong this line away from the origin as t \037 +00,to thenorthwest if CI > 0 and to the southeast if Cl <O.As indicated in Fig.5.6.1,eachsolution curve with C2 -:F 0 is tangent to the line Xl == -X2 at the origin; the point

(Xl (t),X2(t)) approachesthe origin as t \037 -00and approaches+00along thesolution curve as t \037 +00. .)

GeneralizedEigenvectorsThe vector V2 in Eq. (16)is an exampleof a generalizedeigenvector. If A is an

eigenvalueof the matrix A, then a rank r generalizedeigenvectorassociatedwith

A is a vector V such that)

\037) (A -AI)r v ==0 but (A -AI)r-iv -:FO.) (21))

If r == 1,then (21)simply means that v is an eigenvectorassociatedwith A (recallingthe convention that the Oth powerof a squarematrix is the identity matrix). Thus arank 1 generalizedeigenvectoris an ordinary eigenvector.The vector V2 in (16)isarank 2 generalizedeigenvector(and not an ordinary eigenvector).

The multiplicity 2 methoddescribedearlierboilsdown to finding a pair {VI, V2}

of generalizedeigenvectors,one of rank 1 and one of rank 2,such that (A -AI)v2 ==

V1.Highermultiplicity methods involve longer \"chains\" of generalizedeigenvec-tors.A length k chainof generalizedeigenvectorsbasedon the eigenvectorVI isa set {VI, V2, ..., Vk} of k generalizedeigenvectorssuch that

(A -AI)vk ==Vk-I,

(A -AI)Vk-I ==Vk-2,)\037) (22))

(A -AI)v2 ==VI.)))

Example4)


BecauseVI is an ordinary eigenvector,(A -AI)VI = O.Therefore, it followsfrom

(22)that)

(A -AI)k vk = O.) (23))

If {vI,V2, V3} is a length 3 chain of generalizedeigenvectorsassociatedwith

the multiple eigenvalueA of the matrix A, then it is easyto verify that three linearly

independent solutionsof Xl = Ax are given by

xI (t) = v 1 eAt,

X2(t) = (VIt +V2)eAt

,

X3(t) =(\037VIt2 +V2 t +V3) eAt.)

\037) (24))

For instance, the equations in (22)give)

AV3 = V2 +AV3, AV2 = VI +AV2, AVI = AVI,)

so)

AX3 =[\037AVIt2 +AV2t +AV3] eAt

=[\037AVIt2 + (VI + AV2)t + (V2 +AV3)] eAt

= (VIt +V2)eAt +A

(\037VIt2 +V2 t +V3) eAt)

I= x3 .)

Therefore, X3 (t) in (24)does,indeed,definea solutionof Xl = Ax.

Consequently, in orderto \"handle\" a multiplicity 3 eigenvalueA, it sufficesto find a length 3 chain {VI, V2, V3} of generalizedeigenvaluesassociatedwith A.

Lookingat Eq.(23),we seethat we needonly find a solution V3 of

(A -AI)3 v3 = 0)

such that the vectors)

V2 = (A -AI)v3 and VI = (A -AI)v2)

are both nonzero (although, aswe will see,this is not alwayspossible).).......... ...........

Find three linearly independentsolutionsof the system)

[0 1 2

]x/= -i -\037 -6

x.) (25))

Solution The characteristicequation of the coefficientmatrix in Eq.(25)is)

[-A 1 2

]IA -All = -5 -3-A -7

1 0 -A)

= 1 .[-7-2.(-3-A)] + (-A)[(-A)(-3-A) +5]

= -A3 - 3A 2- 3A - 1 = -(A + 1)3= 0,)))


and thus A has the eigenvalue 'A = -1of multiplicity 3.The eigenvectorequation(A - 'AI)v = 0 for an eigenvectorv = [a b c]Tis)

(A +I)v=[-\037 -\037

-r] [\037]

=[\037]

.)

The third row a + c = 0 gives c = -a,then the first row a + b + 2c= 0 givesb = a.Thus, to within a constant multiple, the eigenvalue 'A = -1has only the

singleassociatedeigenvectorv = [a a _a]Twith a i= 0,and so the defectof 'A = -1is 2.

To apply the method describedhere for triple eigenvalues,we first calculate)

(A +1)2=[-\037

1-2

0

and

(A +1)3=[-\037

1-2

0)

-r][-\037-\037 -r]=[=\037=:=\037])

2

] [-2 -1 -3

] [0 0 0

]-i -; -\037 -\037

=\037 \037 \037

\302\267)

Thus any nonzero vector V3 will be a solution of the equation (A + I)3v3 o.Beginning with V3 = [1 0 O]T,for instance, we calculate)

V2 = (A + l)v3 =[-\037

1 -r][g]

=[-\037]

,-20

VI = (A + l)v2 =[-\037

1

-r][-\037]=[=\037].-2

0)

Note that VI is the previously found eigenvector v with a = -2;this agreementservesas a checkof the accuracyof our matrix computations.

Thus we have found a length 3 chain {VI, V2, V3} of generalizedeigenvectorsassociatedwith the triple eigenvalue 'A = -1.Substitution in (24)now yieldsthelinearly independent solutions)

XI(t) = vle-t =[=\037]

e-t ,)

[-2t + 1

]X2(t) = (VIt +v2)e-t = -2t-5 e-t ,

2t + 1)

[-t2+ t + 1

]X3(t) =

(\037vIt2 +V2 t +V3) e-t = -t2-5t e-tt 2 + t)

of the system Xl = Ax.) .)))


The GeneralCase)A fundamental theorem of linear algebra states that every n x n matrix A has n

linearly independent generalizedeigenvectors.Thesen generalized eigenvectorsmay be arranged in chains,with the sum of the lengths of the chains associatedwith a given eigenvalueA equal to the multiplicity of A. But the structure of thesechains dependson the defect of A, and can be quite complicated.For instance,amultiplicity 4 eigenvaluecan correspondto). Four length 1 chains (defect0);. Two length 1 chains and a length 2 chain (defect 1);. Two length 2 chains (defect2);. A length 1 chain and a length 3 chain (defect2);or. A length 4 chain (defect3).)

Note that, in each of thesecases,the length of the longestchain is at most d + 1,where d is the defect of the eigenvalue. Consequently,oncewe have found all the

ordinary eigenvectorsassociatedwith a multiple eigenvalueA, and thereforeknow

the defect d of A, we can beginwith the equation\037 (A -AI)d+IU = 0 (26))

to start building the chains of generalizedeigenvectorsassociatedwith A.)

ALGORITHM Chainsof GeneralizedEigenvectorsBeginwith a nonzero solution UI of Eq. (26)and successivelymultiply by thematrix A -AI until the zero vector is obtained.If)

(A -AI)UI = U2 i= 0,)

(A -AI)Uk-l = Uk i= 0,)

but (A -AI)Uk = 0,then the vectors)

{VI, V2,...,Vk} = {Uk, Uk-I,...,U2,UI})

(listedin reverseorderof their appearance)form a length k chain of generalizedeigenvectorsbasedon the (ordinary)eigenvectorVI.)

Each length k chain {VI, V2, ..., Vk} of generalizedeigenvectors(with VI an

ordinary eigenvectorassociatedwith A) determinesa setof k independentsolutionsof Xl = Ax correspondingto the eigenvalueA:

Xl (t) = VleAt

,

X2(t) = (VIt +V2)eAt

,

X3(t) =(\037Vlt2 +V2 t +V3) eAt,)

\037) (27))

(t k-I 2

)VI Vk-2 t

AtXk (t) = + ...+ +Vk- I t +Vk e .

(k - I)! 2!)))


Example5)

c)

FIGURE5.6.2.Therailway carsofExample 6.)

Note that (27)reducesto Eqs. (18)through (19)and (24) in the casesk = 2 andk = 3, respectively.

Toensurethat we obtain n generalizedeigenvectorsof the n x n matrix A that

are actually linearly independent,and thereforeproducea completesetof n linearlyindependentsolutionsof x'= Ax when we amalgamateall the \"chains of solutions\"

correspondingto different chains of generalizedeigenvectors,we may rely on thefollowing two facts:)

. Any chain of generalizedeigenvectorsconstitutes a linearly independentsetof vectors.. If two chains of generalizedeigenvectors are basedon linearly independenteigenvectors,then the union of thesetwo chains is a linearly independentsetof vectors (whether the two base eigenvectors are associatedwith differenteigenvaluesor with the sameeigenvalue).)

Supposethat the 6 x 6 matrix A has two multiplicity 3 eigenvaluesAl = -2andA2 = 3 with defects1 and 2,respectively.Then Al must have an associatedeigen-vector ulanda length 2 chain {V1,V2} of generalizedeigenvectors(with the eigen-vectorsUl and VI beinglinearly independent),whereasA2 must havea length 3 chain{WI, W2, W3}of generalizedeigenvectorsbasedon its singleeigenvectorWI. The sixgeneralizedeigenvectorsUl, VI, V2, WI, W2, and W3 are then linearly independentand yield the following sixindependent solutionsof x'= Ax:

Xl (t) = Ul e-2t,

x2(t) = VIe-2t

,

x3(t) = (VI t +V2)e-2t,

,X4(t) = Wle3t

,

xs(t)= (WIt +W2)e3t

,

X6(t) =(\037Wlt2 + W2t +W3) e3t. .

As Example5 illustrates, the computation of independent solutions corre-spondingto different eigenvaluesand chains of generalizedeigenvaluesis a routine

matter. The determination of the chain structure associatedwith a given multipleeigenvaluecan be more interesting (as in Example6).)

An ApplicationFigure 5.6.2showstwo railway cars that are connectedwith a spring (permanentlyattached to both cars)and with a damper that exertsoppositeforceson the two cars,of magnitude c(x\037

-x\037) proportionalto their relativevelocity.The two carsare also

subjectto frictional resistanceforcesClx\037and

C2X\037 proportional to their respectivevelocities.An application of Newton'slaw ma = F (as in Example1 of Section5.1)yieldsthe equations of motion)

mlX\037'= k(X2 -Xl) - ClX\037

-c(x\037

-x\037),

m2x{= k(Xl -X2) - C2X\037-

c(x\037-

x\037).

In terms of the positionvector x(t) = [Xl(t) X2(t) ]T, theseequations can bewritten in the matrix form)

(28))

Mx\" = Kx+Rx',) (29))))

Example6)


where M and K are massand stiffnessmatrices (as in Eqs.(2)and (3)of Section5.5),and)

R= [-(C

C+CI) C

]-(c+C2))

is the resistancematrix. Unfortunately, becauseof the presenceof the term involv-

ing x',the methodsof Section5.5cannot beused.Instead,we write (28)as a first-order system in the four unknown functions

Xl (t),X2(t), X3(t) =X\037

(t),and X4(t) =x\037(t). If ml = m2 = 1 we get)

x'= Ax (30)

where now x= [ Xl X2 X3 X4]T and

0 0 1 0

A= 0 0 0 1(31)-k k -(c+CI) C

k -k C -(c+C2))

With ml = m2 = C= 1 and k = CI = C2 = 2,the system in Eq.(30)is)

x' =)

o 0 1 0000 1-2 2 -3 1

2 -2 1-3)x.) (32))

It is not too tediousto calculate manually-althougha computer algebra systemsuch as Maple,Mathematica, or MATLAB is useful here-thecharacteristicequa-tion)

A4 + 6A 3 + 12A2 + 8A = A(A +2)3 = 0)

of the coefficientmatrix A in Eq. (32).Thus A has the distinct eigenvalueAQ = 0and the triple eigenvalueAl = -2.)

CASE1:AQ = O. The eigenvalueequation (A - AI)v = 0 for the eigenvectorv = [a b C d]Tis)

Av=)

o 0 1 0000 1-2 2 -3 1

2 -2 1-3) Cd)

ab)

oooo)

The first two rows give C= d = 0,then the last two rows yield a = b.Thus)

vQ = [1 1 0 O]T)

is an eigenvectorassociatedwith AQ= O.)))


CASE2:Al = -2. The eigenvalueequation (A -AI)v = 0 is)

(A +2I)v=)

2 0 1 0020 1-2 2 -1 1

2 -2 1-1)ab)

oooo)

cd)

The third and fourth scalarequationshere are the differencesof the first and secondequations, and therefore are redundant. Hencev is determined by the first two

equations,)2a +c = 0 and 2b+d = O.)

We can choosea and b independently,then solve for c and d.Thereby we obtaintwo eigenvectorsassociatedwith the triple eigenvalueAl = -2.The choicea = 1,b = 0 yieldsc = -2,d = 0 and thereby the eigenvector)

UI = [ 1) o -2) O]T.)

The choicea = 0,b = 1 yieldsc = 0,d = -2 and thereby the eigenvector)

U2 = [ 0) 1) o _2]T.)

BecauseAl = -2has defect 1,we needa generalizedeigenvectorof rank 2,and hencea nonzero solution V2 of the equation)

2 2 1 1

(A +2I)2v2 = 2 2 1 1V2 = o.0 0 0 0

0 0 0 0

Obviously,V2 = [ 0 0 1 -1]T

is such a vector,and we find that

2 0 1 0 0 1

(A +2I)v2 = 0 2 0 1 0 -1-2 2 -1 1 1 -2 = VI

2 -2 1 -1 -1 2)

is nonzero, and therefore is an eigenvectorassociatedwith Al = -2.Then {VI, V2}

is the length 2 chain we need.The eigenvector VI just found is neither of the two eigenvectors Ul and U2

found previously, but we observethat VI = UI -U2. For a length 1 chain WI to

completethe picture we can chooseany linear combination of Ul and U2 that isindependent of VI. For instance, we couldchooseeither WI = Ul or WI = U2.However,we will seemomentarily that the particularchoice)

WI = UI +U2 = [ 1) 1 -2 _2]T)

yieldsa solutionof the system that is of physical interest.)))


Finally, the chains {va},{WI}, and {VI, V2} yield the four independentsolutions)

XI(t) ==voeO.t ==[1 1 0 O]T,X2(t) ==WI e-2t == [ 1 1 -2 -2]Te-2t

,

x3(t) == V I e-2t == [ 1 -1 -2 2 ]Te-2t

, (33)

X4(t) == (Vlt +v2)e-2t

== [ t -t -2t + 1 2t - 1 ]Te-2t)

of the systemx' ==Ax in (32).The four scalarcomponentsof the general solution)

X(t) ==CIXI (t)+C2X2(t) +C3X3(t) +C4X4(t))

are describedby the equations

Xl (t) ==CI+e-2t(C2+C3+C4t ),

X2(t) ==CI+e-2t(C2-C3-C4t ),

X3(t) ==e-2t(-2C2- 2C3+C4-2C4t),

X4(t) ==e-2t(-2C2 +2C3-C4+ 2C4t).)

(34))

Recall that Xl (t) and X2(t) are the position functions of the two masses,whereasX3 (t) ==

x\037 (t) and X4 (t) ==x\037 (t) are their respectivevelocityfunctions.

For instance, supposethat Xl (0) == X2 (0) == 0 and thatx\037

(0) ==x\037 (0)= va.

Then the equations)Xl (0)==CI + C2 + C3 == 0,X2(0) ==CI + C2- C3 == 0,x\037 (0)== - 2C2- 2C3+ C4== VA,

x\037(O)== - 2C2+ 2C3- C4==

VA)

are readily solvedfor CI ==\037

VA, C2 ==-\037

va, and C3 ==C4==0,so)

Xl (t) ==X2(t) ==\037vo (1-e-2t

) ,,( )

,( )

-2tX I t ==X2 t ==voe .)

In this casethe two railway carscontinue in the samedirection with equal but ex-ponentially dampedvelocities,approaching the displacementsXl = X2 = !va ast ---+ +00.

It is of interest to interpret physically the individual generalizedeigenvectorsolutionsgiven in (33).The degenerate(AO ==0)solution)

Xl (t) == [1 1 0 O]T)

describesthe two massesat rest with position functions Xl (t) = 1 and X2(t) = 1.The solution

X2(t) == [ 1 1 -2 _2]Te-2t

correspondingto the carefully choseneigenvector WI describesdamped motionsXl (t) == e-2t and X2 (t) == e-2t of the two masses,with equal velocities in the

samedirection.Finally, the solutions X3 (t) and X4 (t) resulting from the length 2chain {vI,V2} both describedampedmotion with the two massesmoving in oppositedirections. .)))


_ Pr?blems)

The methods of this sectionapply to complexmultiple eigenvaluesjust as toreal multiple eigenvalues(although the necessarycomputationstend to besomewhatlengthy). Given a complexconjugate pair a ::f:fJi of eigenvaluesof multiplicity k,we work with one of them (say, a - fJi) as if it werereal to find k independentcomplex-valuedsolutions.The real and imaginary parts of thesecomplex-valuedsolutions then provide2k real-valuedsolutionsassociatedwith the two eigenvaluesA = a - fJi and A = a + fJi each of multiplicity k.SeeProblems33 and 34.)

-1]1 x

-1]5 x

-4]9 x)

17.x'=[

1\037

-27

18.x'=[

\037

-2

19.x'=[\037

20.x =[\037

[

-121.x'=

i

22.X/=[\037)

Find generalsolutions of the systems in Problems1through22.In Problems1through 6, usea computer system or graph-ing calculatorto construct a direction fieldand typical solutioncurvesfor the given system.)

2.x'= [i4. x'= [i6.x'= [ \037)

[-2 1

]1.x'= -1 -4 x

[1 -2

]3.x'= 2 5 x

5.x'= [ _\037 n x

7. x'=[-; \037 \037

]x002

8.x'=[-

i\037 2; \037

]x

6 6 13

9.X/=[

-1\037 1; 8\037

]x

-8 4 33

10.x'=[-2i \037\037

=\037\037]

x

[-3 0 -4

]11.x'= -

\037

-b

-\037

x

[-1 0 1

]12.x'= 0 -1 1 x

1 -1 -1

[-1 0 1

]13.x'= 0 1 -4 x

o 1-314.x'=

[-\037 -

\037 -\037

]x

4 1-2

[-2 -9 0

]15.x'=

\037 \037 \037

x

16.x'= [-i -\037 -\037 ]x)

o7

-9o3

-4-4 0

1 0-12 -1-4 01 02 1o 2o 0-4 0

3 02 11 0

3 7-1 -41 3

-6 -14)

\037

]x

-5

\037

]x

-1)

-\037

]-6 x

-1

i]x

\037]x

\037]x)

In Problems23 through 32 the eigenvaluesof the coefficientmatrix A are given. Find a generalsolution of the indicated

system x'= Ax. Especiallyin Problems29 through 32,useofa computer algebrasystem may be useful.)

[39

23.x'= -3672

[28

24. x'= 15-15

[-2

25.x'= -b

26.x'=[

i-3

[-3

27.x =\037)

8 -16

]-5 16 x; A=-I,3,316 -2950 100

]33 60 x; A = -2,3,3

-30 -571

\037 \037

]x; A = 2,2,2

1 2

-1 1

]3 0 x; A = 3,3, 32 1

5 -5]

-1 3 x; A = 2,2,2-8 10)))

5.7 Matrix Exponentialsand LinearSystems 407)

28. x'=[-H

\037

\037 -I!]x; A = 2,2,2

[

-1 1 1 -2

]7 -4 -6 1129. x'= 56

-1 1 3 x; )...= -1,-1,2,2

-2 -2 6

30.X/=[

\037 _j 2\037 _l\037

]X; )\",=-1,-1,2,2

o -27 45 -25

31.x'=[

_\037\037

-\037\037 \037

i\037

]x; )...= 1,1,1,1

-27 9 -3 -2311 -1 26 6 -3o 3 000

32. x'= -9 0 -24 -6 3 x;3 0 9 5-1

-48 -3 -138-30 18)...= 2,2,3,3,3

33. Thecharacteristicequation of the coefficientmatrix A ofthe system)

parts of the complex-valuedsolutions

vleAt and (Vlt + v2)eAt)

to find four independent real-valued solutions ofx'= Ax.

34. Thecharacteristicequation of the coefficientmatrix A ofthe system)

I

[-1\037

x = -933)

o-1-310)

-8o

-2590)

-\037

]-9 x

32)

IS)

(jJ()...)= ()...2_ 4)\", + 13)2=o.Therefore,A has the repeatedcomplex conjugate pair2 :I::3i of eigenvalues. First show that the complexvec-tors)

VI = [ -i) 3 + 3i) O .]

T-l ,)

v2=[3 -10+9i-l O]T)

[

3I 4

x =\037)

-43oo)

1o34) _\037]

x)

form a length 2 chain {VI, V2} associatedwith the eigen-value)...= 2 + 3i.Then calculate(asin Problem 33)four

independent real-valuedsolutions ofx'= Ax.

35.Find the position functions Xl (t) and X2(t) of the railwaycarsofFig. 5.6.1if the physical parameters are given by)

ml = m2 = CI = C2 = C= k = 1)

IS)

(jJ()...)= ()...2_ 6)\", + 25)2= O.Therefore,A has the repeatedcomplexconjugate pair3 :I::4i of eigenvalues.First show that the complexvec-tors)

and the initial conditions are)

Xl (0)= X2(0) = 0,x\037 (0)= x\037(O)

= Va.)

VI = [1 l 0 0]T and V2 = [ 9 0 1 i]T)

How far do the carstravel beforestopping?

36.RepeatProblem 35 under the assumption that car 1 isshieldedfrom air resistanceby car2,sonow CI = O.Show

that, beforestopping, the carstravel twice as far as thoseofProblem 35.)

form a length 2 chain {vI, V2} associatedwith the eigen-value)...= 3 -4i. Then calculatethe real and imaginary)

\037MatrixExponentialsand LinearSystems\037.\"\" h.,.wmn..,..m.\".\"'.W.'...'.'m-'m.,.\"..'.,,,,,m,,,,,,. . \"\" \"\"\"'''_\037''_ -\"'\"\"\"'\",,\" '\" \"\",,_ \" \"\",\037,,\"h\"\"\". ,m\037\" \"\"\

The solution vectorsof an n x n homogeneouslinear system)

\037) x' ==Ax) (1))

can beusedto constructa squarematrix X == \037(t) that satisfiesthe matrix differen-tial equation)

X' ==AX) (1'))))


associatedwith Eq.(1).Supposethat Xl (t),X2(t), ..., Xn (t) are n linearly indepen-dent solutionsof Eq.(1).Then the n x n matrix)

\037) \037(t) = Xl (t) X2(t)) Xn (t )) (2))

having thesesolutionvectorsas its column vectors, iscalleda fundamental matrixfor the system in (1).)

FundamentalMatrix Solutions)Becausethe column vector X = Xj (t)of the fundamental matrix \037 (t) in (2)satisfiesthe differential equation x' = Ax, it follows (from the definition of matrix multi-

plication) that the matrix X = \037 (t) itself satisfies the matrix differentialequationX' = AX. Becauseits column vectors are linearly independent, it alsofollows that

the fundamental matrix \037(t) is nonsingular, and therefore has an inverse matrix

\037(t)-l. Conversely,any nonsingular matrix solution \\II(t) of Eq.(1')has linearlyindependentcolumn vectors that satisfy Eq.(1),so \\II (t) is a fundamental matrix forthe system in (1).

In terms of the fundamental matrix \037(t) in (2),the general solution)

\037) X(t) = CIXI (t)+C2X2(t) + ...+ cnxn(t)) (3))

of the systemx' = Ax can be written in the form)

\037) X(t) = \037(t)c) (4))

where c = [CI C2 cn]Tis an arbitrary constant vector. If \\II(t) is any otherfundamental matrix for (1),then each column vectorof \\II(t) is a linear combinationof the column vectors of \037(t), so it follows from Eq.(4) that)

\\II(t) = \037(t)C) (4'))

for somen x n matrix C of constants.In orderthat the solutionx(t) in (3)satisfy a given initial condition)

X(O) = Xo,) (5))

it sufficesthat the coefficientvectorc in (4)be such that \037 (O)c= Xo; that is, that)

c = \037(O)-IXO.) (6))

When we substitute (6)in Eq.(4),we get the conclusionof the following theorem.)

THEOREM1 FundamentalMatrixSolutions)Let \037(t) be a fundamental matrix for the homogeneouslinear systemx' = Ax.Then the [unique]solution of the initial valueproblem)

\037) x' = Ax, x(O)= xo) (7))

is given by)

\037) x(t) = \037(t)\037(O)-lXO.) (8))))

Example1)


Section5.4tellsus how to find a fundamental matrix for the system)

\037) X' = Ax) (9))

with constant n x n coefficientmatrix A, at least in the casewhere A has a com-pletesetof n linearly independenteigenvectorsVI, V2, ..., Vn associatedwith the

(not necessarilydistinct) eigenvaluesAI, A2, ..., An, respectively.In this event the

correspondingsolutionvectorsof Eq.(9)are given by)

( )A\"t

Xi t = Vie l)

for i = 1,2,..., n. Therefore, the n x n matrix)

\037) ()(t) = vleA}t V2 eA2t) VneAn t) (10))

having the solutions Xl, X2, ..., Xn as column vectors is a fundamental matrix forthe systemx'= Ax.

In order to apply Eq. (8),we must be able to compute the inverse matrix

()(O)-l.The inverseof the nonsingular 2 x 2 matrix)

A=[\037 \037])

IS)

A-I = \037

[d -b

]\037 -c a') (11))

where \037 = det(A) = ad-bc i= O. The inverseof the nonsingular 3 x 3 matrix

A = [ai}]is given by)

T+All -A12 +A13

1A-I = _ -A21 +A22 -A23 (12)

\037

+A31 -A32 +A33)

where \037 = det(A) i= 0 and Ai} denotesthe determinant of the 2 x2 submatrix of A

obtainedby deleting the ith row and jth column of A. (Donot overlookthe symbolT for transposein Eq.(12).)The formula in (12)isalsovalid upon generalizationton x n matrices, but in practiceinversesof larger matricesare usually computedin-steadby row reductionmethods(seeany linear algebra text) or by using a calculatoror computer algebra system.)

......

Find a fundamental matrix for the system)

x'=4x+2y,y' = 3x - y,)

(13))

then use it to find the solution of (13)that satisfiesthe initial conditionsx(0)= 1,y(O)=-l.)))


Solution The linearly independent solutions)

[-2t

]XI (t) = _3:-21) [2e5t

]and X2 (t) == e5t)

found in Example1 of Section5.4yield the fundamental matrix)

[e-2t 2e5t

]()(t) == -3e-2t e5t .) (14))

Then)

.(0)=[-\037 ;l)

and the formula in (11)gives the inversematrix)

.(0)-1=\037

[\037-;J.)

(15))

Hencethe formula in (8)gives the solution)

X(t)=[_3:=\037:2:\037:](\037 )[\037 -7][-\037]=(\037 )[-3:=\037:2:\037:][;l)

and so)

1

[3e-2t+4e5t

]x(t) ==7 _ge-2t +2e5t .)

Thus the solution of the original initial value problem is given by)

x(t) ==\037e-2t +

\037e5t, yet) ==_\037e-2t +

\037e5t.).)

Remark:An advantageof the fundamental matrix approach is this: Oncewe know the fundamental matrix ()(t)and the inversematrix ()(O)-l,we can calcu-late rapidly by matrix multiplication the solutionscorrespondingto different initial

conditions.For example,supposethat we seekthe solution of the system in (13)satisfying the new initial conditions x(0) == 77,y (0) == 49.Then substitution of(14)and (15)in (8)gives the new particular solution)

X(t) =\037 [-3:=\037:2:\037:][\037 -;][\037\037]

1

[e-2t 2e5t

] [-21

]_

[-3e-2t+80e5t

].==

7 -3e-2t e5t 280 - ge-2t +40e5t .)))

Example2)


ExponentialMatricesWe now discussthe possibilityof constructinga fundamental matrix for the constant-coefficient linear system x' = Ax directly from the coefficientmatrix A-that is,without first applying the methodsof earlier sectionsto find a linearly independentsetof solutionvectors.

We have seenthat exponentialfunctions playa central role in the solution oflinear differentialequations and systems,ranging from the scalar equationx' = kxwith solution x(t) = xoekt to the vector solution x(t) = veAt of the linear systemx' = Ax whosecoefficientmatrix A has eigenvalueA with associatedeigenvectorv. We now defineexponentialsof matrices in such a way that)

X(t) = eAt)

is a matrix solutionof the matrix differentialequation)

X/=AX)

with n x n coefficientmatrix A-inanalogy with the fact that the ordinary expo-nential function x(t) = eat is a scalarsolutionof the first-orderdifferential equationx' = ax.

The exponential eZ of the complexnumber z may be defined (as in Section2.3)by means of the exponentialseries)

Z2 Z3 zneZ = 1 + z +-+-+ ...+-+ ....2! 3! n!)(16))

Similarly, if A is an n xn matrix, then the exponentialmatrixeA is the n xn matrix

definedby the series)

\037)

A 2 AneA = I+A +-+ ...+-+ ...2! n!)(17))

where I is the identity matrix. The meaning of the infinite serieson the right in (17)is given by)

00An

(k A n

)-- lim -

\037 n!-hoc

\037 n!)(18))

where A 0 = I,A 2 = AA, A 3 = AA 2, and soon; inductively, An+l = AAn if n > o.It can be shown that the limit in (18)existsfor every n x n square matrix A. Thatis, the exponentialmatrix eA is defined (by Eq.(17))for everysquarematrix A.)

Considerthe 2 x 2 diagonal matrix)

A=[\037 \037J.)

Then it is apparent that)

An = [a; \037])))


for each integer n > 1.It therefore follows that

A2eA = I+A +-+ ...2!

=[\037 \037]

+[\037 \037]

+[

a2

b2!

b2\0372! ]

+..\302\267

=[

1 +a +ao2/2!+...

\037 ].

1 +b +b j2!+ ...)

Thus)

A

[ea 0

]e =0 eb ,

so the exponentialof the diagonal2 x 2 matrix A is obtained simply by exponenti-ating eachdiagonal element of A. .

The n x n analog of the 2 x 2 result in Example2 is establishedin the sameway. The exponentialof the n x n diagonal matrix)

al 00 a2D=0 0

is the n x n diagonal matrix

ea} 0

eD =0 ea2

0 0)

oo)

(19))

an)

oo)

(20))

ean)

obtained by exponentiatingeach diagonal element of D.The exponential matrix eA satisfiesmost of the exponential relations that are

familiar in the caseof scalarexponents.For instance, if 0is the n x n zero matrix,then Eq.(17)yields)

eO= I,) (21))

the n x n identity matrix. In Problem 31we askyou to show that a useful law ofexponentsholdsfor n x n matrices that commute:)

If AB = BA, then eA+B = eAeB.In Problem 32we askyou to concludethat

(eA)-1= e-A

.)

(22))

(23))

In particular, the matrix eA is nonsingular for every n x n matrix A (reminiscentof the fact that eZ i= 0 for all z). It follows from elementary linear algebra that thecolumn vectors of eA are always linearly independent.

If t is a scalarvariable, then substitution of At for A in Eq.(17)gives

At 2 t 2n t n

e = I+At +A -+ ...+A -+ ... . (24)2! n!(Of course,At is obtained simply by multiplying eachelement of A by t.))))

Example3)

Example4)


If

A=[g

3

\037],

00

then

[0\302\260

18] [0\302\260

g],A 2 = 0 0 0 and A 3 = 0 0

o 0 0 o 0)

soAn = 0for n >3.It thereforefollows from Eq.(24)that)

eAt = 1+A t + 4A 2 t 2

=[g \037

\037]

+[g g

\037]

t + 1[g g Ig}2;)

that is,)

[1 3t 4t +9t2

]eAt = 0 1 6t .

o 0 1)

.)

Remark:If An = 0for somepositiveintegern, then the exponentialseriesin (24) terminates after a finite number of terms, so the exponentialmatrix eA (oreAt) is readilycalculatedas in Example3.Sucha matrix-witha vanishing power-is saidto benilpotent. .)

If

[2

3

\037],

A= 0 2o 0

then

A=[g

\302\260

0] [0

3

4]2 0 + 0 o 6 =D+B020o 0)

where D = 21is a diagonal matrix and B is the nilpotent matrix of Example3.Therefore, (20)and (22)give)

eAt = e(D+B)t = eDteBt =[

e

o\037t e0

0

2t\037

] [b

31t 4t

\037t9t2

].

e2t 0 0 1')

thus)

[e2t

eAt =\037)

(4t +9t2)e2t

]6te2t .e2t)

.)3t e2t

e2t

o)))


Example5)

Matrix ExponentialSolutionsIt happensthat term-by-term differentiationof the seriesin (24)is valid, with theresult)

d 2

(2

)At 2 3 t 2t-

(e ) ==A +A t +A -+ ...==A I+At +A -+ ... .dt 2! 2!')

that is,)

\037)

d_(eAt) ==AeAt

,dt)

(25))

in analogy to the formula Dt (ekt

) == kekt from elementary calculus.Thus thematrix-valuedfunction)

\037) X(t) ==eAt)

satisfiesthe matrix differentialequation)

\037) X' ==AX.)

Becausethe matrix eAt is nonsingular, it follows that the matrix exponentialeAt isa fundamental matrix for the linearsystem x' == Ax. In particular, it is the funda-mental matrix X(t) such that X(O)==I.Therefore,Theorem 1 impliesthe followingresult.)

THEOREM2 MatrixExponentialSolutionsIf A is an n x n matrix, then the solution of the initial value problem)

\037) x' ==Ax,) x(O)==Xo) (26))

is given by)

\037) x(t) ==eAtxo,) (27))

and this solution is unique.)

Thus the solution of homogeneouslinear systemsreducesto the task of com-puting exponentialmatrices.Conversely,if we already know a fundamental matrix

()(t) for the linear system x' == Ax, then the facts that eAt == ()(t)C (by Eq. (4/))

and eAoO ==eO==I (the identity matrix) yield)

eAt == ()(t)()(O)-l.) (28))

Sowe can find the matrix exponentialeAt by solving the linear systemx' ==Ax.)

..........

In Example1 we found that the systemx' ==Ax with)

A= [j -i])))


has fundamental matrix)

\302\253)(t)=

[-3:=\037:2:\037:]with \302\253)(0)-1

=\037

[\037 -i].)HenceEq.(28)gives)

eAt =\037 [-3:=\037:2:\037:][\037 -i])1

[e-2t+6e5t

=\"7 -3e-2t +3e5t)

Example6)

-2e-2t +2e5t

] .6e-2t +e5t).)

Usean exponentialmatrix to solve the initial value problem)

[2 3 4

]x'= 0 2 6 x,002) x(O)= [\037\037 ]

.) (29))

Solution The coefficientmatrix A in (29)evidently has characteristicequation (2-A)3 == 0and thus the triple eigenvalue A = 2,2,2. It is easy to seethat the eigenvectorequation)

(A -2I)v=[g g

\037] [\037 ]=

[g])has (to within a constant multiple) the singlesolutionv = [1 0 O]T.Thus thereis only a singleeigenvectorassociatedwith the eigenvalueA = 2,and sowe do not

yet have the three linearly independent solutionsneededfor a fundamental matrix.But we note that A is the samematrix whosematrix exponential)

eAt =)e2t 3te2t

o e2t

o 0)

(4t +9t2)e2t

6te2t

e2t)

was calculated in Example4.Hence,using Theorem 2,the solution of the initial

value problemin (29)is given by)

e2t 3t e2t (4t +9t2)e2t 19x(t) = eAtx(O) = 0 e2t 6te2t 29

0 0 e2t 39

(19+243t +351t2)e2t

- (29+234t)e2t .3ge2t)))


Remark:The sameparticular solutionx(t)as in Example6 couldbefound

using the generalizedeigenvectormethodof Section5.6.Onewould start by findingthe chain of generalizedeigenvectors)

VI =[lg],

V2 =[i], V3 =

[\037])

correspondingto the triple eigenvalueA = 2 of the matrix A. Then one would-using Eqs.(27)in Section5.6-assemblethe linearly independentsolutions)

Xl (t) = Vle2t , X2(t) = (Vlt +V2)e

2t, X3(t) = (4Vlt2 +V2t +V3) e2t)

of the differential equation x' = Ax in (29). The final step would be to deter-mine values of the coefficients Cl, C2, C3 so that the particular solution x(t) =CIXI(t)+ C2X2(t) + C3X3(t) satisfies the initial condition in (29). At this point it

shouldbeapparentthat-especiallyif the matrix exponentialeAt is readilyavailable(for instance, from a computer algebra system)-themethod illustrated in Exam-ple 6 can well be more \"computation ally routine\" than the generalizedeigenvectormethod.)

GeneralMatrix ExponentialsThe relatively simplecalculation of eAt carriedout in Example4 (and usedin Ex-ample 6)was basedon the observation that if)

[2 3 4

]A= 0 2 6 ,002)

then A -21is nilpotent:)

[0 3 4

]3

(A -21)3= 0 0 6 -000) [g g g]= o.) (30))

A similar result holdsfor any 3 x 3 matrix A having a triple eigenvaluer, in

which caseits characteristic equation reducesto (A - r)3 = O. For such a matrix,an explicitcomputationsimilar to that in Eq.(30)will show that)

(A - r 1)3= o.) (31))

(Thisparticular result isa specialcaseof the Cayley-Hamilton theoremof advancedlinear algebra,accordingto which everymatrix satisfiesits own characteristicequa-tion.)Thus the matrix A - r1is nilpotent, and it follows that)

eAt = e(rI+A-rI)t = erIt .e(A-rI)t = ertl.[I+ (A - r I)t + 4 (A - r 1)2t2], (32))

the exponential serieshere terminating becauseof Eq. (31).In this way, we canrather easilycalculate the matrix exponentialeAt for any square matrix having onlya singleeigenvalue.)))

Example7)


The calculation in Eq.(32)motivates a methodof calculatingeAt for any n xn

matrix A whatsoever.As we saw in Section5.6,A has n linearly independentgen-eralizedeigenvectorsU1,U2, ..., Un. Each generalizedeigenvectorU is associatedwith an eigenvalueA of A and has a rank r > 1 such that)

(A -AI)r U ==0) (A -AI)r-lu\037 o.) (33))but)

(If r == 1,then u is an ordinary eigenvectorsuch that Au ==AU.)Even if we donot yet know eAt explicitly,we can considerthe function x(t)==

eAt u, which is a linear combinationof the column vectors of eAt and is thereforeasolution of the linear system x' == Ax with x(O) == u. Indeed,we can calculatex

explicitly in terms of A, u, A, and r:)

x(t) ==eAtu ==e(AI+A-AI)t u ==eAIt e(A-AI)t u

[r-l

]==eAtl I+ (A -AI)t + ...+ (A -AI)r-1

t + ...u,(r - I)!)

so)

[2

At 2 tx(t) ==e u + (A -AI)ut + (A -AI) u-+ ...2!

t r-l

]+ (A -AI)r-lu ,(r - I)!)

(34))

using (33)and the fact that eAIt ==eAtl.

If the linearly independent solutions Xl (t), X2(t), ...,xn(t) of x' == Ax arecalculatedusing (34)with the linearly independentgeneralizedeigenvectorsUI, U2,..., Un, then the n x n matrix)

()(t) == [Xl(t) X2(t) ...xn(t)]) (35))

is a fundamental matrix for the system x' == Ax. Finally, the specificfundamental

matrix X(t) == ()(t)()(O)-1satisfies the initial condition X(O) ==I,and thus is the

desiredmatrix exponentialeAt. We have thereforeoutlined a proof of the followingtheorem.)

THEOREM3 Computationof eAt)

Let U1, U2, ...,Un be n linearly independent generalizedeigenvectorsof the

n x n matrix A. For each i,1< i <n, let Xi (t)be the solutionof x'==Ax givenby (34),substituting u == Ui and the associatedeigenvalueA and rank r of the

generalizedeigenvectorUi. If the fundamental matrix ()(t) is definedby (35),then)

\037) eAt == ()(t)()(O)-l.) (36))

Find eAt if)

A=[g g :].) (37))))


Solution Theorem 3 would apply even if the matrix A werenot upper triangular. But be-causeA is uppertriangular, this fact enablesus to seequickly that its characteristicequation is)

(5 -A)(3 -A)2 = o.)

Thus A has the distinct eigenvalueAl = 5 and the repeatedeigenvalueA2 = 3.)

CASE1:Al = 5. The eigenvectorequation (A -AI)u = 0 for u = [a b e]T)IS)

[-2

(A -51)u=\037) g -\037] [\037]

=[g]

.)

The last two scalarequations 4e = 0 and -2e= 0 give e = O. Then the first

equation-2a+4b = 1 is satisfiedby a = 2 and b = 1.Thus the eigenvalueAl = 5has the (ordinary) eigenvector UI = [2 1 0 ]T.The correspondingsolution ofthe systemx'= Ax is)

XI (t) = e5tu 1 = e5t[ 2 1 0 ]

T.) (38))

CASE2:A2 = 3. The eigenvectorequation (A -AI)u = 0 for u = [a b e]T)IS)

(A -31)u=[g \037

\037] [\037 ]=

[g].

The first two equations 4b +5e= 0 and 2b +4e = 0 imply that b = e = 0,but

leave a arbitrary. Thus the eigenvalueA2 = 3 has the single(ordinary)eigenvectorU2 = [1 0 0]T. The correspondingsolutionof the systemx'= Ax is)

X2 (t) = e3tU2 = e3t [1 0 0]T .) (39))

To look for a generalizedeigenvectorof rank r = 2 in Eq. (33),we considertheequation)

(A _ 31)2u=[g \037

1\037] [\037]

=[g]

.

The first two equations 8b + 16e= 0 and 4b + 8e = 0 are satisfied by b = 2and e = -1,but leave a arbitrary. With a = 0 we get the generalizedeigenvectorU3 = [0 2 -1]T of rank r = 2 associatedwith the eigenvalueA = 3.Because(A - 31)2U= 0,Eq.(34)yieldsthe third solution)

X3 (t) = e3t[U3 + (A - 31)U3t]

= e3t

([_\037]

+[g \037

\037] [_\037 ]t)

= e3t

[\037i]

. (40))

With the solutions listedin Eqs.(39)and (40),the fundamental matrix)

()(t) = [XI(t) X2(t) X3(t)])))

definedby Eq.(35)is)

[2e5t

\037(t) =e5\037)


3t e3t

]o 2e3t

o _e3t)

4>(0)-1==

[\037 -\037 -\037

].

o 0-1)e3t)

with)

HenceTheorem 3 finally yields)

eAt == 4>(t)4>(O)-1

[2e5t e3t 3te3t

] [0

1

-\037]

e5t 0 2e3t 1 -20 0 _e3t 0 0 -1

[e3t 2e5t -2e3t 4e5t - (4+3t)e3t

]== 0 e5t 2e5t -2e3t .

0 0 e3t .)

Remark:As in Example7,Theorem 3 sufficesfor the computation of eAt

provided that a basisconsistingof generalizedeigenvectorsof A can be found. .)

_ Problems)

Find afundamental matrix ofeachof the systems in Problems1through 8, then apply Eq. (8)to find a solution satisfying the

given initial conditions.)

1.x'= [i2.x'=

[_\037

-\037 ]x,

[2 -5

]3.x'==4 _ 2 x,

[3 -1

]4. x'==1 1 x,

5.x'= [-

\037 -;]x, x(O)= [_\037 ]

6.x'=[\037 -;]x, x(O)=

[\037 ]7. x'==

[; -

\037 =\037

]x, x(O)==

[i]4 -2 -4 0

8.x'==

[-; -\037 -;

]x, x(O)==

[\037

]5 5 3 -1)

\037 ]x,) x(O)= [_\037 ]

x(O)= [_i]x(O)= [ \037 ]x(O)= [ \037 ])

Compute the matrix exponential eAt for eachsystem x' == Ax

given in Problems9 through 20.)

9.x\037

== 5XI - 4X2,x\037

== 2XI -X2

10.x\037

== 6XI - 6X2,x\037

== 4XI - 4X211.

x\037

== 5XI - 3X2, x\037

== 2XI

12.x\037

== 5XI - 4X2,x\037

== 3XI - 2X2)

13.x\037

== 9XI - 8X2, x\037

== 6XI - 5X214.

x\037

== 10XI - 6X2, x\037

== 12xI- 7X2

15.x\037

== 6XI - 10x2,x\037

== 2XI - 3X2

16.x\037

== 11xI- 15x2,x\037

== 6XI - 8X2

17.x\037

== 3XI + X2, x\037

== Xl + 3X218.

x\037

== 4XI + 2X2,x\037

== 2XI + 4X219.

x\037

== 9XI + 2X2, x\037

== 2XI + 6X220.

x\037

== 13xI+ 4X2,x\037

== 4XI + 7X2)

In Problems21through 24, show that the matrix A is nilpo-tent and then usethis fact to find (asin Example 3) the matrix

exponential eAt.)

21.A =[ \037

23.A =[1)

-1]-1

=1 -1])

22. A =[ _\037

24.A=[\037)

-:]o -3

]o 7o -3)

Each coefficient matrix A in Problems 25 through 30 is the

sum ofa nilpotent matrix and a multiple ofthe identity matrix.

Usethis fact (asin Example 6) to solvethe given initial value

problem.

25.x'=[\037 ;]x, x(O)= [ \037 ]

26.x'= [ 1\037

\037]

x, x(O)= [-1\037 ]27.x'=

[\037!

\037]

x, x(O)=[ \037

])))


[5 0 0

]28.x'= 10 5 0 x,

20 30 5

[

1 2 3 4

]I 0 1 6 329.x =

0 0 1 2 x,000 1

[

3 0 0 0

] [

1

],6300 130.x =

9 6 3 0 x, x(O)= 112 9 6 3 131.Supposethat the n x n matrices A and B commute; that

is, that AB = BA. Prove that eA+B = eAeB. (Suggestion:Group the terms in the product of the two serieson the

right-hand sideto obtain the serieson the left.)32.Deducefrom the result of Problem 31that, for ev-ery square matrix A, the matrix eA is nonsingular with

(eA

)-I = e-A.

33.Supposethat)

x(O)=[ \037\037 ]

x(O)=[!])

A=[\037 \037J.

Show that A 2n = I and that A 2n+I = A if n is a positiveinteger. Concludethat

eAt = Icosht + A sinh t,)

_ Nonhomogeneous:LinearSystems)

and apply this fact to find a general solution of x'= Ax.

Verify that it is equivalent to the generalsolution found bythe eigenvalue method.

34. Supposethat)

A = [ -\037 \037 J.Show that eAt = Icos2t +

\037

A sin 2t. Apply this fact tofind a general solution of x' = Ax, and verify that it isequivalent to the solution found by the eigenvalue method.)

Apply Theorem 3 to calculatethe matrix exponential eAt foreachof the matrices in Problems35 through 40.

35.A =[ \037

\037] 36.A=[\037

2

1]10

37.A =[\037

3

\037] 38.A=[\037

20 30

]1 10 200 0 5

39.A =[\037

3 3

!]40. A =[\037

4 4

4]1 3 2440 2 0240 0 003)

In Section2.5we exhibitedtwo techniques for finding a singleparticular solutionof a singlenonhomogeneousnth-order linear differentialequation-themethod ofundeterminedcoefficientsand the method of variation of parameters.Each of thesemay be generalizedto nonhomogeneouslinear systems. In a linear system mod-eling a physical situation, nonhomogeneousterms typically correspondto externalinfluences, such as inflow of liquid to a cascadeof brine tanks or an external forceacting on a mass-and-springsystem.

Given the nonhomogeneousfirst-orderlinear system)

\037) X' ==Ax + f(t)) (1))

where A is an n x n constant matrix and the \"nonhomogeneousterm\" f(t) is a givencontinuous vector-valuedfunction, we know from Theorem 4 of Section5.3that ageneral solution of Eq.(1)has the form)

\037)

where)

x(t) ==Xc (t)+xp (t ) ,) (2))

. xc(t)==CIXI (t) +C2X2(t) + ...+ cnxn(t) is a general solution of the associ-ated homogeneoussystemx' ==Ax, and. xp (t) is a singleparticular solutionof the originalnonhomogeneoussystem in

(1).)

Precedingsectionshave dealt with xc(t),soour task now is to find xp(t).)))

5.8NonhomogeneousLinearSystems 421)

UndeterminedCoefficients)First we supposethat the nonhomogeneousterm f(t) in (1)is a linear combination(with constant vector coefficients)of products of polynomials, exponentialfunc-tions, and sinesand cosines.Then the method of undetermined coefficientsfor

systemsis essentiallythe sameas for a singlelinear differentialequation.We makean intelligent guessas to the generalform of a particular solution xp, then attemptto determine the coefficientsin xp by substitution in Eq.(1).Moreover,the choiceof this general form is essentiallythe sameas in the caseof a singleequation (dis-cussedin Section2.5);we modify it only by using undeterminedvectorcoefficientsrather than undeterminedscalars.We will therefore confine the presentdiscussionto illustrative examples.)

Example1)

,... \037 \"\"\",... ,., '\" \" ,,\037'\" '\" ,.... \037.,.....,,.,.,,....,.... y\" '\" \"\"\" '\" \". '\037\"\"\"\"\"\"\" \"\"\"\",.,,...,, '\" '\" \"\"\" \037 \"\"',.... ,.... '\" \037 \"\"\"\"\"\"\037,,,.... \"\"\" ,.... \"\"\",A-'N \"\"\" ,... .....,.... ,....,....

Find a particular solutionof the nonhomogeneoussystem)

XI=[\037 ;]X+[2\037J.)(3))

Solution The nonhomogeneousterm f == [3 2t]T is linear, so it is reasonableto selecta

linear trial particular solutionof the form)

Xp(t) = at +b = [ \037\037 ] t + [ t\037 ] .) (4))

Uponsubstitution of x ==xp in Eq.(3),we get)

[al

]==

[3 2

] [alt +b l

] + [3

]a2 7 5 a2t +b2 2t)

_[

3al +2a2] + [

3bl +2b2 +3

]- 7al+5a2+2t 7bl +5b2 .)

We equate the coefficientsof t and the constant terms (in both XI- and X2-compon-ents)and thereby obtain the equations)

3al +2a2 ==0,7al +5a2+2 ==0,3bl +2b2 +3 ==aI,

7bl +5b2 ==a2.)

(5))

We solve the first two equationsin (5)for al ==4 and a2 ==-6.With thesevalueswecan then solve the last two equations in (5)for b l == 17and b2 ==-25.Substitution

of thesecoefficientsin Eq. (4)gives the particular solution x == [xI X2]T of (3)

describedin scalarform by)

Xl (t) == 4t + 17,X2(t) ==-6t-25.) .)))

422 Chapter5 LinearSystemsof Differential Equations)\037\" ,.,. \037

Figure 5.8.1showsthe systemof three brine tanks investigated in Example2 ofSection5.4.The volumes of the three tanks are VI == 20,V2 == 40, and V3 == 50(gal),and the common flow rate is r == 10(gal/min).Supposethat all three tanks

r (gal/min) containfresh water initially, but that the inflow to tank 1 isbrine containing2 poundsof salt per gallon, so that 20pounds of salt flow into tank 1 per minute. Referringto Eq.(18)in Section5.4,we seethat the vectorx(t) == [ Xl (t) X2(t) X3(t)]Tofamounts of salt (in pounds)in the three tanks at time t satisfiesthe nonhomogeneous

r initial value problem)

Example2)

dx _[-0.5-- 0.5

dt 0)

o-0.25

0.25)\037

]x+

[2\037

],

-0.2 0)x(O)=

[\037]

.) (6))

\037r

FIGURE5.8.1.Thethree brinetanks ofExample 2.)

The nonhomogeneous term f == [20 0 0]T here correspondsto the 20 lb/mininflow of salt to tank 1,with no (external) inflow of salt into tanks 2 and 3.

Becausethe nonhomogeneousterm is constant,we naturally selecta constanttrial function xp == [al a2 a3]T,for which

x\037= O.Then substitution of x ==xp

in (6)yieldsthe system)

[

0

] [-0.5

\037

=\037.5)

o 0

] [a I

] [

20

]-0.25 0 a2 + 00.25-0.2 a3 0)

that we readily solve for al == 40, a2 == 80,and a3 == 100in turn. Thus our

particular solution is xp (t) == [40 80 100]T.

In Example2 of Section5.4we found the general solution)

xc(t)= Cl

[-\037]

e-t /2 +C2

[-r]e-t /4 +C3

[\037]

e-t / 5)

of the associatedhomogeneoussystem,so a general solution x == Xc + xp of the

nonhomogeneoussystemin (6)is given by)

x(t) = Cl

[-\037]

e-t/2 +C2

[-r]e-t /4 +C3

[\037]

e-t/5 +[ li\037 ]

. (7))

When we apply the zero initial conditions in (6),we get the scalarequations)

3CI + 40 ==0,-6CI+ C2 + 80 ==0,

5CI- 5C2+ C3 + 100==0)

that are readily solved for CI == -\037o

, C2 == -160,and C3 == -25300. Substituting

thesecoefficients in Eq. (7),we find that the amounts of salt in the three tanks attime t are given by)

Xl (t) ==40-40e-t/2,

X2(t) ==80 +80e-t/2- 160e-t /4,

X3(t) == 100+1\0370 (_2e-t/2 +24e-t/4 -25e-t/5

) .)

(8))))

Example3)


As illustrated in Fig.5.8.2,we seethe salt in eachof the three tanks approaching,ast --++00,a uniform density of 2Ib/gal-thesame as the salt density in the inflow

to tank 1. .)

120)

100) X3(t)\037 100)

\037 60)

80)X2(t)

\037 80)

40) Xl(t) \037 40)

20)

oo 10 20 30 40 50 60

t)

FIGURE5.8.2.Thesalt amountsolution curvesdefined in (8).)

In the caseof duplicate expressionsin the complementary function and the

nonhomogeneous terms, there is one difference between the method of undeter-mined coefficientsfor systemsand for singleequations (Rule2 in Section2.5).Fora system,the usual first choicefor a trial solution must be multiplied not only bythe smallestintegral powerof t that will eliminate duplication,but alsoby all lower(nonnegativeintegral) powersof t as well, and all the resulting terms must be in-cludedin the trial solution.)

........ ......

Considerthe nonhomogeneoussystem)

I

[4 2

] [15

]-2t

X ==3 -1 x - 4 te .) (9))

In Example1 of Section5.4we found the solution)

[1

]-2t

[2

]5t

Xc (t) ==C1 _ 3 e +C2 1e) (10))

of the associatedhomogeneoussystem.A preliminary trial solutionxp(t) ==ate-2t+be-2t exhibits duplication with the complementary function in (10).We would

therefore selectxp(t)==at2e-2t +bte-2t +ce-2t

as our trial solution, and we would then have sixscalarcoefficientsto determine.Itis simplerto use the method of variation of parameters,our next topic. .)

VariationofParametersRecall from Section2.5that the method of variation of parametersmay be appliedto a linear differential equation with variable coefficientsand is not restricted to

nonhomogeneous terms involving only polynomials, exponentials,and sinusoidalfunctions. The methodof variation of parametersfor systemsenjoys the sameflexi-bility and has a concisematrix formulation that is convenientfor both practicalandtheoreticalpurposes.)))


We want to find a particular solutionxp of the nonhomogeneouslinear system)

\037) X' ==P(t )x+ f(t ) ,) (11))

given that we have already found a general solution)

\037) Xc(t) ==CIXI (t)+C2X2(t) + ...+ cnxn(t)) (12))

of the associatedhomogeneoussystem)

\037) x' ==pet)x.) (13))

We first use the fundamental matrix 4>(t) with column vectors Xl, X2, ..., Xn

to rewrite the complementaryfunction in (12)as)

Xc (t) == 4>(t )c,) (14))

where cdenotesthe column vector whoseentries are the coefficientsCI,C2,..., Cn.Our ideais to replacethe vector \"parameter\" cwith a variablevectoru(t).Thus weseeka particular solution of the form)

\037) Xp(t) == 4>(t)u(t).) (15))

We must determineu(t) so that xp does,indeed,satisfy Eq.(11).The derivativeof Xp (t) is (by the product rule))

X\037(t) == 4>'(t)u(t)+ 4>(t)u'(t).) (16))

Hencesubstitution of Eqs.(15)and (16)in (11)yields)

4>'(t)u(t)+ 4>(t)u'(t) ==P(t)4>(t)u(t)+ f(t).) (17))

But)

4>/(t) ==P(t)4>(t)) (18))

becauseeachcolumn vector of 4>(t) satisfiesEq.(13).Therefore, Eq.(17)reducesto)

4>(t)u'(t) ==f(t).) (19))

Thus it sufficesto chooseu(t) so that)

u'(t) == 4>(t )-1

f( t ) ;) (20))

that is, so that)

u(t) = f \037(t)-lf(t) dt.) (21))

Uponsubstitution of (21)in (15),we finally obtain the desiredparticular solution,as stated in the following theorem.)))


THEOREM1 Variationof Parameters)If 4>(t) is a fundamental matrix for the homogeneoussystem x' == P(t)xonsomeinterval where pet) and f(t) are continuous, then a particular solution ofthe nonhomogeneoussystem)

x' ==P(t)x+ f(t))

is given by)

\037) Xp(t) = \037(t) f \037(t)-lf(t) dt.) (22))

This is the variation of parametersformula for first-orderlinearsystems.Ifwe add this particular solution and the complementaryfunction in (14),we get the

general solution)

\037) X(t) = \037(t)c + \037(t) f \037(t)-lf(t) dt) (23))

of the nonhomogeneoussystem in (11).The choiceof the constant of integration in Eq. (22)is immaterial, for we

needonly a singleparticular solution. In solving initial value problemsit often isconvenientto choosethe constantof integration sothat xp(a)==0, and thus integratefrom a to t:)

Xp(t) = \037(t) 1t

\037(s)-lf(s)ds.) (24))

If we add the particular solutionof the nonhomogeneousproblem)

x' ==P(t)x+f(t), x(a) ==0)

in (24)to the solutionxc(t)== 4>(t)4>(a)-lxa of the associatedhomogeneousprob-lem x' ==pet)x,x(a) ==Xa, we get the solution)

X(t) = \037(t)\037(a)-lXa + \037(t) 1t

\037(s)-If(s)ds) (25))

of the nonhomogeneousinitial value problem)

x' ==P(t )x+ f(t ) , x(a) ==xa.) (26))

Equations (22)and (25)hold for any fundamental matrix 4>(t) of the homo-geneoussystem x' == P(t)x.In the constant-coefficientcasepet) = A we can usefor 4>(t) the exponentialmatrix eAt-that is, the particularfundamental matrix suchthat 4>(0)== I. Then, because(eAt)-l == e-At , substitution of 4>(t) == eAt in (22)yieldsthe particular solution)

Xp(t) = eAt f e-Atf(t) dt) (27))))


of the nonhomogeneoussystemx'= P(t)x+f(t). Similarly, substitution of 4>(t) =eAt in Eq.(25)with a = 0 yieldsthe solution)

\037) x(t) = eAtxo + eAt it e-Atf(t) dt) (28))

of the initial value problem)

\037) x'= P(t)x+f(t), x(O)= xo.) (29))

Remark:If we retain t asthe independentvariablebut uses for the variableof integration, then the solutions in (27)and (28)can be rewritten in the forms)

xp(t)= f e-A(s-t)f(s)ds and x(t)= eAt:xo +it e-A(s-t)f(s)ds. .)

Example4).\".,_.\",\037 \"',\",,0'. \"0\"_ \037 ... \"\"n., \"\"' \037\037v\"'''\"''' \" ,,,...


,

[4 2

] [15

]-2t

X =3 -1 x- 4 te ,) x(O)=

[\037J.)(30))

Solution The solutionof the associatedhomogeneoussystemisdisplayedin Eq.(10).It givesthe fundamental matrix)

[e-2t 2e5t

]4>(t) = -3e-2t e5t)with \037(O)-l=

\037

[\037-i] .)

It follows by Eq. (28)in Section5.7that the matrix exponential for the coefficientmatrix A in (30)is)

At -1

[e 2t 2e5t

]. 1-

[1 -2

]e = \037(t)\037(O) = -3e=2t e5t 7 3 1)

1

[e-2t+6e5t

=\"7 -3e-2t +3e5t)

-2e-2t +2e5t

] .6e-2t + e5t)

Then the variation of parameters formula in Eq.(28)gives)

e-Atx(t)= Xo +it e-ASf(s)ds)

[7

]{t1

[e2s+6e-5s

=3 +10'7 -3e2s+3e-5s)

-2e2s+2e-5s

] [-15se-2s

] ds6e2s+ e-5s -4se-2s)

=[

7

] +it

[-s- 14se=7S] ds3 3s- tse 7s

o)

[7

]1

[-4- 7t 2 +4e-7t +28te-7t

]=

3 + 14 -2+21t2 +2e-7t + 14te-7t .)))


Therefore,)-At 1

[94 - 7t2 +4e-7t +28te-7t

]e x(t)= 14 40+21t2 +2e-7t + 14te-7t .

Uponmultiplication of the right-hand sidehere by eAt, we find that the solution ofthe initial value problem in (30)is given by)

[e-2t+6e5t

x t -.!( ) - 7 -3e-2t +3e5t)

-2e-2t +2e5t

]1

[94 - 7t 2 +4e-7t + 28te-7t

]6e-2t + e5t .14 40+21t2 +2e-7t + 14te-7t)

1

[(6+28t -7t 2)e-2t +92e5t

]= 14 (-4+14t+21t2)e-2t+46e5t.) .)In conclusion,let us investigatehow the variation of parameters formula in

(22)\"reconciles\"with the variation of parameters formula in Theorem 1 of Sec-tion 2.5for the second-orderlinear differentialequation)

y\" + Py' + Qy = f(t).) (31))

If . I I \"\" I h h.

I..

(31).we wrIte y = Xl, Y =

Xl= X2, Y =

Xl= X2, t en t e sIng eequatIonIn IS

equivalentto the linear system X\037

= X2,X\037

= -QXI -PX2 +f (t), that is,)

x'= P(t)x+ r(t),) (32))

where)

x=[\037\037 ]

=[\037I J. P(t) = [-3 -\037 J. and f(t) = [ f\037t) J.)

Now two linearly independentsolutionsYI and Y2 of the homogeneoussystemy\" +Py' +Qy = 0 associatedwith (31)provide two linearly independentsolutions

Xl =[\037;]

and X2 =[ \037\037 ])

of the homogeneoussystem x' = P(t)x associatedwith (32). Observethat the

determinant of the fundamental matrix 4> = [ Xl X2] is simply the Wronskian)

w = YI Y2

Y\037 Y\037)

of the solutions YI and Y2, so the inversefundamental matrix is)

14>-1= _

W)

Y2'

Y- 2

-Y\037 YI)

Therefore the variation of parametersformula xp = 4>f 4>-lrdt in (22)yields)

[ y\037 ]=

[ Y\037 Y\037 ]f\037

[ _Y\037

-Y2

] [0

]dt

Yp YI Y2 W YI YI f

=[\037; ;\037]f \037 [ -\037:;]

dt.)))


The first componentof this column vector is)

Yp= [ Yl) ] J

I

[-

Y2f] J Y2f J Y 1f

Y2- dt = -Yl -dt+ Y2 -dt.W Yl f W W)

If, finally, we supply the independent variable t throughout, the final result on the

right-hand sidehere is simply the variation of parameters formula in Eg.(33)ofSection2.5(where,however,the independentvariable is denotedby x).)

_ Problems)

Apply the method of undetermined coefficientsto find a par-ticular solution ofeachof the systems in Problems1through14.If initial conditions are given, find the particular solutionthat satisfiestheseconditions. Primes denotederivatives with

respectto t.)

1.x'= x + 2y + 3,y' = 2x + y-2

2.x'= 2x + 3y + 5, y' = 2x + y- 2t

3.x'= 3x + 4y, y' = 3x + 2y + t 2; x(O)= y(O)= 04. x'= 4x + y + et , y' = 6x- y

-et ; x(O)= y(O)= 15.x'= 6x-7y + 10,y' = x - 2y -2e-t6.x'= 9x+ y + 2et , y' = -8x-2y + tet7. x'= -3x+4y + sint, y' = 6x-5y; x(O)= 1,y(O)= 08.x'= x -5y + 2 sin t, y' = X -

Y- 3cost

9.xl=x-5y+cos2t,y'=X-y10.x'= x -2y, y' = 2x - y + et sin t11.x'= 2x+ 4y + 2,y' = X + 2y + 3;x(O)= 1,y(O)= -112.x'= x + y + 2t, y' = X + y

-2t13.x'= 2x+ y + 2et , y' = X + 2y - 3et

14.x'= 2x + y + 1,y' = 4x + 2y + e4t)

Problems 15and 16are similar to Example 2, but with twobrine tanks (having volumes VI and V2 gallonsas in Fig. 5.8.2)instead ofthree tanks. Eachtank initially containsfreshwater,and the inflow to tank 1at the rate ofr gallonsperminute hasa salt concentration of Copounds per gallon. (a) Find theamounts Xl (t) and X2(t) of salt in the two tanks after t min-utes. (b) Find the limiting (long-term) amount ofsalt in eachtank. (c) Find how long it takesfor eachtank to reacha saltconcentration of 1 lb /gal.15.VI = 100,V2 = 200,r = 10,Co= 216.VI = 200,V2 = 100,r = 10,Co= 3)

In Problems17through 34,usethe method ofvariation ofpa-rameters (andperhapsa computer algebrasystem) to solvetheinitial value problem

x'= Ax + r(t), x(a)= Xa.

In eachproblem weprovide the matrix exponential eAt aspro-vided by a computer algebrasystem.

17.A =[\037

=\037 1f(t) =[ \037\037 1x(O)= [ \037 1

eAt = 1.[-e-t+ 7e5t 7e-t -7e5t

]6 -e-t + e5t 7e-t _ e5t)

18.RepeatProblem 17,but with f(t) replacedwith [ \037\037\037 119.A =

[\037_;1f(t) =

[ 18g\037 1x(O)=[\037l

At I

[e-3t+ 4e2t -2e-3t + 2e2t

]e =:5 _2e-3t + 2e2t 4e-3t + e2t

20.RepeatProblem 19,but with f(t) replacedwith[75e\0371

21.A =[\037

=\037 1f(t) =D\037:\037: 1x(O)=

[\037l

[-e-t+ 5e3t e-t _ e3t

]eAt = 1.4 -5e-t + 5e3t 5e-t -e3t

22.RepeatProblem21,but with f(t) replacedwith[;\037::t].

23.A =[\037 =n,f(t) =

U1x(O)= U1eAt =

[1+ 3t -t

]9t 1- 3t

24. RepeatProblem 23, but with f(t) =[t\0372 ] and x(I)=

Ul25.A = [ \037 =;1f(t) = [4:1x(O)= [ \037 1

eAt =[

cost\037

2sin t -5 sint. ]SIn t cost -2SIn t

26.RepeatProblem25,but with f(t) = [ \037 \037\037:: ] and x(O)=

Ul27.A =

[\037=\037 1f(t) =

[3\037:2lx(O)=

[\037l

eAt =[

1+ 2t -4t]t 1-2t

28.RepeatProblem27,but with f(t) =[ 4t\037\037t ] and x(1)=

[ -\0371)))


[0 -1

] [seet

] [0]29.A =

1 0'f(t) =0 'x(O)= 0 '

At _[

eost - sin t

]e - .SIn t eost

[0 -2

] [t eos2t

] [0]30.A =

2 0'f(t) =t sin 2t 'x(O)= 0 '

eAt =[ e\037s

2t - sin 2t

]SIn 2t eos2t

31.A =[

\037 i \037

], f(t) =

[\037

], x(O)=

[\037

],

o 0 1 6et 0

[et 2tet (3t + 2t2)et

]eAt = 0 et 2tet

o 0 et

32.A =[

\037 \037 \037

], f(t) =

[\037

], x(O)=

[\037

],

o 0 2 2e2t 0)

[et 3tet

eAt = 0 et

o 0

33.A =[

\037 \037 \0370001 4t 8t + 6t2 32t2 + 8t3

o 1 3t 8t + 6t2eAt =

o 0 1 4to 0 0 1

34. A =[

\037 \037 \037 \037

], f(t) =

[6\037

], x(O)=

[i]

,

o 0 0 2 e2t 1

1 4t 4(-1+ e2t ) 16t(-1+ e2t )o 1 0 4(-1+ e2t )

eAt =o 0 e2t 4te2t

o 0 0 e2t)

(-13-9t)et + 13e2t

]-3et + 3e2t

e2t

\037 l f(f) = 30

[ \037 lx(O)=[\037l)))

NulllericalMethods)

11:IIN\037\037erical\037J?!\037\037.imation:Euler'sMet\037\"\037d)

It is the exceptionrather than the rule when a differential equation of the generalform)

\037)

dydx

==I(x,y))

can be solved exactly and explicitlyby elementarymethods likethose discussedinChapter 1.For example,considerthe simpleequation)

dy -x2--edx- .) (1))

A solution of Eq. (1)is simply an anti derivativeof e-x2.But it is known that everyanti derivative of I(x) == e-x2 is a nonelementaryfunction-onethat cannot beexpressedas a finite combination of the familiar functions of elementarycalculus.Henceno particular solutionof Eq.(1)is finitely expressiblein terms of elementaryfunctions. Any attempt to usethe symbolictechniquesof Chapter 1 to find a simpleexplicitformula for a solution of (1)is therefore doomedto failure.

As a possiblealternative,an old-fashionedcomputerplotter-onethat usesanink pento draw curvesmechanically-canbeprogrammedto draw a solutioncurvethat starts at the initial point (xo,Yo) and attempts to thread its way through the slopefieldof a given differentialequation y' ==I(x,y). The procedurethe plottercarriesout can be describedas follows.). The plotter pen starts at the initial point (xo,Yo) and moves a tiny distance

along the slopesegment though (xo,Yo). This takesit to the point (Xl, YI).. At (Xl, YI) the pen changesdirection, and now moves a tiny distancealongthe slopesegmentthrough this new starting point (Xl, YI). This takesit to thenext starting point (X2, Y2).)

430)))

y)

FIGURE6.1.1.Thefirst few

steps in approximating a solutioncurve.)

(Xn+ 1,Yn+ 1))

hf(xn, Yn))

(Xn, Yn))h) (Xn+ 1,Yn))

FIGURE6.1.2.Thestep from(X n , Yn) to (Xn+l,Yn+l).)

6.1NumericalApproximation:Euler'sMethod 431)

\302\267 At (X2, Y2) the pen changesdirection again, and now moves a tiny distancealong the slopesegment through (X2, Y2). This takes it to the next starting

point (X3, Y3).)

x)

Figure 6.1.1illustrates the result of continuing in this fashion-by a sequenceof discretestraight-line stepsfrom one starting point to the next. In this figure weseea polygonalcurveconsistingof line segmentsthat connect the successivepoints(xo,Yo), (Xl, YI), (X2, Y2), (X3, Y3), ....However,supposethat each \"tiny distance\"the pen travels along a slopesegment-beforethe midcoursecorrection that sendsit along a fresh new slopesegment-isso very small that the naked eye cannotdistinguish the individual line segmentsconstituting the polygonalcurve. Then the

resulting polygonal curve lookslikea smooth, continuously turning solution curveof the differential equation. Indeed,this is (in essence)how most of the solution

curves shown in the figuresof Chapter 1 werecomputergenerated.Leonhard Euler-thegreat 18th-centurymathematician for whom so many

mathematicalconcepts,formulas, methods, and results are named-didnot have acomputerplotter,and his ideawas to do all this numerically rather than graphically.In orderto approximatethe solutionof the initial valueproblem

dydx

= f(x,y), Y(XO) = Yo,)\037) (2))

we first choosea fixed (horizontal) step size h to use in making each step fromone point to the next. Supposewe've started at the initial point (xo,Yo) and aftern stepshave reachedthe point (xn, Yn). Then the step from (xn, Yn) to the next

point (Xn+I, Yn+I) is illustrated in Fig. 6.1.2.The slopeof the direction segmentthrough (xn, Yn) is m = f(xn,y,J. Hencea horizontal change of h from Xn toXn+I correspondsto a vertical change of m .h = h .f (xn, y,J from Yn to Yn+ 1.Therefore the coordinatesof the new point (xn+I,Yn+ I)are given in terms of the oldcoordinatesby)

Xn+1 = Xn +h, Yn+1 = Yn +h .f(xn,Yn).

Given the initial value problem in (2),Euler'smethodwith stepsizeh con-sistsof starting with the initial point (xo,Yo) and applying the formulas)

XI = Xo + h

X2 = XI + h

X3 = X2 + h)

YI = Yo + h .f(xo,Yo)

Y2 = Y I + h .f (x1 , Y 1)

Y3 = Y2 + h .f(X2,Y2))

to calculate successivepoints (XI, YI), (X2, Y2), (X3, Y3), ...on an approximateso-lution curve.

However,we ordinarily do not sketchthe correspondingpolygonalapproxi-mation. Instead,the numerical result of applying Euler'smethod is the sequenceofapproximations)

Y I , Y2, Y3, ..., Yn, ...)

to the true values)

Y(XI), Y(X2), Y(X3), ...,Y(Xn), ...at the points Xl, X2, X3, ..., Xn , ...of the exact (though unknown) solution y(x) ofthe initial valueproblem.Theseresultstypically are presentedin the form of a tableof approximatevaluesof the desiredsolution.)))

432 Chapter6 NumericalMethods)

Example1)

ALGORITHM TheEulerMethodGiven the initial value problem)

\037)

dy-= f(x,y), y(xo) = Yo,dx)

(2))

Euler'smethodwith stepsizeh consistsof applying the iterative formula)

\037) Yn+l = Yn + h .f(xn, Yn) (n >0)) (3))

to calculate successiveapproximationsYl, Y2, Y3, ...to the [true] values Y(XI),Y(X2), Y(X3), ...of the [exact] solution Y == y(x) at the points Xl, X2, x3, ...,respectively.)

The iterative formula in (3)tellsus how to make the typical stepfrom Yn to

Yn+l and is the heart of Euler'smethod. Although the most important applicationsof Euler'smethod are to nonlinear equations, we first illustrate the method with asimpleinitial value problemwhoseexact solution is available,just for the purposeof comparisonof approximateand actual solutions.)

o 0

Apply Euler'smethod to approximatethe solutionof the initial valueproblem

dy 1

dx ==x+5 y , y(O)==-3,

(a) first with stepsizeh == 1 on the interval [0,5],(b) then with stepsizeh ==0.2on the interval [0,1].)

(4))

Solution (a) With Xo ==0,Yo == -3,f(x,y) == X +\037Y,

and h == 1 the iterative formula in

(3)yieldsthe approximatevalues

YI == Yo + h .[xo+\037Yo]

==(-3)+ (1)[0+ \037(-3)]==-3.6,Y2 == YI + h . [Xl +

\037YI]==(-3.6)+ (1)[1+ \037(-3.6)]==-3.32,

Y3 == Y2 + h . [X2 +\037Y2]

==(-3.32)+ (1)[2+ \037(-3.32)]==-1.984,Y4 == Y3 + h . [X3 +

\037Y3]==(-1.984)+ (1)[3+ \037(-1.984)]==0.6192,and

Ys == Y4 + h . [X4 +\037Y4]

==(0.6912)+ (1)[4+ \037(0.6912)]\037 4.7430at the points Xl == 1,X2 == 2,X3 == 3,X4 == 4, and Xs == 5. Note how the result ofeachcalculation feedsinto the next one.The resulting table of approximatevaluesIS)

X 0 1 2 3 4 5

Approx.Y -3 -3.6 -3.32 -1.9840.69124.7430)

Figure 6.1.3showsthe graph of this approximation,together with the graphsof the Euler approximationsobtained with stepsizesh == 0.2and 0.05,as well asthe graph of the exactsolution

y(x) ==22ex/s -5x- 25)

that is readily found using the linear-equationtechniqueof Section1.5.We seethat

decreasingthe stepsizeincreasesthe accuracy,but with any singleapproximation,the accuracydecreaseswith distancefrom the initial point.)))

Example2)


10)

5)

Exact solution

\037)

y)

o)

-3)

-5o) 4)2) 3) 5)

x)

FIGURE6.1.3.Graphs of Euler approximations with step sizesh = 1,h = 0.2,and h = 0.05.)

(b) Starting afresh with Xo = 0,Yo = -3,f(x,y) = x +\037y,

and h = 0.2,we getthe approximatevalues)

YI = Yo + h .[xo+\037Yo]

= (-3)+(0.2)[0+ \037(-3)]= -3.12,Y2 = YI + h . [Xl +

\037YI]= (-3.12)+ (0.2)[0.2+ \037(-3.12)]\037 -3.205,

Y3 = Y2 + h . [X2 +\037

Y2] \037 (-3.205)+ (0.2)[0.4+\037(-3.205)]\037 -3.253,

Y4 = Y3 + h .[X3 +\037Y3]

\037 (-3.253)+ (0.2)[0.6+ \037(-3.253)]\037 -3.263,Y5 = Y4 +h.[X4 +

\037Y4]\037 (-3.263)+ (0.2)[0.8+ \037(-3.263)]\037 -3.234)

at the points Xl = 0.2,X2 = 0.4,X3 = 0.6,X4 = 0.8,and X5 = 1.The resultingtable of approximatevalues is)

X 0 0.2 0.4 0.6 0.8 1

Approx.Y -3 -3.12 -3.205-3.253-3.263-3.234) .)

High accuracywith Euler'smethod usually requiresa very small stepsizeand

hencea larger number of steps than can reasonably be carried out by hand. Theapplicationmaterial for this sectioncontains calculator and computerprogramsfor

automating Euler'smethod. Oneof theseprograms was usedto calculatethe tableentries shown in Fig.6.1.4.We seethat 500Euler steps(with stepsizeh == 0.002)from X = 0 to X = 1 yield values that are accurate to within 0.001.)

Supposethe baseballof Example3 in Section1.3is simply dropped(insteadofbeingthrown downward)from the helicopter.Then its velocityvet) after t secondssatisfiesthe initial valueproblem)

dv-= 32 -0.16v,v(O)= O.dt)

(5))

We useEuler'smethod with h = 1 to track the ball'sincreasingvelocityat I-secondintervals for the first 10secondsof fall. With to = 0,vo = 0,F(t, v) == 32-0.16v,)))

.... PD..........\037.... .......... :,;,,;l:::':\\;\037DP\037()\037;:\037:<

D\302\273\037\037i\037\03714\037:il,>:!V\037)\037i>\037i\037i,jil\037'iiQ\037;:>:'

.>)

..... .....Approx:ywithh <........ O.002)

Actualvalueof y)x)

o0.20.40.60.81)

-3.000-3.120-3.205-3.253-3.263-3.234)

-3.000-3.104-3.172-3.201-3.191-3.140)

-3.000-3.102-3.168-3.196-3.184-3.130)

-3.000-3.102-3.168-3.195-3.183-3.129)

FIGURE6.1.4.Euler approximations with step sizesh = 0.2,h = 0.02,and h = 0.002.)

and h = 1 the iterative formula in (3)yieldsthe approximatevalues)

VI = Vo + h .[32-0.16vo]= (0)+ (1)[32-0.16(0)]= 32,V2 = VI + h .[32-0.16vI]= (32)+(1)[32-0.16(32)]= 58.88,V3 = V2 + h .[32-0.16v2]= (58.88)+ (1)[32-0.16(58.88)]\037 81.46,V4 = V3 +h.[32-0.16v3]= (81.46)+ (1)[32-0.16(81.46)]\037 100.43,and

V5 = V4 + h .[32-0.16V4] = (100.43)+ (1)[32-0.16(100.43)]\037 116.36.)Continuing in this fashion, we completethe h = 1 column of v-valuesshown in thetable of Fig.6.1.5-wherewe have rounded off velocity entries to the nearest footpersecond.The valuescorrespondingto h = 0.1werecalculatedusing a computer,and we seethat they are accurate to within about 1 ftls. Note also that after 10secondsthe falling ball has attained about 80% of its limiting velocity of 200ftls.

.)

'i!)*!i'I\037\037ltl']\037!!l:!'\" 2LL.I'I]'!:\037!>I\037\037.>\037

'/\037jtgj.h';:::\302\245'l:' '\037itli.ft;::::iO.l)

Actualvalue of v)t .)

123456789

10)

325981

100116130141150158165)

305577

95111124135145153160)

30557695110123135144153160)

FIGURE6.1.5.Euler approximations in Example 2 with stepsizesh = 1 and h = 0.1.)

LocalandCumulativeErrors)There are severalsourcesof error in Euler'smethod that may make the approxima-tion Yn to y(xn) unreliablefor large valuesof n, those for which Xn isnot sufficientlycloseto Xo. The error in the linear approximationformula)

Y(Xn+l) \037 Yn + h .f(xn, Yn) = Yn+I) (6))))

Y)

Xn)

I }Local errorI (Xn + I'Yn + I)I

I

I

I

I

I)

Xn+1)

FIGURE6.1.6.Thelocalerrorin Euler'smethod.)


x)

is the amount by which the tangent line at (xn, Yn) departs from the solution curvethrough (xn, Yn), as illustrated in Fig.6.1.6.This error, introducedat each step in

the process,is calledthe localerrorin Euler'smethod.The local error indicated in Fig.6.1.6would be the total error in Yn+ 1 if the

starting point Yn in (6)werean exactvalue, rather than merely an approximationto the actual value y(xn). But Yn itself suffers from the accumulatedeffectsof allthe localerrorsintroduced at the previous steps.Thus the tangent line in Fig.6.1.6is tangent to the \"wrong\" solution curve-theone through (xn, Yn) rather than the

actual solution curve through the initial point (xo,Yo). Figure 6.1.7illustrates this

cumulative errorin Euler'smethod; it is the amount by which the polygonalstep-wisepath from (xo,Yo) departs from the actual solutioncurve through (xo,Yo).)

Y)

Cumulative error)

I

I (xI'YI)I

I

I

I

I

I)

Approximatevalues)

Xo)x)

XI) X3) Xn)X2)

FIGURE6.1.7.Thecumulative error in Euler'smethod.)

The usual way of attempting to reducethe cumulative error in Euler'smethodis to decreasethe stepsizeh. The table in Fig.6.1.8showsthe results obtainedin

approximatingthe exactsolution Y (x)= 2eX -x-Iof the initial value problem)

dydx

= x + Y, y(O)= 1,)

using the successivelysmaller step sizesh = 0.1,h = 0.02,h = 0.005,andh = 0.001.We show computed valuesonly at intervals of \037x = 0.1.For instance,with h = 0.001,the computation required1000Euler steps,but the value Yn isshown only when n is a multiple of 100,so that Xn is an integral multiple of 0.1.

By scanning the columns in Fig.6.1.8we observethat, for each fixedstepsizeh, the error Yactual

-Yapprox increasesasxgetsfarther from the starting point Xo = O.

But by scanningthe rowsof the table we seethat for each fixedx,the errordecreasesas the stepsizeh is reduced.The percentageerrorsat the final point x = 1 rangefrom 7.25%with h = 0.1down to only 0.08%with h = 0.001.Thus the smallerthe stepsize,the more slowlydoesthe error grow with increasingdistancefrom thestarting point.)))


ywith y:Wiflt YWitlt y with Actualx Ji-? ...... O.:I. ,-.JI...O.O.2 Ji>O.005 Ji =0.001 y

0.1 1.1000 1.1082 1.1098 1.1102 1.11030.2 1.2200 1.2380 1.2416 1.2426 1.24280.3 1.3620 1.3917 1.3977 1.3993 1.39970.4 1.5282 1.5719 1.5807 1.5831 1.58360.5 1.7210 1.7812 1.7933 1.7966 1.79740.6 1.9461 2.0227 2.0388 2.0431 2.04420.7 2.1974 2.2998 2.3205 2.3261 2.32750.8 2.4872 2.6161 2.6422 2.6493 2.65110.9 2.8159 2.9757 3.0082 3.0170 3.01921.0 3.1875 3.3832 3.4230 3.4238 3.4266)

FIGURE6.1.8.Approximating the solution ofdy/dx = x + y, y(O)= 1 with

successivelysmaller step sizes.)

The column of data for h = 0.1in Fig.6.1.8requiresonly 10steps,soEuler'smethod can be carriedout with a hand-held calculator. But 50stepsare requiredto reach x = 1 with h = 0.02,200stepswith h = 0.005,and 1000stepswith

h = 0.001.A computer is almost always usedto implement Euler'smethod when

more than 10or 20stepsare required.Oncean appropriatecomputerprogram hasbeenwritten, one stepsizeis-inprinciple-justasconvenientasanother;after all,the computerhardly careshow many stepsit is askedto carry out.

Why, then, do we not simply choosean exceedinglysmall stepsize(suchash = 10-12),with the expectation that very great accuracywill result? There are two

reasonsfor not doing so.The first isobvious:the time requiredfor the computation.For example,the data in Fig.6.1.8wereobtained using a hand-heldcalculator that

carriedout nine Euler stepsper second.Thus it requiredslightly over one secondto approximate y(l) with h = 0.1and about 1 min 50s with h = 0.001.But with

h = 10-12it would require over 3000years!The secondreasonis more subtle.In addition to the local and cumulative er-

rorsdiscussedpreviously,the computeritself will contributeroundoff errorat eachstagebecauseonly finitely many significant digitscan be usedin each calculation.An Euler'smethod computation with h = 0.0001will introduce roundoff errors1000times as often as one with h = 0.1.Hencewith certain differentialequations,h = 0.1might actually producemore accurate resultsthan those obtained with

h = 0.0001,becausethe cumulative effect of roundofferror in the latter casemightexceedcombinedcumulative and roundofferror in the caseh = 0.1.

The \"best\" choiceof h is difficult to determine in practiceaswell as in theory.It dependson the nature of the function f (x,y) in the initial valueproblemin (2),onthe exactcodein which the program is written, and on the specificcomputerused.With a step size that is too large, the approximations inherent in Euler'smethod

may not be sufficiently accurate,whereasif h is too small, then roundofferrorsmayaccumulate to an unacceptabledegreeor the program may require too much time tobe practical.The subjectof errorpropagationin numericalalgorithms is treated in

numerical analysis coursesand textbooks.)))

Example3)

Example4)


The computations in Fig.6.1.8illustrate the common strategy of applying anumericalalgorithm, suchasEuler'smethod,severaltimes in succession,beginningwith a selectednumber n of subintervalsfor the first application,then doubling n foreachsucceedingapplicationof the method. Visual comparisonof successiveresultsoften can provide an \"intuitive feel\" for their accuracy.In the next two exampleswepresentgraphically the resultsof successiveapplicationsof Euler'smethod.)

The exact solutionof the logisticinitial value problem)

dy 1

dx= 3y (8 -y), y(O)= 1)

is y(x) = 8/(1+ 7e-8x /3). Figure 6.1.9showsboth the exact solution curve and

approximate solution curves obtained by applying Euler'smethod on the interval

o <x <5 with n = 5,n = 10,and n = 20subintervals.Each of these \"curves\" ac-tually consistsof line segmentsjoining successivepoints (xn , Yn) and (Xn+l, Yn+l).The Euler approximationwith 5 subintervalsispoor,and the approximationwith 10subintervalsalsoovershootsthe limiting value Y = 8 of the solution before levelingoff, but with 20subintervalswe obtain fairly goodqualitative agreementwith the

actual behaviorof the solution. .)

12

10

8

;>-.. 6

4

2

00 1 2 3 4 5)

3 Exact)

2)

;>-..)

o)

o)

x)

10x)

5) 15)

FIGURE6.1.9.Approximating alogistic solution using Euler'smethod with n = 5,n = 10,andn = 20 subintervals.)

FIGURE6.1.10.Approximatingthe exactsolution y = esin x

usingEuler'smethod with 50,100,200,and 400subintervals.)

\"\"'-

The exact solutionof the initial value problem)

dydx

= ycosx, y(O)= 1)

is the periodicfunction y (x) = esinx. Figure 6.1.10showsboth the exact solutioncurve and approximatesolutioncurves obtained by applying Euler'smethod on the

interval 0 < x < 6n with n = 50,n = 100,n = 200,and n = 400subintervals.Even with this many subintervals,Euler'smethod evidently has considerablediffi-

culty keepingup with the oscillationsin the actual solution. Consequently,the moreaccuratemethodsdiscussedin succeedingsectionsare neededfor seriousnumericalinvestigations. .)))


A Word ofCaution)The data shown in Fig.6.1.8indicate that Euler'smethodworkswell in approximat-ing the solution of dy/dx ==x+y, y(O) == 1 on the interval [0,1].That is, for eachfixedx it appearsthat the approximatevalues approach the actual value of y(x) asthe stepsizeh is decreased.For instance, the approximatevalues in the rowscorre-spondingto x ==0.3and x ==0.5suggestthat y(0.3)\037 1.40and y(0.5)\037 1.80,in

accordwith the actual values shown in the final column of the table.Example5,in contrast, showsthat someinitial valueproblemsare not sowell

behaved.)

Example5)\037_'NNN\037NNNN\037-\037_'N.',-.'\037_-\037.'''Nn_'Nn;,.... UN..N....\"..;..\"'\" ,.... .\" \"\"\" ____ ..., \"...\" .... \037.... \037 '\" ,....

UseEuler'smethod to approximatethe solutionof the initial value problem)

dy-==x2 + y2, y(O) == 1dx)

(7))

on the interval [0,1].)

Solution Heref (x,y) ==x2 + y2, so the iterative formula of Euler'smethod is)

Yn+l == Yn + h .(x\037 + y\037).) (8))

With stepsizeh ==0.1we obtain)

Yl == 1 + (0.1).[(0)2+ (1)2]== 1.1,Y2 == 1.1+ (0.1).[(0.1)2+ (1.1)2]== 1.222,Y3 == 1.222+ (0.1).[(0.2)2+(1.222)2]\037 1.3753,)

and so forth. Rounded to four decimalplaces,the first ten values obtained in this

manner are)

Y1 == 1.1000Y2 == 1.2220Y3 == 1.3753Y4 == 1.5735Ys == 1.8371)

Y6 ==2.1995Y7 ==2.7193Ys ==3.5078Y9 ==4.8023

YIO ==7.1895)

But instead of naively acceptingtheseresultsas accurate approximations,wedecidedto usea computer to repeat the computationswith smaller valuesof h. Thetable in Fig.6.1.11showsthe resultsobtained with stepsizesh == 0.1,h == 0.02,and h ==0.005.Observethat now the \"stability\" of the procedurein Example1 ismissing.Indeed,it seemsobvious that somethingis going wrong near x == 1.

Figure 6.1.12providesa graphicalclueto the difficulty. It showsa slopefieldfor dy/dx == x2 + y2, together with a solution curve through (0,1)plotted usingone of the more accurate approximation methods of the following two sections.It appearsthat this solution curve may have a vertical asymptote near x == 0.97.Indeed,an exactsolutionusing Besselfunctions (seeProblem 16in Section3.6)canbe used to show that y(x) ---* +00as x ---* 0.969811(approximately). AlthoughEuler'smethod gives values (albeit spuriousones)at x == 1,the actual solutiondoesnot existon the entire interval [0,1].Moreover,Euler'smethod is unable to\"keepup\" with the rapid changesin y (x) that occuras x approachesthe infinite

discontinuity near 0.969811. .)))

X\":')

.......'.1.:.....\037.....1\037\037:..;.....'.;..........,...,.........,;.:....'......:........;.:..\037.....Wi.t\037.:.............

..,

.:,\037.:\037'.Q\037t\":',:'

, ..1t';'.Q\037Q2..... -,' -------\037.:'>-, -, ',-, >: :-, --'-'.-'-' -- -- ,)

..:....>y...\"\"itl1

,...\"....:::....iO.:005)

0.10.20.30.40.50.60.70.80.91.0)

1.10001.22201.37531.57351.83712.19952.71933.50784.80237.1895)

1.10881.24581.42431.66582.00742.52013.36124.96019.0000

30.9167)

1.11081.25121.43571.68822.05122.61043.57065.576312.2061

1502.2090)

8)

6)

4)

\0372)

o)

-2)

-4-2.0-1.0) 0.0x)

1.0) 2.0)

FIGURE6.1.11.Attempting to approximate thesolution ofdy/dx = x2 + y2, y(O)= 1.)

FIGURE6.1.12.Solution ofdy/dx = x2 + y2, y(O) = 1.)

The moral of Example5 is that there are pitfalls in the numericalsolution ofcertain initial value problems.Certainly it'spointlessto attempt to approximateasolutionon an interval where it doesn'tevenexist(orwhere it isnot unique, in which

casethere'sno generalway to predictwhich way the numericalapproximationswill

branch at a point of nonuniqueness).Oneshouldneveracceptasaccuratethe resultsof applying Euler'smethod with a singlefixed stepsizeh. A second\"run\" with

smaller stepsize(hI2,say, or h15,or hllO)may give seeminglyconsistentresults,thereby suggestingtheir accuracy,or it may-asin Example5-revealthe presenceof somehidden difficulty in the problem.Many problemssimply require the moreaccurate and powerful methods that are discussedin the final two sectionsof this

chapter.)

11II..J>'E\037I?\037\037\037\037..\037_ .._.....)In Problems1through 10,an initial value problem and its ex-actsolution y(x) are given. Apply Euler'smethod twice to ap-proximate to this solution on the interval [0,4],first with stepsizeh = 0.25,then with stepsizeh = 0.1.Comparethe three-decimal-placevalues of the two approximations at x = 4 with

the value y (4) of the actualsolution.

1.y' = -y,y(O)= 2;y(x) = 2e-x2.y' = 2y, y(O)= 4;y(x) = 4e2x3.y' = y + 1,y(O)= 1;y(x) = 2ex - 14. y'=x-y,y(0)=I;y(x)=2e-x +x-l5. y' = y

-x-I,y(O)= 1;y(x) = 2 + x - eX26.y' = -2xy,y(O)= 2;y(x) = 2e-x

7. y' = -3x2y,y(O)= 3;y(x) = 3e-x3

8.y' = e-Y , y(O)= 0; y(x) = In (x + 1)9.y' = *(1+ y2), y(O)= 1;y(x) = tan *(x + n)

10.y' = 2xy2,y(O)= 1;y(x) = 12I-x

Note: The application following this problem set lists illus-trative calculator/computerprograms that can be usedin the

remaining problems.)

A programmable calculatoror a computer will be useful forProblems 11through 16.In eachproblem find the exactso-lution of the given initial value problem. Then apply Euler'smethod twice to approximate (tofour decimalplaces)this so-lution on the given interval, first with step sizeh = 0.01,then

with step sizeh = 0.005.Make a table showing the approxi-mate values and the actual value, together with the percentageerror in the more accurateapproximation, for x an integralmultiple of 0.2.Throughout, primes denote derivatives with

respectto x.)

11.y' = y-2, y(O)= 1;0 < x < 1

12.y' = 4(y - 1)2,y(O)= 2;0 < x < 113.y y' = 2x3, Y(1)= 3; 1 < x < 214.xy' = y2, Y(1)= 1;1 < x < 215.xy'= 3x -2y, y(2) = 3;2 < x < 316.y2y'=2x5,y(2)=3;2< x < 3)

A computer with a printer is required for Problems 17through24. In these initial value problems, use Euler'smethod with

step sizesh = 0.1,0.02,0.004,and 0.0008to approximate to

four decimalplacesthe values of the solution at ten equally)))

spacedpoints of the given interval. Print the results in tabular

form with appropriate headings to make it easy to gauge the

effectof varying the step sizeh. Throughout, primes denotederivatives with respectto x.17.y' = x2 + y2, Y(0)= 0; 0 < x < 118.y' =x2 - y2, Y(0) = 1;0 < x < 219.y'=x+\037,y(0)=1;0< x < 220.y'=x+\037,y(0)=-1;0< x < 221.y' = In y, y(l) = 2; 1 < x < 222.y' = x2/3 + y2/3, Y(0) = 1;0 < x < 223.y' = sin x + cosy, y (0)= 0; 0 < x < 1

x24. y' =1+ y2

'Y(-1)= 1;-1< x < 1

25.You bailout of the helicopterof Example 2 and immedi-

ately pull the ripcord of your parachute. Now k = 1.6in Eq. (5),so your downward velocity satisfies the initial

value problem)

dv-= 32-1 6v V (O) = 0dt

. ,)

(with t in secondsand v in ft/sec). UseEuler'smethodwith a programmable calculator or computer to approx-imate the solution for 0 < t < 2, first with step sizeh = 0.01and then with h = 0.005,rounding off approx-imate v-values to onedecimal place.What percentageofthe limiting velocity 20 ft/sechas been attained after 1second?After 2 seconds?

26.Supposethe deerpopulation P (t) in a small forest initiallynumbers 25and satisfiesthe logistic equation)

dP-= 0.0225P-0.0003p2dt)

(with t in months). UseEuler's method with a pro-grammable calculatoror computer to approximate the so-lution for 10years, first with step sizeh = 1and then with

h = 0.5,rounding off approximate P-valuesto integralnumbers ofdeer.What percentageof the limiting popula-tion of 75 deerhas beenattained after 5 years? After 10years?)

UseEuler'smethod with a computer system to find the desiredsolution values in Problems 27 and 28. Start with step size)

6.1Application)

h = 0.1,and then use successivelysmaller step sizesuntil suc-cessiveapproximate solution values at x = 2 agree rounded

off to two decimalplaces.27.y' = x2 + y2 - 1,y(O)= 0; y(2) = ?28.y' = x + !y2,y(-2)= 0; y(2) = ?29.Considerthe initial value problem)

dy7x-+y=0, y(-I)=1.dx

(a) Solvethis problem for the exactsolution)

1y (x) = -

X 1/7'

which has an infinite discontinuity at x = O. (b) ApplyEuler'smethod with step sizeh = 0.15to approximatethis solution on the interval -1< x < 0.5.Note that,from thesedata alone,you might not suspectany difficultynear x = O.Thereasonis that the numerical approxima-tion \"jumps acrossthe discontinuity\" to another solution

of 7xy' + y = 0 for x > O. (c) Finally, apply Euler'smethod with step sizesh = 0.03and h = 0.006,but still

printing results only at the original points x = -1.00,-0.85,-0.70,..., 1.20,1.35.and 1.50.Would you now

suspecta discontinuity in the exactsolution?

30.Apply Euler'smethod with successivelysmaller stepsizeson the interval [0,2] to verify empirically that the solution

of the initial value problem)

dy

dx= x2 + y2, y(O)= 0)

has a vertical asymptote near x = 2.003147.(Contrastthis with Example 2, in which y (0)= 1.)31.Thegeneral solution of the equation)

dy = (1+ y2)cosxdx

is y(x) = tan(C + sin x). With the initial condition

y(O) = 0 the solution y(x) = tan(sinx) is well behaved.But with y(O) = 1 the solution y(x) = tan (in+ sinx)has a vertical asymptote at x = sin-1(nj4) \037 0.90334.UseEuler'smethod to verify this fact empirically.)

ImplementingEuler'sMethod\" \" \037 \"\".N\"''''.',_''\",,,,,,,\037 nn '-' '-' \"'_\"n.\",\", n ..... ...\" n \037 N '\" n ,,-. '\" \"'''''' ,...,'\" ,.....,..,.... .....\037... '-' \"\"........ \"\" ... '-' \037....,.. '\" '\" \"''')

Construction of a calculator or computer program to implement a numerical algo-rithm can sharpen one'sunderstanding of the algorithm. Figure 6.1.13listsTI-85and BASICprograms implementing Euler'smethod to approximatethe solutionofthe initial value problem)

dydx

= x + y, y(O)= 1)

consideredin this section.The commentsprovided in the final column shouldmaketheseprograms intelligible even if you have little familiarity with the BASICand)))


'......:TI\0378!).........

..BASIC) Comment)

PROGRAM: EULER:10\037N

:O\037X

:l\037Y:1\037X1:(Xl-X)/N\037H

:For(I,l,N):X+Y\037F:Y+H*F\037 Y

:X+H\037X

:DispX,Y

:End)

ProgramEULER

N - 10X = 0Y = 1Xl = 1H = (X1-X)/NFOR 1=1TO N

F - X + Y

Y = Y + H*F

X = X + H

PRINT X,Y

NEXT I)

ProgramtitleNumber of stepsInitial xInitial Y

Finalx

stepsizeBeginloopFunctionvalueEuleriterationNew x

DisplayresultsEnd loop)

FIGURE6.1.13.TI-85and BASICEuler'smethod programs.)

TI calculator programming languages.Indeed,the BASIClanguage is no longerwidelyusedfor programming computersbut is still useful (asin Fig.6.1.13and sub-sequentonesin this text) for brief descriptionof mathematicalalgorithms in a trans-

parent form intermediatebetweenEnglishand higher programming languages.(Ap-propriately, the name BASICis an acronymdescribingthe Beginner'sAll-purposeSymbolicInstruction Codeintroduced in 1963,intially for instructional useat Dart-mouth College.)

To increasethe number of steps(and therebydecreasethe stepsize)you needonly change the valueof N specifiedin the first line of the program. Toapply Euler'smethod to a different equation dyjdx = I(x,y), you needchange only the singleline that calculatesthe function value F.

Any other proceduralprogramming language (suchas FORTRANor Pascal)wouldfollow the pattern illustrated by the parallel linesofTI-85and BASICcodein

Fig.6.1.13.Someof the modem functional programming languagesmirror standard

mathematical notation even more closely.Figure 6.1.14showsa MAT LAB imple-mentation of Euler'smethod. The eulerfunction takes as input the initial value x,the initial value y, the final valuexlof x,and the desirednumber n of subintervals.)

functionyp - f(x,y)yp = x + Yi) % yp = y')

function[X,Y] -h - (xl - x)/niX = XiY = Yifor i = l:n)

Y - Y + h*f(x,Y)ix = x + hiX - [XiX]iY - [YiY]iend)

euler(x,y,x1,n)% stepsize% initialx% initialY

% beginloop% Euleriteration)% new x% update x-column% update y-column% end loop)

FIGURE6.1.14.MATLAB implementation ofEuler'smethod.)))

For instance, the MATLAB command

[X, Y] = euler(O,1, 1, 10)then generatesthe xn - and Yn-data shown in the first two columnsof the table of Fig.6.1.8.

You shouldbeginthis projectby implementing Euler'smethod with your owncalculator or computer system.Test your program by first applying it to the initial

value problemin Example1,then to someof the problemsfor this section.)

FamousNumbersInvestigation)The following problemsdescribethe numbers e \037 2.71828,In2 \037 0.69315,andJT \037 3.14159as specificvalues of solutions of certain initial value problems.Ineachcase,apply Euler'smethod with n = 50,100,200,...subintervals(doublingn eachtime).Howmany subintervalsare neededto obtain-twicein succession-the correctvalue of the target number rounded to three decimalplaces?)

1.The number e = y(I), where y(x) is the solution of the initial value problemdyldx = y, y(O)= 1.

2.The number In 2 = y (2),where y (x) is the solution of the initial value prob-lem dyldx = llx,y(l) = O.

3.The number JT = y(I), where y(x) is the solutionof the initial valueproblemdyldx = 41(1+x2), y(O)= O.)

Also explain in eachproblemwhat the point is-why the indicated famousnumber is the expectednumerical result.)

lIBA CloserLookat the EulerMethod)

The Eulermethod as presentedin Section6.1is not often usedin practice,mainlybecausemore accuratemethodsare available.But Euler'smethodhas the advantageof simplicity, and a careful study of this method yieldsinsights into the workings ofmore accuratemethods,becausemany of the latter are extensionsor refinementsofthe Eulermethod.

Tocomparetwo differentmethodsof numericalapproximation,we needsomeway to measurethe accuracy of each.Theorem 1 tellswhat degreeof accuracywecan expectwhen we useEuler'smethod.)

THEOREM1 TheError in theEulerMethod)

Supposethat the initial value problem)

dydx

= I(x,y), y(xo)= Yo) (1))

has a unique solution y(x) on the closedinterval [a,b] with a = Xo, and assumethat y(x) has a continuous secondderivativeon [a,b].(This would follow fromthe assumptionthat I,Ix,and Iy are all continuousfor a <x <band c <

y<d,

where c <y(x) <d for all x in [a,b].)Then there existsa constant C such that)))

6.2A CloserLookat the EulerMethod 443)

the followingis true: If the approximationsYl, Y2, Y3, ..., Yk to the actual valuesY(Xl), Y(X2), Y(X3), ...,Y(Xk) at points of [a,b] are computed using Euler'smethod with stepsizeh >0,then)

IYn- y(xn)1 <Ch) (2))

for eachn = 1,2,3,...,k.)

Remark:The error)

Yactual-

Yapprox== Y (xn ) - Yn)

in (2)denotesthe [cumulative]error in Euler'smethod after n stepsin the approxi-mation, exclusiveof roundofferror (asthough we wereusing a perfectmachinethat

made no roundofferrors).The theoremcan be summarizedby saying that the errorin Euler'smethod is of orderh; that is, the error is boundedby a [predetermined]constant C multiplied by the stepsizeh. It follows, for instance, that (on a givenclosedinterval) halving the stepsizecuts the maximum error in half; similarly, with

stepsizeh/l0we get 10times the accuracy (that is, 1/10the maximum error) aswith stepsizeh. Consequently,we can-inprinciple-getany degreeof accuracywe want by choosingh sufficiently small. .)

We will omit the proof of this theorem, but one can be found in Chapter7 ofG.Birkhoff and G.-C.Rota, Ordinary DifferentialEquations, 4th ed.(New York:John Wiley, 1989).The constant C deservessomecomment. BecauseC tends toincreaseas the maximum valueof

I y\" (x)Ion [a,b] increases,it followsthat C must

dependin a fairly complicatedway on Y, and actual computationof a value of Csuch that the inequality in (2)holdsisusually impractical. In practice,the following

type of procedureis commonlyemployed.

1.Apply Euler'smethod to the initial value problemin (1)with a reasonablevalue of h.

2.Repeat with h12,h14,and so forth, at each stagehalving the stepsizefor the

next applicationof Euler'smethod.3.Continueuntil the resultsobtainedat onestageagree-toan appropriatenum-

berof significant digits-withthose obtained at the previousstage.Then the

approximatevalues obtained at this stageare consideredlikely to beaccurateto the indicated number of significant digits.)

Example1) Carry out this procedurewith the initial value problem)

dydx)

2xy1 +x2 ') y(O) == 1) (3))

to approximateaccurately the value Y (1)of the solutionat x == 1.)

Solution Using an Euler method program, perhapsone of those listedin Figs.6.1.13and6.1.14,we beginwith a stepsizeh ==0.04requiring n == 25 steps/toreachx == 1.The table in Fig.6.2.1showsthe approximatevalues of y(l) obtainedwith succes-sivelysmallervaluesof h. The data suggestthat the true valueof y(l) isexactly0.5.Indeed,the exact solution of the initial value problemin (3)is y(x) == 1/(1+x2),so the true value of Y (1)is exactly 4. .)))


y)

(x + h, y(x +h\302\273

i }Error

,. ./ PredIcted

// y-value

//

//

// Slopey'(x)

//

//

//)

x x+h)

FIGURE6.2.2.True and

predictedvalues in Euler'smethod.)

Ii) \037JJP..ij\037ijbateJ'(l ))Actual y(l)

0.500000.500000.500000.500000.500000.500000.500000.50000)

IErrorl/h

0.110.110.110.110.110.100.110.10)

0.040.020.010.0050.00250.001250.0006250.0003125)

0.504510.502200.501090.500540.500270.500130.500070.50003)

FIGURE6.2.1.Tableofvalues in Example 1.)

The final column of the table in Fig.6.2.1displaysthe ratio of the magnitudeof the error to h; that is, I Yactual

-Yapprox I/h. Observehow the data in this column

substantiateTheorem I-inthis computation,the error bound in (2)appearsto holdwith a value of C slightly larger than 0.1.)

An ImprovementinEuler'sMethod)

As Fig.6.2.2shows,Euler'smethod is rather unsymmetrical. It usesthe predictedslopek == f(xn , Yn) of the graph of the solution at the left-hand endpoint of theinterval [xn, Xn + h] as if it were the actual slopeof the solution over that entireinterval. We now turn our attention to a way in which increasedaccuracycan easilybe obtained; it is known as the improvedEulermethod.

Given the initial value problem)

dydx

==f(x,y), Y(Xo) ==Yo,) (4))

x)

supposethat after carrying out n stepswith stepsizeh we have computed the ap-proximation Yn to the actual value y(xn ) of the solution at Xn == Xo +nh. We canusethe Euler method to obtain a first estimate-whichwe now call Un+I rather than

Yn+ I-ofthe value of the solution at Xn+ I ==Xn +h. Thus)

\037) Un+I == Yn +h.f(xn , Yn) == Yn +h.k l .)

Now that Un+I \037 Y(Xn+I) has beencomputed,we can take)

\037) k2 == f(Xn+I,Un+I))

as a secondestimate of the slopeof the solutioncurve Y == Y (x)at x ==Xn+1.Of course,the approximateslopek I == f(xn, Yn) at x == Xn has alreadybeen

calculated.Why not average thesetwo slopesto obtain a more accurateestimateofthe average slopeof the solution curve over the entire subinterval [xn, Xn+I]?Thisideais the essenceof the i\037proved Eulermethod. Figure 6.2.3showsthe geometrybehind this method.)))


Y)

Slopek 2=f(xn+ \\,un+ \\)

\037)

(Xn,Yn)I

I

I

I

I

I

I

I

I)

Xn)

Xn+ 1)Xn+2)

x)

FIGURE6.2.3.Theimproved Euler method: Average the slopesof the tangent linesat (Xn, Yn) and (Xn+l, Un+l).)

ALGORITHM TheImprovedEulerMethodGiven the initial valueproblem)

dy-= f(x,y), y(xo)= Yo,dx)

the improvedEulermethod with step sizeh consistsin applying the iterativeformulas)

k1 = f(xn , Yn),

Un+l = Yn + h .k1,k2 = f(Xn+l, Un+l),

Yn+l = Yn +h.\037(kl

+k2))

(5))

to compute successiveapproximationsYl, Y2, Y3, ...to the [true] valuesY(XI),

Y(X2), Y(X3), ...of the [exact] solution Y = y(x) at the points Xl, X2, X3, ...,respectively.)

Remark:The final formula in (5)takes the \"Euler form\

Yn+l = Yn + h .k)

if we write)

k = k l +k22

for the approximateaverage slopeon the interval [xn, Xn+l].) .)The improvedEuler method is one of a classof numericaltechniquesknown

aspredictor-correctormethods.First a predictorUn+l of the next y-valueis com-puted;then it is usedto correctitself. Thus the improvedEulermethod with step)))

Example2)

sizeh consistsof using the predictor)

\037) Un+l == Yn + h .I(xn , Yn)) (6))

and the corrector)

\037) Yn+l == Yn + h .\037 [/(xn , Yn) + I(Xn+l,Un+l)]) (7))

iteratively to compute successiveapproximationsYl, Y2, Y2, ...to the valuesY(Xl),Y (X2), Y (X3), ...of the actual solutionof the initial valueproblem in (4).

Remark: Each improvedEuler steprequirestwo evaluationsof the func-tion I(x,y), as comparedwith the singlefunction evaluationrequiredfor an ordi-nary Euler step. We naturally wonder whether this doubledcomputationallabor isworth the trouble.

Answer: Underthe assumption that the exactsolution Y == y(x) of the

initial value problemin (4)has a continuous third derivative,it can beproved-seeChapter 7 of Birkhoff and Rota-thatthe error in the improved Euler method isof orderh 2. This means that on a given boundedinterval [a,b], each approximatevalue Yn satisfiesthe inequality)

\037) Iy(xn ) - Ynl<Ch2

,) (8))

where the constant C doesnot dependon h. Becauseh 2 is much smaller than h ifh itself is small, this means that the improvedEuler method is more accurate than

Euler'smethod itself.This advantage is offset by the fact that about twice as many

computationsare required.But the factor h 2 in (8)means that halving the stepsizeresultsin 1/4the maximum error, and with step sizeh/l0we get 100times the

accuracy (that is, 1/100the maximum error) as with stepsizeh. .)...

Figure 6.1.8showsresultsof applying Euler'smethod to the initial value problem)

dydx ==x+y, y(O)==1) (9))

with exactsolution y(x) ==2ex -x-I.With I(x,y) ==x + Y in Eqs.(6)and (7),the predictor-correctorformulas for the improvedEulermethod are)

Un+l == Yn + h .(xn + Yn),

Yn+l == Yn +h.\037 [(xn + Yn) + (Xn+l +Un+l)].)

With stepsizeh ==0.1we calculate)

Ul == 1 + (0.1).(0+ 1)== 1.1,Yl == 1 + (0.05).[(0+1)+ (0.1+ 1.1)]== 1.11,U2 == 1.11+ (0.1).(0.1+ 1.11)== 1.231,Y2 == 1.11+ (0.05).[(0.1+ 1.11)+(0.2+ 1.231)]== 1.24205,)

and so forth. The table in Fig.6.2.4comparesthe resultsobtained using the im-

proved Eulermethod with thoseobtainedpreviouslyusing the \"unimproved\" Eulermethod. When the samestepsizeh == 0.1is used,the error in the Euler approxi-mation to y(l) is 7.25%,but the error in the improvedEulerapproximationis only0.24%.)))

x)

0.00.10.20.30.40.50.60.70.80.91.0)

\037ml\037()ved

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1.000001.110341.242801.399711.583641.797442.044232.327492.651073.019193.43654)

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1.000001.110341.242811.399721.583651.797442.044242.327512.651083.019213.43656)

FIGURE6.2.5.Improved Euler

approximation to the solution ofEq. (9)with step sizeh = 0.005.)

Example3)


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Actual

y

1.11031.24281.39971.58361.79742.04422.32752.65113.01923.4366)

Improved Euler,h =0.1

Values of y)

0.10.20.30.40.50.60.70.80.91.0)

1.10001.22001.36201.52821.72101.94312.19742.48722.81593.1875)

1.10981.24161.39771.58071.79332.03882.32052.64223.00823.4230)

1.11001.24211.39851.58181.79492.04092.32312.64563.01243.4282)

FIGURE6.2.4.Euler and improved Euler approximations to the solution ofdyjdx = x + y, y(O)= 1.)

Indeed,the improved Euler method with h = 0.1is more accurate (in this

example)than the original Euler method with h = 0.005.The latter requires200evaluations of the function I(x,y), but the former requiresonly 20 such evalua-tions, so in this casethe improvedEuler method yieldsgreater accuracywith onlyabout one-tenth the work.

Figure 6.2.5showsthe resultsobtained when the improvedEuler method isappliedto the initial value problemin (9)using stepsizeh = 0.005.Accuracyoffive significant figures is apparent in the table. This suggeststhat, in contrastwith

the original Eulermethod, the improved Euler method is sufficiently accurateforcertain practical applications-suchas plotting solutioncurves. .)

An improvedEuler program (similar to the oneslistedin the project materialfor this section)was usedto compute approximationsto the exact value y(l) = 0.5of the solution y(x) = 1/(1+x2) of the initial value problem)

dydx)

2xy1 +x2 ') y(O)= 1) (3))

of Example1.The resultsobtained by successivelyhalving the stepsizeappearin

the table in Fig.6.2.6.Note that the final column of this table impressivelycor-roboratesthe form of the error bound in (8),and that eachhalving of the step sizereducesthe error by a factor of almost exactly 4, as should happen if the error isproportional to h 2.

In the following two exampleswe exhibit graphical results obtainedby em-ploying this strategy of successivelyhalving the stepsize, and thus doubling the

number of subintervalsof a fixed interval on which we are approximating a solu-ti on.)

...... ..... .....

In Example3 of Section6.1we appliedEuler'smethod to the logistic initial valueproblem)

dy 1

dx= 3y (8 -y), y(O)= 1.)))

12

10

8

\037 6

4

2

00 2 3 4 5)

x)

FIGURE6.2.7.Approximatinga logisticsolution using the

improved Euler method withn = 5,n = 10,n = 20,andn = 40 subintervals.)

0.040.020.010.0050.00250.001250.0006250.0003125)

0.5001959030.5000494940.5000124370.5000031170.5000007800.5000001950.5000000490.500000012)

-0.000195903-0.000049494-0.000012437-0.000003117-0.000000780-0.000000195-0.000000049-0.000000012)

0.120.120.120.120.120.120.120.12)

FIGURE6.2.6.Improved Euler approximation to y(l) fordyjdx = -2xyj(1+ x2),y(O)= 1.)

Figure 6.1.9shows an obvious difference between the exact solution y(x) =8/(1+ 7e-8x /3) and the Eulerapproximationon 0 < x < 5 using n = 20subin-tervals. Figure 6.2.7showsapproximatesolutioncurvesplotted using the improvedEuler'smethod.

The approximation with five subintervalsis still bad-perhapsworse!It ap-pearsto level off considerablyshort of the actual limiting population M = 8.You

should carry out at least the first two improved Euler steps manually to seeforyourself how it happensthat, after increasingappropriatelyduring the first step,the

approximatesolutiondecreasesin the secondsteprather than continuing to increase(as it should).In the projectfor this sectionwe askyou to showempiricallythat theimprovedEulerapproximatesolution with stepsizeh = 1 levelsoff at y \037 4.3542.

In contrast, the approximate solution curve with n = 20subintervals tracksthe exactsolution curve rather closely,and with n = 40 subintervals the exactand approximate solution curves are indistinguishable in Fig.6.2.7.The table in

Fig.6.2.8indicatesthat the improvedEulerapproximationwith n = 200subinter-vals is accurate rounded to three decimalplaces(that is, four significant digits)onthe interval 0 < x < 5. Becausediscrepanciesin the fourth significant digit arenot visually apparentat the resolutionof an ordinary computerscreen,the improvedEulermethod (usingseveralhundred subintervals)is consideredadequatefor many

graphicalpurposes. .)

o12345)

1.00005.38227.73857.98137.99877.9999)

1.00005.38097.73797.98127.99877.9999)

FIGURE6.2.8.Using the improved Eulermethod to approximate the actual solution of theinitial value problem in Example 3.)))

Example4)

3)n= 50)

2)

;::.....)

1)

o)

o) 5) 10x)

15)

FIGURE6.2.9.Approximatingthe exactsolution y = esinx

usingthe improved Euler method withn = 50,100,and 200subintervals.)

6.2A CloserLookat the Euler Method 449)

In Example4 of Section6.1we appliedEuler'smethod to the initial value problem)

dy-= y cosx, y(O)= 1.dx)

Figure 6.1.10showsobvious visual differencesbetween the periodicexact solution

y(x) = esinx and the Euler approximationson 0 <x <6n with asmany as n = 400subintervals.

Figure 6.2.9showsthe exact solution curve and approximatesolution curvesplotted using the improved Euler method with n = 50,n = 100,and n = 200subintervals. The approximationobtained with n = 200is indistinguishable fromthe exact solutioncurve, and the approximationwith n = 100is only barelydistin-

guishablefrom it. .Although Figs.6.2.7and 6.2.9indicate that the improvedEuler method can

provide accuracy that sufficesfor many graphicalpurposes,it doesnot providethe

higher-precisionnumerical accuracy that sometimesis neededfor more careful in-

vestigations.For instance,consideragain the initial value problem)

dydx)

2xy1 +x2 ') y(O)= 1)

of Example1.The final column of the table in Fig.6.2.6suggeststhat, if the im-

provedEulermethod is usedon the interval 0 <x < 1 with n subintervals and stepsize h = Iln, then the resulting error E in the final approximationYn \037 y(l) isgiven by)

A hand-heldcalculatorwill suffice for Problems1through 10,where an initial value problem and its exactsolution aregiven.Apply the improved Euler method to approximate this solutionon the interval [0,0.5]with stepsizeh = 0.1.Construct atable showing four-decimal-placevalues of the approximatesolution and actual solution at the points x = 0.1,0.2,0.3,0.4,0.5.1.y' = -y,y(O)= 2;y(x) = 2e-x2.y' = 2y, y(O)= 4; y(x) = 4e2X

3.y' = y + 1,y(O)= 1;y(x) = 2eX - 14. y' = x -y, y (0)= 1;y (x) = 2e-x + x-I5.y'=y-x-l,y(0)=I;y(x)=2+x-ex)

2 0.12E = ly(l)- Ynl \037 (0.12)h=\037.n

If so,then 12-placeaccuracy (for instance) in the value y(l) would require that

(0.12)n-2 < 5 x 10-13, which means that n > 489,898.Thus, roughly half amillion stepsof length h \037 0.000002would be required.Asidefrom the possi-ble impracticalityof this many steps(using availablecomputationalresources),theroundoff error resulting from so many successivestepsmight well overwhelm the

cumulative error predictedby theory (which assumesexact computationsin eachseparatestep).Consequently,still more accurate methods than the improved Eulermethod are neededfor such high-precisioncomputations. Sucha method is pre-sentedin Section6.3.)

\037ProbleIns\037_ ___\037 \037 'mmmm.n.M N N ,,,___ _____ \037mm\037\037. \037 N m._.n_ ,y__\037,.. m..,..m_u_. ..mn 'mm....'._\037m.' \037_. ____,. N ..,m__\037__\037.'. __ 0 ..y., ,.y)

6.y' = -2xy,y(O)= 2;y(x) = 2e-x2

7. y' = -3x2y,y(O)= 3;y(x) = 3e-x3

8.y' = e-Y , y(O)= 0; y(x) = In(x + 1)9.y' =

\037(1+ y2), Y(0)= 1;y (x) = tan

\037(x + Jr)

, 110.y = 2xy 2,y(O)= 1;y(x) =2I-x

Note: Theapplication following this problem setlists illustra-

tive calculator/computer programs that can be usedin Prob-lems11through 24.A programmable calculatoror a computer will be useful forProblems 11through 16.In eachproblem find the exactso-lution of the given initial value problem. Then apply the im-)))

proved Euler method twice to approximate (to five decimalplaces)this solution on the given interval, first with stepsizeh =0.01,then with stepsizeh = 0.005.Make a table showingthe approximate values and the actual value, together with the

percentageerror in the more accurateapproximations, for xan integral multiple of0.2.Throughout, primes denotederiva-tives with respectto x.)

11.y' = y -2,y(O)= 1;0 < x < 112.y' =

\037(y - 1)2,y(O)= 2;0 < x < 1

13.yy' = 2x3, y(l) = 3; 1 < x < 214.xy' = y2, y(l) = 1;1 < x < 215.xy' = 3x -2y, y(2) = 3;2 < x < 316.y2 y '=2x5,y(2)=3;2< x < 3)

A computer with a printer is requiredfor Problems17through24. In these initial value problems, use the improved Eulermethod with step sizesh = 0.1,0.02,0.004,and 0.0008to

approximate to five decimalplacesthe values of the solutionat ten equally spacedpoints of the given interval. Print theresults in tabular form with appropriateheadings to make it

easyto gauge the effectofvarying the step sizeh. Throughout,primes denotederivatives with respectto x.17.y' = x2 + y2, y(O)= 0; 0 < x < 118.y' = x2 - y2, Y(0)= 1;0 < x < 219.y' = x + -JY, y(O)= 1;0 < x < 220.y'=x+,yJ,y(0)=-1;0< x < 221.y' = In y, y(l) = 2; 1 < x < 222.y' = X213+ y213, Y(0)= 1;0 < x < 223.y' = sin x +cosy,y(O)= 0; 0 < x < 1

x24. y' =1+ y2

'Y(-1)= 1;-1< x < 1

25.As in Problem 25 of Section6.1,you bailout of a he-licopter and immediately open your parachute, so yourdownward velocity satisfiesthe initial value problem

dv

dt= 32-1.6v, v(O) = 0)

(with t in secondsand v in ftls). Usethe improved Eu-ler method with a programmable calculator or computerto approximate the solution for 0 < t < 2, first with stepsizeh = 0.01and then with h = 0.005,rounding offapproximate v-values to three decimal places.What per-centageof the limiting velocity 20 ftls has beenattainedafter 1second?After 2 seconds?

26.As in Problem 26of Section6.1,supposethe deerpopu-lation pet) in a small forest initially numbers 25 and sat-isfiesthe logisticequation)

dPdt

= 0.0225P-0.0003p2

(with t in months). Usethe improved Euler method with aprogrammable calculatoror computer to approximate thesolution for 10years, first with step sizeh = 1 and then)

with h = 0.5,rounding off approximate P-valuesto threedecimal places.What percentageof the limiting popula-tion of 75 deerhas beenattained after 5 years?After 10years?)

Usethe improved Euler method with a computer system to findthe desiredsolution values in Problems27and 28. Start with

step sizeh = 0.1,and then usesuccessivelysmaller step sizesuntil successiveapproximate solution values at x = 2 agreerounded off to four decimalplaces.27.y' = x2 + y2 -1,y(O)= 0; y(2) = ?28.y' = x +

\037y2, y(-2)= 0; y(2) =?29.Considerthe crossbowbolt of Example 2 in Section1.8,

shot straight upward from the ground with an initial veloc-ity of49mls.Becauseof linear air resistance,its velocityfunction vet) satisfiesthe initial value problem)

dv-= -(0.04)v-9.8, v(O) = 49dt)

with exactsolution vet) = 294e-t/25-245.Usea calcu-lator or computer implementation of the improved Eulermethod to approximate v (t) for 0 < t < lOusingbothn = 50 and n = 100subintervals. Display the resultsat intervals of 1 second. Do the two approximations-each rounded to two decimal places-agreeboth with

eachother and with the exactsolution? If the exactso-lution wereunavailable, explain how you could use the

improved Euler method to approximate closely (a) thebolt's time of ascentto its apex(given in Section1.8as4.56s)and (b) its impact velocity after 9.41s in the air.

30.Considernow the crossbowbolt of Example 3 in Section1.8.It still is shot straight upward from the ground with

an initial velocity of49mis,but becauseof air resistanceproportional to the squareof its velocity, its velocity func-tion v (t) satisfiesthe initial value problem)

dv-= -(O.OOll)vlvl-9.8, v(O) = 49.dt)

The symbolic solution discussedin Section1.8requiredseparateinvestigations of the bolt'sascentand its descent,with vet) given by a tangent function during ascentand

by a hyperbolic tangent function during descent. Butthe improved Euler method requires no such distinction.Usea calculator or computer implementation of the im-

proved Euler method to approximate v (t) for 0 < t < 10using both n = 100and n = 200 subintervals. Dis-play the results at intervals of 1 second. Do the two

approximations-eachrounded to two decimal places-agree with each other? If an exact solution were un-

available, explain how you could use the improved Eulermethod to approximate closely (a) the bolt'stime ofas-cent to its apex(given in Section1.8as4.61s)and (b) its

impact velocity after 9.41s in the air.)))

6.2Application)

6.2A CloserLookat the Euler Method 451)

I\037.p_\037\037\037\037d \037\037!.\037,\037., I\037.e.!.\037_!!!\037\037.\037ation)

Figure 6.2.10listsTI-85and BASICprograms implementing the improved Eulermethod to approximatethe solutionof the initial valueproblem)

dydx

= x+ y, y(O)= 1)

consideredin Example2 of this section.The commentsprovidedin the final columnshould make theseprograms intelligibleeven if you have little familiarity with the

BASICand TI programming languages.)

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,-, -\", :.- :\037-;.:i,-:-: :.,_ .->.-_'.-,.. :-,: \"-'-,_ '-,:':,- -

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ir!.BA.SIC') Comment),/:,' ::-.;: ',_: '::;'-

'->?:::?.;:\302\273:C::;:/h2\\,:;{\037:;)

':\037:': :.:: :\037:-:: -:;- :;-'- ;-' -. -:-,

;:':;':'-;-':-::,/::-::,:\037-:'_'..;._:',}:c)

PROGRAM:IMPEULER

:F=X+Y:10-+N:O-+X:l-+Y:1-+X1:(X1-X)/N-+H:For(I,l,N):Y-+YO:F-+K1:YO+H*K1-+Y:X+H-+X:F-+K2:(K1+K2)/2-+K:YO+H*K-+Y

:DispX,Y

:End)

ProgramIMPEULER

DEF FN F(X,Y) = X + Y

N - 10X = 0Y = 1Xl = 1H = (X1-X)/NFOR 1=1TO N

YO = Y

K1 = FNF(X,Y)Y = YO + H*K1X = X + H

K2 = FNF(X,Y)K = (K1+ K2)/2Y = YO + H*K

PRINT X,Y

NEXT I)

ProgramtitleDefine functionfNo. of stepsInitial xInitial yFinalxstepsizeBeginloopSave previousyFirst slopePredictorNew xSecondslopeAverage slopeCorrectorDisplayresultsEnd loop)

FIGURE6.2.10.TI-85and BASICimproved Euler programs.)

Toapply the improvedEulermethodto a differentialequationdy/dx = I(x,y),one needonly change the initial line of the program, in which the function I isde-fined. To increasethe number of steps(and therebydecreasethe stepsize)oneneedonly change the value of N specifiedin the secondline of the program.

Figure 6.2.11exhibits one MAT LAB implementation of the improved Eulermethod. The impeulerfunction takes as input the initial value x,the initial valuey, the final value xlof x, and the desirednumber n of subintervals. As output it

producesthe resulting column vectors X and Y of x-and y-values. For instance,the

MAT LAB command)

[X, Y] = impeuler(O,1, 1, 10)then generatesthe first and fourth columnsof data shown in Fig.6.2.4.

You should beginthis projectby implementing the improvedEuler methodwith your own calculator or computer system. Test your program by applying it

first to the initial valueproblemof Example1,then to someof the problemsfor this

section.)))

functionyp - f(x,y)yp = x + Yi) % yp = y')

function[X,Y] -h - (xl - x)/niX = XiY = Yifor i = l:ni)

kl = f(x,Y)ik2 = f(x+h,y+h*kl)ik - (kl + k2)/2iix = x + hiY - Y + h*kiX - [XiX]iY - [Yi y] iend)

impeuler(x,y,xl,n)% stepsize% initialx% initialy% beginloop% first slope% secondslope% average slope% new x% new y% update x-column% update y-column% end loop)

FIGURE6.2.11.MATLAB implementation of improved Euler method.)

FamousNumbersRevisited)

The following problemsdescribethe numbers e \037 2.7182818,In 2 \037 0.6931472,and Jr \037 3.1415927asspecificvaluesof certain initial valueproblems.In eachcase,apply the improvedEuler method with n == 10,20,40,...subintervals(doublingn

each time).How many subintervalsare neededto obtain-twicein succession-thecorrectvalue of the target number rounded to five decimalplaces?)

1.The number e == y (1),where y (x)is the solutionof the initial value problemdy/dx ==y, y(O) == 1.

2.The number In 2 == y(2), where y(x) is the solution of the initial value prob-lem dy /dx == l/x,y(l) ==O.

3.The number Jr ==Y (1),where y (x)is the solutionof the initial valueproblemdy/dx ==4/(1+x2), y(O) == O.)

LogisticPopulationInvestigation)

Apply your improvedEulerprogram to the initial valueproblemdy/dx ==\037y(8-y),

y(O) == 1 of Example3. In particular, verify (as claimed)that the approximatesolution with stepsizeh == 1 levels off at y \037 4.3542rather than at the limitingvalue y == 8 of the exact solution. Perhaps a table of values for 0 <x < 100will

make this apparent.For your own logisticpopulation to investigate,considerthe initial valueprob-)

lem)

dy 1

dx==

ny(m -y), y(O) == 1,)

where m and n are (for instance) the largest and smallestnonzero digits in yourstudent ID number. Doesthe improved Euler approximationwith stepsizeh == 1level off at the \"correct\"limiting value of the exact solution? If not, find a smallervalue of h so that it does.)))

2.0

1.5

1.0

;>-.. 0.5

0.0

-0.5-1.0

0 1 2 3 4 5t)

FIGURE6.2.12.Solution curvesofdy /dt = Y(1-y) -sin 2:rrt.)

6.3The Runge-KuttaMethod 453)

PeriodicHarvestingandRestockingThe differentialequation)

dy .(

2nt)-==ky(M-y)-hsln-dt P)

modelsa logisticpopulation that isperiodicallyharvestedand restockedwith periodP and maximal harvesting/restockingrate h. A numerical approximationprogramwas usedto plot the typical solution curves for the casek == M ==h == P == 1 that

are shown in Fig.6.2.12.This figure suggests-althoughit doesnot prove-theexistenceof a threshold initial populationsuch that). Beginning with an initial population above this threshold, the population os-

cillates(perhapswith periodP?) around the (unharvested)stable limiting

population yet) = M, whereas. The populationdiesout if it beginswith an initial populationbelowthis thresh-old.)

Usean appropriateplotting utility to investigateyour own logisticpopulation with

periodicharvesting and restocking(selectingtypical valuesof the parametersk, M,h, and P).Do the observationsindicatedhere appear to hold for your population?)- \037\037_!!.\037\037g\037-=\037\037!!\037__\037\037\037!t !?_\037________...___.__._ ________...__......_..... ..)

We now discussa method for approximating the solution y == y(x) of the initial

value problem)

dydx

==I(x,y), y(xo)==Yo) (1))

that is considerablymore accurate than the improved Euler method and is morewidely usedin practicethan any of the numericalmethodsdiscussedin Sections6.1and 6.2.It is calledthe Runge-Kuttamethod,after the Germanmathematicians who

developedit, Carl Runge (1856-1927)and Wilhelm Kutta (1867-1944).With the usual notation, supposethat we have computed the approximations

Yl, Y2, Y3, ...,Yn to the actual values Y(XI), Y(X2), Y(X3), ...,y(xn ) and now want

to compute Yn+l \037 Y(Xn+l). Then)

lxn +1 lxn+h

Y(Xn+l) -y(xn ) == y'(x)dx== y'(x)dxXn Xn)

(2))

by the fundamental theorem of calculus.Next, Simpson'srule for numerical inte-

gration yields)

h

[, ,

(h

),

]Y(Xn+l) -y(xn ) \037

6 Y (xn ) +4y Xn +2

+ y (Xn+l) .) (3))

Hencewe want to define Yn+ 1 so that)

h

[, ,

(h

),

(h

),

]Yn+l \037 Yn +6 Y (xn) +2y Xn +

2+2y Xn +

2+ y (Xn+l) ;) (4))))

we have split 4y' (xn + !h)into a sum of two terms becausewe intend to approxi-mate the slopey' (Xn +!h) at the midpoint Xn +!h of the interval [xn, Xn+1]in twodifferent ways.

On the right-hand side in (4), we replacethe [true] slope values y'(xn),y' (Xn +!h ), y' (xn +!h ), and y'(xn+1),respectively,with the followingestimates.)

\037) k l ==f(xn, Yn)) (5a))

. This is the Euler method slopeat Xn.)

\037) k2 ==f (Xn + !h,Yn + !hk1))

(5b)). This is an estimate of the slopeat the midpoint of the interval [Xn, Xn+1]usingthe Euler method to predictthe ordinate there.)

\037) k3 ==f(xn + !h,Yn + !hk2))(5c)). This is an improvedEuler value for the slopeat the midpoint.)

\037) k4 ==f (Xn+l, Yn + hk 3 )) (5d))

. This is the Eulermethod slopeat Xn+1,using the improved slopek3 at the

midpoint to stepto Xn+1.)

When thesesubstitutions are made in (4),the result is the iterative formula)

\037)

hYn+l ==Yn +-(k l +2k2 + 2k3 + k4).

6)(6))

The useof this formula to compute the approximationsYl, Y2, Y3, ...successivelyconstitutes the Runge-Kuttamethod.Note that Eq.(6)takes the \"Euler form\

Yn+l ==Yn + h .k)

if we write)

1k == -(k1 + 2k2 + 2k3 +k4)

6)(7))

for the approximateaverage slopeon the interval [xn, Xn+l].The Runge-Kutta method is afourth-order method-itcan beproved that the

cumulative error on a boundedinterval [a,b] with a ==Xo is of orderh 4. (Thus the

iteration in (6)is sometimescalledthe fourth-order Runge-Kutta method becauseit is possibleto developRunge-Kutta methodsof other orders.)That is,)

\037) Iy(xn) -Ynl

<Ch4,) (8))

where the constant C dependson the function f(x,y) and the interval [a,b], but

does not dependon the step size h. The following exampleillustrates this high

accuracy in comparisonwith the lower-orderaccuracy of our previous numericalmethods.)))

Example1)


We first apply the Runge-Kutta method to the illustrative initial value problem)

dydx ==x+y, y(O) == 1) (9))

that we consideredin Fig.6.1.8of Section6.1and again in Example 2 of Section6.2.The exact solutionof this problemis y (x) ==2ex -x-I.To make a point weuseh ==0.5,a larger stepsizethan in any previousexample,so only two stepsarerequiredto gofrom x ==0 to x == 1.

In the first stepwe usethe formulas in (5)and (6)to calculate)

k 1 ==0 + 1 == 1,k2 ==(0+0.25)+ (1+ (0.25).(1\302\273

== 1.5,k3 == (0+0.25)+(1+ (0.25). (1.5\302\273

== 1.625,k4 == (0.5)+ (1+ (0.5). (1.625\302\273

==2.3125,)

and then)

0.5Yl == 1 +-[1+2.(1.5)+2.(1.625)+2.3125]\037 1.7969.

6)

Similarly, the secondstepyieldsY2 \037 3.4347.Figure 6.3.1presentstheseresultstogether with the results (from Fig.6.2.4)

of applying the improvedEulermethod with stepsizeh == 0.1.We seethat evenwith the larger stepsize,the Runge-Kutta method gives (for this problem) four tofive times the accuracy(in terms of relativepercentageerrors)of the improved Eulermethod. .)

'-.;\";:-,;;-,-;:-.:;':;-,:;-.;:-:-'::':::':::-'::',::,.:',,',:,;,,

:il:\037II\037\037\037\0374ll\037\037i';[,;'

x.;','.:..i\037...\037i.,.c\037\037i1i..:'\037i,1.\037.':ijl\037.\037.'.r':.;!'.:.\037I\037I\037*,\037.,_\037i.jr....';)

Runge-Kutta:'1V,i'Wi\037hi1z

--n.5 PercentError Actual y)

0.00.51.0)

0.00%0.03%0.05%)

1.00001.79743.4366)

1.00001.79493.4282)

0.00%0.14%0.24%)

1.00001.79693.4347)

FIGURE6.3.1.Runge-Kutta and improved Euler results for the initial value problemdy/dx = x + y, y(O)= 1.)

It is customary to measure the computational labor involved in solvingdy/dx == I(x,y) numerically by counting the number of evaluationsof the func-tion I(x,y) that are required. In Example1,the Runge-Kutta method requiredeight evaluations of I(x,y) == x + y (four at each step),whereas the improvedEulermethod required20such evaluations (two for each of 10steps). Thus the

Runge-Kutta method gaveover four times the accuracywith only 40% of the labor.Computer programs implementing the Runge-Kutta method are listed in the

projectmaterialfor this section.Figure6.3.2showsthe resultsobtainedby applyingthe improved Euler and Runge-Kutta methods to the problem dy/dx == x + y,y (0)== 1 with the samestepsizeh ==0.1.The relativeerror in the improved Eulervalueat x == 1 isabout0.24%,but for the Runge-Kutta value it is0.00012%.In this

comparisonthe Runge-Kutta method is about 2000times as accurate,but requiresonly twice as many function evaluations,as the improvedEuler method.)))

asx approachesthe infinite discontinuity near x = 0.969811(seeFig.6.3.3).Nowwe apply the Runge-Kutta method to this initial valueproblem.

Figure 6.3.4showsRunge-Kutta resultson the interval [0.0,0.9],computedwith stepsizesh = 0.1,h = 0.05,and h = 0.025.There is still somedifficulty near

2.0 x = 0.9,but it seemssafe to concludefrom thesedata that y(0.5)\037 2.0670.)

Example2)

8)

6)

\0372)

o)

-2)

-4-2.0 -1.0 0.0x)

1.0)

FIGURE6.3.3.Solutions ofdy/dx = x2 + y2, y(O)= 1.)

0.10.20.30.40.50.60.70.80.91.0)

1.11001.24211.39851.58181.79492.04092.32312.64563.01243.4282)

1.1103421.2428051.3997171.5836481.7974412.0442362.3275032.6510793.0192033.436559)

1.1103421.2428061.3997181.5836491.7974432.0442382.3275052.6510823.0192063.436564)

FIGURE6.3.2.Runge-Kutta and improved Euler results for the initial valueproblem dy/dx = x + y, y(O)= 1,with the samestep sizeh = 0.1.)

The error bound)

Iy(xn ) -Ynl<Ch4)

(8))

for the Runge-Kutta method resultsin a rapid decreasein the magnitude of errorswhen the stepsizeh is reduced(exceptfor the possibilitythat very small stepsizesmay result in unacceptableroundofferrors).It followsfrom the inequality in (8)that

(on a fixed boundedinterval) halving the stepsizedecreasesthe absolute error by

a factor of(\037)

4 = /6. Consequently,the common practiceof successivelyhalvingthe stepsizeuntil the computed results \"stabilize\" is particularly effectivewith theRunge-Kutta method.)

In Example5 of Section6.1we saw that Euler'smethod is not adequate to approx-imate the solution Y (x)of the initial value problem)

dy-=x2+y2, y(O)= 1dx)

(10))

x),; \" ':\" -,: ::: ,: --c' ':',-::-:.:: .';' ;.:' -: \037.:.: ..::. :. -\"\037

.1ti\037i\037ti./i\037j;\037,!tQ\037;I.:H):\":,i,\037,j.'t1;'\037,:)\037:\\:\037\037:JJ\037(l:.',- -\".'.' '\037.:: ;.\"

-..,.-,-,)

,.,

..................,....\037........tti.....h..;j#,...fJ..f.)\037

1.11151.43972.06703.6529

14.3021)

0.10.30.50.70.9)

1.11151.43972.06703.6522

14.0218)

1.11151.43972.06703.6529

14.2712)

FIGURE6.3.4.Approximating the solution of the initial value problem in Eq.(10).)))

Example3)


We therefore beginanew and apply the Runge-Kutta method to the initial

value problem)

dydx

= x2 +l, y(0.5)= 2.0670. (11)Figure 6.3.5showsresults on the interval [0.5,0.9],obtained with step sizesh ==

0.01,h = 0.005,and h ==0.0025.We now concludethat y(0.9)\037 14.3049.)

;:X'i;':'),.

...,.;..........,.,

'l;..\037:\"i.:.n\037\"\"li\037J.I\037,\037\037::::;\037\".;ft\037.\037\037,i..:.....:.,) y with h -0.0025

2.06702.64403.65295.8486

14.3049)

...;..........'..;l;....\037.....\037tl1..h.......==.....O..OO5

2.06702.64403.65295.8486

14.3049)

:.?:::?::,::;;;;;::;;;:\037?:::',>;::\037;;:;::':

:'-',

'';)

0.50.60.70.80.9)

2.06702.64403.65295.8486

14.3048)

FIGURE6.3.5.Approximating the solution of the initial value problem in Eq. (11).)

Finally, Fig.6.3.6showsresultson the interval [0.90,0.95]for the initial valueproblem)

dydx

= x2 +l, y(0.9)= 14.3049, (12)

obtained using step sizesh == 0.002,h == 0.001,and h == 0.0005.Our final

approximate result is y(0.95)\037 50.4723.The actual value of the solution at x =0.95is y(0.95)\037 50.471867.Our slight overestimateresults mainly from the factthat the four-place initial value in (12)is (in effect) the result of rounding up the

actual value y(0.9) \037 14.304864;such errorsare magnified considerablyas weapproach the vertical asymptote. .)\037:<'\037::-.-:: <c. ::\037:\037>:'::'-\">'

>\",x.)

..:;\037;\037:\037i\\.fml_\037\\\037lli!:;:\037\0371;\037tlla;\037tl\037QfJi)ywith h.=O.0005

14.304916.702420.061725.107333.536350.4723)

0.900.910.920.930.940.95)

14.304916.702420.061725.107333.536350.4722)

14.304916.702420.061725.107333.536350.4723)

FIGURE6.3.6.Approximating the solution of the initial value problem in Eq. (12).)

\"...\037 .'...

A skydiver with a massof 60kg jumps from a helicopterhovering at an initial

altitude of 5 kilometers.Assume that she falls vertically with initial velocityzeroand experiencesan upward force FR of air resistancegiven in terms of her velocityv (in meters persecond)by

FR ==(0.0096)(100v+ 10v2 + v3))

(in newtons, and with the coordinate axisdirecteddownward so that v > 0 duringher descentto the ground). If she doesnot openher parachute, what will be herterminal velocity? How fast will shebe falling after 5 s have elapsed?After 10s?After 20s?)))


Solution)60)

40)

20)

S' 0\037)

-20)

-40)

-6\03760-40-20 0 20 40 60 80 100v)

FIGURE6.3.7.Graph of I(v)=9.8-(0.00016)(100v+ 10v2+ v 3).)

t{s)..,v(mls)

o 01 9.6362 18.3863 25.2994 29.9495 32.6786 34.1377 34.8758 35.2399 35.41510 35.500)

t{s)........v(lt1Is))

11 35.54112 35.56013 35.56914 35.57415 35.57616 35.57717 35.57818 35.57819 35.57820 35.578)

FIGURE6.3.8.Theskydiver'svelocity data.)

Example4)

Newton'slaw F ==ma gives)dv

mdt

==mg -FR ;)

that is,)dv60dt

= (60)(9.8)- (0.0096)(100v+ lOv2 + v3) (13)

becausem == 60and g == 9.8.Thus the velocity function vet) satisfies the initial

value problem)dv

dt==I(v), v(O)==0,) (14))

where)

I(v) ==9.8- (0.00016)(100v+ 10v2 + v3).) (15))

The skydiver reachesher terminal velocity when the forcesof gravity and airresistancebalance,soI(v) == O. We can therefore calculate her terminal velocityimmediatelyby solving the equation)

I(v) ==9.8- (0.00016)(100v+ 10v2 + v3) ==O.) (16))

Figure 6.3.7showsthe graph of the function I(v) and exhibits the singlereal so-lution v \037 35.5780(found graphicallyor by using a calculatoror computerSolveprocedure).Thus the skydiver='s terminal speedis approximately35.578mjs,about128kmjh (almost 80 mijh).

Figure 6.3.8showsthe resultsof Runge-Kutta approximationsto the solutionof the initial value problemin (14);the stepsizesh == 0.2and h == 0.1yield thesameresults(to three decimalplaces).Observethat the terminal velocity is effec-tively attained in only 15s.But the skydiver'svelocity is 91.85%of her terminal

velocity after only 5 s, and 99.78%after 10s. .)The final exampleof this sectioncontains a warning: For certain types of

initial value problems,the numerical methods we have discussedare not nearly sosuccessfulas in the previousexamples.)

Considerthe seeminglyinnocuous initial valueproblem

dy x

dx==5y -6e-, y(O) == 1) (17))

whoseexactsolution is y(x) == e-x . The table in Fig.6.3.9showsthe results ob-tained by applying the Runge-Kutta method on the interval [0,4] with stepsizesh == 0.2,h == 0.1,and h == 0.05.Obviously theseattempts are spectacularlyun-

successful.Although y (x) ==e-x ---+ 0 asx ---+ +00,it appearsthat our numerical

approximationsare headedtoward -00rather than zero.The explanationliesin the fact that the general solutionof the equationdyjdx =

5y -6e-x is)

y(x) ==e-x +Ce5x.) (18))

The particular solution of (17)satisfying the initial condition y(O) == 1 is ob-tained with C == O. But any departure, however small, from the exact solution

y(x) == e-x-evenif due only to roundoff error-introduces[in effect] a nonzerovalue of C in Eq. (18).And as indicated in Fig.6.3.10,all solution curves of the

form in (18)with C f::0 diverge rapidly away from the one with C == 0,even iftheir initial values are closeto 1. .)))

1,::ii:i!:iiil\"1i;t\037,\037::,(';,;\\I(\037rl\037l\037l\037YMw.

\037iii\302\245!!!rrt_lI1\037.li\037\\;\\ii ,i\"'<i\\iWjptl7ii\037'If.l)

Runge--Kuttay

Withh = 0.05) Actual y)

0.40.81.21.62.02.42.83.23.64.0)

0.668800.437130.21099

-0.46019-4.72142

-35.53415-261.25023

-1,916.69395-14059.35494

-103,126.5270)

0.670200.448330.293760.14697

-0.27026-2.90419-22.05352

-163.25077-1205.71249-8903.12866)

0.670310.449260.300670.198020.10668

-0.12102-1.50367

-11.51868-85.38156

-631.03934)

0.670320.449330.301990.201900.135340.090720.060810.040760.027320.01832)

FIGURE6.3.9.Runge-Kutta attempts to solvenumerically the initial value problem in

Eq.(17).)

2.5

2.0

1.5

1.0)\037)

0.5

0.0

-0.5

-1.0o 0.5 1.0 1.5 2.0 2.5 3.0

x)

FIGURE6.3.10.Directionfield andsolution curvesfor dy /dx = 5y

-6e-x .)

Difficultiesof the sort illustrated by Example4 sometimesare unavoidable,but one can at leasthopeto recognizesuch a problem when it appears.Approxi-mate valueswhoseorderof magnitude varies with changingstepsizeare a commonindicator of such instability. Thesedifficultiesare discussedin numerical analysistextbooksand are the subjectof current researchin the field.)

_ Problems)26.y' = -2xy,y(O)= 2;y(x) = 2e-x

7. y' = -3x2y,y(O)= 3;y(x) = 3e-x3

8.y' = e-Y , y(O)= 0; y(x) = In(x + 1)9.y' = *(1+ y2), Y(0)= 1;y (x) = tan *(x + Jr)

110.y' = 2xy 2,y(O)= 1;y(x) =1 2-x)

A hand-held calculatorwill suffice for Problems1through 10,where an initial value problemand its exactsolution aregiven.Apply the Runge-Kulla method to approximate this solution onthe interval [0,0.5]with stepsizeh = 0.25.Construct a tableshowing five-decimal-placevalues ofthe approximate solutionand actual solution at the points x = 0.25and 0.5.)

1.y' = -y,y(O)= 2;y(x) = 2e-X2.y' = 2y, y(O)= 4; y(x) = 4e2x

3.y' = y + 1,y(O)= 1;y(x) = 2ex - 14. y' = x -y, y (0) = 1;y (x) = 2e-x + x-I5.y'=y-x-1,y(0)=1;y(x)=2+x-ex)

Note: The application following this problem set lists illus-trative calculator/computer programs that can be usedin the

remaining problems.)))

A programmable calculatoror a computer will be useful forProblems11through 16.In eachproblem find the exactsolu-tion ofthe given initial value problem. Then apply the Runge-Kulla method twice to approximate (tofive decimalplaces)thissolution on the given interval, first with step sizeh = 0.2,thenwith step sizeh = 0.1.Make a table showing the approxi-mate values and the actual value, together with the percentageerror in the more accurateapproximation, for x an integralmultiple of0.2.Throughout, primes denotederivatives with

respectto x.)

11.y' = y -2,y(O)= 1;0 < x < 112.y' = 4(y - 1)2,Y(0) = 2;0 < x < 1

13.yy' = 2x3, Y(1)= 3; 1 < x < 214.xy' = y2, y(l) = 1;1 < x < 215.xy' = 3x - 2y, y(2) = 3;2 < x < 316.y2y'=2x5,y(2)=3;2< x < 3)

A computer with a printer is requiredforProblems17through24. In these initial value problems, use the Runge-Kullamethod with step sizesh =0.2,0.1,0.05,and 0.025to approx-imate to six decimalplacesthe values of the solution at fiveequally spacedpoints of the given interval. Print the resultsin tabular form with appropriateheadings to make it easyto

gauge the effectofvarying the stepsizeh. Throughout, primesdenotederivatives with respectto x.17.y' = x2 + y2, y(O)= 0;0 < x < 118.y' = x2 - y2, Y(0)= 1;0 < x < 219.y'=x+,JY,y(0)=1;0< x < 220.y'=x+\037,y(0)=-1;0< x < 221.y' = In y, y(l) = 2; 1 < x < 222.y' = x2/3 + y2/3, y(O)= 1;0 < x < 223.y' = sin x + cosy, y(O)= 0; 0 < x < 1

24. Y'= x

Y(-1) = 1.-1< x < 11+ y2

' , --25.As in Problem 25 of Section6.2,you bail out of a he-

licopter and immediately open your parachute, so yourdownward velocity satisfiesthe initial value problem)

dv

dt= 32-1.6v, v(O) = 0)

(with t in secondsand v in ftls). Usethe Runge-Kuttamethod with a programmable calculator or computer to

approximate the solution for 0 < t < 2, first with stepsizeh =0.1and then with h = 0.05,rounding off approx-imate v-values to three decimal places.What percentageof the limiting velocity 20 ftls has been attained after 1second?After 2 seconds?

26.As in Problem 26of Section6.2,supposethe deerpopu-lation pet) in a small forest initially numbers 25 and sat-isfiesthe logistic equation)

dPdt = 0.0225P-0.0003p2)

(with t in months). Usethe Runge-Kutta method with aprogrammable calculatoror computer to approximate thesolution for 10years, first with step sizeh = 6 and thenwith h = 3, rounding off approximate P-valuesto fourdecimal places.What percentageof the limiting popula-tion of 75 deerhas beenattained after 5 years?After 10years?)

Usethe Runge-Kulla method with a computer system to findthe desiredsolution values in Problems27and 28. Start with

step sizeh = 1,and then usesuccessivelysmaller step sizesuntil successiveapproximate solution values at x = 2 agreerounded off to five decimalplaces.27.y' = x2 + y2 - 1,y(O)= 0; y(2) =?28.y' = x + 4y2,y(-2)= 0; y(2) =?

Velocity-AccelerationProblemsIn Problems 29 and 30, the linear accelerationa = dv/dtof a moving particle is given by a formula dv/dt = f(t, v),where the velocity v = dy/dt is the derivative of the functiony = yet) giving the position of the particleat time t. Supposethat the velocity vet) is approximated using the Runge-Kuttamethod to solvenumerically the initial value problem

dv

dt= f(t,v), v(O) = Vo.) (19))

That is, starting with to = 0 and vo, the formulas in Eqs.(5)and (6) are applied-with t and v in place of x and y-tocalculatethe successiveapproximate velocity values VI, V2,

V3, ...,vm at the successivetimes tI, t2, t3, ...,tm (with

tn+I = tn + h). Now supposethat we alsowant to approximatethe distanceyet) traveled by the particle. We can do this by

beginning with the initial position y (0)= Yo and calculating)

Yn+l = Yn + vnh + 4an h 2)

(20))

(n = 1,2, 3, ...),where an = f (tn, vn ) \037 v'(tn ) is the parti-cle'sapproximate accelerationat time tn. Theformula in (20)would give the correctincrement (from Yn to Yn+ 1)if the accel-eration an remained constant during the time interval [tn, tn+d.

Thus, oncea table of approximate velocitieshas beencalculated,Eq. (20)providesa simple way to calculatea table

of correspondingsuccessivepositions. This processis illus-trated in the projectfor this section,by beginning with the ve-locity data in Fig. 6.3.8(Example 3) and proceedingto followthe skydiver's position during her descentto the ground.

29.Consideragain the crossbowbolt of Example 2 in Sec-tion 1.8,shot straight upward from the ground with an ini-

tial velocity of 49 m/s.Becauseof linear air resistance,its velocity function v = dy/dt satisfies the initial value

problem)

dv

dt= -(0.04)v-9.8, v(O) = 49

with exactsolution vet) = 294e-t125-245. (a) Usea cal-culator or computer implementation of the Runge-Kuttamethod to approximate v (t) for 0 < t < lOusing both)))

n = 100and n = 200subintervals. Display the results atintervals of 1 second.Do the two approximations-eachrounded to four decimal places-agreeboth with eachother and with the exactsolution? (b) Now usethe veloc-ity data from part (a) to approximate yet) for 0 < t < 10using n = 200 subintervals. Display the results at inter-vals of 1 second.Dotheseapproximate position values-eachrounded to two decimalplaces-agreewith the exactsolution)

Beginning with this initial value problem, repeatparts (a)through (c) of Problem 25 (exceptthat you may needn = 200 subintervals to get four-placeaccuracy in part(a) and n = 400 subintervals for two-placeaccuracyin

part (b\302\273. According to the results of Problems 17and 18in Section1.8,the bolt's velocity and position functions

during ascentand descentaregiven by the following for-

mulas.)

yet) = 7350(1-e-t/25) -245t?

(c) If the exactsolution wereunavailable, explain how

you could use the Runge-Kutta method to approximatecloselythe bolt'stimes ofascentand descentand the max-imum height it attains.

30.Now consideragain the crossbowbolt of Example 3 in

Section1.8.It still is shot straight upward from the groundwith an initial velocity of49m/ s,but becauseof air resis-tance proportional to the squareof its velocity, its velocityfunction vet) satisfiesthe initial value problem

dv

dt= -(O.OOll)vlvl-9.8, v(O) = 49.)

Ascent:vet) = (94.388)tan(0.478837- [0.103827]t),yet) = 108.465

+ (909.091)In (cos(0.478837- [O.103827]t\302\273 ;)

Descent:vet) = -(94.388)tanh(0.103827[t-4.6119]),yet) = 108.465- (909.091)In (cosh(0.103827[t-

4.6119]\302\273 .)

6.3Application)Runge-KuttaImplementation)

Figure 6.3.11lists TI-85and BASICprograms implementing the Runge-Kuttamethod to approximate the solutionof the initial value problem)

dydx =x+ y , y(O)= 1)

consideredin Example1 of this section.The commentsprovidedin the final columnshould make theseprograms intelligible even if you have little familiarity with the

BASICand TI programming languages.To apply the Runge-Kutta method to a different equationdy/dx = I(x,y),

one needonly change the initial line of the program, in which the function I isdefined.To increasethe number of steps(and thereby decreasethe stepsize),oneneedonly change the value of N specifiedin the secondline of the program.

Figure 6.3.12exhibits a MATLAB implementation of the Runge-Kuttamethod. Supposethat the function I describingthe differential equation y' =f (x,y) has beendefined. Then the rk function takes as input the initial value x,the initial value y, the final valuexlof x,and the desirednumber n of subintervals.As output it producesthe resulting column vectors X and Y of x-and y-values. Forinstance, the MATLAB command

[X, Y] = rk(O, 1, 1, 10)then generatesthe first and third columnsof data shown in the table in Fig.6.3.2.

You shouldbeginthis projectby implementing the Runge-Kuttamethod with

your own calculator or computer system.Test your program by applying it first tothe initial valueproblemin Example1,then to someof the problemsfor this section.)))

TI.;85<)

Cjmli1.,nt).'...;;......,........//;.'0.

.

.

.

...... ......

.

...

.....lit,c...

.

...

I....\037..........i..j...G......!......'.....,;/>..<::::-::;:,//::::::,::::\"';,,,,;/-'

\"\"->;-<:_:',:;:\037-;:\037....:-.-.-; ,,\"-' -\"'-'\" ,';,., -,)

PROGRAM:RK

:F=X+Y:10---+N:0---+X:1---+Y:1---+Xl:(Xl-X)/N---+H

:For(I,l,N):X---+ XO:Y---+ YO:F---+K1:XO+H/2---+X:YO+H*K1/2---+Y:F---+ K2:YO+H*K2/2---+Y

:F---+K3:XO+H---+ X:YO+H*K3---+Y:F---+ K4:(K1+2*K2+2*K3+K4)/6---+K:YO+H*K---+ Y

:DispX,Y

:End)

ProgramRK

DEF FN F(X,Y) - X + Y

N - 10X - 0Y - 1Xl = 1H = (X1-X)/NFOR 1=1TO N

XO - X

YO = Y

K1 = FNF(X,Y)X = XO + H/2Y = YO + H*K1/2K2 = FNF(X,Y)Y = YO + H*K2/2K3 = FNF(X,Y)X = XO + H

Y = YO + H*K3

K4 = FNF(X,Y)K = (K1+2*K2+2*K3

+K4)/6Y = YO + K*K

PRINT X,Y

NEXT I)

Program'ti'tleDefinefunc'tionfNo. of s'tepsIni'tialxIni'tialyFinalx

S'tepsizeBeginloopSave previousxSave previousyFirs'tslopeMidpoin'tMidp't predic'torSecondslopeMidp't predic'torThirdslopeNew x

Endp't predic'torFour'thslopeAverage slope)

Correc'torDisplayresul'tsEnd loop)

FIGURE6.3.11.TI-85and BASICRunge-Kutta programs.)

func'tionyp - f(x,y)yp = x + y;)

func'tion[X,Y] -h - (xl - x)/n;X = x;Y = y;for i =)

rk(x,y,x1,n))

l:nk1- f(x,y);k2 - f(x+h/2,y+h*k1/2);k3 - f(x+h/2,y+h*k2/2);k4 - f(x+h,y+h*k3);k - (k1+2*k2+2*k3+k4)/6;- x + h;- y + h*k;

[XiX];[Y;y];)

x)

yX -Y -end)

% yp - y')

% s'tepsize% ini'tialx% ini'tialy% beginloop% firs't slope% secondslope% 'thirdslope% four'th slope% average slope% new x% new y% upda'tex-column% upda'tey-column% end loop)

FIGURE6.3.12.MATLAB implementation of the Runge-Kutta method.)))

t($)

o2468101214161820)

V....\037rrVS\037:.............!..;...i.;;\037'\037.j;L.,'-,:'-'-:.: :,::-,--_:>\037_::,:\037_..:/, ,::\037

,:;,)

o18.38629.94934.13735.23935.50035.56035.57435.57735.57835.578)

o18.98468.825

133.763203.392274.192345.266416.403487.555558.710629.866)

FIGURE6.3.13.Theskydiver'svelocity and position data.)


FamousNumbersRevisited,OneLastTime)

The followingproblemsdescribethe numbers)

e \037 2.71828182846,In 2 \037 0.69314718056,and Jr \037 3.14159265359)

as specificvalues of certain initial value problems.In each case,apply the Runge-Kutta method with n == 10,20,40, ...subintervals(doublingn each time). Howmany subintervalsare neededto obtain-twicein succession-thecorrectvalue ofthe target number rounded to nine decimalplaces?)

1.The number e == y (1),where y (x)is the solutionof the initial value problemdy/dx ==y, y(O) == 1.

2.The number In 2 == y(2), where y(x) is the solutionof the initial value prob-lem dy/dx == l/x,y(1)==O.

3.The number Jr == Y(1),where y (x)is the solutionof the initial value problemdy/dx ==4/(1+x2), y(O) ==O.)

The Skydiver'sDescent)The following MATLAB function describesthe skydiver'saccelerationfunction in

Example3.)

function vp = f(t,v)vp = 9.8- 0.00016*(100*v+ 10*vA 2 + v A 3);

Then the commands)

k = 200[t,v] = rk(O, 20,0, k);[t(1:10:k+1);v(1:10:k+1)])

% 200subintervals% Runge-Kutta approximation% Displayevery 10thentry)

producethe table of approximatevelocities shown in Fig.6.3.8.Finally, the com-mands)

y - zeros(k+1,1):h - 0.1;for n = l:k

a = f(t(n),v(n\302\273:y(n+1) = y(n) + v(n)*h + 0.5*a*hA 2;

end[t(1:20:k+1),v(1:20:k+1),y(1:20:k+1)])

% initializey% stepsize% for n = 1 to k% acceleration% Equation (20)% end loop% each 20thentry)

carry out the position function calculations describedin Eg.(20)in the instruc-tions for Problems29 and 30. The resultsof these calculations are shown in

the table in Fig.6.3.13.It appearsthat the skydiver falls 629.866m during herfirst 20s of descent,and then free falls the remaining 4370.134meters to the

ground at her terminal speedof 35.578m/s. Henceher total time of descentis20+ (4370.134/35.578)\037 142.833s, or about 2 min 23 s.

For an individual problemto solve after implementing thesemethods usingan availablecomputer system, analyze your own skydive (perhapsfrom a different

height), using your own massm and a plausibleair-resistanceforce of the formFR ==av +bv 2 +cv3

.)))


11IINumericalMethodsfor Systems)We now discussthe numericalapproximationof solutionsof systemsof differentialequations.Our goal is to apply the methodsof Sections6.1through 6.3to the initial

value problem)

\037) X' ==f(t, x), x(to) ==Xo) (1))

for a system of m first-orderdifferentialequations.In (1)the independentvariableis the scalart, and)

x ==(XI,X2,...,xm ) and f== (/1,/2,...,/m))

are vector-valuedfunctions. If the component functions of f and their first -orderpartial derivatives are all continuous in a neighborhood of the point (to, xo),then

Theorems3 and 4 of the Appendix guarantee the existenceand uniquenessof asolutionx ==x(t)of (1)on somesubinterval [of the t-axis]containing to. With this

assurancewe can proceedto discussthe numericalapproximationof this solution.Beginning with stepsizeh, we want to approximate the values of x(t) at the

points tI, t2, t3, ..., where tn+I == tn + h for n > O. Supposethat we have alreadycomputed the approximations)

Xl, X2, x3, ..., xn

to the actual values

X(tI) , X(t2) , X(t3) , ..., x(tn ))

of the exactsolution of the system in (1).We can then make the stepfrom Xn tothe next approximationXn+I \037 X(tn+l) by anyone of the methods of Sections6.1through 6.3.Essentiallyall that is requiredis to write the iterative formula of the

selectedmethod in the vector notation of the presentdiscussion.)

EulerMethodsforSystemsFor example,the iterative formula of Euler'smethod for systemsis)

\037) Xn+I ==Xn +hf(t, xn ).) (2))

Toexamine the casem ==2 of a pair of first-orderdifferentialequations,let us write)

x =[\037 ] and f =

[ \037 J.)

Then the initial value problemin (1)is)

x'==f (t,x,y),y' ==g(t,x,y),)

x(to)==xo,

y(to) == Yo,)(3))

and the scalarcomponentsof the vector formula in (2)are)

Xn+I ==Xn +hf(tn , xn , Yn),

Yn+I ==Yn +hg(tn , Xn , Yn).)(4))))

Example1)

6.4NumericalMethodsfor Systems 465)

Note that eachiterative formula in (4)has the form of a singleEuler iteration, but

with Yn insertedlikea parameter in the first formula (forxn+I)and with Xn insertedlikea parameter in the secondformula (for Yn+l). The generalizationto the systemin (3)of eachof the other methods in Sections6.1through 6.3follows a similar

pattern.The improved Euler method for systemsconsistsat each stepof calculating

first the predictor)

\037) Un+l ==Xn +hf(t n , xn )) (5))

and then the corrector)

\037)

hXn+l ==Xn + -[f(tn , xn ) + f(tn+l, Un+l)].

2)(6))

For the caseof the two-dimensionalinitial value problem in (3),the scalarcompo-nents of the formulas in (5)and (6)are)

Un+l ==Xn +hf(tn , Xn , Yn),

Vn+l ==Yn +hg(tn , Xn , Yn))(7))

and)

hXn+l ==Xn +

2 [f(tn , Xn , Yn) + f(tn+l, Un+l, Vn+l)],)

(8))h

Yn+l ==Yn +2 [g(tn , Xn , Yn) +g(tn+l,Un+l, Vn+l)].)

...... ..... ......

Considerthe initial valueproblem)

x'==3x-2y,y' ==5x-4Y ,)

X(O) ==3;y(O) ==6.)

(9))

The exactsolutionof the system in (9)is

x(t) ==2e-2t +et,) yet) ==5e-2t +et

.) (10))

Herewe havef (x,y) ==3x-2Y and g(x,y) ==5x-4Y in (3),sothe Euler iterativeformulas in (4)are)

Xn+l ==Xn + h .(3xn - 2Yn), Yn+l == Yn + h .(5xn -4Yn).)

With stepsizeh ==0.1we calculate)

Xl ==3 + (0.1).[3 .3 -2 .6] ==2.7,YI ==6 + (0.1).[5 .3 -4 .6] ==5.1)

and)

X2 ==2.7+ (0.1).[3 .(2.7)-2.(5.1)]==2.49,Y2 ==5.1+ (0.1).[5 .(2.7)-4 .(5.1)]==4.41.)

The actual valuesat t2 ==0.2given by (10)arex(0.2)\037 2.562and y(0.2)\037 4.573.)))


To compute the improved Euler approximationsto x(0.2)and y(0.2)with asinglestepof sizeh = 0.2,we first calculate the predictors)

UI = 3 + (0.2).[3 .3-2.6]= 2.4,VI = 6 + (0.2).[5 .3-4.6] = 4.2.)

Then the correctorformulas in (8)yield)

Xl = 3 + (0.1).([3.3 - 2 .6] + [3 .(2.4)- 2 .(4.2)])= 2.58,YI = 6 + (0.1).([5.3-4.6]+ [5 .(2.4)-4.(4.2)])= 4.62.)

As we would expect,a singleimproved Euler stepgives better accuracy than two

ordinary Eulersteps. .)

The Runge-KuttaMethodandSecond-OrderEquations)The vector version of the iterative formula for the Runge-Kutta method is)

\037)

hxn+l = Xn + -(kl +2k2 +2k3+k4)

6)(11))

where the vectors k l , k2, k3, and k4 are defined (by analogy with Eqs.(5a)-(5d)ofSection6.3)as follows:)

\037)

k l = f(tn , xn ),k2 = f(tn + 4h , Xn + 4hk l)'k3 = f(tn + 4h , Xn + 4hk 2)'k4 = f(tn +h, Xn +hk 3).)

(12))

To describein scalar notation the Runge-Kutta method for the two-dimensional initial value problem)

X' = f(t,x,y),y' = get,x,y),)

x(to)= Xo,

y(to) = Yo,)(3))

let us write)

x =[\037l f =

[ \037 land k j = [ \037: l)Then the Runge-Kutta iterative formulas for the stepfrom (xn , Yn) to the next ap-proximation (Xn+l, Yn+l) \037 (X(tn+l), y(tn+l))are)

hXn+l = Xn +

6 (F1 +2F2 +2F3 + F4),

hYn+l = Yn +

6 (GI +2G2+2G3 + G4),)

(13))))

t)


where the values F1 , F2, F3, and F4 of the function J are)

Fl = J(tn , xn , Yn),

F2 = J (tn + 4h , Xn + 4hFI,Yn + 4hGI)'F3 = J(tn + 4h , Xn + 4hF2, Yn + 4hG2) ,

F4 = J(tn +h, Xn +hF3, Yn +hG3);)

(14))

0.51.01.52.02.53.03.54.04.55.0)

G 1 , G2, G3,and G4 are the similarly definedvaluesof the function g.Perhapsthe most common application of the two-dimensionalRunge-Kutta

method is to the numerical solution of second-orderinitial value problemsof the

form)

x\" = get,x,x'),x(to)= Xo, x'(to) = Yo.)

.......x:::..sint\";.}':#:i\037\037st;\037)

+0.47943+0.84147+0.99749+0.90930+0.59847+0.14112-0.35078-0.75680-0.97753-0.95892)

+0.87758+0.54030+0.07074-0.41615-0.80114-0.98999-0.93646-0.65364-0.21080+0.28366)

FIGURE6.4.1.Runge-Kuttavalues (with h = 0.05)for the

problem in Eg.(18).)

Example2)

Example3)

(15))

If we introduce the auxiliary variable Y = x',then the problem in (15)translatesinto the two-dimensionalfirst-orderproblem)

IX = Y,

y'=g(t,x,y),)x(to) = Xo,

y(to) = Yo.)(16))

This is a problemof the form in (3)with J(t,x,y) = y.If the functions J and g are not too complicated,then it is feasible to carry

out manually a reasonablenumber of stepsof the two-dimensionalRunge-Kuttamethod describedhere. But the first operating electroniccomputers were con-structed (during World War II) specifically to implement methods similar to the

Runge-Kutta method for the numericalcomputationof trajectoriesof artillery pro-jectiles.The applicationmaterial for this sectionlistsTI-85 and BASICversionsofProgramRK2DIMthat can be usedwith two-dimensionalsystems.)

The exactsolutionof the initial valueproblem)

x\" = -x; x(O)= 0, x'(0)= 1) (17))

is x(t) = sin t. The substitution y = x' translates (17)into the two-dimensionalproblem)

IX = y,

y' = -x,)x(O)= 0;y (0)= 1,)

(18))

which has the form in (3)with J(t,x,y) = y and get,x,y) = -x. The tablein Fig.6.4.1showsthe resultsproducedfor 0 < t < 5 (radians) using ProgramRK2DIMwith stepsizeh = 0.05.The values shown for x = sin t and y = cost

are all accurate to five decimalplaces. .)In Example4 of Section1.8we considereda lunar lander that initially is falling

freely toward the surface of the moon. Its retrorockets,when fired, provide a de-celeration of T = 4 mls2. We found that a soft touchdown at the lunar surfaceis achieved by igniting theseretrorockets when the lander is at a height of 41,870meters Gust over 26 miles)above the surface and is then descendingat the rate of450mls.)))


Now we want to compute the descenttime of the lunar lander.Letthe distancex(t) of the lander from the center of the moon be measuredin meters and measuretime t in seconds.According to the analysis in Section1.8(where we usedr(t)instead of x(t)),x(t) satisfiesthe initial valueproblem)

d2x GM 4.9044x 1012-==T--==4-dt2 x2 x2)

x(O)== R +41870== 1,781,870,x'(0)== -450)(19))

where G \037 6.6726X 10-11N.(m/kg)2is the universal gravitational constant andM ==7.35x 1022

kg and R == 1.74x 106 m are the massand radius of the moon.We seekthe value of t when x(t) == R = 1,740,000.

The problemin (19)is equivalentto the first-ordersystem)

dxdt

= y, x(0)== 1,781,870;)

dy 4.9044X 1012-=4-dx x2)y(O)= -450.)

(20))

The table in Fig.6.4.2showsthe result of a Runge-Kutta approximationwith stepsizeh = 1 (the indicated data agreeing with those obtained with stepsizeh = 2).Evidently, touchdown on the lunar surface (x = 1,740,000)occursat sometimebetweent = 180and t = 190seconds.The table in Fig.6.4.3showsa secondRunge-Kutta approximationwith t(O) = 180,x(O)= 1,740,059,y(O) = -16.83,and h = 0.1.Now it is apparent that the lander'stime of descentto the lunar surfaceis very closeto 187seconds;that is, 3 min 7 s. (The final velocity terms in thesetwo tablesare positive becausethe lander would beginto ascendif its retrorocketswerenot turned off at touchdown.) .)

t(s)) ,x'(mJ) '.\037(m)) vfmls,)

o20406080

100120140160180200)

1,781,8701,773,3601,765,8261,759,2641,753,6671,749,0331,745,3571,742,6371,740,8721,740,0591,740,199)

......:.......:..i\037.:...flZ\037)

-450.00-401.04-352.37-303.95-255.74-207.73-159.86-112.11-64.45-16.83

30.77)

t:;ts\037;\

180181182183184185186187188189190)

1,740,0591,740,0441,740,0301,740,0191,740,0111,740,0051,740,0011,740,0001,740,0011,740,0041,740,010)

-16.83-14.45-12.07-9.69-7.31-4.93-2.55-0.172.214.596.97)

FIGURE6.4.2.Thelander'sdescentto the lunar surface.)

FIGURE6.4.3.Focusing on thelunar lander'ssoft touchdown.)))

Example4)


Higher-OrderSystemsAs we saw in Section5.1,any system of higher-orderdifferential equationscan bereplacedwith an equivalent system of first-orderdifferentialequations. For exam-ple,considerthe system)

\" F(, '

)x = t,x,y,x,y,\" G(

, ')y = t,x,y,x,y) (21))

of second-orderequations.If we substitute)

x = Xl,), ,

X = X3 = Xl'), ,

y = X4 = X2,)y = X2,)

then we get the equivalentsystem)

,xl = X3,,

X2 = X4,

X\037

= F(t,Xl, X2, X3, X4),

X\037

= G(t,XI,X2,X3,X4))

(22))

of four first-orderequations in the unknown functions Xl (t) = X(t), X2(t) = yet),X3(t), and X4(t).It would be a routine (if slightly tedious)matter to write a four-dimensional version of program RK2DIMfor the purposeof solving such a sys-tem. But in a programming language that accommodatesvectors,an n-dimensionalRunge-Kutta program is scarcelymore complicatedthan a one-dimensionalpro-gram. For instance, the applicationmaterial for this sectionliststhe n-dimensionalMATLAB program rkn that closelyresemblesthe one-dimensionalprogram rk ofFig.6.3.11.)

_.,,\037.\037,.....-.... '\" \"'\037 .....\"...'0. \" \"\"'''''. V\"\"',,, \037U\"'\"'''' \" ....,.. .............\037 ..\".....vu.w.\"...... '\" .y....

Supposethat a batted ball starts at Xo = 0,Yo = 0 with initial velocity Vo = 160ftls and with initial angle of inclination () =

30\302\260. If air resistanceis ignored,wefind by the elementarymethodsof Section1.2that the baseballtravels a [horizontal]distanceof 400,J3\"ft (approximately693 ft) in 5 s before striking the ground. Now

supposethat in addition to a downward gravitational acceleration(g = 32 ftj s2),the baseballexperiencesan acceleration due to air resistanceof (0.0025)v2 feetper secondper second,directedoppositeto its instantaneousdirection of motion.Determine how far the baseballwill travel horizontally under theseconditions.)

Solution According to Problem 30 of Section5.1,the equationsof motion of the baseballare)

d2x dx-= -ev-,dt2 dt)

d2y dy-= -ev--gdt2 dt)

(23))

where v = J(x')2+ (y')2 is the speedof the ball, and where e 0.0025and

g = 32 in fps units. We convert to a first-ordersystemas in (22)and thereby obtainthe system)

,xl = X3,,x2 = X4,)

(24)),vi

2 2X3 = -ex3 X3 +X4'

,vi

2 2X4 = -eX4 X3 +X4 - g)))

of four first-orderdifferentialequations with)

Xl (0)= X2(0) = 0,X3(0) = 80v'3,X4(0) = 80.)

(25))

Note that X3(t) and X4(t ) are simply the x- and y-components of the baseball'svelocityvector,so v = Jx1+xi.We proceedto apply the Runge-Kutta method toinvestigate the motion of the batted baseballdescribedby the initial value problemin (24)and (25),first taking c = 0 to ignoreair resistanceand then using c = 0.0025to take air resistanceinto account.)

WITHOUTAIR RESISTANCE:Figure6.4.4showsthe numericalresultsobtainedwhen a Runge-Kutta programsuchasrkn isappliedwith stepsizeh = 0.1and with

c = 0 (no air resistance).For convenience in interpreting the results,the printedoutput at eachselectedstep consistsof the horizontal and vertical coordinates X

and y of the baseball,its velocity v, and the angle of inclination a of its velocityvector (in degreesmeasuredfrom the horizontal).Theseresultsagreewith the exactsolution when c = O. The ball travels a horizontal distanceof 400,J3\"\037 692.82ft

in exactly5 s, having reacheda maximum height of 100ft after 2.5s.Note alsothat

the ball strikesthe ground at the sameangle and with the same speedas its initial

angle and speed.)

\", .,. -.--._-,- -,-\";.:. \037:-; :., --: :.;

,,'-;'

-;.'

--,\" .' '-' -\" ,

,'.',...--..,.....--....-.---.-.-..'..'-\"-.,

t oX ....1)', ...\037/ ......(i..........\"'.a:...<.;.'\" ':. ,.;;....;' ;.' :-- .,. .\"\037

0.0 0.00 0.00 160.00 +30 ,',\",-\"..--..'------.--- - - -.-,

;1:(......;;.....:;;v\037:....... y v IX0.5 69.28 36.00 152.63 +251.0 138.56 64.00 146.64 +19 0.0 0.00 0.00 160.00+301.5 207.85 84.00 142.21+13 0.5 63.25 32.74 127.18 +242.0 277.13 96.00 139.48 +7 1.0 117.11 53.20 104.86 +172.5 346.41 100.00 138.56 +0 1.5 164.32 63.60 89.72 +83.0 415.69 96.00 139.48 -7 2.0 206.48 65.30 80.17 -33.5 484.97 84.00 142.21 -13 2.5 244.61 59.22 75.22 -154.0 554.26 64.00 146.64 -19 3.0 279.29 46.05 73.99 -274.5 623.54 36.00 152.63 -25 3.5 310.91 26.41 75.47 -375.0 692.82 0.00 160.00 -30 4.0 339.67 0.91 78.66 -46)

FIGURE6.4.4.Thebatted baseballwith no airresistance(c = 0).)

FIGURE6.4.5.Thebatted baseballwith airresistance(c = 0.0025).)

WITH AIR RESISTANCE:Figure6.4.5showsthe resultsobtainedwith the fairlyrealisticvalue of c = 0.0025for the air resistancefor a batted baseball.To within ahundredth of a foot in either direction, the sameresultsare obtained with stepsizesh = 0.05and h = 0.025.We now seethat with air resistancethe ball travels adistancewell under 400 ft in just over 4 s.The more refineddata in Fig.6.4.6showthat the ball travels horizontally only about 340 ft and that its maximum height isonly about 66 ft. As illustrated in Fig.6.4.7,air resistancehas converteda massivehome run into a routine fly ball (if hit straightaway to center field). Note alsothat

when the ball strikesthe ground, it has slightly under half its initial speed(onlyabout 79 ft/s) and is falling at a steeperangle (about 46\302\260). Every baseballfan hasobservedempirically theseaspectsof the trajectoryof a fly ball.)))

1.5 164.32 63.60 89.72 +81.6 173.11 64.60 87.40 +51.7 181.72 65.26 85.29 +31.8 190.15 65.60 83.39 +11.9 198.40 65.61 81.68 +--Apex-12.0 206.48 65.30 80.17 -3)

3.83.94.04.14.2)

328.50334.14339.67345.10350.41)

11.776.450.91-4.84-10.79)

77.2477.9378.6679.4380.22)

-42-44-46+--Impact-47-49)

The massive home run)

FIGURE6.4.6.Thebatted ball'sapexand its impact with the

ground.)

FIGURE6.4.7.An \"easy out\" ora home run?)

VariableStepSizeMethods)The Runge-Kutta methodfor a large systemrequiresan appreciableamount ofcom-putational labor, even when a computer is employed.Therefore, just as the stepsizeh should not be so large that the resulting error in the solution is unaccept-able, h ought not to be so small that too many stepsare needed,hence requiringan unacceptableamount of computation. Thus the practical numericalsolution ofdifferentialequations involvesa tradeoffbetweenaccuracyand efficiency.

To facilitate this tradeoff, modemvariable step sizemethods vary the stepsizeh as the solution processproceeds.Largestepsare taken in regions wherethe dependentvariables are changing slowly; smaller stepsare taken when thesevariablesare changing rapidly, in orderto prevent largeerrors.

An adaptableor variable stepsizeRunge-Kutta methodemploysboth a pre-assignedminimum error toleranceMinToland a maximum error toleranceMaxTolto attempt to ensurethat the error made in the typical stepfrom Xn to Xn+l is nei-ther too large (and henceinaccurate)nor too small (and henceinefficient).A fairly

simpleschemefor doing this may be outlined as follows:). Having reachedXn with a Runge-Kutta stepof length tn- tn-l = h, let x(l)

denotethe result of a further Runge-Kutta stepof length h and let X(2) denotethe result of two successiveRunge-Kutta stepseach of length h/2..On the grounds that x(2) shouldbea more accurateapproximationto x(tn + h)than is x(1) take,)

Err= Ix(1)-X(2)I)

as an estimate of the error in X(l).. If MinTol<Err<MaxTol,then let Xn+l = x(l),tn+l = tn + h, and proceedto the next step.. If Err <MinTol,then the error is too small!Hencelet Xn+l = x(l),tn+l =tn +h, but doublethe stepsizeto 2h before making the next step.. If Err>MaxTol,then the error is too large.Hencerejectx(l) and start afreshat Xn with the halved stepsizeh /2.)))


/////)

X2 S(x1'x2)/\037/ \\

rE / \\/\\

// \\

/ \\/ \\rM

\\

\\

\\

\\

\\

M(1- \037, 0)X

1)E(-\037, 0))

FIGURE6.4.8.TheEarth-Moon center-of-masscoordinate system.)

The detailedimplementation of such a schemecan be complicated.For amuch more completebut readablediscussionof adaptive Runge-Kutta methods,seeSection15.2of William H.Presset aI.,NumericalRecipes:TheArt ofScientificComputing (New York: CambridgeUniversity Press,1986).

Severalwidely availablescientificcomputingpackages(suchasMaple,Math-ematica, and MATLAB) include sophisticatedvariable stepsizeprograms that will

accommodate an essentiallyarbitrary number of simultaneous differential equa-tions. Sucha general-purposeprogram might be used,for example,to model nu-

merically the major componentsof the solarsystem:the sun and the nine (known)major planets. If mi denotesthe massand ri = (Xi, Yi, Zi) denotesthe positionvector of the ith one of these10bodies,then-by Newton'slaws-theequationofmotion of mi is)

\" \037 Gmimjmi ri =

\037 3 (rj- ri),ji=i (rij ))

(26))

where rij = Irj - r;j denotesthe distancebetween mi and mj. For each i = 1,2,...,10,the summation in Eq. (26)is over all values of j i= i from 1 to 10.The 10vector equations in (26)constitute a systemof 30 second-orderscalarequa-tions, and the equivalent first-order system consistsof 60 differential equations in

the coordinatesand velocitycomponentsof the 10major bodiesin the solarsystem.Mathematical modelsthat involve this many (or more) differentialequations-andthat require sophisticatedsoftware and hardware for their numerical analysis-arequite common in science,engineering,and appliedtechnology.)

Earth-MoonSatelliteOrbits)For an exampleof a program whoseefficient solution requiresadaptive stepsizemethods,we consideran Apollo satellite in orbit about the Earth E and Moon M.Figure 6.4.8showsan xlx2-coordinatesystem whoseorigin liesat the center ofmassof the Earth and the Moonand which rotates at the rate of one revolution per\"moon month\" of approximately T = 27.32days,so the Earth and Moon remainfixed in their positionson the xl-axis.If we take asunit distancethe distance(about384,000kilometers,assumedconstant) betweenthe Earth and Moon centers,then

their coordinatesare E(-f1,0) and M(1- f1,0),where f1 = mMj(mE +mM) in

terms of the Earth massmE and Moonmassm M. If we take the total massmE +m Mas the unit of massand Tj(2n) \037 4.35days as the unit of time, then the gravitationalconstant is G = 1 in Eq. (26),and the equations of motion of the satellite positionS(XI,X2) are)

X\" -X 2x'_ (1- f1)(XI + f1) _ f1(XI- 1 + f1)

1- 1 + 2 (rE )3 (rM)3

')

(27))

X\" = X _ 2x'_ (1- f1)X2 _ f1X22 2 I

(rE)3 (rM)3')

where rE and rM denotethe satellite'sdistanceto the Earth and Moon (indicatedin Fig.6.4.8).The initial two terms on the right-hand sideof each equation resultfrom the rotation of the coordinatesystem.In the systemof units describedhere,thelunar massis approximatelymM = 0.012277471.The second-ordersystem in (27)can be converted to an equivalentfirst-ordersystem (of four differentialequations)by substituting)

,xl = X3,)\" ,x2 = x4.)

,x2 = X4,) so that) X\" -x,I-

3')))

X2)

FIGURE6.4.9.Apollo Moon-Earth bus orbit with insertion

velocity Vo = 7476km/h.)

FIGURE6.4.10.Apollo Moon-Earth bus orbit with insertion

velocity Vo = 7365kmlh.)


Supposethat the satellite initially is in a clockwisecircular orbit of ra-dius about 2400kilometersabout the Moon. At its farthest point from the Earth

(Xl = 0.994)it is \"launched\" into Earth-Moonorbit with initial velocity Vo. Thecorrespondinginitial conditionsare)

Xl (0)= 0.994,X2(0) = 0, X3(0) = 0, X4(0) = -Vo.)

Xl)

An adaptive stepsizemethod (ode4S)in the MATLAB software system was usedto solve numerically the system in (27).The orbits in Figs.6.4.9and 6.4.10wereobtained with)

Vo = 2.031732629557and Vo = 2.001585106379,)

Xl)

respectively.[In the system of units used here,the unit of velocity is approxi-mately 3680km/h.]In each casea closedbut multiloopedperiodictrajectoryaboutthe Earth and the Moon-aso-calledbusorbit-isobtained, but a relatively smallchange in the initial velocity changesthe number of loops!For more information,seeNASA ContractorReport CR-61139,\"Study of the Methodsfor the NumericalSolutionof OrdinaryDifferentialEquations,\" preparedby O.B.Francis,Jr. et al. forthe NASA-GeorgeC.Marshall SpaceFlight Center,June 7, 1966.

So-calledMoon-Earth\"bus orbits\" are periodic-thatis, are closedtrajecto-riestraversedrepeatedlyby the satellite-onlyin a rotating xlx2-coordinatesystemas discussedabove. The satellite of Fig.6.4.9traverses its closedorbit and returns

to rendezvous with the Moon about 48.4days after its insertion into orbit. Fig-ures6.4.11and 6.4.12illustrate the motion of the samesatellite-butin an ordinary

nonrotating xy-coordinatesystem centeredat the Earth, in which the Moon encir-clesthe Earth counterclockwisein a near-circularorbit, completingone revolutionin about 27.3days.The Moonstarts at point S,and after 48.4days it has completeda bit over 1.75revolutionsabout the Earth and reachesthe point R at which its ren-dezvous with the satellite occurs.Figure 6.4.11showsthe positionsof Moon and

satellite a day and a half after the satellite'sinsertion into its orbit, each traveling in

a generally counterclockwisedirection around the Earth. Figure 6.4.12showstheir

positionsa day and a half before their rendezvous at point R, the satellitemean-while having encircledthe Earth about 2.5times in an orbit that (in the indicatedxy-coordinatesystem)appearsto resemblea slowly varying ellipse.)

y) y)--.,,\"\"\"

\"\"

\"/IIII,I\\

\\

\\

\\

\\ ,....,,........-)

--.,,\"\"\"

\"\"

\"/IIII,I\\

\\

\\

\\

\\ ,....,'....)

-............,....,\\\\

Moon

\\

ISX)

R)

-........,,....,\\

\\

\\

\\

\\

ISX)

FIGURE6.4.11.Themoon andsatellite in a nonrotating coordinatesystem, 1.5days after orbital insertionof the satelliteat starting point S.)

FIGURE6.4.12.Themoon andsatellite in a nonrotating coordinate

system, 1.5days beforetheirrendezvous at point R.)))


III]Problems)A hand-held calculatorwill suffice for Problems1through 8.In eachproblem an initial value problem and its exactsolu-tion are given. Approximate the values ofx(0.2)and y (0.2)in three ways: (a) by the Euler method with two stepsofsizeh =0.1;(b) by the improved Euler method with a single stepofsizeh = 0.2;and (c) by the Runge-Kutta method with asingle step ofsizeh = 0.2.Comparethe approximate valueswith the actual values x(0.2)and y(0.2).1.x'= x + 2y, x(O)= 0,

y' = 2x + y, y(O)= 2;x(t)= e3t -e-t , yet) = e3t + e-t

2.x'= 2x + 3y, x(O)= 1,y' = 2x + y, y (0)= -1;x(t) =e-t , yet) = -e-t

3.x'= 3x + 4y, x(O)= 1,y' = 3x +2y, y(O)= 1;x(t) =

\037(8e6t -e-t),yet) =

\037(6e6t + e-t )

4. x'= 9x+ 5y, x (0)= 1,y' = -6x-2y, y(O)= 0;x(t) = -5e3t + 6e4t , yet) = 6e3t -6e4t

5. x'= 2x -5y, x(O)= 2,y' = 4x-2y, y(O)= 3;x(t) = 2cos4t - Ii sin4t, yet) = 3cos4t+

\037

sin4t6.x'=x -2y,x(0)= 0,

y' = 2x + y, y(O)= 4;x(t) = -4et sin 2t, yet) = 4et cos2t

7. x'= 3x - y, x(O)= 2,y' =x + y, y(O)= 1;x(t) = (t + 2)e2t , yet) = (t + l)e2t

8.x'= 5x -9y, x (0)= 0,y' = 2x - y, y(O) = -1;x(t) = 3e2t sin3t, yet) = e2t (sin3t-cos3t))

A computer will be requiredfor the remaining problems in thissection.In Problems 9 through 12,an initial value problemand its exactsolution are given. In eachof thesefour prob-lems, use the Runge-Kutta method with step sizesh = 0.1andh =0.05to approximate to five decimalplacesthe values x(l)and y(I). Comparethe approximations with the actual values.

9.x'= 2x- y, x (0)= 1,y' = x + 2y, y(O)= 0;x(t) = e2t cost, yet) = e2t sin t

10.x'=x + 2y, x(O)= 0,y' =x + e-t , y (0)= 0;x(t) =

\037(2e2t -2e-t + 6te-t),

yet) =\037(e2t -e-t + 6te-t )

11.x'= -x-y- (1+ t 3)e-t , x(O)= 0,

y' = -x-y- (t - 3t2)e-t , y(O)= 1;

x(t)= e-t (sin t - t), yet) = e-t (cost + t 3)12.x\" + x = sin t, x(O)= 0;x(t) =

\037(sint - t cost)

13.Supposethat a crossbowbolt is shot straight upward with

initial velocity 288 ftl s.If its decelerationdue to air re-sistanceis (0.04)v, then its height x(t) satisfiesthe initial)

value problem)

x\" = -32- (0.04)x';x(O)= 0, x'(0)= 288.)

Find the maximum height that the bolt attains and the time

required for it to reachthis height.14.RepeatProblem 13,but assume instead that the decelera-

tion of the bolt due to air resistanceis (0.0002)v2.15.Supposethat a projectileis fired straight upward with ini-

tial velocity Va from the surface of the earth. If air resis-tance is not a factor, then its height x(t)at time t satisfiesthe initial value problem)

d2xdt 2)

gR2 .(x + R)2

') x(O)= 0, x'(0)= Va.)

Usethe values g = 32.15ft/s 2\037 0.006089mi/s2 for the

gravitational accelerationof the earth at its surface and

R = 3960mi as the radius of the earth. If Va = 1 mils,find the maximum height attained by the projectileand its

time ofascentto this height.)

Problems 16through 18deal with the batted baseballofEx-ample 4, having initial velocity 160ftIsand air resistanceco-efficient c = 0.0025.16.Find the range-thehorizontal distancethe ball travels be-

fore it hits the ground-andits total time of flight with

initial inclination angles 40\302\260 , 45\302\260 , and 50\302\260 .17.Find (to the nearest degree)the initial inclination that

maximizes the range. If there were no air resistanceitwould beexactly 45\302\260,

but your answer should be lessthan

45\302\260 .18.Find (to the nearest half degree)the initial inclination an-

glegreater than 45\302\260 for which the range is 300ft.

19.Find the initial velocity of a baseballhit by BabeRuth

(with c = 0.0025and initial inclination40\302\260)

if it hit the

bleachersat a point 50ft high and 500horizontal feet from

home plate.20.Considerthe crossbowbolt of Problem 14,fired with the

sameinitial velocity of288ftls and with the air resistancedeceleration(0.0002)v2 directedoppositeits direction ofmotion. Supposethat this bolt is fired from ground levelat

an initial angle of 45\302\260. Find how high vertically and how

far horizontally it goes,and how long it remains in the air.

21.Supposethat an artillery projectileis fired from groundlevel with initial velocity 3000ftls and initial inclination

angle 40\302\260. Assume that its air resistancedecelerationis(0.0001)v2. (a) What is the range of the projectileand

what is its total time of flight? What is its speedat impactwith the ground? (b) What is the maximum altitude ofthe projectile,and when is that altitude attained? (c) You

will find that the projectileis still losing speedat the apexof its trajectory. What is the minimum speedthat it attains

during its descent?)))

6.4Application)


.Comets.andSp\037\037ecraft)

Figure 6.4.13listsTI-85and BASICversionsof the two-dimensionalRunge-Kuttaprogram RK2DIM.You should note that it closelyparallelsthe one-dimensionalRunge-Kutta program listedin Fig.6.3.11,with a singleline there replaced(whereappropriate) with two lineshere to calculate a pairof x- and y-values or slopes.Note also that the notation used is essentiallythat of Eqs. (13)and (14)in this

section.The first several linesdefine the functions and initial data neededfor Ex-ample 1.

Figure 6.4.14exhibits an n-dimensional MATLAB implementation of the

Runge-Kutta method. The MATLAB function f defines the vector of right-handsidesof the differentialequations in the system x' = f(t, x) to be solved.The rknfunction then takesas input the initial t-value t, the column vector x of initial x-values, the final t-value t1,and the desirednumber n of subintervals. As outputit producesthe resulting column vector T of t-values and the matrix x whoserowsgive the correspondingx-values.For instance, with f as indicated in the figure, theMATLAB command)

[T,X] = rkn(O, [0;1],5, 50)then generatesthe data shown in the table of Fig.6.4.1(which listsonly every fifth

value of eachvariable).You can use Examples1 through 3 in this sectionto test your own imple-

mentation of the Runge-Kutta method. Then investigate the comet and space-craft problemsdescribednext. Additional application material at the Web sitewww.prenhall.com/edwardsdescribesadditional numerical ODE investiga-tions ranging from batted baseballsto the Apollo orbits shown in Figs.6.4.9and6.4.10.)

Your SpacecraftLandingYour spacecraft is traveling at constant velocity V, approachinga distant earthlike

planet with massM and radius R. When activated,your decelerationsystem pro-vides a constant thrust T until impact with the surface of the planet. During the

periodof deceleration,your distancex(t) from the center of the planet satisfiesthedifferentialequation)

d2X GM--T--dt2 - x2 ') (1))

where G \037 6.6726X 10-11N.(m/kg)2as in Example3. Your question is this:At what altitude above the surface should your decelerationsystembe activated in

orderto achievea soft touchdown?For a reasonableproblem,you can take)

M = 5.97X 1024(kg),R = 6.38X 106 (m),V = p X 104 (km/h),T = g +q (m/s2))

where g = GM/R2 is the surface gravitational accelerationof the planet. Choosepto be the smallestnonzero digit and q the next-to-smallestnonzerodigit in your IDnumber. Find the \"ignition altitude\" accurate to the nearest meter and the resulting\"descenttime\" accurate to the nearest tenth of a second.)))


TI.8S) ...... Comment), ..,...........\"\302\273\"\037I\037:st\037:)

PROGRAM:RK2DIM)

:F=Y)

:G=-X:50---+N:0---+T:0---+X

:1---+Y:5---+Tl:(TI-T)/N---+H

:For(I,I,N):T---+TO:X---+XO

:Y---+YO

:F---+Fl:G---+Gl:TO+H/2---+T:XO+H*Fl/2---+X:YO+H*Gl/2---+Y

:F---+F2

:G---+G2:XO+H*F2/2---+X:YO+H*G2/2---+Y:F---+F3:G---+G3:TO+H---+T:XO+H*F3---+X

:YO+H*G3---+Y:F---+F4

:G---+ G4:(Fl+2*F2+2*F3

+F4)/6---+FA

:(Gl+2*G2+2*G3+G4)/6---+GA:XO+H*FA---+X

:YO+H*GA---+ Y

:DispT,X,Y:End)

ProgramRK2DIM

DEF FN F(T,X,Y) - Y

DEF FN G(T,X,Y)- -XN - 50T - 0X - 0Y - 1Tl = 5H = (TI-T)/NFOR 1=1TO N

TOXO

YO

FlGl)

ProgramtitleDefine functionfDefine functiongNo. of stepsInitial tInitial x

Initial yFinaltStepsizeBeginloopSave previoustSave previousxSave previousyFirst f-slopeFirstg-slopeMidpointtMidpt x-predictorMidpt y-predictorSecondf-slopeSecondg-slopeMidpt x-predictorMidpt y-predictorThird f-slopeThirdg-slopeNew tEndpt x-predictorEndpt y-predictorFourth f-slopeFourth g-slopeAverage f-slope)

T -X -Y -)

- T- X- Y- FNF(T,X,Y)- FNG(T,X,Y)TO + H/2XO + H*Fl/2YO + H*Gl/2)

F2 = FNF(T,X,Y)G2 = FNG(T,X,Y)X - XO + H*F2/2Y = YO + H*G2/2F3 = FNF(T,X,Y)G3 = FNG(T,X,Y)

TO + H

XO + H*F3YO + H*G3- FNF(T,X,Y)- FNG(T,X,Y)

FA - (Fl+2*F2+2*F3+F4)/6

GA - (Gl+2*G2+2*G3+G4)/6

X - YO + H*FA

Y - YO + H*GA

PRINT T,X,YNEXT I)

T -X -Y -F4G4)

Average g-slope)

x-correctory-correctorDisplayresultsEnd loop)

FIGURE6.4.13.TI-85and BASICtwo-dimensional Runge-Kutta programs.)))


functionxp = f(t,x)xp = X;xp(l)- x(2);xp(2)= -x(l);)function[T,Y] - rkn(t,x,tl,n)h - (tl - t)/n;T = t;X = x';for i = l:n)

kl - f(t,x);k2 - f(t+h/2,x+h*kl/2);k3 - f(t+h/2,x+h*k2/2);k4 - f(t+h ,x+h*k3 ) ;k - (kl+2*k2+2*k3+k4)/6;t - t + h;x - x + h*k;T - [T;t];X - [XiX'];end)

% stepsize% initialt% initialx-vector% beginloop% first k-vector% secondk-vector% third k-vector% fourthk-vector% average k-vector% new t% new x% update t-column% update x-matrix% end loop)

FIGURE6.4.14.MATLAB implementation of the Runge-Kutta method.)

Kepler'sLawofPlanetary(orSatellite)Motion)

Considera satellite in elliptical orbit around a planet of massM, and supposethat

physical units are so chosenthat GM= 1 (where G is the gravitational constant).If the planet is locatedat the origin in the xy-plane,then the equationsof motion ofthe satellite are)

d2xdt 2)

X d2y

(x2 +y2)3/2'

dt2)

Y

(x2+y2)3/2.) (2))

Let T denote the periodof revolution of the satellite.Kepler'sthird law says that

the squareof T is proportional to the cubeof the major semiaxisa of its ellipticalorbit. In particular, if GM = 1,then)

T2 = 4n2a3.) (3))

(For details,seeSection11.6of Edwardsand Penney, Calculus:Early Transcen-dentals,7th ed.(UpperSaddleRiver, NJ:Prentice Hall,2008).)If the satellite'sx-and y-componentsof velocity,X3 = x' =

x\037and X4 = y' =

x\037,are introduced,then

the systemin (2)translates into a system of four first-orderdifferential equationshaving the form of those in Eq.(22)of this section.

(a) Solvethis 4 x 4 system numerically with the initial conditions)

x(O)= 1, y(O)= 0, x'(O)= 0, y'(O)= 1)

that correspondtheoreticallyto a circularorbit of radius a = 1,soEq.(3)givesT = 2n.Isthis what you get?)))


z)

-5)

-10)

FIGURE6.4.15.yz-projectionof the orbit ofHalley'scomet.)

(b) Now solve the systemnumericallywith the initial conditions)

x(O)= 1, y(O)= 0, x'(0)= 0, y'(0)= \037,J6)

that correspondtheoretically to an elliptical orbit with major semiaxisa = 2,soEq.(3)gives T = 4n,J2.Isthis what you get?)

Halley'sCometHalley'scomet last reachedperihelion (its point of closestapproachto the sun at theorigin) on February 9,1986.Itsposition and velocitycomponentsat that time were)

Po = (0.325514,-0.459460,0.166229)and

Vo = (-9.096111,-6.916686,-1.305721))(respectively),with positionin AU (astronomicalunits, in which the unit of distanceis the major semiaxisof the earth'sorbit) and time in years.In this system,the three-dimensionalequations of motion of the comet are)

d2x f1X d2y f1Y d2z f1Z- -- - -- - --

dt2 3 ' dt2 3 ' dt2 r 3r r

where)

(4))

f1 = 4n2 and r = Jx2 + y2 +Z2.

Solvethe equations in (4)numerically to verify the appearanceof the yz-projectionof the orbit of Halley'scomet shown in Fig.6.4.15.Plot the xy- and xz-projectionsaswell.

Figure 6.4.16showsthe graph of the distancer(t) of Halley'scomet fromthe sun. Inspectionof this graph indicatesthat Halley'scomet reachesa maximumdistance(at aphelion) of about 35 AU in a bit lessthan 40 years and returns toperihelion after about three-quarters of a century. The closerlook in Fig.6.4.17indicatesthat the periodof revolution of Halley'scomet is about76years.Useyournumerical solution to refine theseobservations.What is your bestestimate of thecalendar date of the comet'snext perihelionpassage?)

r)

30

20

10)

25) 200 t)50) 150) 175)75) 100) 125)

FIGURE6.4.16.200-yearplot of the distancer(t)ofHalley'scometfrom the sun.Is there a cuspnear t = 75?)

Your Own Comet)The night before your birthday in 2007you setup your telescopeon a nearbymoun-

taintop. It was a clearnight, and you had a strokeof luck:At 12:30A.M. you spotteda new comet.After repeating the observationon successivenights, you were ableto calculate its solarsystemcoordinatesPo = (xo,Yo, zo) and its velocity vectorVo = (vxo, vyo, vzo) on that first night. Usingthis information, determine the fol-lowing:)))


r)

5)

15)

10)

74) 75) 76) 77) 78 t)

FIGURE6.4.17.A closerlook at Halley'sperihelionpassageafter about 76 years.)

. the comet'sperihelion(point nearest the sun) and aphelion(point farthest fromthe sun),. the comet'svelocity at perihelion and at aphelion,. the comet'speriodof revolution around the sun, and. the comet'snext two datesof perihelionpassage.)

Usingunits of length in AU and time in earth years,the equationsof motionof your comet are given in (4).For your personalcomet, begin with random initial

position and velocityvectors with the sameorderof magnitude as those of Halley'scomet.Repeat the random selectionof initial position and velocityvectors,if nec-essary,until you get a plausibleeccentricorbit that ranges well outside the earth'sorbit (asmost real cometsdo).)))

NonlinearSysteIllsand PhenoIllena)

1fII..\037_\037l1!\037!!\037!.\037!-!.\".:1Soll!.\037.i\037.\037\037.and Stabil!\037....)

In previouschapterswe haveoftenusedexplicitsolutionsof differentialequationstoanswer specificnumericalquestions.But even when a given differentialequationisdifficult or impossibleto solveexplicitly,it often ispossibleto extractqualitative in-formation about general propertiesof its solutions.For example,we may beabletoestablishthat every solutionx(t) growswithout bound as t --++00,or approachesa finite limit, or is a periodicfunction of t.In this sectionwe introduce-mainlybyconsideration of simpledifferential equations that can be solved explicitly-someof the more important qualitative questionsthat can sometimesbe answeredfor

equations that are difficult or impossibleto solve.)

Example1)

\037\037\037\037\037N\037 \037A o....,\037\037 ..... \"'\037

Let x(t) denote the temperature of a body with initial temperaturex(O)= Xo. At

time t = 0 this body is immersedin a medium with constant temperature A. As-

suming Newton'slaw of cooling,dxdt

= -k(x- A)) (k >0 constant),) (1))

we readily solve (by separationof variables)for the explicitsolution

x(t) = A + (xo -A)e-kt.)

It follows immediately that)

lim x(t) = A,t ---HX))

(2))t)

FIGURE7.1.1.Typicalsolutioncurves for the equation ofNewton's law ofcooling,dx/dt = -k(x- A).)

so the temperature of the body approachesthat of the surrounding medium (as isevident to one'sintuition). Note that the constant function x(t) = A is a solution ofEq.(1);it correspondsto the temperatureof the body when it is in thermal equilib-rium with the surrounding medium. In Fig.7.1.1the limit in (2)means that everyother solution curve approachesthe equilibrium solution curve x = A asymptoti-cally as t --++00. .)

480)))

x'>0) x'<0).I .x<A x=A x>A)

FIGURE7.1.2.Phasediagramfor the equationdx/dt = I(x)= k(A -x).)

Example2)

7.1Equilibrium Solutionsand Stability 481)

Remark:The behaviorof solutionsof Eq.(1)is summarizedbriefly by thephasediagramin Fig.7.1.2-whichindicates the direction (or \"phase\") of changein x asa function of x itself. The right-hand sideI(x)= -k(x-A) = k(A -x) ispositive if x < A, negative if x > A. This observationcorrespondsto the fact that

solutions starting above the line x = A and those starting belowit both approachthe limiting solutionx(t) = A as t increases(asindicatedby the arrows). .)

In Section1.7we introducedthe generalpopulationequation)

dxdt

= (fJ -8)x,) (3))

where fJ and 8 are the birth and death rates, respectively, in births or deaths perindividual perunit of time. The questionof whether a populationx(t) isboundedorunbounded as t --++00is of evident interest. In many situations-likethe logisticand explosion/extinctionpopulations of Section1.7-thebirth and death rates areknown functions of x.Then Eq.(3)takes the form)

dx-= I(x).dt)

(4))

This is an autonomousfirst-orderdifferentialequation-onein which the indepen-dent variable t doesnot appearexplicitly (the terminologyhere stemming from the

Greekword autonomosfor \"independent,\" e.g.,of the time t). As in Example1,the

solutions of the equation I(x) = 0 play an important roleand are calledcriticalpointsof the autonomousdifferentialequationdx/dt = I(x).

If x = c is a critical point of Eq. (4),then the differential equationhas the

constant solution x(t) = c.A constant solution of a differential equationis some-times calledan equilibriumsolution(onemay think of a population that remainsconstantbecauseit is in \"equilibrium\" with its environment).Thus the criticalpointx = c, a number, correspondsto the equilibrium solution x(t) = c, a constant-valued function.

Example2 illustrates the fact that the qualitative behavior (as t increases)ofthe solutionsof an autonomousfirst-orderequation can be describedin terms of its

critical points.)

Considerthe logisticdifferentialequation)

dx-= kx (M-x)dt)

(5))

(with k >0 and M >0).It has two criticalpoints-thesolutionsx = 0 and x = Mof the equation)

I(x)= kx(M-x) = o.)

In Section1.7we discussedthe logistic-equationsolution)

Mxox(t) =

Xo + (M-xo)e-kMt) (6))

satisfying the initial condition x(0) = Xo. Note that the initial valuesXo = 0 andXo = M yield the equilibrium solutionsx(t) = 0 and x(t) = M ofEq.(5).)))

482 Chapter7 NonlinearSystemsand Phenomena)

t)

FIGURE7.1.3.Typicalsolutioncurves for the logisticequationdx/dt = kx(M -x).)

76543

\037 21)

-1-2-3 20 4 6

t)

8 10 12)

FIGURE7.1.4.Solution curves,funnel, and spout fordx/dt = 4x-x2

.)

X'<0 x'>0 x'<01 t t I 1

x=O x=MUnstable Stable)

FIGURE7.1.5.Phasediagramfor the logisticequationdx/dt = f(x) = kx(M -x).)

We observedin Section1.7that if Xo >0,then x(t) --+ M as t --++00.Butif Xo <0,then the denominator in Eq.(6)initially is positive,but vanisheswhen)

1 M -Xo

t = tl =-In >O.kM -Xo)

Becausethe numerator in (6)is negative in this case,it follows that)

lim x(t) = -00 if Xo <O.t\037tl)

It follows that the solutioncurvesof the logisticequation in (5)lookas illustrated in

Fig.7.1.3.Herewe seegraphically that every solution either approachesthe equi-librium solutionx(t) = M as t increases,or (in a visually obvious sense)divergesaway from the other equilibrium solutionx(t) = O. .)

StabilityofCriticalPointsFigure 7.1.3illustrates the conceptof stability. A critical point x = c of an au-tonomous first-orderequation is saidto be stableprovided that, if the initial valueXo is sufficiently closeto c,then x(t) remainscloseto cfor all t >O.Moreprecisely,the critical point c is stableif, for each E >0,there exists8 >0 such that)

Ixo -cl <8 impliesthat Ix(t)-cl < E) (7))

for all t >O.The critical point x = c is unstableif it is not stable.Figure7.1.4showsa \"wider view\" of the solutioncurvesof a logisticequation

with k = 1 and M = 4. Note that the strip 3.5< x < 4.5enclosingthe stableequilibrium curve x = 4 acts likeafunnel-solutioncurves (moving from left to

right) enter this strip and thereafter remain within it. By contrast, the strip -0.5<x < 0.5enclosingthe unstable solution curve x = 0 acts like a spout-solutioncurves leave this strip and thereafterremain outsideit. Thus the criticalpoint x = Mis stable,whereasthe critical point x = 0 is unstable.

Remark 1:We can summarize the behavior of solutions of the logisticequation in (5)-interms of their initial values-bymeans of the phasediagramshown in Fig.7.1.5.It indicatesthat x(t) --+ M as t --+ +00if either Xo > Mor 0 < Xo < M, whereasx(t) --+ -00as t increasesif Xo < O. The fact that Mis a stablecritical point would be important, for instance, if we wishedto conductan experiment with a population of M bacteria.It is impossibleto count preciselyM bacteria for M large,but any initially positive population will approach M as t

Increases.)

Remark 2: Related to the stability of the limiting solution M = alb ofthe logisticequation)

dx 2-= ax-bxdt)

(8))

is the \"predictability\" of M for an actual population. The coefficientsa and b areunlikely to beknown preciselyfor an actualpopulation. But if they are replacedwith

closeapproximationsa* and b*-derivedperhapsfrom empiricalmeasurements-then the approximate limiting population M* = a*Ib* will be closeto the actual

limiting populationM = alb.We may thereforesay that the valueM of the limiting

population predictedby a logisticequation not only is a stablecritical point of the)))

Example3)

\037)

x=o)

t)

FIGURE7.1.6.Typicalsolutioncurves for the explosion/extinctionequation dx/dt = kx(x -M).)

x'>0 x'<0 x'>0..I . . ..x=O x=MStable Unstable)

FIGURE7.1.7.Phasediagramfor the explosion/extinctionequationdx/dt = f(x) = kx(x -M).)

Example4)


differential equation; this value alsois \"stable\" with respectto small perturbationsof the constant coefficientsin the equation. (Note that one of thesetwo statementsinvolveschangesin the initial valueXo; the other involveschangesin the coefficientsa and b.) .)

Considernow the explosion/extinctionequation)

dx-= kx (x-M)dt)

(9))

of Eq. (10)in Section1.7.Likethe logisticequation, it has the two criticalpointsx = 0 and x = M correspondingto the equilibrium solutionsx(t) = 0 and x(t)=M. According to Problem 33 in Section1.7,its solution with x(0)= Xo isgiven by)

Mxox(t) =

Xo + (M-xo)ekM1) (10))

(with only a singledifference in sign from the logisticsolution in (6)).If Xo < M,then (becausethe coefficientof the exponential in the denominator is positive)it

followsimmediatelyfrom Eq.(10)that x(t) --+0 as t --++00.But if Xo >M, then

the denominator in (10)initially is positive,but vanisheswhen)

1 Xot = tl =-In >O.

kM Xo-M)

Becausethe numerator in (10)is positive in this case,it follows that)

lim x(t) = +00 if Xo > M.1-+11)

Therefore, the solution curves of the explosion/extinctionequation in (9)look asillustrated in Fig.7.1.6.A narrow band along the equilibrium curve x = 0 (as in

Fig.7.1.4)would serve as a funnel, while a band along the solution curvex == Mwould serve as a spout for solutions.The behavior of the solutions of Eq. (9) issummarizedby the phasediagram in Fig.7.1.7,where we seethat the criticalpointx = 0 is stableand the critical point x = M is unstable. .)

Harvestinga LogisticPopulationThe autonomousdifferentialequation)

dx 2-= ax-bx - hdt)

(11))

(with a, b, and h all positive) may be consideredto describea logisticpopulationwith harvesting. For instance, we might think of the population of fish in a lakefrom which h fish peryear are removedby fishing.)

Let us rewriteEq.(11)in the form)

dx-= kx (M-x)-h,dt)

(12))))

t)

FIGURE7.1.8.Typicalsolutioncurves for the logisticharvestingequationdx/dt = keN -x)(x-H).)

x'<O) x'>O) x'<O). . .I .x=H x=N

Unstable Stable)

FIGURE7.1.9.Phasediagramfor the logisticharvesting equationdx/dt = f(x) = k(N-x)(x-H).)

Example5)

which exhibits the limiting population M in the caseh = 0 of no harvesting. As-suming hereafter that h >0,we can solve the quadraticequation-kx2+kMx-h=o for the two critical points)

H,N = kM ::1:.J(\037\037)2

-4hk =\037 (M::1:.JM2-4hjk ) , (13))

assuming that the harvesting rate h is sufficiently small that 4h < kM2, so bothroots HandN are real with 0 < H < N <M. Then we can rewriteEq.(12)in theform)

dxdt = keN-x)(x-H).

For instance, the number of critical points of the equation may change abruptly asthe valueof a parameter ischanged.In Problem24 we askyou to solve this equationfor the solution)

(14))

N(xo-H)-H(xo-N)e-k(N-H)fx(t) =

(xo -H)- (xo -N)e-k(N-H)fin terms of the initial valuex(0)= Xo.

Note that the exponent -keN-H)t is negativefor t > O. If Xo > N, then

eachof the coefficientswithin parenthesesin Eq.(15)is positive; it follows that)

(15))

(16))If Xo > N then x(t) --+N as t --++00.In Problem 25 weaskyou to deducealsofrom Eq.(15)that

If H <Xo < N then x(t) --+N as t --++00, whereas (17)if Xo < H then x(t) --+-00 as t --+ tl (18))

for a positive value tl that dependson Xo. It follows that the solution curves ofEq. (12)-stillassuming that 4h < kM2-100kas illustrated in Fig.7.1.8.(Canyou visualize a funnel along the line x = N and a spout along the line x = H?)Thus the constant solution x(t) = N is an equilibrium limiting solution, whereasx(t) -H is a threshold solution that separatesdifferent behaviors-thepopulationapproachesN if Xo > H,while it becomesextinctbecauseof harvesting if Xo < H.Finally, the stablecritical point x = N and the unstable critical point x = H areillustrated in the phasediagram in Fig.7.1.9. .)

For a concreteapplication of our stability conclusionsin Example4, supposethat

k = 1 and M = 4 for a logisticpopulation x(t) of fish in a lake, measured in

hundreds after t years.Without any fishing at all, the lakewould eventually contain

nearly 400 fish, whatever the initial population. Now supposethat h = 3, so that

300fish are \"harvested\" annually (at a constant rate throughout the year).Equation

(12)is then dx/dt = x(4-x)-3, and the quadraticequation

-x2 +4x -3 = (3 -x)(x- 1)= 0)

has solutions H = 1 and N = 3.Thus the thresholdpopulation is 100fish and the

(new) limiting population is 300fish. In short, if the lakeis stockedinitially with

more than 100fish, then as t increases,the fish population will approacha limitingvalue of 300fish. But if the lakeis stockedinitially with fewer than 100fish, then

the lakewill be \"fished out\" and the fish will disappearentirely within a finite periodof time. .)))

Example6)

5)

3)

\037)

-1)

o) 2t)

4)

FIGURE7.1.10.Solution curvesof the equation x'= x(4-x)- hwith critical harvesting h = 4.)

5)

3)

\037)

-1)

o) 2t)

4)

FIGURE7.1.11.Solution curvesof the equation x'= x(4-x)- hwith excessiveharvesting h = 5.)


BifurcationandDependenceonParametersA biologicalor physical system that is modeledby a differential equationmay de-pend crucially on the numerical values of certain coefficientsor parameters that

appear in the equation. For instance, the number of critical points of the equationmay change abruptly as the value of a parameter is changed.)

,,,\037..y

The differentialequation)

dx-=x(4-x)-hdt)

(19))

(with x in hundreds) modelsthe harvesting of a logisticpopulation with k = 1and limiting population M = 4 (hundred).In Example5 we consideredthe caseof harvesting level h = 3, and found that the new limiting population is N = 3hundred and the threshold population is H = 1 hundred. Typical solution curves,including the equilibrium solutions x(t) = 3 and x(t) = 1,then look like thosepictured in Fig.7.1.8.

Now let'sinvestigatethe dependenceof this picture upon the harvesting levelh. According to Eq. (13)with k = 1 and M = 4, the limiting and threshold

populationsNand H are given by)

H,N =\037 (4::1::\037 16-4h)

= 2 ::I::\037 4 - h.) (20))

If h < 4-wecan considernegative values of h to describestocking rather than

harvesting the fish-then there are distinct equilibrium solutions x(t) = Nandx(t) = H with N > H as in Fig.7.1.8.

But if h = 4, then Eq. (20)gives N = H = 2,so the differential equationhas only the singleequilibrium solutionx(t) = 2.In this casethe solution curvesof the equation looklikethoseillustrated in Fig.7.1.10.If the initial number Xo (inhundreds) of fish exceeds2, then the population approachesa limiting populationof 2 (hundred fish). However,any initial populationXo < 2 (hundred) results in ex-tinction with the fish dying out as a consequenceof the harvesting of 4 hundred fish

annually. The critical point x = 2 might thereforebe describedas \"semistable\"-itlooksstableon the sidex > 2 where solution curves approach the equilibrium so-lution x(t) = 2 as t increases,but unstable on the sidex <2 where solution curvesinstead divergeaway from the equilibrium solution.

If, finally, h > 4, then the quadratic equation correspondingto (20)has noreal solutionsand the differentialequation in (19)has no equilibrium solutions.Thesolution curves then look like those illustrated in Fig.7.1.11,and (whatever the

initial number of fish) the populationdiesout as a result of the excessiveharvesting..)If we imagine turning a dial to gradually increasethe value of the parameterh

in Eq.(19),then the picture of the solution curveschangesfrom one like Fig.7.1.8withh < 4, to Fig.7.1.10withh= 4, to one likeFig.7.1.11withh > 4. Thus the

differentialequation has). two critical points if h < 4 ;. one critical point if h = 4 ;. no critical point if h > 4.)))

486 Chapter7 NonlinearSystems and Phenomena)

c)

FIGURE7.1.12.Theparabola(c-2)2= 4- h is the bifurcation

diagram for the differential

equation x'= x(4-x)- h.)

_ Problems)

The value h = 4-forwhich the qualitative nature of the solutions changesas h

increases-iscalleda bifurcationpoint for the differentialequation containing the

parameter h. A common way to visualize the corresponding\"bifurcation\" in thesolutions is to plot the bifurcationdiagramconsistingof all points (h,c),wherecis a critical point of the equation x' = x(4-x) + h . For instance, if we rewriteEq.(20)as)

h) c=2::i::,J4-h,(c-2)2 = 4 - h,)

whereeither c = N or c = H,then we get the equationof the parabola that is shownin Fig.7.1.12.This parabola is then the bifurcation diagram for our differentialequation that modelsa logisticfish populationwith harvesting at the level specifiedby the parameter h.)

In Problems 1through 12first solvethe equation f (x)= 0to find the criticalpoints of the given autonomous differentialequation dx/dt = f (x).Then analyze the sign off (x)to de-termine whether eachcriticalpoint is stableor unstable, andconstruct the correspondingphase diagram for the differen-tial equation. Next, solvethe differential equation explicitlyfor x (t) in terms of t. Finally, use either the exactsolutionor a computer-generatedslopefield to sketch typical solutioncurves for the given differential equation, and verify visuallythe stability ofeachcriticalpoint.

dx1.-= x -4dtdx3.-=x2 -4xdtdx 25.-= x -4dtdx7.-= (x - 2)2dtdx9.-= x2 -5x + 4dtdx11.-= (x - 1)3dt)

dx2.-= 3 -xdtdx4. -= 3x -x2dtdx6.-= 9-x2dtdx 28.dt

= -(3-x)dx 2-= 7x -x - 10dtdx12.-= (2-x)3dt)

10.)

In Problems13through 18,usea computer system orgraphingcalculatorto plot a slopefield and/or enough solution curvesto indicate the stability or instability of eachcriticalpoint ofthe given differential equation. (Someof thesecriticalpointsmay besemistablein the sensementioned in Example 6.)

dx dx13.-= (x + 2)(x-2)2 14.-= x(x2 -4)dt dtdx dx15.-= (x2 -4)2 16.-= (x2 -4)3dt dtdx dx17.-= x2(x2 -4) 18.-= X3(x2 -4)dt dt

19.Thedifferential equation dx/dt = /ox(IO-x)-hmodelsa logisticpopulation with harvesting at rate h. Determine)

(asin Example 6) the dependenceof the number ofcritical

points on the parameter h, and then construct a bifurcation

diagram like Fig. 7.1.12.20.Thedifferential equation dx/dt =

I\037Ox(x-5)+ s models

a population with stocking at rates.Determine the depen-denceof the number ofcritical points c on the parameter s,and then construct the correspondingbifurcation diagramin the sc-plane.

21.Considerthe differential equation dx/dt = kx - x3.(a) If k < 0, show that the only critical value c = 0 ofx is stable.(b) If k > 0, show that the critical point c = 0is now unstable, but that the critical points c = \037,Jk arestable.Thus the qualitative nature of the solutions changesat k = 0 as the parameter k increases,and so k = 0 is abifurcation point for the differential equation with param-eterk. Theplot of all points of the form (k, c) where c is acriticalpoint of the equation x'= kx -x3 is the \"pitchforkdiagram\" shown in Fig. 7.1.13.)

FIGURE7.1.13.Bifurcation diagram fordx/dt = kx -x3

.)

22.Considerthe differential equation dx/dt = x + kx 3 con-taining the parameter k. Analyze (as in Problem 21)the

dependenceof the number and nature of the critical pointson the value of k, and construct the correspondingbifur-cation diagram.)))

23.Supposethat the logistic equation dx/dt = kx(M -x)modelsa population x(t) of fish in a lake after t months

during which no fishing occurs. Now supposethat, be-causeof fishing, fish areremovedfrom the lake at the rateof hx fish per month (with h a positive constant). Thusfish are \"harvested\" at a rate proportional to the existingfish population, rather than at the constant rate of Exam-

ple4. (a) If 0 < h < kM, show that the population isstill logistic.What is the new limiting population ? (b) Ifh > kM, show that x(t) ---+ 0 aret ---+ +00,so the lake iseventually fished out.

24. Separatevariables in the logistic harvesting equationdx/dt = keN -x)(x- H) and then usepartial fractionsto derive the solution given in Eq.(15).

25.Usethe alternative forms

N(xo-H)+ H(N-xo)e-k(N-H)tx(t)=(xo-H)+ (N -xo)e-k(N-H)t

H(N-xo)e-k(N-H)t -N(H-xo)(N -xo)e-k(N-H)t - (H-xo)

of the solution in (15)to establish the conclusionsstatedin (17)and (18).)

Example 4 dealt with the case4h > kM 2 in the equationdx/dt = kx(M -x)- h that describesconstant-rate har-vesting ofa logisticpopulation. Problems26and 27dealwith

the other cases.)

26.If 4h = kM2, show that typical solution curves lookas illustrated in Fig. 7.1.14.Thus if Xo > M/2, then

x(t) ---+ M/2 as t ---+ +00.But if Xo < M/2, thenx (t ) = 0 after a finite periodof time, so the lake isfished out. Thecritical point x = M/2 might be calledsemistable,becauseit looksstable from oneside,unstablefrom the other.

27.If 4h > kM 2, show that x(t)= 0 after a finite periodoftime, so the lake is fished out (whatever the initial popula-)

\037)

t)

FIGURE7.1.14.Solutioncurves for harvesting a logisticpopulation with 4h = kM 2

.)


tion). [Suggestion:Completethe square to rewrite the dif-ferential equation in the form dx/dt = -k[(x-a)2+ b2].Then solveexplicitly by separation of variables.]There-sults of this and the previous problem (together with Ex-ample 4) show that h = ikM2 is a critical harvesting ratefor a logisticpopulation. At any lesserharvesting rate the

population approachesa limiting population N that is lessthan M (why?), whereasat any greaterharvesting rate the

population reachesextinction.28.This problem dealswith the differential equation dx/dt =

kx (x-M)-h that modelsthe harvesting ofan unsophisti-catedpopulation (such asalligators). Show that this equa-tion can be rewritten in the form dx/dt = k(x -H)(x-K), where)

H =\037 (M + JM2 + 4hjk ) > 0,

K =\037 (M -JM2 + 4hjk)

< O.)

Show that typical solution curves look as illustrated in

Fig.7.1.15.29.Considerthe two differential equations

dx-= (x -a)(x -b)(x -c)dt)

(21))

and)dx-= (a -x)(b-x)(c-x),dt)

(22))

eachhaving the critical points a, b, and c;supposethat

a < b < c. For oneof theseequations, only the criti-

calpoint b is stable; for the other equation, b is the onlyunstable critical point. Construct phasediagrams for the

two equations to determine which is which. Without at-

tempting to solveeither equation explicitly, make roughsketchesof typical solution curves for each. You should

seetwo funnels and a spout in onecase,two spouts and afunnel in the other.)

\037)

t)

FIGURE7.1.15.Solutioncurves for harvesting apopulation of alligators.)))

488 Chapter7 NonlinearSystemsand Phenomena)..:IStabilityand the PhasePlane\037\",..,.....,.... .....d..\"'\" \" ,,. .._..\".,\"_\",.,.\"'\"\037\037,._ ,\037....'.'dd.'\"\".'\" _\"_'\"'''.'d_,.,.._,.\037_.d ,_,.,\",.\037.,.\" '\"d_'''\" \

Awidevariety of natural phenomenaare modeledby two-dimensionalfirst -ordersystemsof the form)

\037)

dxdt

= F(x,y),

dy-= G(x,y)dt)

(1))

in which the independent variable t doesnot appear explicitly. We usually think

of the dependentvariables x and y as position variables in the xy-planeand of t

as a time variable. We will seethat the absenceof t on the right-hand sidesin (1)makesthe system easierto analyze and its solutionseasierto visualize. Usingthe

terminology of Section7.1,such a systemof differential equations in which thederivative values are independent (or \"autonomous\") of time t is often calledanautonomoussystem.

We generally assumethat the functions F and G are continuously differen-tiable in some region R of the xy-plane. Then accordingto the existenceanduniquenesstheoremsof the Appendix,given to and any point (xo, Yo) of R, there isa unique solution x = x(t), y = yet) of (1)that is definedon someopeninterval

(a,b) containing to and satisfiesthe initial conditions)

x(to)= Xo, y(to) = Yo.) (2))

The equations x = x(t), y = yet) then describea parametrized solution curve in

the phaseplane. Any such solution curve is calleda trajectory of the system in

(1),and preciselyone trajectorypassesthrough each point of the region R (Problem29).A criticalpoint of the system in (1)is a point (x*, y*) such that)

\037) F(x*,y*) = G(x*,y*) = o.) (3))

If (x*, y*) is a critical point of the system,then the constant-valuedfunctions)

x(t) = x*, yet) = y*) (4))

have derivativesx'(t) = 0 and y'(t) = 0,and therefore automatically satisfy the

equations in (1).Sucha constant-valuedsolution is calledan equilibriumsolutionof the system.Note that the trajectory of the equilibrium solution in (4)consistsofthe singlepoint (x*, y*).

In somepractical situations thesevery simplesolutions and trajectories arethe onesof greatest interest.For example,supposethat the system x' = F(x,y),y' = G(x,y) modelstwo populationsx(t) and y (t) of animals that cohabit the sameenvironment, and perhapscompetefor the samefood or prey on one another;x(t)might denote the number of rabbits and yet) the number of squirrelspresentat timet. Then a critical point (x*, y*) of the system specifiesa constant populationx* ofrabbits and a constant population y* of squirrelsthat can coexistwith one anotherin the environment. If (xo,Yo) is not a critical point of the system, then it is not

possiblefor constant populations of Xo rabbits and Yo squirrelsto coexist;one orboth must change with time.)))

7.2Stability and the PhasePlane 489)

Example1)

......Find the critical points of the system

dx 2-= 14x-2x -xy,dt

dy 2dt

= 16y-2y -xy.)

(5))

Solution When we lookat the equations

14x-2x2-xy = x(14-2x- y) = 0,16y-2y

2-xy = y(16-2y -x) = 0)

that a critical point (x,y) must satisfy, we seethat either

x = 0 or 14-2x - y = 0,) (6a))

and either)

y=O or 16-2y-x=0.) (6b))

If x = 0 and y i= 0,then the secondequation in (6b)gives y = 8.If y = 0 and

x i= 0,then the secondequation in (6a) givesx = 7. If both x and y are nonzero,then we solve the simultaneousequations)

2x + y = 14, x+2y = 16for x = 4, y = 6.Thus the system in (5)has the four critical points (0,0),(0,8),(7,0),and (4,6).If x(t) and y(t) denote the number of rabbits and the numberof squirrels,respectively,and if both populations are constant, it follows that theequations in (5)allow only three nontrivial possibilities:either no rabbits and 8squirrels,or 7 rabbits and no squirrels,or 4 rabbits and 6 squirrels.In particular,the critical point (4,6)describesthe only possibilityfor the coexistenceof constantnonzero populationsof both species. .)

PhasePortraits)If the initial point (xo,Yo) is not a critical point, then the correspondingtrajectory isa curve in the xy-planealong which the point (x(t),y(t)) moves as t increases.Itturns out that any trajectorynot consistingof a singlepoint isa nondegeneratecurvewith no self-intersections(Problem 30).We can exhibit qualitatively the behaviorof solutions of the autonomous system in (1)by constructinga picture that showsits critical points together with a collectionof typical solutioncurvesor trajectoriesin the xy-plane.Sucha picture is calleda phaseportrait(or phaseplanepicture)becauseit illustrates \"phases\" or xy-statesof the system, and indicates how they

change with time.Another way of visualizing the system is to construct a slopefield in the xy-

phaseplane by drawing typical line segmentshaving slope

dy y' G(x,y)---- ,dx x' F(x,y)

or a directionfield by drawing typical vectors pointing the same directionat eachpoint (x,y) asdoesthe vector (F(x,y), G(x,y)).Sucha vectorfieldthen indicateswhich direction along a trajectory to travel in orderto \"go with the flow\" describedby the system.)))


(0,8)(4,6))

\037)

2 4) 8 10)

x)

FIGURE7.2.1.Direction field and

phaseportrait for the rabbit- squirrelsystem x'= 14x- 2X2 -xy,y' = 16y-2y

2 -xy of Example 1.)

Example2)

y)

G(x,y) = 1 -x2=0x=+l)

x=-lG(x, y) = 1 -x2=0)

FIGURE7.2.4.Thetwo critical

points (-1,-1)and (+1,+1)in

Example 2 as the intersection ofthe curves F (x,y) = x -

y = 0and G(x,y) = 1-x2 = o.)

Remark:It is worth emphasizing that if our system of differential equa-tions werenot autonomous, then its critical points, trajectories, and direction vec-tors would generallybechanging with time. In this event, the concretevisualizationafforded by a (fixed)phaseportrait or direction field would not be available to us.Indeed,this is a principal reasonwhy an introductory study of nonlinear systemsconcentrateson autonomousones. .)

Figure 7.2.1showsa direction field and phaseportrait for the rabbit-squirrelsystem of Example1.The direction fieldarrows indicate the directionof motion ofthe point (x(t),y(t)). We seethat, given any positive initial numbers Xo i= 4 andYo i= 6 of rabbits and squirrels,this point moves along a trajectoryapproachingthecritical point (4,6)as t increases.)

3 It!:' .j. J2 J; J,

J,. -t.

1 t. \037\037

t (1,1)\037 \\

\037 0,l1I

\":Ii

,.\037\037\\i\\,

-1 /'\037\037'\\a\\t/f-:)\037\037\\r-2 /t4\037\037'\\i\037\037-)\037\037-3 -1 0 1 2 3-3

x

FIGURE7.2.2.Direction fieldfor the system in Eq.(7).)

4321

\037 0-1-2)

o 1 234x)

FIGURE7.2.3.Phaseportraitfor the system in Eq.(7).)

For the system)

IX = X - y,

y' = 1-x2)(7))

we seefrom the first equation that x = y and from the secondthat x = ::f:1 ateachcritical point. Thus this systemhas the two criticalpoints (-1,-1)and (1,1).The directionfield in Fig.7.2.2suggeststhat trajectoriessomehow\"circulate\" coun-terclockwisearound the critical point (-1,-1),whereas it appearsthat sometra-

jectoriesmay approach, while others recedefrom, the critical point (1,1).Theseobservations are corroboratedby the phraseportrait in Fig.7.2.3for the system

x in (7). .)Remark:One couldcarelesslywrite the critical points in Example2 as

(::f:1,::f:1)and then jump to the erroneousconclusionthat the system in (7)has four

rather than just two critical points.When feasible,a sure-fireway to determine the

number of critical points of an autonomoussystem is to plot the curves F(x,y) = 0and G(x,y) = 0 and then note their intersections,eachof which representsa criticalpoint of the system. For instance, Fig.7.2.4showsthe curve (line) F(x,y) =x -

y = 0 and the pair of linesx = +1 and x = -1that constitute the \"curve\"

G(x,y) = l-x2 = O.The (only) two points of intersection(-1,-1)and (+1,+1)are then visually apparent. .)))

Example3)

2

1

;:..... 0

-1-2

-2 -1 0 1 2x)

FIGURE7.2.5.A proper node;the trajectoriesapproach the

origin, so it is a nodal sink.)

2)

1)

;:..... 0)

-1)

-2-2 -1) o

x)

1) 2)

FIGURE7.2.6.An impropernodebecauseall trajectoriesaretangent to a single line; theyapproach the origin, so it is a nodalsink.)


CriticalPointBehavior)The behavior of the trajectories near an isolatedcritical point of an autonomoussystemis of particular interest. In the remainder of this sectionwe illustrate with

simpleexamplessomeof the most commonpossibilities.)

Considerthe autonomouslinear system)

dx-==-x,dt

dy-==-ky (k a nonzeroconstant),dt)

(8))

which has the origin (0,0)as its only critical point. The solution with initial point

(xo,Yo) is)

x(t) ==xoe-t , yet) ==yoe-kt.) (9))

If Xo i= 0,we can write)

y = yoe-k1 =y\037

(xoe-1)k= bxk ,Xo)

(10))

where b == yo/x\037.The nature of the critical point (0,0) dependson whether the

nonzero parameter k is positiveor negative.)

CASE1:k > O. If k is positive, then we seefrom (9)that the point (x(t),yet))approachesthe origin along the curve y ==bxk as t --++00.The appearanceof this

curve dependson the magnitude of k:). If k == 1,then y == bx with b == yo/xo is a straight line through the point

(xo,Yo). Thesestraight-linetrajectoriesare illustrated by the phaseportrait in

Fig.7.2.5.. If k > 1 and neither Xo nor Yo in Eq. (10)is zero, then the curve y == bxk

is tangent to the x-axisat the origin. This caseis illustrated by the phaseportrait in Fig.7.2.6,where k == 2 and the trajectories are parabolas.Moreprecisely,the trajectoriesare the semiaxesand the right and left halvesof theseparabolas.. If 0 <k < 1 and neither Xo nor Yo is zero, then the phaseportrait is similar toFig.7.2.6,exceptthat each curve y ==bxk is tangent to the y-axis(rather than

to the x-axis)at the origin.)

The type of criticalpoint illustrated in Figs.7.2.5and 7.2.6iscalleda node.Ingeneral,the critical point (x*, y*) of the autonomoussystem in (1)is calleda nodeprovided that). Either every trajectory approaches(x*, y*) as t --+ +00orevery trajectory

recedesfrom (x*, y*) as t --++00,and. Every trajectory is tangent at (x*, y*) to somestraight line through the criticalpoint.)

A nodeis saidto beproperprovided that no two different pairsof \"opposite\"

trajectories are tangent to the samestraight line through the criticalpoint. This isthe situation in Fig.7.2.5(in which the trajectoriesarestraight lines, not merely)))

492) Chapter7 NonlinearSystemsand Phenomena)

;::.... 0)

-1)

-1) ox)

1)

FIGURE7.2.7.A saddlepointwith trajectoriesresembling thecontour curvesofa saddlepoint ona surface.)

Example3)

Continued)

tangent to straight lines).A propernodemight becalleda \"star point.\" In Fig.7.2.6all trajectoriesexceptfor a singleoppositepair are tangent to a singlestraight line

through the critical point. This type of nodeis saidto be improper.A nodeis alsocalleda sink if all trajectories approachthe critical point, a

sourceif all trajectories recede(or emanate) from it. Thus the origin in Fig.7.2.5is a propernodal sink, whereasin Fig.7.2.6it is an improper nodal sink. If the

direction fieldarrows in eachfigure werereversed,then the origin would bea nodalsourcein eachfigure.)

CASE2:k < O. If k is negative, then the trajectories resemblethose for the

casek == -1,which is illustrated in Fig.7.2.7.If neither Xo nor Yo is zero, then

the correspondingtrajectory in Fig.7.2.7is one branch of the rectangularhyperbolaxy ==b, and Iy(t) I

--++00as t --++00.If eitherXo or Yo iszero, then the trajectoryis a semiaxisof the hyperbola.The point (x(t),y(t)) approachesthe origin alongthe x-axis,but recedesfrom it along the y-axis,as t --+ +00.Thus there are two

trajectories that approach the critical point (0,0), but all others are unboundedast --+ +00.This type of critical point, illustrated in Fig.7.2.7,is calleda saddlepoint. .)

StabilityA criticalpoint (x*, y*) of the autonomoussystem in (1)is saidto bestableprovidedthat if the initial point (xo,Yo) is sufficiently closeto (x*, y*), then (x(t),y(t)) re-mains closeto (x*, y*) for all t >O.In vectornotation, with x(t) == (x(t),y(t)),the

distancebetweenthe initial point Xo == (xo,Yo) and the critical point x* == (x*, y*)IS)

Ixo -x*1 ==)(xo -x*)2 + (Yo-y*)2.

Thus the critical point x* is stableprovided that, for eachE >0,there exists8 >0such that)

\037) Ixo - x*1 <8 impliesthat Ix(t) -x*1 < E) (11))

for all t > O. Note that the condition in (11)certainly holdsif x(t) --+ x* as t \037

+00,as in the caseof a nodal sink. Thus the nodal sinks illustrated in Figs.7.2.5and 7.2.6can alsobe describedasstablenodes.

The criticalpoint (x*, y*) iscalledunstableif it isnot stable.The saddlepointat (0,0) in Fig.7.2.7is an unstablecritical point becausethe point (x(t),y(t)) goesto infinity as t --++00(assumingthat XoYo i= 0),and hencethe condition in (11)isnot satisfied.)

If the signson the right-hand sidein (8)are changed to obtain the system)

dx-==x,dt

dy-==ky (k a nonzero constant),dt)

(12))

then the solution is x(t) == xoet , y(t) == yoekt . Then with k == 1 and k == 2, the

trajectories are the sameas those shown in Figs.7.2.5and 7.2.6,respectively,but

with the arrows reversed,so that the point (x(t),y(t)) goesto infinity as t --+00.The result in eachcaseis a nodal source-thatis,an unstable node-at(0,0). .)))

Example4)

54321

;:..., 0-1-2-3-4-5

-5-4-3-2-10 1 2 3 4 5x)

FIGURE7.2.8.Direction fieldand elliptical trajectoriesfor the

system x'= y, y' = -*x.Theorigin is a stable center.)


If (x*, y*) is a critical point, then the equilibrium solutionx(t) = x*, yet) =y* is calledstableor unstable dependingon the nature of the critical point. Inapplications the stability of an equilibrium solution is often a crucial matter. Forinstance, supposein Example1 that x(t) and yet) denote the rabbit and squirrelpopulations, respectively, in hundreds.We will seein Section7.4that the criticalpoint (4,6) in Fig.7.2.1is stable. It follows that if we begin with closeto 400rabbits and 600squirrels-ratherthan exactly theseequilibrium values-thenfor allfuture time there will remain closeto 400 rabbits and closeto 600squirrels.Thusthe practical consequenceof stability is that slight changes(perhapsdue to randombirths and deaths)in the equilibrium populations will not so upset the equilibriumas to result in large deviationsfrom the equilibrium solutions.

It is possiblefor trajectories to remain near a stablecritical point without ap-proaching it, asExample4 shows.)

...... ...... ......

Considera massm that oscillateswithout damping on a spring with Hooke'scon-stant k, so that its position function x(t) satisfiesthe differentialequationx\" +w2x =o (where w 2 = kjm).If we introduce the velocity y = dxjdtof the mass,we getthe system)

dxdt

= y,

dy 2-= -w xdt)

(13))


x(t) = Acoswt + Bsinwt,yet) = -Awsinwt+ Bwcoswt.)

(14a)(14b))

With e = .JA2 + B2,A = ecosa,and B = e sin a,we can rewrite the solutionin (14)in the form)

x(t) = e cos(wt-a),yet) = -wCsin(wt -a),)

(15a)(15b

))

soit becomesclearthat eachtrajectoryother than the criticalpoint (0,0) isan ellipsewith equation of the form)

x2 y22+ 2 2= 1.ewe) (16))

As illustrated by the phaseportrait in Fig.7.2.8(where w = 4),each point (xo,Yo)other than the origin in the xy-plane lieson exactly one of theseellipses,and eachsolution (x(t),yet)) traverses the ellipsethrough its initial point (xo,Yo) in the

clockwisedirection with periodP = 2Jrjw. (It is clearfrom (15)that x(t + P) =x(t) and yet + P) = yet) for all t.) Thus eachnontrivial solution of the systemin (13)is periodicand its trajectory is a simpleclosedcurve enclosingthe criticalpoint at the origin. .)))

y)

e)

x)

-- -)

FIGURE7.2.9.If the initial

point (xo,Yo) lieswithin distance8of the origin, then the point(x(t),y(t)) stays within distanceE

of the origin.)

Figure 7.2.9showsa typical elliptical trajectory in Example4, with its minorsemiaxisdenotedby 8 and its major semiaxisby E. We seethat if the initial point(xo,Yo) lieswithin distance8 of the origin-sothat its ellipticaltrajectoryliesinsidethe one shown-thenthe point (x(t),y(t)) always remains within distanceE ofthe origin.Hencethe origin (0,0) is a stablecritical point of the system x' = y,y' = -w2X . Unlikethe situation illustrated in Figs.7.2.5and 7.2.6,however,nosingletrajectory approachesthe point (0,0).A stablecritical point surroundedby

simpleclosedtrajectoriesrepresentingperiodicsolutionsis calleda (stable)center.)

AsymptoticStability)The critical point (x*, y*) is calledasymptoticallystableif it is stable and, more-over, every trajectory that begins sufficiently closeto (x*, y*) also approaches(x*, y*) as t --++00.That is, there exists8 >0 such that)

\037) Ix -x* I<8 impliesthat) (17))lim x(t)= x*,

t\037oo)

where Xo = (xo,Yo), x* = (x*, y*), and x(t) = (x(t),y(t)) is a solution with

x(O)= Xo.

Remark:The stablenodesshown in Figs.7.2.5and 7.2.6are asymptoti-cally stablebecauseeverytrajectoryapproachesthe criticalpoint (0,0)as t --++00.The center (0,0) shown in Fig.7.2.8is stablebut not asymptoticallystable,becausehowever small an elliptical trajectory we consider,a point moving around this el-lipsedoesnot approach the origin.Thus asymptotic stability is a strongerconditionthan mere stability. .)

Now supposethat x(t) and y(t) denote coexistingpopulations for which

(x*, y*) is an asymptotically stablecritical point. Then if the initial populationsXo and Yo are sufficiently closeto x* and y*, respectively,it follows that both)

lim x(t) = x* and lim y(t) = y*.t\037oo t\037oo)

(18))

That is, x(t) and y(t) actually approach the equilibrium populations x* and y* ast --++00,rather than merely remaining closeto those values.

For a mechanical system as in Example4, a critical point representsan equi-librium state of the system-ifthe velocity y = x' and the acceleration y' = x\"

vanish simultaneously, then the massremains at rest with no net force acting onit. Stability of a critical point concernsthe question whether, when the mass is

displacedslightly from its equilibrium, it

1.Movesbacktoward the equilibrium point as t --++00,2.Merelyremains near the equilibrium point without approachingit, or3.Movesfarther away from equilibrium.)

In Case1 the critical [equilibrium]point is asymptoticallystable;in Case2it is stablebut not asymptotically so;in Case3 it is an unstable critical point. A

marble balancedon the top of a soccerball is an exampleof an unstable criticalpoint. A masson a spring with damping illustrates the caseof asymptotic stabilityof a mechanical system.The mass-and-springwithout damping in Example4 is an

exampleof a systemthat is stablebut not asymptoticallystable.)))

Example5)

y)

FIGURE7.2.10.A stable spiralpoint and onenearby trajectory.)

Example6)


.....

Supposethat m 1 and k 2 for the massand spring of Example4 and that

the mass is attached also to a dashpot with damping constant c = 2. Then its

displacementfunction x(t) satisfiesthe second-orderequation)

x\" (t)+2X'(t) +2x(t)= O.) (19))

With y = x'we obtain the equivalentfirst-ordersystem)

dxdt

= y,

dy-= -2x-2ydt)

(20))

with critical point (0,0).The characteristicequation r 2 + 2r+2 = 0 of Eg.(19)has roots -1+ i and -1- i,so the general solution of the system in (20)is givenby)

x) x(t) = e-t (A cost + B sin t) = Ce-t cos(t-a),yet) = e-t [(B-A) cost - (A +B)sin t]

= -Che-t sin (t-a + in),

where C = v'A2 + B2 and a = tan- I (B/A).We seethat x(t) and yet) oscillatebetweenpositive and negative values and that both approach zero as t -* +00.Thus a typical trajectoryspiralsinward toward the origin, as illustrated by the spiralin Fig.7.2.10. .)

(21a))

(21b))

It is clearfrom (21)that the point (x(t),yet)) approachesthe origin as t -*+00,so it follows that (0,0)is an asymptoticallystablecriticalpoint for the systemx' = y, y' = -2x- 2y of Example5. Suchan asymptotically stable criticalpoint-aroundwhich the trajectories spiral as they approachit-iscalleda stablespiralpoint (or a spiralsink). In the caseof a mass-spring-dashpotsystem, aspiral sink is the manifestation in the phaseplane of the damped oscillationsthat

occurbecauseof resistance.If the arrows in Fig.7.2.10werereversed,we would seea trajectoryspiraling

outward from the origin.An unstable critical point-aroundwhich the trajectoriesspiral as they emanate and recedefrom it-iscalledan unstablespiral point (ora spiralsource).Example6 showsthat it alsois possiblefor a trajectoryto spiralinto a closedtrajectory-asimpleclosedsolutioncurve that representsa periodicsolution (likethe elliptical trajectories in Fig.7.2.8).)

Considerthe system)

dx 2 2dt

= -ky +x(1-x -y ),

dy 2 2dt

= kx +y(1-x -y ).)

(22))

In Problem 21we askyou to show that (0,0) is its only critical point. This systemcan be solved explicitly by introducing polar coordinatesx = r cos0,y = r sin 0,as follows.First note that)

dO =!!.-(arctan Y

)= xy' -x'Y .

dt dt X x2 + y2)))

2)

1)

;::....0)

-1)

-2-2) -1) ox)

1)

FIGURE7.2.11.Spiraltrajectories of the system in

Eq. (22)with k = 5.)

Then substitute the expressionsgiven in (22)for x' and y' to obtain

dO k(x2 + y2)- -kdt

-x2 + y2

- .)

It follows that)

O(t) = kt +00,) (23))where 00= 0(0).)

Then differentiationof r 2 = x2 + y2 yieldsdr dx dy2rdt

= 2xdt +2y dt

= 2(x2 +y2)(1-x2 _ y2) = 2r2(1_ r 2),so r = r (t) satisfiesthe differentialequation

dr 2dt

= r(1- r ).

In Problem 22 we askyou to derive the solution

ror(t) = , (25)

JrJ+ (1-rJ)e-21

where ro = reO).Thus the typical solutionofEq.(22)may beexpressedin the form)

(24))

2)

x(t) = r(t) cos(kt+(0),yet) = r(t) sin(kt +(0).

If ro = 1,then Eq. (25)gives r(t) = 1 (the unit circle).Otherwise,if ro > 0,then Eq. (25)impliesthat r(t) --+ 1 as t --+ +00.Hencethe trajectory definedin (26)spiralsin toward the unit circleif ro > 1 and spiralsout toward this closedtrajectory if 0 < ro < 1.Figure7.2.11showsa trajectoryspiralingoutward from theorigin and four trajectories spiraling inward, all approaching the closedtrajectoryr(t) = 1. .)

(26))

Underrather general hypothesesit can be shown that there are four possibili-tiesfor a nondegeneratetrajectoryof the autonomoussystem)

dx-= F(x,y),dtThe four possibilitiesare these:)

dy-= G(x,y).dt)

1.(x(t), y (t))approachesa critical point as t --++00.2. (x(t),yet)) is unbounded with increasingt.3.(x(t), y (t)) is a periodicsolution with a closedtrajectory.4. (x(t),yet)) spiralstoward a closedtrajectoryas t -+ +00.)

As a consequence,the qualitative nature of the phaseplane picture of the

trajectories of an autonomous system is determined largely by the locations of its

critical points and by the behavior of its trajectoriesnear its critical points.We will

seein Section7.3that, subjectto mild restrictions on the functions F and G,eachisolatedcritical point of the systemx'= F(x,y), y' = G(x,y) resemblesqualita-tively one of the examplesof this section-itis either a node(properor improper),a saddlepoint, a center, or a spiral point.)))

7.2Stability and the PhasePlane) 497)

lIB)Problems)

In Problems1through 8,find the criticalpoint orpoints ofthe dx dy3.-= x -2y + 3, -=x-y+2given autonomous system, and thereby match eachsystem with dt dtits phaseportrait among Figs. 7.2.12through 7.2.19.

dx dydx dy 4. -= 2x -2y -4, -=x+4y+31.dt

= 2x - y, -= x - 3y dt dtdt

2.dx dy dx1- y2,

dy-= x - y, -= x + 3y -4 5.- -= x + 2ydt dt dt dt)

54321

;::.... 0-1-2-3-4-5-5-4 .....3-2-1 0 1 2 3 4 5

x)

4321

;::.... 0)

-1-2)

o 1 234x)

4321

;::.... 0-1-2-3

\\ \\ \\ I I r.-4-4 -3-2-1 0 1 2 3 4x)

FIGURE7.2.12.Spiral point(-2, 1)and saddlepoint (2,-1).)

FIGURE7.2.13.Spiral point(1,-1).)

FIGURE7.2.14.Saddlepoint(0,0).)

4321

;::.... 0-1-2)

4321

;::.... 0-1-2)

o 1 234x)

3

2

1

;::.... 0

-1-2-3 -2 -1 0 1 2 3-3

x)

o 1

x)

234)FIGURE7.2.15.Spiral point(0,0);saddlepoints (-2,-1)and

(2, 1).)

FIGURE7.2.16.Node(1,1).) FIGURE7.2.17.Spiral point(-1,-1),saddlepoint (0,0),andnode (1,-1).)

6

4

2

;::.... 0

-2-4-6

-6 -4 -2 0 2 4 6x)

4321)

;::.... 0-1-2-3-4-4 -3-2-1 0 1 2 3 4

x)

FIGURE7.2.18.Spiral point

(-2,\037)

and saddlepoint (2,-\037 ) .)

FIGURE7.2.19.Stablecenter(-1,1).)))

6.dx = 2 _ 4x- 15y,dy = 4 _ x2

dt dtdx dy 37.-= x -2y, -= 4x-xdt dtdx 2 dy 28.dt

= x -y-x + xy, dt

= -y-x

In Problems 9 through 12,find each equilibrium solution

x(t) = Xo of the given second-orderdifferential equationx\" + f(x,x') = o. Usea computer system or graphing cal-culator to construct a phaseportrait and direction fieldfor the

equivalent first-ordersystem x'= y, y' = -f (x,y). Therebyascertainwhether the criticalpoint (xo,0) looks like a center,a saddlepoint, or a spiralpoint of this system.

9.xl!+ 4x-x3 = 010.xl!+ 2x'+ x + 4x3 = 011.xl!+ 3x'+ 4 sin x = 012.xl!+ (x2 - I)x'+ x = 0)

Solveeachof the linear systems in Problems13through 20 todetermine whether the criticalpoint (0,0) is stable, asymptot-ically stable, or unstable. Usea computer system or graphingcalculatorto construct a phaseportrait and direction fieldforthe given system. Thereby ascertainthe stability or instabil-

ity ofeachcriticalpoint, and identify it visually as a node, asaddlepoint, a center, or a spiralpoint.

dx dy13.-= -2x, -= -2ydt dtdx dy14.dt = 2x,

dt= -2y

dx dy15.-= -2x, -= -ydt dtdx dy16.dt =x, dt = 3y

dx dy17.-= y, -= -xdt dtdx dy18.dt = -y, dt

= 4x

dx dy19.-= 2y, -= -2xdt dtdx dy20.-= y, -= -5x-4ydt dt

21.Verify that (0, 0) is the only critical point of the system in

Example 6.22. Separatevariables in Eq.(24) to derive the solution in

(25).)

In Problems 23 through 26, a system dxjdtdyjdt = G(x,y) is given. Solvethe equation)

F(x,y),)

dy G(x,y)dx F(x,y))

to find the trajectoriesofthe given system. Usea computer sys-tem or graphing calculatorto construct a phaseportrait and)

direction fieldfor the system, and thereby identify visually the

apparent characterand stability of the criticalpoint (0,0) ofthe given system.)

23.)dx dy

dt= y, dt

= -xdx-= y(l +x2+ y2),dtdx-= 4y(1+ x2 + y2),dt)

dy-= x(I+ x2 + y2)dt

dy-= -x(l+ x2 + y2)dt)

24.)

25.)

dx dy26.-= y3 ex+y , -= -x3ex+y

dt dt27.Let (x(t),yet\302\273\037

be a nontrivial solution of the nonau-tonomous system)

dx-=y,dt)

dy-= tx.dt)

Supposethat </J(t) = x(t + y) and 1fr(t) = yet + y),where y =I- O.Show that (</J(t),1fr(t\302\273

is not a solution ofthe system.)

Problems28 through 30dealwith the system)

dxdt

= F(x,y),)dy

dt =G(x,y))

in a region where the functions F and G are continuously dif-ferentiable, sofor eachnumber a and point (xo, Yo), there is aunique solution with x(a)= Xo and yea)= Yo.

28.Supposethat (x(t), Y (t\302\273is a solution of the autonomous

system and that y =I- O. Define</J (t) = x (t + y) and

1fr(t) = yet + y). Then show (in contrast with the sit-uation in Problem 27) that (</J(t), 1fr(t\302\273

is alsoa solutionof the system. Thus autonomous systems have the simplebut important property that a \"t-translate\" of a solution isagain a solution.

29.Let (XI (t), YI(t\302\273and (X2(t), Y2(t\302\273

be two solutions hav-

ing trajectories that meet at the point (xo,Yo); thus

XI (a) = x2(b) = Xo and YI (a) = Y2(b) = Yo for somevalues a and b oft.Define)

X3(t) = X2(t + y) and Y3(t) = Y2(t + y),

where y = b - a, so (X2(t), Y2(t\302\273and (X3(t), Y3(t\302\273

have the same trajectory. Apply the uniqueness theoremto show that (XI (t), YI (t\302\273

and (X3(t), Y3(t\302\273are identical

solutions. Hencethe original two trajectoriesare identi-cal. Thus no two different trajectoriesof an autonomous

system can intersect.30.Supposethat the solution (XI (t), YI (t\302\273

is defined for all tand that its trajectory has an apparent self-intersection:)

XI (a) = x I (a + P)= Xo, YI (a) = YI (a + P)=Yo)

for someP > O.Introduce the solution)

X2(t) = XI(t + P), Y2(t) = YI(t + P))))

7.2Stability andthe PhasePlane 499)

and then apply the uniqueness theorem to show that

Xl (t + P) = Xl (t) and YI (t) = YI (t + P)for all t.Thus the solution (Xl (t), YI (t)) is periodicwith)

periodP and has a closedtrajectory. Consequently a so-lution of an autonomous system either is periodic with

a closed trajectory, or else its trajectory never passesthrough the samepoint twice.)

7.2Application)\"wm\037_\037'\037\"\037\037\037\037.!,\037\037\037\037\037 ,,\037\037\037\037_\037\037.!\"\037\037\"\"'\037!!\037,,_\037 \037\037\037.\037:,Q.,\037,\037\037\037 _,,\037\037l\037\037\037!\037!ls)

Considera first-orderdifferentialequationof the form)

dy G(x,y)-- ,dx F(x,y))(1))

which may be difficult or impossibleto solve explicitly. Its solution curves canneverthelessbeplotted as trajectoriesof the correspondingautonomoustwo-dimen-sional system)

dxdt

= F(x,y),)dy-= G(x,y).dt)

(2))

Most ODE plotters can routinely generate phase portraits for autonomous sys-tems. Many of those appearing in this chapter wereplotted using (as illustrated

in Fig.7.2.20)John Polking'sMATLAB-basedpplaneprogram that is availablefree for educational use (mat.h.rice.edu/-dfield).Another freely availableand user-friendlyMATLAB-basedODEpackagewith similar graphicalcapabilitiesis lode(www.mat.h.uiuc.edu/iode).)

FIGURE7.2.20.MATLAB pplanemenu entries to plot a direction field and

phaseportrait for the system x'= -X,y' = -2y(asshown in Fig. 7.2.6).)

For example,to plot solutioncurves for the differentialequation)

dy 2xy -y2-- ,dx x2-2xy)

(3))))

500 Chapter7 NonlinearSystemsandPhenomena)

432)

;>.,0-1-2)

FIGURE7.2.21.Phaseportraitfor the system in Eq.(4).)

4321

;>.,0)

-1-2-3-4-4 -3-2-1 0 1 2 3 4

x)

FIGURE7.2.22.Phaseportraitfor the system correspondingtoEq. (5).)

weplot trajectoriesof the system

dx 2-=x -2xy,dt)

dy 2-= 2xy - y .dt)

(4))

The result is shown in Fig.7.2.21.Plot similarly somesolutioncurves for the followingdifferentialequations.)

dy 4x -5y1.-=dx 2x+ 3ydy 4x -5y2. dx 2x-3ydy 4x -3y3.dx 2x-5ydy 2xy4. dx x2- y2

dy x2 +2xy5.dx y2 +2xy

Now construct someexamplesof your own. Homogeneousfunctions likethosein Problems1 through 5-rationalfunctions with numerator and denominatorof the samedegreein x and y-workwell.The differentialequation)

dydx)

25x+y(1-x2-y2)(4-x2 _ y2)

-25y+x(1-x2-y2)(4-x2 - y2))(5))

of this form generalizesExample6 in this sectionbut would beinconvenientto solveexplicitly.Its phaseportrait (Fig.7.2.22)showstwo periodicclosedtrajectories-the circlesr = 1 and r = 2.Anyone want to try for three circles?)

__\037!.\037e;;1r\0371!\037_.\037\037O\037\037 Li_!,ea.!._\037I\037!\037_\037.\037_______.____._._...)

We now discussthe behaviorof solutionsof the autonomoussystem)

\037)

dxdt

= f(x,y),)dydt

= g(x,y)) (1))

near an isolatedcritical point (xo,Yo) where f(xo,Yo) = g(xo,Yo) = O. A criticalpoint is calledisolatedif someneighborhoodof it contains no other critical point.We assumethroughout that the functions f and g are continuouslydifferentiablein

a neighborhoodof (xo,Yo).We can assumewithout loss of generality that Xo = Yo = O. Otherwise,

we make the substitutions u = x - Xo, v = y-

Yo. Then dxjdt= dujdt and

dyjdt = dvjdt, so (1)is equivalentto the system

dudt

= f(u +xo,v + Yo) = ft(u, v),)

(2))dv-= g(u+xo,v +Yo) = gt(u,v)dt)

that has (0,0)as an isolatedcritical point.)))

Example1)

(1,2))

1

x)

FIGURE7.3.1.Thesaddlepoint(1,2) for the systemx'= 3x -x2 -xy,y' = y + y2 - 3xyofExample 1.)

1)

(0,0))

-1-1) ou)

FIGURE7.3.2.Thesaddlepoint(0,0) for the equivalent system

I 2u = -u- v - u - uv,v' = -6u+ 2v + V

2 - 3uv.)

7.3Linearand Almost LinearSystems 501)

The system)

dx 2-= 3x-x -xy = x(3-x-y),dt

dy = y +l-3xy = y(1-3x+y)dt)

(3))

has (1,2)as oneof its critical points.We substitute u = x-I,v = y-2;that is,

x = u + 1,y = v +2.Then)

3 -x-y = 3 - (u + 1)- (v +2) = -u- v)

and)

1-3x+ y = 1 -3(u + 1)+ (v +2)= -3u+ v,

so the systemin (3)takesthe form)

2) du 2-= (u + 1)(-u- v) = -u- v - u - uv,dt

dv-= (v +2)(-3u+ v) = -6u+2v + v2- 3uvdt)

(4))

and has (0,0)as a critical point. If we can determine the trajectoriesof the systemin (4)near (0,0), then their translationsunder the rigid motion that carries(0,0) to(1,2)will be the trajectoriesnear (1,2)of the original system in (3).This equiva-lenceis illustrated by Fig.7.3.1(which showscomputer-plottedtrajectoriesof the

systemin (3)near the critical point (1,2)in the xy-plane)and Fig.7.3.2(whichshowscomputer-plottedtrajectoriesof the system in (4)near the criticalpoint (0,0)in the uv-plane). .)

Figures7.3.1and 7.3.2illustrate the fact that the solution curves of the xy-system in (1)are simply the imagesunder the translation (u,v) \037 (u +xo, v + Yo)of the solution curves of the uv-system in (2).Near the two correspondingcriticalpoints-(xo,Yo) in the xy-planeand (0,0) in the uv-plane-thetwo phaseportraitstherefore lookpreciselythe same.)

1)

LinearizationNear a CriticalPoint)

Taylor'sformula for functions of two variablesimpliesthat-ifthe function f (x,y)is continuouslydifferentiablenear the fixedpoint (xo,Yo)-then)

f(xo+u, Yo + v) = f(xo,Yo) + fx(xo,yo)u + fy(xo, yo)v +r(u, v))

where the \"remainder term\" r (u,v) satisfiesthe condition)

1. r(u, v) -01m - .

(u,v)-+(O,O)v'u 2 + v 2)

(Note that this condition would not be satisfied if r(u, v) were a sum containingeither constants or terms linear in u or v. In this sense,r(u, v) consistsof the\"nonlinear part\" of the function f(xo+u, Yo + v) of u and v.))))

Example1)

Continued)

If we apply Taylor'sformula to both Iand g in (2)and assumethat (xo,Yo)is an isolatedcritical point so I(xo,Yo) = g(xo,Yo) = 0,the result is

dudt

= Ix(xo,yo)u + fy(xo, yo)v + r(u, v),)

dv

dt= gx(xo,Yo)u +gy(xo,yo)v +s(u, v))

(5))

where r(u, v) and the analogousremainder term s(u, v) for g satisfy the condition)

1. r(u, v)1m

(u,v)-+(O,O) -vu 2 + v 2)

. s(u, v) 0hm = .(u,v)-+(O,O) -vu 2 + v 2)

(6))

Then, when the values u and v are small, the remainder terms r(u, v) and s(u,v)are very small (beingsmall even in comparisonwith u and v).

If we dropthe presumablysmallnonlinear terms r(u, v) and s(u,v) in (5),theresult is the linearsystem)

\037)

dudt

= fx(xo,yo)u + Iy(xo,yo)v,

dv

dt= gx(xo,yo)u +gy(xo,yo)v)

(7))

whoseconstant coefficients (of the variables u and v) are the values Ix(xo,Yo),

Iy(xo,Yo) and gx(xo,Yo), gy(xo,Yo) of the functions I and g at the critical point(xo,Yo). Because(5)is equivalent to the original (and generally)nonlinear systemu' = I(xo+u, Yo + v), v' = g(xo+u, Yo + v) in (2),the conditions in (6)suggestthat the linearizedsystem in (7)closelyapproximates the given nonlinear systemwhen (u,v) is closeto (0,0).

Assuming that (0,0) is also an isolatedcritical point of the linear system,and that the remainder terms in (5)satisfy the condition in (6),the original systemx'= f (x,y), y' = g(x,y) is saidto be almostlinearat the isolatedcritical point(xo,Yo). In this case,its linearizationat (xo,Yo) is the linear systemin (7). In

short, this linearization is the linear systemu' = Ju (where u = [u v]T)whosecoefficientmatrix is the so-calledJacobianmatrix)

\037) J(xo,Yo) =[

Ix(xo,Yo) Iy (xo,Yo)

]gx(xo,yo) gy(xo,yo))(8))

of the functionsIand g, evaluatedat the point (xo,Yo).)

\"\"\037 c<''-' \"\" \037 \037.\"\" v\"\"\"'\" \"\" .-. \037 \037 \037 _ \"\"'.. \037.-. ,.\",,,..' \"\" '\" \"\" n\"'\037\"Y \037 _ \037...n ....... \"..y............ \"............, \"'\" \"\"\037v \" ,,,,,,.. ......., '\" ,... ..... \037 \"\"... . ,... '\" '\" ,.d.\".\"..... ..... . ..In (3)wehave I(x,y) = 3x-x2 -xy and g(x,y) = y + y2 -3xy. Then)

[3-2X-yJ(x,y) =

-3y) 1 + 2yx-3x] ,) [-1 -1

]so J(I,2)= -6 2.)Hencethe linearizationof the systemx'= 3x-x2 -xy, y' = y + y2 -3xy at its

critical point (1,2) is the linear system)

,u = -u- v,v' = -6u+ 2v)

that we get when we dropthe nonlinear (quadratic)terms in (4).) .)))

__

u____

__

--)

FIGURE7.3.3.Theobliqueu v-coordinatesystem determin-edby the eigenvectorsv 1 and v2.)


It turns out that in most (though not all) cases,the phaseportrait of an al-most linear system near an isolatedcritical point (xo,Yo) strongly resembles-qualitatively-thephaseportrait near the origin of its linearization. Consequently,the first step toward understanding generalautonomoussystemsis to characterizethe criticalpointsof linearsystems.)

CriticalPointsofLinearSystemsWe can use the eigenvalue-eigenvectormethod of Section5.4to investigate the

critical point (0,0)of a linear system)

\037)

[\037;]

=[\037 \037] [\037 ])

(9))

with constant-coefficientmatrix A. Recall that the eigenvaluesAl and A2 of A arethe solutionsof the characteristicequation)

\037)

a -A bdet(A -AI) = c d _ A

= (a -A)(d -A) -bc ==O.)

We assumethat (0,0) is an isolatedcritical point of the system in (9),so it followsthat the coefficientdeterminant ad-bc of the system ax+by ==0, cx+ dy == 0is nonzero. This impliesthat A == 0 is not a solution of (9),and hence that both

eigenvaluesof the matrix A are nonzero.The nature of the isolatedcriticalpoint (0,0)then dependson whether the two

nonzero eigenvaluesAl and A2 of A are)

\302\267 real and unequal with the samesign;\302\267 real and unequal with oppositesigns;\302\267 real and equal;\302\267 complexconjugates with nonzero real part; or\302\267 pure imaginary numbers.)

Thesefive casesare discussedseparately. In eachcasethe critical point (0,0) re-semblesone of thosewe saw in the examplesof Section7.2-anode (properorimproper),a saddlepoint, a spiral point, or a center.)

UNEQUAL REAL EIGENVALUES WITH THESAME SIGN: In this casethe ma-trix A has linearly independent eigenvectors VI and V2, and the general solution

x(t) == [ x(t) y(t)]T of (9)takes the form)

x(t)= cIvIeA1t + c2v2eA2t.) (10))

x)This solution is most simply describedin the obliqueuv-coordinate system indi-catedin Fig.7.3.3,in which the u- and v-axesare determinedby the eigenvectorsVI

and V2. Then the uv-coordinatefunctions u(t)and vet) of the moving point x(t)aresimply its distancesfrom the originmeasuredin the directionsparallelto the vectorsVI and V2, so it follows from Eq.(10)that a trajectoryof the system isdescribedby)

u(t) ==uoeA1t , vet) ==voeA2t) (11))

where Uo = u(O) and Vo = v(O). If Vo = 0,then this trajectory lieson the u-axis,whereasif Uo == 0,then it lieson the v-axis. Otherwise-ifUo and Vo are)))

Example2)

x)

FIGURE7.3.4.Theimpropernodal sourceofExample 2.)

54321

;;:.... 0-1-2-3-4 '\" '\\ t-5-5-4 -3 -2 -1 0 1 2 3 4 5

x)

FIGURE7.3.5.ThesaddlepointofExample 3.)

Example3)

both nonzero-theparametric curve in (11)takesthe explicitform v = Cuk wherek = A2/AI >O.Thesesolutioncurves are tangent at (0,0)to the u-axisif k > 1,tothe v-axis if 0 <k < 1.Thus wehave in this casean impropernodeas in Example3 of Section7.2.If Al and A2 are both positive, then we seefrom (10)and (11)that

thesesolution curves \"depart from the origin\" as t increases,so (0,0) is a nodalsource.But if Al and A2 are both negative, then thesesolutioncurves approach the

origin as t increases,so (0,0) is a nodal sink.)

(a) The matrix)

I

[7

A = 8 -3) 1\037 ])

has eigenvalues Al = 1 and A2 = 2 with associatedeigenvectorsVI = [3 I]Tand V2 = [1 3]T.Figure 7.3.4showsa direction field and typical trajectoriesofthe correspondinglinear systemx' = Ax. Note that the two eigenvectorspoint in

the directionsof the linear trajectories.As is typical of an impropernode,all othertrajectoriesare tangent to one of the obliqueaxesthrough the origin.In this examplethe two unequal real eigenvaluesare both positive, so the critical point (0,0) is an

impropernodal source.(b) The matrix)

I

[-7B = -A= 8 3)

-3]-17)

has eigenvalues A I = -1and A2 = -2 with the sameassociatedeigenvectorsVI = [3 I]Tand V2 = [1 3]T.The new linear system x' = Bx has the samedirection field and trajectoriesas in Fig.7.3.4exceptwith the direction fieldarrowsnow all reversed,so (0,0) is now an impropernodal sink. .)

UNEQUAL REAL EIGENVALUES WITH OPPOSITESIGNS: Herethe situation

is the sameas in the previouscase,exceptthat A2 <0 <Al in (11).The trajectorieswith Ua = 0 or Va = 0 lieon the u- and v-axes through the critical point (0,0).Thosewith Ua and Va both nonzero are curves of the explicitform v = Cuk, wherek = A2/AI < O. As in the casek < 0 of Example3 in Section7.2,the nonlineartrajectoriesresemblehyperbolas,and the criticalpoint (0,0)is thereforean unstablesaddlepoint.)

The matrix)

I

[5A= 4 3)

-3]-5)

has eigenvaluesA I = 1 and A2 = -1with associatedeigenvectorsV I = [3 1]Tand V2 = [1 3]T.Figure 7.3.5showsa direction field and typical trajectoriesof the correspondinglinear system x' = Ax. Note that the two eigenvectorsagainpoint in the directionsof the linear trajectories.Herek = -1and the nonlineartrajectories are (true) hyperbolasin the obliqueuv-coordinate system,so we havethe saddlepoint indicated in the figure. Note that the two eigenvectorspoint in thedirectionsof the asymptotes to thesehyperbolas. .)))

Example4)

54321

;>., 0-1-2-3-4 t '\\

-5-5-4 -3 -2 -1 0 1 2 3 4 5x)

FIGURE7.3.6.Theimpropernodal sink ofExample 4.)


EQUALREAL ROOTS: In this case,with A = Al = A2 i= 0,the characterof the critical point (0,0) dependson whether or not the coefficientmatrix A hastwo linearly independent eigenvectorsVI and V2. If so,then we have oblique uv-coordinatesas in Fig.7.3.3,and the trajectoriesare describedby)

u(t) = uoeAt, vet) = voeAt) (12))

as in (11).But now k = A2/AI = 1,so the trajectories with Uo i= 0 are all of theform v = Cu and hencelieon straight linesthrough the origin. Therefore,(0,0)isa propernode(orstar)as illustrated in Fig.7.2.4,and is a sourceif A >0, a sinkif A <O.

If the multiple eigenvalueA i= 0 has only a singleassociatedeigenvectorVI,then (aswe saw in Section5.6)there neverthelessexistsa generalizedeigenvectorV2 such that (A -AI)v2 = VI,and the linear system x' = Ax has the two linearly

independent solutions)

Xl (t) = vle At and X2(t) = (Vlt +v2)eAt.) (13))

We can still use the two vectors VI and V2 to introduceobliqueuv-coordinatesas in

Fig.7.3.3.Then it follows from (13)that the coordinate functions u(t) and vet) ofthe moving point x(t)on a trajectoryare given by)

u(t) = (uo + vot)eAt, vet) = voeAt

,) (14))

where Uo = u(O) and Vo = v(O). If Vo = 0 then this trajectory lieson the u-axis.Otherwisewe have a nonlinear trajectory with)

dv

du)

A voeAt-voeAt +A(UO + vot)eAt)

AVO-Vo +A(UO + vot))

dv/dtdu/dt)

We seethat dv/du \037 0 as t \037 :f::oo,so it follows that each trajectoryis tangent tothe u-axis.Therefore, (0,0)is an impropernode.If A <0,then we seefrom (14)that this nodeis a sink, but it is a sourceif A >O.)

The matrix)

A_1.[-II 9

]-8 -1 -5

has the multiple eigenvalue A = -1with the singleassociatedeigenvectorVI

[ 3 1]T. It happensthat V2 = [1 3]T is a generalizedeigenvectorbasedonVI, but only the actual eigenvectorshowsup in a phaseportrait for the linear sys-tem x' = Ax. As indicated in Fig.7.3.6,the eigenvectorVI determinesthe u-axisthrough the improper nodal sink (0,0),this axisbeingtangent to each of the non-linear trajectories. .)COMPLEXCONJUGATE EIGENVALUES: Supposethat the matrix A has eigen-values A = P +qi and A = P -qi (with p and q both nonzero)having associatedcomplexconjugateeigenvectorsV = a +bi and V = a-bi.Then we saw in Section5.4-seeEq. (22)there-thatthe linear system x' = Ax has the two independentreal-valued solutions)

Xl (t) = ept (acosqt -b sin qt) and X2(t) = ept (bcosqt +a sin qt). (15))))


Example5)

x)

FIGURE7.3.7.Thespiral sinkofExample 5.)

Example6)

54321

;>., 0-1-2-3-4

It::\" 1C:'tL\"t.C-5-5-4 -3 -2 -1 0 1 2 3 4 5x)

FIGURE7.3.8.ThestablecenterofExample 6.)

Thus the componentsx(t) and yet) of any solutionx(t)= CIXI (t) +C2X2(t)oscillatebetweenpositive and negative values as t increases,so the critical point (0,0) is aspiralpoint as in Example5 of Section7.2.If the real part p of the eigenvaluesis negative, then it is clearfrom (15)that x(t) \037 0 as t \037 +00,so the origin is aspiral sink.But if p is positive, then the critical point is a spiral source.)

The matrix)

I

[-10

A = 4 -15) 1\037 ])

has the complexconjugateeigenvaluesA = -\037

:f::3iwith negativereal part, so(0,0)is a spiralsink. Figure 7.3.7showsa direction field and a typical spiral trajectoryapproaching the origin as t \037 +00. .)

PURE IMAGINARY EIGENVALUES: If the matrix A has conjugate imaginaryeigenvalues A = qi and A = -qiwith associatedcomplexconjugate eigenvec-tors v = a + bi and v = a -bi,then (15)with p = 0 gives the independentsolutions)

5)XI(t) = acosqt-bsinqt and X2(t) = bcosqt+asinqt) (16))

of the linear systemx' = Ax. Just as in Example4 of Section7.2,it follows that

any solution x(t) = CIXI (t) + C2X2(t) describesan ellipsecenteredat the origin in

the xy-plane.Hence(0,0) is a stablecenterin this case.)

The matrix)

I

[-9

A = 4 -15) 1\037 ])

has the pure imaginary conjugate eigenvalues A = :f::3i,and therefore (0,0)is astablecenter.Figure 7.3.8showsa direction field and typical elliptical trajectoriesenclosingthe critical point. .

For the two-dimensionallinear system x' = Ax with det A i= 0,the table in

Fig.7.3.9lists the type of critical point at (0,0)found in the five casesdiscussedhere,accordingto the nature of the eigenvaluesA I and A2 of the coefficientmatrix A.Our discussionof the variouscasesshowsthat the stability of the criticalpoint (0,0)is determined by the signsof the real parts of theseeigenvalues,as summarizedin

Theorem 1.Note that if Al and A2 are real, then they are themselvestheir real parts.)

Real,unequal, samesign

Real,unequal, oppositesign

Realand equalComplexconjugate

Pure imaginary)

Improper nodeSaddlepoint

Properor improper node

Spiral point

Center)

FIGURE7.3.9.Classificationof the critical point (0,0)of the two-dimensional system x'= Ax.)))

Al = qi)

11-1=r+si)

A2=-qi)

11-2= r -si)

FIGURE7.3.10.Theeffectsofperturbation ofpure imaginaryroots.)

y)Complexconjugateroots)

11-2\302\260)

FIGURE7.3.11.Theeffectsofperturbation of real equal roots.)


THEOREM1 Stabilityof LinearSystemsLet A 1 and A2 be the eigenvalues of the coefficient matrix A of the two-dimensionallinear system)

dx-= ax+by,dt

dy-=ex+dydt)

(17))

with ad-be :f= O.Then the critical point (0,0)is

1.Asymptotically stableif the real parts of A1 and A2 are both negative;2.Stablebut not asymptoticallystableif the real parts of At and A2 are both

zero(sothat At, A2 = -::!::.qi);3.Unstableif either A1or A2 has a positivereal part.)

x)

It is worthwhile to considerthe effectof smallperturbationsin the coefficientsa, b, c, and d of the linear system in (17),which result in small perturbations ofthe eigenvaluesA 1 and A2. If theseperturbationsare sufficiently small, then positivereal parts (of Al and A2) remain positive and negative real parts remain negative.Hencean asymptoticallystablecritical point remains asymptoticallystable and an

unstable critical point remains unstable.Part 2 of Theorem 1 is therefore the onlycasein which arbitrarily small perturbations can affect the stability of the criticalpoint (0,0).In this casepure imaginary roots AI, A2 = :!:::.qiof the characteristicequation can be changedto nearby complexroots /11,/12 = r :!:::.si,with r eitherpositive or negative (seeFig.7.3.10).Consequently, a small perturbation of the

coefficientsof the linear system in (7)can change a stablecenter to a spiral pointthat is either unstable or asymptoticallystable.

There is one other exceptionalcasein which the type, though not the stability,of the critical point (0,0) can be altered by a small perturbation of its coefficients.This is the casewith Al = A2, equal roots that (under a small perturbation of the

coefficients)can split into two roots /11and /12,which are eithercomplexconjugatesor unequal real roots (seeFig.7.3.11).In either case,the signof the real parts of the

roots is preserved,so the stability of the critical point is unaltered. Its nature maychange,however; the table in Fig.7.3.9showsthat a nodewith Al = A2 can eitherremain a node(if /11and /12are real) or change to a spiral point (if /11and J.l2 arecomplexconjugates).

Supposethat the linear system in (17)is usedto model a physical situation. Itis unlikely that the coefficientsin (17)can be measuredwith total accuracy,so letthe unknown preciselinear model be)

dx-= a*x +b*y,dt

dy-= c*x+d*y.dt)

(17*))

If the coefficientsin (17)are sufficiently closeto those in (17*),it then followsfromthe discussionin the precedingparagraph that the origin (0,0) is an asymptoticallystablecritical point for (17)if it is an asymptoticallystablecriticalpoint for (17*),and is an unstable critical point for (17)if it is an unstable critical point for (17*).Thus in this casethe approximate model in (17)and the precisemodel in (17*))))

predictthe samequalitative behavior (with respectto asymptotic stability versusinstability).)

AlmostLinearSystemsRecall that we first encountered an almost linear system at the beginning of this

section,when we usedTaylor'sformula to write the nonlinear system (2) in the

almost linear form (5)which led to the linearization (7)of the original nonlinearsystem. In casethe nonlinear systemx'= f(x,y), y' = g(x,y) has (0,0)as an

isolatedcritical point, the correspondingalmost linear system is

dxdt =ax+by+r(x,y),

dy-=cx+dy+s(x,y)dt)

(18))

where a = fx(O,0),b = fy(O,0) and c = gx(O,0),d = gy(O,0);we assumealsothat ad - bc i= O. Theorem 2,which we state without proof, essentiallyimpliesthat-with regard to the type and stability of the critical point (0,O)-theeffectofthe small nonlinear terms r (x,y) and s(x,y) is equivalent to the effect of a small

perturbation in the coefficientsof the associatedlinearsystem in (17).)

THEOREM2 Stabilityof AlmostLinearSystems)Let A 1 and A2 be the eigenvaluesof the coefficientmatrix of the linear system in

(17)associatedwith the almost linear system in (18).Then

1.If Al = A2 are equal real eigenvalues,then the criticalpoint (0,0)of (18)iseither a nodeor a spiral point, and is asymptoticallystableif Ai = A2 <0,unstable if Al = A2 >O.

2.If Al and A2 are pure imaginary, then (0,0) is either a center or a spiralpoint, and may be either asymptoticallystable,stable,or unstable.

3.Otherwise-thatis, unless Al and A2 are either real equal or pureimaginary-thecritical point (0,0) of the almost linear system in (18)isof the sametype and stability as the critical point (0,0)of the associatedlinear systemin (17).)

Thus, if Al i= A2 and Re(AI) i= 0,then the type and stability of the criticalpoint of the almost linear systemin (18)can bedeterminedby analysisof its associ-ated linear systemin (17),and only in the caseof pure imaginary eigenvaluesis the

stability of (0,0)not determinedby the linear system.Exceptin the sensitivecasesA I = A2 and Re(Ai) = 0,the trajectories near (0,0) will resemblequalitativelythoseof the associatedlinear system-they enter or leave the critical point in the

sameway, but may be \"deformed\" in a nonlinear manner. The table in Fig.7.3.12summarizes the situation.

An important consequenceof the classificationof casesin Theorem 2 is that

a criticalpoint of an almost linearsystemis asymptoticallystableif it isan asymp-totically stablecriticalpoint of the linearization of the system. Moreover,a criticalpoint of the almost linear system is unstable if it is an unstable critical point of the

linearized system.If an almost linear systemis usedto modela physical situation,

then-apartfrom the sensitivecasesmentionedearlier-itfollows that the qualita-tive behavior of the systemnear a critical point can bedeterminedby examining its

linearization.)))

f\037\037IA\037f.\"_)TypeofCriticalPoint of

the Almost. LinearSystem)

Al < A2 < 0Al = A2 < 0Al < 0 < A2

Al = A2 > 0Al > A2 > 0AI, A2 = a -:f::.bi (a < 0)AI, A2 = a -:f::. bi (a > 0)AI, A2 = -:f::.bi)

Stableimproper nodeStablenode or spiral point

Unstable saddlepoint

Unstable node or spiral point

Unstable improper nodeStablespiral point

Unstable spiral point

Stableor unstable, centeror spiral point)

FIGURE7.3.12.Classificationof critical points of an almost linear system.)

Example7)- -

Determine the type and stability of the critical point (0,0) of the almost linear

system)dx-= 4x +2y +2x2-3y2,dt

dy-= 4x -3y +7xy.dt)

(19))

Solution The characteristic equation for the associatedlinear system (obtained simply by

deleting the quadratic terms in (19))is)

(4-A)( -3-A) - 8 = (A -5)(A +4) = 0,)

so the eigenvaluesAl = 5 and A2 = -4are real, unequal, and have oppositesigns.By our discussionof this casewe know that (0,0) is an unstable saddlepoint ofthe linear system,and henceby Part 3 of Theorem 2,it is also an unstable sad-dle point of the almost linear system in (19).The trajectoriesof the linear systemnear (0,0)are shown in Fig.7.3.13,and those of the nonlinear system in (19)areshown in Fig.7.3.14.Figure 7.3.15showsa phaseportrait of the nonlinear sys-tem in (19)from a \"wider view.\" In addition to the saddlepoint at (0,0),there arespiralpoints near the points (0.279,1.065)and (0.933,-1.057),and a node near

(-2.354,-0.483). .)

0.4) 0.4)3)

\037 0.0) \037 0.0) \037 0)

2)

0.2) 0.2)1)

-0.2) -0.2) -1)

-2)-0.4) -0.4)

-0.4 -0.2 0.0 0.2 0.4x)

-0.4 -0.2 0.0 0.2 0.4x)

ox)

2 3)

FIGURE7.3.13.Trajectoriesofthe linearizedsystem ofExample 7.)

FIGURE7.3.14.Trajectoriesofthe original almost linear system ofExample 7.)

FIGURE7.3.15.Phaseportraitfor the almost linear system in

Eq.(19).)))


We have seenthat the systemx' = f(x,y), y' = g(x,y) with isolatedcriticalpoint (xo,Yo) transforms via the substitution x = u +xo, y = v + Yo to an equiv-alent u v-system with correspondingcritical point (0,0) and linearizationu' = Ju,whosecoefficientmatrix J is the Jacobian matrix in (8)of the functions f and g at

(xo,Yo).Consequentlywe neednot carry out the substitution explicitly;instead,wecan proceeddirectly to calculate the eigenvaluesof J preparatory to applicationofTheorem 2.)

Example8)............... ........

Determine the type and stability of the critical point (4,3) of the almost linearsystem)

dx 2dt

= 33 - lOx-3y +x ,

dydt

= -18+6x+2y-xy .)

(20))

Solution With f(x,y) = 33 - lOx- 3y +x2, g(x,y) = -18+6x+2y -xy and Xo = 4,Yo = 3 wehave)

J( ) - [-10+2xx, y -

6 -y)-3

]2 -x ') [-2 -3

]so J(4,3) =3 -2 .)

The associatedlinear system)

du-= -2u- 3v,dt

dv-= 3u - 2vdt)

(21))

has characteristic equation (A + 2)2 + 9 = 0,with complexconjugate roots A =-2 ::f::3i.Hence(0,0)is an asymptoticallystablespiral point of the linear system in

(21),soTheorem 2 impliesthat (4,3) is an asymptoticallystablespiral point of the

original almost linear systemin (20).Figure 7.3.16showssometypical trajectoriesof the linear system in (21),and Fig.7.3.17showshow this spiral point fits into the

phaseportrait for the original almost linear system in (20). .)

2 1210

1 86

:::.0 \037 42

-1)

-2-2) -1) ou)

1) 2)

x)

FIGURE7.3.16.Spiral trajectoriesof the linear system in Eq.(21).)

FIGURE7.3.17.Phaseportrait forthe almost linear system in Eq.(20).)))

_ Problems)In Problems1through 10,apply Theorem 1to determine the

type of the criticalpoint (0,0) and whether it is asymtoticallystable, stable,or unstable. Verify your conclusionby using acomputer system or graphing calculatorto construct a phaseportrait for the given linear system.)

1.dx dy-= -2x+ y-= x -2ydt ' dt

2.dx dy-= 4x- y, -=2x+ydt dt

3.dx dy

dt= x + 2y, -=2x+ydt

4. dx dy-= 3x + y, -= 5x- Ydt dtdx dy5.-= x - 2y, -= 2x- 3ydt dtdx dy6.-= 5x-3y, -= 3x -Ydt dtdx dy7.-= 3x - 2y, -= 4x- Ydt dt

8.dx dy-= x - 3y, -= 6x-5ydt dt

9.dx dy-= 2x -2y, -= 4x-2ydt dt

10.dx dy-= x - 2y, -= 5x- Ydt dt)

Eachof the systems in Problems11through 18has a singlecriticalpoint (xo,Yo). Apply Theorem 2 to classifythis criticalpoint as to type and stability. Verify your conclusionby usinga computer system or graphing calculatorto construct aphaseportrait for the given system.)

11.)dx dy

dt= x - 2y, dt

= 3x -4y -2

dx dy

dt= x - 2y - 8,

dt= x + 4y + 10

dx dy-= 2x- Y-2 -= 3x -2y

-2dt '

dtdx dy

dt= x + y

-7,dt

= 3x - y-5

dx dy

dt= x - y, dt

= 5x- 3y -2

dx dy

dt= x -2y + 1,

dt= x + 3y -9

dx dy

dt= x - 5y -5, dt

= x -y- 3

dx dy

dt= 4x-5y + 3, dt

= 5x-4y + 6)

12.)

13.)

14.)

15.)

16.)

17.)

18.)

In Problems19through 28, investigate the type of the criti-calpoint (0,0) of the given almost linear system. Verify yourconclusionby using a computer system or graphing calculatorto construct a phaseportrait. Also, describethe approximate)


locationsand apparent types of any other critical points that

are visible in your figure. Feelfree to investigate theseaddi-tional criticalpoints; you can use the computational methods

discussedin the application material for this section.)

dx19.-= x - 3y + 2xy,dt

dx20.-=6x-5y+x2,dt)

dy-= 4x-6y -xydt

dy-= 2x - y + y2dt)

dx dy21.-= x + 2y + x2 + y2, -= 2x -2y - 3xydt dt

22.dx = x + 4y _ xy2,dy = 2x - y + x2

ydt dt

dx dy23.-=2x-5y+x3, -=4x-6y+y4dt dt

dx dy24.-= 5x -3y + y(x2 + y2), -= 5x + y(x2 + y2)dt dt

dx dy 2 225.dt

= x -2y + 3xy, dt= 2x - 3y -x -

y)

dx 2 226.dt

= 3x - 2y -x -y ,)

dy-= 2x - y- 3xy

dt)

dx27.-= x -y + x4 _ y2,dt

dx28.-= 3x - y + x3 + y3,dt)

dy-= 2x - y + y4-x2

dt

dy-= 13x-3y + 3xydt)

In Problems29 through 32,find all criticalpoints ofthe given

system, and investigate the type and stability ofeach. Verify

your conclusionsby meansofa phaseportrait constructed us-ing a computer system or graphing calculator.)

29.dx dy-= x - y, -= x2 -ydt dt

30.dx dy

dt= Y

-1, -= x2 -ydt

dx dy31.-= y2- 1, -= x3 -

ydt dt

32.dx dy-= xy -2, -= x - 2ydt dt

Bifurcations)

The term bifurcation generally refersto something \"splitting

apart.\" With regard to differential equations or systems involv-

ing a parameter, it refersto abrupt changesin the characterofthe solutions as the parameteris changedcontinuously. Prob-lems 33 through 36 illustrate sensitive casesin which small

perturbations in the coefficientsof a linear or almost linearsystem can change the type or stability (or both) ofa criticalpoint.)))

1.0)

0.5)

\037 0.0)

-0.5)

-1.0-1.0 -0.5) 0.0x)

0.5) 1.0)

FIGURE7.3.18(a).Stablespiralwith E = -0.2.)

1.0) 1.0)

-1.0-1.0-0.5)

0.5) 0.5)

\037 0.0) \037 0.0)

-0.5) -0.5)

-1.0-1.0 -0.5) 0.0x)

0.5) 1.0)

FIGURE7.3.18(c).Stablecenterwith E = O.)

1.0)

0.5)

\037 0.0)

-0.5)

-1.0-1.0-0.5) 0.5) 1.0)0.0x)

FIGURE7.3.18(b).Stablespiralwith E = -0.05.)

1.0)

0.5)

\037 0.0)

-0.5)

0.0x)

-1.0 ....-1.0-0.5 0.0 0.5 1.0

x)

0.5) 1.0)

FIGURE7.3.18(d).Unstable

spiral with E = 0.05.)FIGURE7.3.18(e).Unstable

spiral with E = 0.2.)

33.Considerthe linear system)

dxdt

= EX -y,)

dy-=X+EY.dt)

Show that the critical point (0,0) is (a) a stable spiralpoint if E < 0; (b) a centerif E = 0; (c) an unstable spi-ral point if E > O.Thus small perturbations of the systemx'= -y, y' = x can change both the type and stability ofthe critical point. Figures 7.3.18(a)-(e)illustrate the lossof stability that occursat E = 0 as the parameter increasesfrom E < 0 to E > O.

34. Considerthe linear system)

dxdt

= -x+ EY,)dy-= x - y.dt)

35.)

Show that the critical point (0,0) is (a) a stable spiralpoint if E < 0; (b) a stable nodeif 0 < E < 1.Thussmall perturbations of the system x'= -x,y' = x-y canchange the type of the critical point (0,0)without chang-ing its stability.This problem dealswith the almost linear system)

dxdt

= Y + hx(x 2 + y2),)dy

dt= -x+ hy(x

2 + y2),)

in illustration of the sensitive caseofTheorem2, in which

the theorem providesno information about the stability ofthe criticalpoint (0,0). (a) Show that (0,0) is a centerofthe linear system obtained by setting h = O. (b) Supposethat h i= O.Letr 2 = x2 + y2, then apply the fact that

dx dy drx

dt+ y dt

= rdt

to show that dr/dt = hr 3. (c) Supposethat h = -1.Integrate the differential equation in (b); then show that

r \037 0 as t \037 +00.Thus (0,0) is an asymptotically sta-

ble critical point of the almost linear system in this case.(d) Supposethat h = +1.Show that r \037 +00as t in-

creases,so (0,0) is an unstable critical point in this case.36.This problem presents the famous Hopfbifurcation for the

almost linear system

dxdt

= EX + y-x(x2 + y2),

dy 2 2dt

= -x+ EY- y(x + y ),

which has imaginary characteristicroots A = ::i:iif E =O.(a) Change to polar coordinatesas in Example 6 ofSec-tion 7.2to obtain the system r' = r(E - r 2),e'= -1.)))

7.4EcologicalModels:Predatorsand Competitors513)

37.)

(b) Separatevariables and integrate directly to show thatif E < 0, then r(t) -+ 0 as t -+ +00,so in this casethe

origin is a stable spiral point. (c) Show similarly that ifE > 0, then r(t) -+ ,JEas t -+ +00,so in this casethe

origin is an unstable spiral point. Thecircler(t) = ,JEitself is a closedperiodicsolution or limit cycle.Thus alimit cycleof increasingsizeis spawned as the parameterE increasesthrough the critical value O.In the caseofa two-dimensional system that is not almostlinear, the trajectoriesnear an isolatedcritical point canexhibit a considerably more complicatedstructure than

those near the nodes,centers,saddlepoints, and spiralpoints discussedin this section. For example, considerthe system)

dxdt

= x(x3 -2y3),

dy_ = y(2x3 _ y3)dt

having (0,0)as an isolatedcritical point. This system isnot almost linear because(0,0) is not an isolatedcritical

point of the trivial associatedlinear system x'= 0,y' = O.Solvethe homogeneousfirst-order equation)

FIGURE7.3.19.Trajectoriesof the

system in Eq.(22).)

(22))

D)

dy y(2x3 _ y3)dx x(x3 -2y3)

to show that the trajectoriesof the system in (22)arefoliaofDescartesof the form

x3 + y3 = 3cxy,

where c is an arbitrary constant (Fig.7.3.19).38.First note that the characteristicequation of the 2 x 2 ma-trix A can be written in the form A2 - TA + D = 0,where D is the determinant of A and the traceT of thematrix A is the sum of its two diagonal elements.Thenapply Theorem1 to show that the type of the critical point(0,0)of the system x'= Ax is determined-asindicatedin Fig. 7.3.20-bythe location of the point (T, D) in thetrace-determinant planewith horizontal T-axisand verti-calD-axis.)

Spiralsource)

T)

FIGURE7.3.20.Thecritical point (0,0)of the

system x'= Ax is a. spiral sink or sourceif the point (T,D) liesabove the parabola T2 = 4Dbut off theD-axis;. stablecenterif (T, D)lieson the positiveD-axis;. nodal sink or sourceif (T, D) liesbetween the

parabolaand the T-axis;. saddlepoint if (T, D)liesbeneath the T-axis.)

IIIJEcol\037g\037callV!odels:?redat\037rsan\037 C0l1!pet\037tors)

Someof the most interesting and important applicationsof stability theory involvethe interactions betweentwo or more biologicalpopulations occupying the sameenvironment. We considerfirst a predator-preysituation involving two species.Onespecies-thepredators-feedson the other species-theprey-whichin turn

feedson some third food item readily available in the environment. A standard

exampleis a population of foxesand rabbits in a woodland; the foxes (predators)eat rabbits (the prey),while the rabbitseat certainvegetationin the woodland.Otherexamplesare sharks(predators)and food fish (prey),bass (predators)and sunfish

(prey),ladybugs (predators)and aphids (prey),and beetles(predators)and scaleinsects(prey).)))

Example1)

The classicalmathematicalmodel of a predator-preysituation was developedin the 1920sby the Italian mathematician Vito Volterra (1860-1940)in order toanalyze the cyclicvariationsobservedin the shark and food-fishpopulations in the

Adriatic Sea.To construct such a model,we denote the number of prey at timet by x(t), the number of predatorsby y(t), and make the following simplifying

assumptions.)

1.In the absenceof predators,the prey population would grow at a natural rate,with dxjdt= ax,a >O.

2. In the absenceof prey, the predator populationwoulddeclineat a natural rate,with dyjdt = -by,b >O.

3.When both predatorsand prey are present,there occurs,in combinationwith

thesenatural rates of growth and decline,a declinein the preypopulationand agrowth in the predator population,eachat a rate proportionalto the frequencyof encountersbetweenindividuals of the two species.We assumefurther that

the frequency of such encounters is proportional to the productxy, reasoningthat doubling either population alone shoulddoublethe frequencyof encoun-ters, while doubling both populations ought to quadruple the frequency ofencounters.Consequently,the consumptionof prey by predators results in). an interactionrate of decline-pxy in the prey populationx,and. an interaction rate of growth qxy in the predator population y.)

When wecombinethe natural and interactionrates axand -pxy for the preypopulationx,aswell as the natural and interactionrates -byand qxy for the preda-tor population y, weget the predator-preysystem

dx-= ax-pxy = x(a-py),dt)

\037) (1))dy-= -by+qxy = y(-b+qx),dt)

with the constantsa, b, p,and q all positive.[Note: You may seethe predatorandprey equations written in either order in (1).It is important to recognizethat the

predator equation has negative linear term and positive interaction term, whereasthe prey equation has positive linear term and negativeinteraction term.])

TheCriticalPoints A critical point of the general predator-preysystem in (1)isa solution (x,y) of the equations)

x(a-py) = 0, y(-b+qx) = O.) (2))

The first of these two equations impliesthat either x = 0 or y = ajp =1= 0,and

the secondimpliesthat either y = 0 or x = bjq =1= O. It follows readily that this

predator-preysystemhas the two (isolated)critical points (0,0)and (bjq,ajp).)

THECRITICALPOINT (0,0): The Jacobian matrix of the system in (1)is)

[a -py -px

]J(x,y)=qy -b+qx') so J(O,O)=

[\037-\037 l) (3))

The matrix J(O,0)has characteristicequation (a-A)(-b-A) = 0 and the eigen-values Al = a > 0,A2 = -b < 0 with different signs.Henceit follows from)))


Theorems 1 and 2 in Section7.3that the critical point (0,0)is an unstable saddlepoint, both of the predator-preysystem and of its linearizationat (0,0). The cor-respondingequilibrium solutionx(t) = 0,yet) = 0 merely describessimultaneousextinctionof the prey (x)and predator (y) populations.)

THECRITICALPOINT (blq,alp): The Jacobian matrix)

o)pb--q

o)J(bjq,ajp) =)

aqp)

(4))

has characteristic equation A2 + ab = 0 and the pure imaginary eigenvaluesAI,

A2 = ::f::i,J(ib.It follows from Theorem 1 in Section7.3that the linearization of the

predator-preysystem at (bjq,ajp) has a stable center at the origin. Thus we havethe indeterminatecaseof Theorem 2 in Section7.3,in which casethe criticalpointcan (asidefrom a stablecenter)alsobeeithera stablespiral sink or an unstablespiralsourceof the predator-preysystemitself. Hencefurther investigation is requiredtodetermine the actual character of the critical point (bjq,ajp).The correspondingequilibrium solution x(t) = bjq, yet) - ajp describesthe only nonzeroconstant

prey (x)and predator (y) populations that coexistpermanently.)

THEPHASEPLANE PORTRAIT In Problem 1 we askyou to analyzenumericallya typical predator-preysystem and verify that the linearizationsat its two criticalpoints agreequalitatively with the phaseplane portrait shown in Fig.7.4.I-wherethe nontrivial critical point appearsvisually to be a stablecenter. Of course,onlythe first quadrant of this portrait correspondsto physically meaningful solutions

describingnonnegativepopulationsof prey and predators.In Problem 2 weaskyou to derive an exact implicit solutionof the predator-

prey system of Fig.7.4.I-asolution that can be usedto show that its phaseplanetrajectories in the first quadrant are, indeed,simpleclosedcurves that encirclethe

critical point (75,50)as indicated in the figure. It then follows from Problem30 in)

50 100 150 200x (prey)

FIGURE7.4.2.Thepredator-preyphaseportrait ofExample 2.)

200)

200

-- 150tr.J

'\"'\"0\037

\037

\"'0 100(I.)'\"'\"0..'-\";;..-,

50

0)

100)

;;..-,)

o)

-100)

-100) o) 100) 200) 300)x)

FIGURE7.4.1.Phaseplane portrait for the predator-prey systemx'= 200x-4xy,y' = -150y+ 2xy with critical points (0,0) and (75,50).)

o)))


Example2)

8070

6050

\037 40\037

302010o0 10 20 30 40 50 60 70 80

t)

FIGURE7.4.3.Periodicoscillations of the predator and

prey populations in Example 2.)

Section7.2that the explicitsolution functions x(t) and y(t) are both periodicfunc-tions of t-thusexplaining the periodicfluctuations that are observedempiricallyin predator-preypopulations. .)

OscillatingPopulations Figure 7.4.2showsa computer-generateddirection fieldand phaseportrait for the predator-preysystem)

dx-= (0.2)x- (0.005)xy= (0.005)x(40-y),dt

dydt

= -(0.5)y+ (O.OI)xy= (O.OI)y(-50+x),)

(5))

where x(t) denotesthe number of rabbits and y(t) the number of foxesafter t

months. Evidently the critical point (50,40) is a stablecenter representing equi-librium populations of 50rabbits and 40 foxes.Any other initial point lieson aclosedtrajectory enclosingthis equilibrium point. The direction field indicates that

the point (x(t),Y (t)) traverses its trajectory in a counterclockwisedirection, with

the rabbit and fox populations oscillating periodicallybetweentheir separatemax-imum and minimum values. A drawback is that the phaseplane plot provides noindication as to the speedwith which eachtrajectory is traversed.

This lost \"senseof time\" is recaptured by graphing the two individual popu-lation functions as functions of time t. In Fig.7.4.3we have graphed approximatesolution functions x(t) and y(t) calculated using the Runge-Kutta method of Sec-tion 6.4with initial values x(O) = 70 and y(O) = 40.We seethat the rabbit

population oscillatesbetweenthe extreme values Xmax \037 72and Xmin \037 33,whilethe fox population oscillates(out of phase)between the extreme values Ymax

\037 70and Ymin \037 20.A careful measurement indicatesthat the periodP of oscillationof each population is slightly over 20months. Onecould\"zoom in\" on the maxi-mum/minimum points on eachgraph in orderto refine theseestimatesof the periodand the maximum and minimum rabbit and fox populations.

Any positive initial conditions Xo = x(O)and Yo = y(O) yield a similar pic-ture, with the rabbit and fox populations both surviving in coexistencewith eachother. .)CompetingSpeciesNow we considertwo species(of animals, plants, or bacteria, for instance) with

populationsx(t) and y(t) at time t and which competewith each other for the foodavailable in their common environment. This is in marked contrast to the casein

which one speciespreyson the other. To construct a mathematicalmodel that is asrealisticas possible,let us assumethat in the absenceof either species,the otherwould have a bounded(logistic)population likethoseconsideredin Section1.7.Inthe absenceof any interaction or competitionbetweenthe two species,their popu-lationsx(t) and y (t) would then satisfy the differentialequations)

dx 2dt

= alx-b1x,

dy 2-= a2Y-b2y ,

dt)

(6))

eachof the form of Eq.(2)of Section1.7.But in addition,we assumethat competi-tion has the effectof a rate of declinein eachpopulation that is proportionalto their)))

\037)

product xy. We insert such terms with negativeproportionality constants-Cland-C2in the equations in (6)to obtain the competitionsystem

dx 2-= alx-blx -CIXY = x(al-blx-CIY),dt

dy 2-= a2Y-b2y

-C2X Y = y(a2 -b2y-C2X),

dt)

(7))

where the coefficientsaI,a2, b I , b2, CI,and C2are all positive.The almost linear system in (7)has four criticalpoints.Upon setting the right-

hand sidesof the two equationsequal to zero,we seethat if x = 0,then eithery = 0or y = a2/b2,whereasif y = 0,then eitherx = 0 or x = ai/bl.This givesthe threecritical points (0,0), (0,a2/b2),and (ai/bI , 0).The fourth criticalpoint is obtainedfrom the simultaneoussolutionof the equations)

blx+ClY = aI, C2X + b2Y = a2.) (8))

We assumethat, as in most interesting applications, theseequationshave a singlesolution and that the correspondingcritical point liesin the first quadrant of the xy-plane.This point (xE , YE) is then the fourth criticalpoint of the system in (7),and it

representsthe possibilityof coexistenceof the two species,with constantnonzeroequilibrium populationsx(t) = xE and y(t) = YE .

We are interested in the stability of the critical point (xE , YE ). This turns out

to dependon whether)

CIC2 b l b2.) (9))

Each inequality in (9)has a natural interpretation. Examining the equationsin (6),weseethat the coefficientsb i and b2 representthe inhibiting effectof eachpopula-tion on its own growth (possiblydue to limitations of food or space).On the otherhand, CI and C2representthe effect of competition between the two populations.Thus b l b2 is a measureof inhibition while CIC2is a measure of competition. A

general analysis of the system in (7)showsthe following:)

1.If CIC2 < b i b2, so that competition is small in comparison with inhibition,then (xE , YE ) is an asymptotically stablecritical point that is approachedbyeachsolution as t --+ +00.Thus the two speciescan and do coexistin this

case.2.If CIC2 > b i b2, so that competitionis large in comparisonwith inhibition, then

(xE , YE ) is an unstable critical point, and either x(t) or y(t) approacheszeroas t --+ +00.Thus the two speciescannot coexistin this case;one survivesand the other becomesextinct.)

Rather than carrying out this general analysis, we presenttwo examplesthat

illustrate thesetwo possibilities.)

___\037\037 .__\037 \037 \037 \037_ \037 _ \037 .,._ ..\"\",.,,__ ,_\037__\".\"\",..\"\",.\"\",.\", \"\"\" \037 \"., \037_m\"'\" \037 \037\"m\"\037 ,,, \037 .\037___ \" _\037.\" _\"\037\". \"_,, _ \037_.\" __ \" _\037 _ \037 __..,,_ ,.,,\037\037_o ___,_ _\037 Survival of a SingleSpeciesSupposethat the populationsx(t) and y(t) satisfythe equations)

dx I 2-= 14x- 2x -xy,dt

dy I 2-= 16y- 2Y-xy,

dt)

(10))))


x)

FIGURE7.4.4.Phaseplaneportrait for the linear systemx'== 14x,y' == 16ycorrespondingto the critical point (0,0).)

\037 0)

-1)

-2-2 -1) o

u)

1) 2)

FIGURE7.4.5.Phaseplaneportrait for the linear system in

Eq. (13)correspondingto thecritical point (0,32).)

in which al = 14,a2 = 16,b i = b2 = !,and CI = C2 = 1.Then CIC2= 1 >

\037

= b I b2, so we should expectsurvival of a singlespeciesas predictedin Case2previously. We find readily that the four critical points are (0,0),(0,32),(28,0),and (12,8).We shall investigatethem individually.)

THECRITICALPOINT(0,0): The Jacobian matrix of the system in (10)is)

J(x,y) =[14

-_\037

-y)

[14 0

]so J(O,O)= 0 16 .) (11))-x

]16-y-x')

The matrix J(O,0) has characteristic equation (14-A)(16- A) = 0 and has the

eigenvalues)

Al = 14 with eigenvector VI = [1 0 ]T)

and)

A2 = 16 with eigenvector V2 = [ 0 I]T.)

Both eigenvalues are positive, so it follows that (0,0) is a nodal sourcefor the

system'slinearizationx'= 14x,y' = 16yat (0,0),and hence-byTheorem 2 in

Section7.3-isalsoan unstable nodal sourcefor the original system in (10).Figure7.4.4showsa phaseportrait for the linearized system near (0,0).)

THECRITICALPOINT (0,32): Substitution of x = 0,y = 32 in the Jacobianmatrix J(x,y) shown in (11)yieldsthe Jacobian matrix)

[-18 0

]J(0,32)= -32 -16) (12))

of the nonlinear system (10)at the point (0,32).Comparing Eqs. (7)and (8)in

Section7.3,weseethat this Jacobian matrix correspondsto the linearization)

du-= -18udt 'dv-= -32u- 16vdt)

(13))

of (10)at (0,32).The matrix J(O,32)has characteristicequation(-18-A)(-16-A) = 0 and has the eigenvaluesAl = -18with eigenvectorVI = [1 16]Tand

A2 = -16with eigenvectorV2 = [ 0 1]T. Becauseboth eigenvaluesare negative,it follows that (0,0) is a nodal sink for the linearizedsystem,and hence-byTheo-rem 2 in Section7.3-that(0,32)is alsoa stablenodal sink for the original systemin (10).Figure 7.4.5showsa phaseportrait for the linearized systemnear (0,0).)))

1)

::::> 0)

-1)

-2-2 -1) o

u)

1) 2)


Eq.(15)correspondingto thecritical point (28,0).)

2)

::::> 0)

-1)

-2-2 -1) o

u)

1) 2)


Eq.(17)correspondingto thecritical point (12,8).)


THECRITICALPOINT(28,0): The Jacobian matrix)

J(28,0)= [-16=i\037])(14))

correspondsto the linearization)

du-= -14u-28vdt 'dv-= -12vdt

of (10)at (28,0).The matrix J(28,0)has characteristicequation(-14-A)(-12-A) = 0 and has the eigenvaluesAl = -14with eigenvectorVI = [1 O]Tand A2 =-12with eigenvectorV2 = [-14 1]T. Becauseboth eigenvaluesare negative, it

follows that (0,0)is a nodal sink for the linearizedsystem,and hence-byTheorem2 in Section7.3-that(28,0)is alsoa stable nodal sink for the original nonlinear

system in (10).Figure 7.4.6showsa phaseportrait for the linearizedsystem near(0,0).)

(15))

THECRITICALPOINT(12,8): The Jacobian matrix)

[-6 -12

]J(12,8) = -8 -4) (16))


du-= -6u- 12v,dt

dv-= -8u-4vdt)

(17))

of (10)at (12,8).The matrix J(12,8) has characteristicequation)

(-6-A)(-4-A) - (-8)(-12)= A2 + lOA -72= 0)

and has the eigenvalues)

)1.\\ = -5-v'97<0 with eigenvector VI = [ k(1+ffi) 1 f)

and)

A.2 = -5+v'97>0 with eigenvector V2 = [k(1-ffi) 1f \302\267)

Becausethe two eigenvalues have oppositesigns, it follows that (0,0) is a sad-dle point for the linearized system and hence-byTheorem 2 in Section7.3-that(12,8) is alsoan unstable saddlepoint for the original system in (10).Figure 7.4.7showsa phaseportrait for the linearizedsystemnear (0,0).

Now that our local analysis of each of the four critical points is complete, it

remains to assemblethe information found into a coherent global picture. If weacceptthe facts that)))

Example4)

. Near eachcritical point, the trajectoriesfor the original system in (10)resem-blequalitatively the linearizedtrajectories shown in Figs.7.4.4-7.4.7,and. As t \037 +00each trajectory either approachesa critical point or divergestoward infinity,)

then it wouldappearthat the phaseplane portrait for the original systemmust resem-ble the rough sketchshown in Fig.7.4.8.This sketchshowsa few typical freehandtrajectories connecting a nodal sourceat (0,0),nodal sinks at (0,32)and (28,0),and a saddlepoint at (12,8),with indicateddirectionsof flow along thesetrajecto-riesconsistentwith the known characterof thesecritical points.Figure 7.4.9showsa more precisecomputer-generatedphaseportrait and direction fieldfor the nonlin-

ear systemin (10).)

y

(0,32) Region I)

35(0,32)

302520

\037

15)

(0,0))x)(28,0)) x)

FIGURE7.4.8.Rough sketch consistentwith the analysis in Example 3.)

FIGURE7.4.9.Phaseplane portraitfor the system in Example 3.)

The two trajectories that approach the saddlepoint (12,8),together with that

saddlepoint, form a separatrixthat separatesregionsI and II in Figure 7.4.9.It plays a crucial rolein determining the long-term behavior of the two popula-tions. If the initial point (xo,Yo) liespreciselyon the separatrix, then (x(t),yet))approaches(12,8) as t \037 +00.Of course,random events make it extremelyunlikely that (x(t),yet)) will remain on the separatrix. If not, peaceful coexis-tenceof the two speciesis impossible.If (xo,Yo) liesin Region I above the sep-aratrix, then (x(t),yet)) approaches(0,32)as t \037 +00,so the population x(t)decreasesto zero.Alternatively, if (xo,Yo) liesin Region IIbelowthe separatrix,then (x(t),yet)) approaches(28,0) as t \037 +00,so the population yet) diesout.In short, whicheverpopulationhas the initial competitiveadvantagesurvives,whilethe other facesextinction. .)

................n........... ..... ..................... ,.\", .....

PeacefulCoexistenceof Two SpeciesSupposethat the populationsx(t) and yet)satisfy the competitionsystem

dx 2-= 14x-2x -Xydt '

dy 2dt

= 16y-2y -xy,)

(18))

for which al = 14,a2 = 16,b l = b2 = 2,and CI = C2 = 1.Then CIC2= 1 <4 = b l b2, sonow the effectof inhibition is greater than that of competition.We find

readily that the four critical points are (0,0),(0,8),(7,0),and (4,6).We proceedas in Example3.)))

2

1

::::> 0

-1

-2-2 -1 0 1 2

u)


Eq.(20)correspondingto thecriticalpoint (0,8).)

1)

::::> 0)

-1)

-2-2 -1) o

u)

1)


Eq.(22)correspondingto thecritical point (7, 0).)


THE CRITICALPOINT (0,0):When we drop the quadratic terms in (18),weget the samelinearizationx' = 14x,y' = 16yat (0,0)as in Example3.Thus its

coefficientmatrix has the two positive eigenvaluesAl = 14and A2 = 16,and its

phaseportrait is the sameas that shown in Fig.7.4.3.Therefore,(0,0)isan unstablenodal sourcefor the original system in (18).)

THECRITICALPOINT(0,8):The Jacobian matrix of the system in (18)is)

J(x,y) =[

14-\037\037

-y) -x

]16-4y -x ') so J(O,8) =[_\037 -1\037 l (19))

The matrix J(0,8) correspondsto the linearization)

du-=6udt

'dv-= -8u- 16vdt)

(20))

of (18)at (0,8).It has characteristicequation (6-A)( -16-A) = 0 and has the

positive eigenvalue Al = 6 with eigenvector VI = [11 _4]Tand the negative

eigenvalue A2 = -16with eigenvectorV2 = [0 1]T. It follows that (0,0) is asaddlepoint for the linearized system,and hencethat (0,8) is an unstable saddlepoint for the original system in (18).Figure 7.4.10showsa phaseportrait for the

linearized systemnear (0,0).)

THECRITICALPOINT (7,0): The Jacobian matrix)

J(7,0)= [-16-

\037])

(21))


du-= -14u- 7vdt

'dv-=9vdt)

(22))

of (18)at (7,0).The matrix J(7,0)has characteristicequation (-14-A) (9-A) == 0and has the negativeeigenvalueAl = -14with eigenvectorVI = [1 O]T and the

positive eigenvalue A2 = 9 with eigenvector V2 = [-7 23]T. It follows that

(0,0) is a saddlepoint for the linearizedsystem,and hencethat (7,0)is an unstablesaddlepoint for the original system in (18).Figure7.4.11showsa phaseportrait forthe linearized systemnear (0,0).)

THECRITICALPOINT(4,6):The Jacobian matrix)

[-8 -4

]J(4,6)= -6 -12) (23))))

2)

::::> 0)

-1)

-2-2 -1) o

u)

1) 2)


Eq. (24)correspondingto thecritical point (4,6).)

(0, 8))

;::......)

(0,0))x)

FIGURE7.4.13.Direction fieldand phaseportrait for the

competition systemx'= 14x- 2X2 -xy,y' = 16y-2y

2 - xyofExample 4.)


du-= -8u-4v,dt

dv-= -6u- 12vdt)

(24))

of (18)at (4,6).The matrix J(4,6)has characteristicequation)

(-8-A)(-12-A) - (-6)(-4)= A2 + 20A +72= 0)

and has the two negativeeigenvalues)

)I.J = 2(-5-.J7) with eigenvector VI = [ H-1+.J7) 1rI)

and)

)..2= 2(-5+.J7) with eigenvector V2 = [ H-1-.J7) 1r .)

It follows that (0,0) is a nodal sink for the linearized system,and hencethat (4,6)is a stablenodal sink for the original systemin (18).Figure 7.4.12showsa phaseportrait for the linearized system near (0,0).

Figure 7.4.13assemblesall this localinformation into a global phaseplaneportrait for the original system in (18).The notable feature of this system isthat-forany positive initial population values Xo and Yo-thepoint (x(t),yet))approachesthe singlecritical point (4,6) as t \037 +00.It follows that the two

speciesboth survive in stable(peaceful)existence. .)InteractionsofLogisticPopulationsIf the coefficientsaI,a2, b l , b2 are positivebut CI = C2 = 0,then the equations)

dx 2-= alx-b1x-CIXY,dt

dy 2-= a2Y -b2y-C2XY

dt)

(25))

describetwo separatelogisticpopulationsx(t) and yet) that have no effect on eachother. Examples3 and 4 illustrate casesin which the xy-coefficients CI and C2are both positive. The interaction betweenthe two populations is then describedas competition,becausethe effect of the xy-termsin (25)is to decreasethe ratesof growth of both populations-thatis, each population is \"hurt\" by their mutual

interaction.Suppose,however, that the interaction coefficientsCI and C2 in (25)are both

negative. Then the effect of the xy-termsis to increasethe rates of growth of bothpopulations-thatis, eachpopulation is \"helped\" by their mutual interaction. Thistype of interaction is aptly describedascooperationbetweenthe two logisticpop-ulations.

Finally, the interactionbetweenthe two populations is one of predationif theinteraction coefficientshave different signs.For instance, if CI >0 but C2 < 0,then

the x-population is hurt but the y-population ishelpedby their interaction. We maytherefore describex(t) as a prey population and yet) as a predator population.)))

If either b I or b2 is zero in (25),then the correspondingpopulation would

(in the absenceof the other) exhibit exponentialgrowth rather than logisticgrowth.For instance, supposethat al > 0,a2 < 0,b I = b2 = 0,and CI > 0, C2 < o.Then x(t) is a naturally growing prey populationwhile yet) is a naturally decliningpredator population. This is the originalpredator-preymodel with which we beganthis section.

Problems26through 34 illustrate a variety of the possibilitiesindicatedhere.The problemsand examplesin this section illustrate the power of elementarycritical-pointanalysis.But remember that ecologicalsystemsin nature are seldomso simpleas in theseexamples.Frequently they involve more than two species,and the growth rates of thesepopulationsand the interactionsamong them often aremore complicatedthan those discussedin this section.Consequently,the mathe-matical modeling of ecologicalsystemsremains an activearea of current research.)

_ Problems)Problems1and 2 dealwith the predator-preysystem

dx-= 200x-4xydt 'dy

dt= -150y+ 2xy)

(1))

that correspondsto Fig. 7.4.1.1.Starting with the Jacobianmatrix of the system in (1),de-

rive its linearizations at the two critical points (0,0) and

(75,50).Usea graphing calculator or computer systemto construct phaseplane portraits for thesetwo lineariza-tions that are consistent with the \"big picture\" shown in

Fig. 7.4.1.2.Separatethe variables in the quotient)

dy--dx)

-150y+ 2xy200x- 4xy)

of the two equations in (1),and thereby derive the exactimplicit solution)

200lny + 150lnx-2x-4y = C)

of the system. Usethe contour plot facility of a graphingcalculator or computer system to plot the contour curvesof this equation through the points (75,100),(75,150),(75,200), (75,250),and (75,300) in the xy-plane.Are

your results consistentwith Fig. 7.4.1?3.Letx(t)bea harmful insectpopulation (aphids?) that un-

dernatural conditions is held somewhat in checkby a be-nign predator insectpopulation y(t) (ladybugs?).Assumethat x(t)and y(t) satisfy the predator-preyequations in

(1),sothat the stableequilibrium populations arexE= bjq

and YE= ajp. Now supposethat an insecticideis em-

ployed that kills (perunit time) the samefraction f < aof eachspeciesof insect.Show that the harmful popula-tion xE is increased,while the benign population YE is de-creased,sothe useof the insecticideis counterproductive.)

This is an instance in which mathematical analysis revealsundesirable consequencesof a well-intentioned interfer-encewith nature.)

Problems4 through 7 dealwith the competition system

dx 2-= 60x-4x -3xydt 'dy 2dt

= 42y-2y - 3xy,)

(2))

in which CIC2 = 9 > 8 = blb2, so the effectof competitionshould exceedthat of inhibition. Problems4 through 7 implythat the four criticalpoints (0,0), (0,21),(15,0),and (6,12)of the system in (2) resemblethose shown in Fig. 7.4.9-anodal sourceat the origin, a nodal sink on eachcoordinate

axis, and a saddlepoint interior to the first quadrant. In eachof theseproblems usea graphing calculatoror computer sys-tem to construct a phaseplaneportrait for the linearization atthe indicatedcriticalpoint. Finally, construct afirst-quadrantphaseplaneportrait for the nonlinear system in (2).Doyourlocaland globalportraits lookconsistent?

4. Show that the coefficientmatrix of the linearization x/ =60x, y/ = 42y of (2) at (0,0) has positive eigenvaluesAl = 60 and A2 = 42.Hence(0,0)is a nodal sourcefor

(2).5. Show that the linearization of (2)at (0,21)is u/ = -3u,

v/ = -63u-42v.Then show that the coefficientmatrix ofthis linear system has negative eigenvalues Al = -3and

A2 = -42.Hence(0,21)is a nodal sink for the system in

(2).6.Show that the linearization of(2)at (15,0) is u/ = -60u-

45v, v/ = -3v.Then show that the coefficientmatrix ofthis linear system has negative eigenvalues Al = -60and

A2 = -3.Hence(15,0) is a nodal sink for the system in

(2).7. Show that the linearization of (2) at (6,12) is u/ =

-24u- 18v, v/ = -36u- 24v. Then show that the)))

coefficientmatrix of this linear system has eigenvaluesAl = -24- 18,J2< 0 and A2 = -24+ 18,J2> O.Hence(6, 12)is a saddlepoint for the system in (2).)

Problems 8 through 10dealwith the competition system)

dx 2-= 60x-3x -4xydt 'dy 2dt

= 42y- 3y -2xy,)

(3))

in which CIC2 = 8 < 9 = bl b2, so the effect of inhibitionshould exceedthat ofcompetition. Thelinearization ofthe sys-tem in (3)at (0,0) is the sameas that of(2).This observationand Problems8 through 10imply that the four criticalpoints(0,0), (0,14),(20,0),and (12,6)of(3)resemblethose shownin Fig. 7.4.13-anodal sourceat the origin, a saddlepointon eachcoordinateaxis, and a nodal sink interior to the first

quadrant. In eachoftheseproblemsusea graphing calculatoror computer system to construct a phaseplaneportrait for the

linearization at the indicated criticalpoint. Finally, constructa first-quadrant phaseplaneportrait for the nonlinear systemin (3).Doyour localand globalportraits lookconsistent?

8.Show that the linearization of (3) at (0, 14)is u ' = 4u,v' = -28u -42v. Then show that the coefficientmatrixof this linear system has the positive eigenvalue A I = 4and the negative eigenvalue A2 = -42.Hence(0, 14)is asaddlepoint for the system in (3).

9.Show that the linearization of(3)at (20,0)is u '= -60u-80v,v' = 2v.Then show that the coefficientmatrix of thislinear system has the negative eigenvalue A I = -60andthe positive eigenvalue A2 = 2.Hence(20,0) is a saddlepoint for the system in (3).

10.Show that the linearization of (3) at (12,6)is u ' =-36u- 48v, v' = -12u- 18v. Then show that thecoefficientmatrix of this linear system has eigenvaluesAl = -27+ 3,J73and A2 = -27-3,J73,both ofwhicharenegative. Hence(12,6) is a nodal sink for the systemin (3).)

Problems11through 13dealwith the predator-preysystem)

dx 2-= 5x-x -Xydt 'dy

dt= -2y+xy,)

(4))

in which the prey population x(t) is logistic but the predatorpopulation y(t) would (in the absenceof any prey) declinenaturally. Problems 11through 13imply that the three crit-icalpoints (0,0), (5,0), and (2,3) of the system in (4)are asshown in Fig. 7.4.14-withsaddlepoints at the origin and onthe positive x-axis,and with a spiralsink interior to the firstquadrant. In eachof theseproblems usea graphing calcula-tor or computer system to construct a phaseplaneportrait forthe linearization at the indicatedcriticalpoint. Doyour localportraits lookconsistent with Fig. 7.4.14?)

8

6

\037

4

2

00 2 4 6 8

x)

FIGURE7.4.14.Directionfield and phaseportrai t for the predator-prey system ofProblems 11through 13.)

11.Show that the coefficientmatrix of the linearization x'=5x, y' = -2yof (4) at (0,0)has the positive eigenvalueA I = 5 and the negative eigenvalue A2 = -2.Hence(0,0) is a saddlepoint of the system in (4).

12.Show that the linearization of (4)at (5,0)is u ' = -5u-5v, v' = 3v. Then show that the coefficientmatrix of this

linear system has the negative eigenvalue A I = -5and the

positive eigenvalue A2 = 3.Hence(5,0)is a saddlepointfor the system in (4).

13.Show that the linearization of (4)at (2,3) is u ' = -2u-2v, v' = 3u. Then show that the coefficientmatrix of this

linear system has the complexconjugate eigenvalues AI,A2 = -1:I:i-J5with negative real part. Hence(2,3) is aspiral sink for the system in (4).)

Problems14through 17dealwith the predator-preysystem)

dx 2-=x -2x-xy,dt

dy 2-= Y-4y + xy.

dt)

(5))

Here each population-the prey population x(t) and the

predatorpopulation y(t)-is an unsophisticated population

(like the alligators of Section1.7)for which the only alter-natives (in the absenceof the other population) are doomsdayand extinction. Problems 14 through 17imply that the fourcriticalpoints (0,0), (0,4),(2,0),and (3, 1)of the system in

(5)are as shown in Fig. 7.4.15-anodal sink at the origin, asaddlepoint on eachcoordinateaxis, and a spiralsourcein-

terior to the first quadrant. This is a two-dimensional version

of \"doomsday versus extinction.\" If the initial point (xo, Yo)

liesin Region I, then both populations increasewithout bound

(until doomsday), whereasif it liesin Region II, then both pop-ulations decreaseto zero(and thus both becomeextinct). In

eachoftheseproblemsusea graphing calculatoror computer

system to construct aphaseplaneportrait for the linearizationat the indicated criticalpoint. Do your localportraits lookconsistentwith Fig. 7.4.15?)))


(0,4))

\037)

4 6)x)

FIGURE7.4.15.Direction field and phaseportrait for the predator-preysystem ofProblems 14through 17.)

14.Show that the coefficientmatrix of the linearization x/ =-2x,y/ = -4y of the system in (5)at (0,0)has the neg-ative eigenvaluesA I = -2and A2 = -4.Hence(0,0) isa nodal sink for (5).

15.Show that the linearization of (5) at (0,4)is u/ = -6u,v/ = 4u + 4v. Then show that the coefficientmatrix ofthis linear system has the negative eigenvalue A I = -6and the positive eigenvalue A2 = 4. Hence(0,4) is a sad-dlepoint for the system in (5).16.Show that the linearization of(5)at (2,0) is u/ = 2u -2v,v/ = -2v. Then show that the coefficientmatrix of thislinear system has the positive eigenvalue Al = 2 and the

negative eigenvalue A2 = -2.Hence(2,0) is a saddlepoint for the system in (5).

17.Show that the linearization of(5)at (3, 1)is u/ = 3u -3v,v/ = u + v. Then show that the coefficientmatrix ofthis linear system has complexconjugateeigenvaluesAI,A2 = 2 :I:i-J2with positive realpart. Hence(3, 1)is aspiral sourcefor (5).)

Problems18through 25dealwith the predator-preysystem

dxdt

= 2x- xy + Ex(5-x),

dy-= -5y +xydt ')

(6))

for which a bifurcation occursat the value E = 0 of the pa-rameter E. Problems18and 19deal with the caseE = 0, in

which casethe system in (6)takes the form)

dx-= 2x-Xydt ')

dy-= -5x+ xydt ') (7))

and theseproblems suggest that the two criticalpoints (0,0)and (5,2)of the system in (7) are as shown in Fig. 7.4.16-asaddlepoint at the origin and a centerat (5,2).In eachprob-lem usea graphing calculatoror computer system to constructa phaseplane portrait for the linearization at the indicatedcriticalpoint. Doyour localportraits lookconsistentwith Fig.7.4.16?)

20

16

12\037

8

4

0

(0,0) 0 4 8 12 16 20x)

FIGURE7.4.16.ThecaseE = 0(Problems 18and 19).)

18.Show that the coefficientmatrix of the linearization x/ =2x, y/ = -5y of (7) at (0,0) has the positive eigenvalueAl = 2 and the negative eigenvalue A2 = -5.Hence(0,0)is a saddlepoint for the system in (7).19.Show that the linearization of the system in (7) at (5,2) isu/ = -5v, v/ = 2u.Then show that the coefficient matrixof this linear system has conjugate imaginary eigenvaluesAI, A2 = :l:iJIQ.Hence(0,0) is a stablecenter for thelinear system. Although this is the indeterminate caseofTheorem2 in Section7.3,Fig. 7.4.16suggests that (5,2)alsois a stablecenterfor (7).)

Problems20 through 22dealwith the caseE = -1,for which

the system in (6)becomesdx 2 dy

dt=-3x+x-xy, dt =-5y+xy, (8)

and imply that the three criticalpoints (0,0), (3,0),and (5,2)of(8)areasshown in Fig. 7.4.17-witha nodal sink at the ori-gin, a saddlepoint on the positivex-axis,and a spiralsourceat(5,2).In eachproblem usea graphing calculatoror computersystem to construct aphaseplaneportrait for the linearizationat the indicated criticalpoint. Do your localportraits lookconsistentwith Fig. 7.4.17?)

10)

8)

6)\037)

4)

2)

o

(0,0) 2 4(3,0) x

FIGURE7.4.17.ThecaseE = -1(Problems20 through 22).

20.Show that the coefficientmatrix of the linearization x/ =-3x,y/ = -5yof the system in (8)at (0,0)has the neg-ative eigenvalues Al = -3and A2 = -5.Hence(0,0)isa nodal sink for (8).)))


21.Show that the linearization of the system in (8)at (3,0) isu ' = 3u - 3v, v' = -2v.Then show that the coefficientmatrix of this linear system has the positive eigenvalueA I = 3 and the negative eigenvalue A2 = -2.Hence(3,0) is a saddlepoint for (8).

22.Show that the linearization of(8)at (5,2) is u '= 5u-5v,v' = 2u.Then show that the coefficientmatrix of this lin-ear system has complexconjugate eigenvaluesAI, A2 =4(5 :I:i,JI5)with positive real part. Hence(5,2) is aspiral sourcefor the system in (8).)

Problems23 through 25dealwith the caseE = 1, so that the

system in (6)takes the form)

dx 2-=7x-x -xydt ')

dy-= -5y+xydt ') (9))

and theseproblems imply that the three criticalpoints (0,0),(7,0),and (5,2) of the system in (9) are as shown in

Fig. 7.4.18-withsaddlepoints at the origin and on the pos-itive x-axisand with a spiralsink at (5,2).In eachproblemuse a graphing calculatoror computer system to construct aphaseplaneportrait for the linearization at the indicated crit-icalpoint. Do your localportraits look consistent with Fig.7.4.18?)

10)

8)

6)\037)

4)

2)

o

(0,0) 6 8x (7,0)

FIGURE7.4.18.ThecaseE = +1(Problems23 through 25).)

23.Show that the coefficientmatrix of the linearization x' =7x, y' = -5yof (9) at (0,0)has the positive eigenvalue)

Al = 7 and the negative eigenvalue A2 = -5.Hence(0,0)is a saddlepoint for the system in (9).

24. Show that the linearization of (9)at (7, 0) is u ' = -7u -7v, v' = 2v. Then show that the coefficientmatrix of this

linear system has the negative eigenvalue A I = -7 and the

positive eigenvalue A2 = 2.Hence(7, 0) is a saddlepointfor the system in (9).

25.Show that the linearization of (9)at (5,2)is u ' = -5u-5v, v' = 2u. Then show that the coefficientmatrix of this

linear system has the complexconjugate eigenvalues AI,A2 = 4(-5:1:i,JI5)with negative real part. Hence(5,2)is a spiral sink for the system in (9).)

For each two-population system in Problems 26 through 34,first describethe type ofx-and y-populations involved (ex-ponential or logistic) and the nature of their interaction-competition, cooperation,or predation. Then find and char-acterizethe system's criticalpoints (asto type and stability).Determine what nonzero x-and y-populations can coexist.Finally, construct a phaseplaneportrait that enablesyou to

describethe long-term behavior of the two populations in

terms of their initial populations x(O)and y(O).dx dy26.-= 2x-xy, -= 3y -xydt dtdx dy27.-= 2xy -4x, -= xy - 3ydt dtdx dy28.-= 2xy - 16x,-= 4y -xydt dtdx 2 129.dt

= 3x-x -2xy ,

dx 2 130.-= 3x-x + -xy,dt 2dx 2 131.dt

= 3x -x -4xy ,

dx32.-= 30x-3x2 +xy,dtdx 233.-= 30x-2x -Xydt

'dx34.-= 30x- 2X2 -Xydt

')

_ NonlinearMechanicalSystems\037 \"\" N N _,.,....,.,.... _ _ ,.,....\037 __ \037 ...\",....\037__'\"\"'\"\"\"',... \"\"....,...,..,.. N_\"\"\"\"\"___\037\037\"\" ,...___ N... \"\"\037 N.-..... \"',........,..._ \037.\"_,,.... '\" '\" \037 ..,..,..,\037 \"\"\"\"''''' ....\037....,.,............ \"'\"V.....,...n...-\037.... ....\037\037.....,........ __ ,.... _,...,.,........\037 \037........ ...,,___ \"''\"\"''''\"'' \"',..,.\".,....,.......\",.... '\" \",,_ ,..._\037_,...'\"\"'\" '\" ____,...,.,....,.,,,.....,)

Now we apply the qualitative methods of Sections7.2and 7.3to the analysis ofsimplemechanical systemslike the mass-on-a-springsystem shown in Fig.7.5.1.Let m denotethe mass in a suitable system of units and let x(t) denote the dis-placement of the massat time t from its equilibrium position (in which the springis unstretched).Previouslywe have always assumedthat the force F(x)exertedbythe springon the massis a linearfunction of x:F(x) = -kx(Hooke'slaw). In

reality, however,every spring in nature actually is nonlinear (even if only slightly

so).Moreover,springsin someautomobilesuspensionsystemsdeliberatelyare de-signedto be nonlinear. Here,then, we are interested specifically in the effects ofnonlinearity.)

I)

Equilibrium

positionI)

FIGURE7.5.1.Themass on aspnng.)

dy-= 4y - 2xydtdy 1-= -xy -

Ydt 5dy-= xy -2ydtdy-= 60y- 3y2 +4xydtdy-= 80y -4y2+ 2xydtdy-= 20y-4y2+ 2xydt)))

7.5NonlinearMechanicalSystems 527)

Sonow we allow the force function F(x)to be nonlinear. BecauseF(O)=o at the equilibrium position x = 0,we may assumethat F has a power seriesexpansionof the form)

\037) F(x)= -kx+ax2 + fJx3 + ...

.) (1))

We take k >0 so that the reaction of the spring isdirectedoppositeto the displace-ment when x is sufficiently small.If we assumealsothat the reactionof the springis symmetric with respectto positive and negativedisplacementsby the samedis-tance, then F(-x)= -F(x),so F is an odd function. In this caseit followsthat

the coefficientof x n in Eq. (1)is zero if n is even, so the first nonlinear term is the

one involving x 3.For a simplemathematicalmodel of a nonlinear spring we thereforetake)

\037) F(x)= -kx+ fJx3

,) (2))

ignoring all terms in Eq.(1)of degreegreater than 3.The equationof motion of the

massm is then)

\037) mx\" = -kx+ fJx3

.) (3))

ThePosition-VelocityPhasePlaneIf we introduce the velocity)

yet) = x'(t)) (4))

of the masswith positionx(t), then we get from Eq. (3)the equivalent first-ordersystem)

dxdt

= y,

dy 3m-= -kx+ fJx .dt)

(5))

A phaseplane trajectoryof this system is a position-velocityplot that illustrates the

motion of the masson the spring.We can solve explicitlyfor the trajectoriesof this

systemby writing)

dy dyjdtdx dxjdt)

-kx+ fJx3

my)

whence)my dy + (kx - fJx

3) dx = O.)

Integration then yields)

4my2 + 4kx2- \037fJx4= E) (6))

for the equation of a typical trajectory. We write E for the arbitrary constant ofintegration becauseKE = 4my2 is the kinetic energy of the masswith velocityy,and it is natural to define)

PE=\037kx2

-\037fJx4) (7))

as the potential energy of the spring.Then Eq. (6)takesthe form KE +PE = E,so the constant E turns out to be the total energyof the mass-springsystem.Eq.(6)then expressesconservation of energy for the undamped motion of a mass on aspnng.)))

The behaviorof the massdependson the signof the nonlinear term in Eq.(2).The springis called)

. hard if fJ <0,. soft if fJ >o.)

We considerthe two casesseparately.)

HARD SPRINGOSCILLATIONS:If fJ <0,then the secondequation in (5)takesthe form my' = -x(lfJlx

2 +k), so it follows that the only critical point of the

systemis the origin (0,0).Each trajectory)

4my2 + 4kx2+ \037lfJlx4= E > 0) (8))

is an oval closedcurve like those shown in Fig.7.5.2,and thus (0,0) is a stablecenter.As the point (x(t),yet)) traverses a trajectory in the clockwisedirection,the position x(t) and velocity yet) of the massoscillatealternately, as illustrated in

Fig.7.5.3.The massis moving to the right (with x increasing)when y >0,to the

left when y <O.Thus the behaviorof a masson a nonlinearhard spring resemblesqualitatively that of a masson a linear spring with fJ = 0 (asin Example4 of Section7.2).But one differencebetweenthe linear and nonlinear situations is that, whereasthe periodT = 2ny'm/k of oscillation of a masson a linear spring is independentof the initial conditions,the periodof a masson a nonlinear spring dependson its

initial position x(O)and initial velocity y(O)(Problems21through 26).)

6)6)

4)

>-.2

......

'g 0a:) l'>_ 2 l' l'l'l' l'l'l' l'-4 l'l' l'

-6 l'l' i

-6 -4 -2 0 2Position)

>-.......

'u.9 2(1);;>] 0C\\$

c::o:-8 -2

C/Jo\037 -4) Position x)

-6o) 2 3 456

t)

FIGURE7.5.2.Position-velocityphaseplane portrait for the hard

mass-and-springsystem with

m = k = 2 and ,B = -4< O.)

FIGURE7.5.3.Position and

velocity solution curves for the hard

mass-and-spring system with

m = k = 2 and ,B= -4< O.)

Remark:The hard spring equation mx\" = -kx-IfJlx

3 has equivalentfirst-ordersystem)

,x = y,), k IfJl 3

y = --x--xm m)

with Jacobianmatrix)

o 1

J(x,y) =_\037 _ 31131x2 0

m m)

so J(O,O)=[_\0372 \037])))

Example1)


(writing k/m = w 2 as usual). The latter matrix has characteristicequation A2 +w 2 = 0 and pure imaginary eigenvaluesAI, A2 = :f:wi.Thus the linearizedsystemx/ = y, y/ = -w2x has a stable center at the critical point (0,O)-aswe observedin Example4 of Section7.2.However,the nonlinear cubicterm in the differential

equation has (in effect) replacedthe elliptical trajectories (as in Fig.7.2.7)of thelinear system with the \"flatter\" quartic ovals we seein Fig.7.5.2. .)

SOFTSPRINGOSCILLATIONS:If fJ >0,then the secondequation in (5)takesthe form my/ = x (fJx

2-k),so it follows that the systemhas the two criticalpoints

(:f:,Jk/fJ , 0) in addition to the critical point (0,0).Thesethree criticalpoints yieldthe only solutions for which the masscan remain at rest.The following exampleillustrates the greater range of possiblebehaviorsof a masson a soft spring.)

If m = 1,k = 4, and fJ = 1,then the equationof motion of the massis)

d2x-+4x -x3 = 0dt2 ') (9))

and Eq.(6)gives the trajectories in the form)

\037y2+2x2- ix4 = E.) (10))

After solving for)

y = ::l:.J2E-4x2 + 4x2,) (10/))

we couldselecta fixedvalueof the constantenergy E and plot manually a trajectorylike one of thoseshown in the computer-generatedposition-velocity phaseplaneportrait in Fig.7.5.4.)

543

(-2,0)21

\037 0-1-2 (2,0)-3-4 J

It' \" ....-5-5-4 -3 -2 -1 0 1 2 3 4 5x)

FIGURE7.5.4.Position-velocityphaseplane portraitfor the soft mass-and-spring system with m = 1,k = 4,and f3 = 1 > O.Theseparatricesareemphasized.)

The different types of phaseplane trajectoriescorrespondto different valuesof the energy E. If we substitute x = :f:,Jk/fJ and y = 0 into (6),we get the

energy value E = k2/(4fJ) = 4 (becausek = 4 and fJ = 1)that correspondstothe trajectories that intersect the x-axisat the nontrivial critical points (-2,0)and(2,0).Theseemphasizedtrajectoriesare calledseparatricesbecausethey separatephaseplane regionsof different behavior.

The nature of the motion of the massisdeterminedby which type of trajectoryits initial conditions determine.The simpleclosedtrajectories encircling(0,0) in)))

10)

5)

\037

\037)

o)

-5-1 -0.5) ot)

0.5) 1)

FIGURE7.5.5.Position and

velocity solution curves for thesoft mass-and-spring system with

m = 1,k = 4, f3 = 1 > 0, and

energy E = 8-sufficientlygreatthat the mass approachesthe

origin from the left and continueson indefinitely to the right.)

the region boundedby the separatricescorrespondto energiesin the range 0 < E <4. Theseclosedtrajectories representperiodicoscillationsof the massbackandforth around the equilibrium point x = o.

The unbounded trajectories lying in the regionsabove and belowthe separa-tricescorrespondto values of E greater than 4.Theserepresentmotions in which

the massapproachesx = 0 with sufficient energy that it continues on through the

equilibrium point, never to return again (asindicated in Fig.7.5.5).The unboundedtrajectoriesopeningto the right and left correspondto negative

values of E.Theserepresentmotions in which the massinitially is headedtowardthe equilibrium point x = 0,but with insufficient energy to reach it. At somepointthe massreversesdirection and headsbackwhence it came.

In Fig.7.5.4it appearsthat the critical point (0,0) is a stable center,whereasthe criticalpoints (::f::2,0)looklikesaddlepoints of the equivalentfirst-ordersystem)

x'= y, y' = -4x+x3) (11))

with Jacobian matrix)

J(x,y) =[-4\0373X2 \037J.

To checktheseobservations against the usual critical-point analysis, we note first

that the Jacobian matrix

J(O,O)=[_\037 \037]

at the critical point (0,0)has characteristicequationA2+4 = 0 and pure imaginaryeigenvaluesAI,A2 = ::f::2iconsistentwith a stablecenter.Moreover, the Jacobianmatrix

J(:!:2,O)=[\037 \037]

correspondingto the other two criticalpoints has characteristicequationA2-8 = 0and real eigenvalues AI, A2 = ::f::-J8of oppositesign,consistentwith the saddle-point behavior that we observenear (-2,0)and (+2,0). .

Remark:Figures7.5.2and 7.5.4illustrate a significant qualitative differ-encebetweenhard springswith f3 < 0 and soft springswith f3 >0 in the nonlinear

equation mx\" = kx + f3x3. Whereas the phaseplane trajectories for a hard spring

are all bounded,a soft springhas unbounded phaseplane trajectories (as well asboundedones).However,we should realizethat the unbounded soft-springtrajec-toriesceaseto representphysicallyrealisticmotions faithfully when they exceedthe

spring'scapability of expansionwithout breaking. .)

DampedNonlinearVibrationsSupposenow that the masson a springisconnectedalsoto a dashpot that providesaforce of resistanceproportional to the velocity y = dxjdtof the mass.If the springis still assumednonlinearas in Eq.(2),then the equation of motion of the massis)

\037) mx\" = -ex'-kx + f3x3

,) (12))

where c > 0 is the resistanceconstant. If f3 > 0,then the equivalent first-ordersystem)

dx dydt

= y,dt)

-kx-cy + fix3

= _\037y_ \037x (1 _ f3 x2

)m m m k)(13))))

Example2)


has critical points (0,0)and(\037 -Vk/f3 , 0)and Jacobian matrix)

o 1

J(x,y) = k 3f3 2 e--+-xm m m)

Now the critical point at the origin is the most interestingone.The Jacobianmatrix)

J(O,O)=)

o

k) e)

1)

m m)


(e

)k 1 2(-A) ---A+-=-(mA+eA+k)=O

m m m)

and eigenvalues) -e\037 -ve2-4kmAI, A2 = .

2m

It followsfrom Theorem2 in Section7.3in that the criticalpoint (0,0) of the systemin (13)is). a nodal sink if the resistanceis so great that e2 > 4km (in which casethe

eigenvaluesare negativeand unequal), but is. a spiral sink if e2 < 4km (in which casethe eigenvaluesare complexconju-gateswith negativereal part).)

The followingexampleillustrates the latter case.(In the borderlinecasewith equalnegativeeigenvalues,the origin may beeither a nodal or a spiral sink.))

\037Nm'mS\037pp\037\037\037\037\"th\037t ;;'\037--

1-\037-'\037\037

--\037

-'2\037k\037

\037'\037\037\037

-5\037\037\037-\037\037d\037\037,8\037

\037\037'\"\037- \037

\037\037-:\037'Th\"e\037

the \037onlinear systemin (13)i\037)

dx-=y,dt)

dy 5 3 1 2-= -5x-2y + 4x = -2y-5x(1- 4x ).dt)

(14))

It has critical points (0,0), (\0372, 0)and Jacobian matrix)

J(x,y) = [-5:7x2\0372 J.)

At (0,0):The Jacobian matrix)

J(O,O)=[_\037 _\037])

has characteristicequation A2+2A +5 = 0 and has complexconjugateeigenvaluesAI, A2 = -1\037 2i with negative real part. Hence(0,0) is a spiral sink of the

nonlinear system in (14),and the linearizedposition function of the massis of the

form)

x(t) = e-t (A cos2t + B sin 2t))))

that correspondsto an exponentiallydampedoscillationabout the equilibrium posi-tion x = o.At (:1:2,0):The Jacobian matrix

J(:!:2,0)=[1\037 _\037]

has characteristicequationA2+2A -10= 0 and real eigenvaluesAl = -1-,JI1<o and A2 = -1+ ,JI1>0 with different signs.It follows that (-2,0)and (+2,0)are both saddlepoints of the system in (14).

The position-velocityphaseplane portrait in Fig.7.5.6showstrajectoriesof(14)and the spiral sink at (0,0),aswell as the unstablesaddlepoints at (-2,0)and(2,0).The emphasizedseparatricesdivide the phaseplane into regionsof differentbehavior. The behavior of the massdependson the region in which its initial point(xo,Yo) is located.If this initial point liesin). Region I betweenthe separatrices,then the trajectory spiralsinto the ori-

gin as t \037 +00,and hencethe periodicoscillationsof the undamped case(Fig.7.5.4)are now replacedwith dampedoscillationsaround the stableequi-librium positionx = 0;. Region II,then the masspassesthrough x = 0 moving from left to right (xincreasing);. Region III,then the masspassesthrough x = 0 moving from right to left (xdecreasing);. Region IV, then the massapproaches(but doesnot reach)the unstable equi-librium positionx = -2 from the left, but stopsand then returns to the left;. Region V, then the massapproaches(but doesnot reach)the unstableequilib-rium positionx = 2 from the right, but stopsand then returns to the right.)

If the initial point (xo,Yo) liespreciselyon one of the separatrices,then the corre-spondingtrajectory either approachesthe stablespiral point or recedesto infinity

from a saddlepoint as t \037 +00. .)

4)

;>-.. 0)

-2)

-4)

-4) -2) ox)

2) 4)

FIGURE7.5.6.Position-velocityphaseplane portrait for the soft mass-and-springsystem with m = 1,k = 5, f3 =

\037,

and resistanceconstant c = 2.The(black)separatricesareemphasized.)))


TheNonlinearPendulum)

m)

In Section2.4we derived the equation

d2() g.\037 -+-SIn () ==0 (15)dt2 Lfor the undamped oscillationsof the simplependulum shown in Fig.7.5.7.Therewe usedthe approximation sin () \037 () for () near zero to replaceEq. (15)with thelinear model)

////)

L_---I

I)

d2()_2 +w 2() ==0,dt

where w 2 ==gjL.The general solution

()(t) == A coswt + B sin wt)

(16))FIGURE7.5.7.Thesimplependulum.)

(17))

of Eq.(16)describesoscillationsaround the equilibrium position () ==0 with circu-lar frequencywand amplitude C == (A

2 + B2)1/2.The linear modeldoes not adequately describethe possiblemotions of the

pendulum for large values of ().For instance, the equilibrium solution()(t)= n ofEq.(15),with the pendulum standing straight up, doesnot satisfy the linearequationin (16).Nor doesEq. (17)include the situation in which the pendulum \"goesoverthe top\" repeatedly, so that ()(t) is a steadily increasing rather than an oscillatoryfunction of t.To investigatethesephenomenawe must analyze the nonlinear equa-tion ()// + w 2 sin () == 0 rather than merely its linearization()// +w 2() == O.We alsowant to include the possibilityof resistanceproportionalto velocity,sowe considerthe general nonlinear pendulum equation

d2() d()\037 _

2 +c-+w2sin()==0. (18)dt dtThe casec > 0 correspondsto damped motion in which there actually is

resistanceproportional to (angular) velocity. But we examine first the undampedcasein which c == O. With x(t) == ()(t)and yet) == ()/(t) the equivalent first-ordersystem is)

dx-==y,dt)

dy 2 .-==-w sinx.dt)

(19))

We seethat this system is almost linear by writing it in the form

dxdt

==y,

dy 2dt

==-w x+g(x),)(20))

where)

g(x) ==-w2(sinx_ x) ==w2 (X 3

_ x5 +...)3! 5!has only higher-degreeterms.

The criticalpointsof the system in (19)are the points (nn,0)with n an integer,and its Jacobian matrix is given by

J(x,y) =[_(V2\037osx \037].

(21)

The nature of the critical point (nn,0)dependson whether n is evenor odd.)))


EVEN CASE: If n ==2m is an even integer, then cosnn ==+1,so (21)yieldsthematrix)

J(2mn,0)=[_\0372 \037]

with characteristicequation A2 +w 2 ==0 and pure imaginary eigenvaluesAI, A2 ==

::i::wi.The linearizationof (19)at (nn,0) is therefore the system)

du-== v,dt)

dv 2-==-w udt)

(22))

for which (0,0) is the familiar stablecenter enclosedby elliptical trajectories(asin

Example4 of Section7.2).Although this is the delicatecasefor which Theorem2of Section7.3doesnot settle the matter, we will seepresently that (2mn,0) is alsoa stablecenter for the original nonlinearpendulum system in (19).)

ODDCASE: If n == 2m + 1 is an oddinteger, then cosnn ==-1,so (21)yieldsthe matrix)

J\302\2532m + 1)n,0)=[\0372 \037]

with characteristic equation A2 - w 2 == 0 and real eigenvaluesAI, A2 == ::i::wwith

differentsigns.The linearizationof (19)at\302\2532m + l)n,0) is therefore the system)

du-== vdt ')

dv 2-==wudt)

(23))

for which (0,0) is a saddlepoint. It follows from Theorem 2 of Section7.3that

the critical point \302\2532m + l)n,0) is a similar saddlepoint for the original nonlinear

pendulum systemin (19).)

THE TRAJECTORIES:We can seehow these \"even centers\"and \"odd saddlepoints\" fit together by solving the system in (19)explicitly for the phaseplane tra-

jectories.If we write)

dy dyjdtdx dxjdt)

w 2sin x)

y)

and separatethe variables,)y dy +w 2sin xdx==0,)

then integration from x ==0 to x ==x yields)

1y2+w 2(1-cosx) == E.) (24))

We write E for the arbitrary constant of integration because,if physical units are sochosenthat m == L == 1,then the first term on the left is the kinetic energy and the

secondterm the potential energy of the masson the end of the pendulum. Then Eis the total mechanical energy; Eq. (24)thus expressesconservationof mechanicalenergy for the undampedpendulum.

If we solveEq.(24)for y and usea half-angle identity, we get the equation)

y = \037J2E-4w2sin2!x) (25))))

4)

-It) o) It) 2It)

2)

\037 0)

-2)

-4)

x)

FIGURE7.5.8.Position-velocityphaseplane portrait for the undamped pendulumsystem x'= y, y' = - sin x.The(black)separatricesareemphasized.)

that defines the phaseplane trajectories.Note that the radicand in (25)remainspositive if E > 2w2.Figure 7.5.8shows(along with a directionfield) the results ofplotting thesetrajectoriesfor variousvalues of the energy E.

The emphasizedseparatricesin Fig.7.5.8correspondto the criticalvalue E =2w2 of the energy; they enter and leave the unstable critical points (nn,0)with n

an oddinteger. Following the arrows along a separatrix,the pendulum theoreticallyapproachesa balancedverticalpositione = x = (2m+l)n with just enoughenergyto reach it but not enough to \"go over the top.\" The instability of this equilibriumposition indicatesthat this behaviormay neverbe observedin practice!

The simpleclosedtrajectories encircling the stable critical points-allofwhich correspondto the downwardpositione = 2mn of the pendulum-representperiodicoscillationsof the pendulum backand forth around the stable equilibriumposition e = O.Thesecorrespondto energiesE < 2w2 that are insufficient for the

pendulum to ascendto the vertical upward position-soits back-and-forthmotionis that which we normally associatewith a \"swinging pendulum.\"

The unbounded trajectories with E > 2w2 representwhirling motions of the

pendulum in which it goesover the top repeatedly-ina clockwisedirectionif yet)remains positive, in a counterclockwisedirection if y (t) is negative.)

PeriodofUndampedOscillationIf the pendulum is releasedfrom restwith initial conditions)

x(O)= e(o)= ex, y(O)= e'(0)= 0,) (26))

then Eq.(24)with t = 0 reducesto)

w 2(1-cosex) = E.) (27))

HenceE < 2w2 if 0 < ex < n,so a periodicoscillation of the pendulum ensues.To determine the periodof this oscillation,we subtract Eq.(27)from Eq.(24)and)))

write the result (with x = e and y = dejdt) in the form)

(de

)2

4 dt= Cll(cosB-cosa).) (28))

The periodT of time requiredfor one completeoscillation is four times the

amount of time requiredfor e to decreasefrom e = ex to e = 0,one-fourthof an

oscillation.Hencewe solve Eq.(28)for dtjdeand integrate to get)

4 lotT--w-J2 0 ,Jcose-cosex)

de)(29))

To attempt to evaluate this integral we first use the identity cose = 1 -2 sin2(e/2)and get)

2fot de

T =w 10 Jk2_ sin2(Bj2)

,)

where) . exk = SIn -.

2Next, the substitution u = (ljk)sin(ej2)yields)

4 11T---

w 0 J(1- u 2)(1- k2u 2))

du)

Finally, the substitution u = sin c/J gives)

4 f1C(2 dc/JT =w 10 Jt-k2sin2ifJ

The integral in (30)is the ellipticintegral of the first kind that is often denotedby F(k,nj2).Whereas ellipticintegrals normally cannot be evaluated in closedform, this integral can be approximated numerically as follows. First we use the

binomial series)

(30))

1\037

1 .3 ...(2n - 1)= 1 + \037xn

,J1-x n= 12 .4 ...(2n))

(31))

with x = k2sin2c/J < 1 to expandthe integrand in (30).Then we integratetermwise

using the tabulated integral formula)

l1C(2 . 2n n 1.3 ...(2n - 1)SIn c/J dc/J

= - . .o 2 2 .4 ...(2n))

(32))

The final result is the formula)

2n

[\037

(1 .3 ...(2n - 1)

)2

2

]T=- 1+\037 k n

w 2 .4 ...(2n)n=1

[ ( )2

( )2

( )2

]1 2 1.3 4 1.3.56= To 1 + - k + - k + k +...2 2.4 2.4.6) (33))

for the periodT of the nonlinear pendulum releasedfrom rest with initial anglee(O)= ex, in terms of the linearizedperiodTo= 2njw and k = sin(exj2).)))

.- -. --,-\037 --' _:.:, -.::':.,

-\302\253',)

\": -' ':- :' \" - .,-\"- '--\037

,......1J}f'1i'.,/.\",.0:)<'.------._--- --'--,,---,,-, ---- - '--- - -. -' --,- .-.,-, ,.\037-\".,::;:.

--::;\037- ':::::;::=<:/>.:,:;::-:::<>

'.I:: ::_:::\037-,: -:-

-;: -,:',)

10\302\260

20\302\260

30\302\260

40\302\260

50\302\260

60\302\260

70\302\260

80\302\260

90\302\260)

1.00191.00771.01741.03131.04981.07321.10211.13751.1803)

FIGURE7.5.9.Dependenceofthe periodT of a nonlinear

pendulum on its initial angle a.)


The infinite serieswithin the secondpair of bracketsin Eq. (33)gives the

factor T/To by which the nonlinear periodT is longer than the linearizedperiod.The table in Fig.7.5.9,obtained by summing this seriesnumerically, showshow

T/To increasesas ex is increased.Thus T is 0.19%greater than To if ex ==10\302\260,

whereasT is 18.03%greater than Toif ex ==90\302\260. But even a 0.19%discrepancyis

significant-thecalculation

seconds hours days(0.0019)x 3600 x 24 x 7 \037 1149(seconds/week)hour day week

showsthat the linearizedmodelisquite inadequatefor a pendulum clock;a discrep-ancy of 19min 9 s after only one weekis unacceptable.)

DampedPendulumOscillationsFinally, we discussbriefly the dampednonlinearpendulum. The almostlinearfirst-

ordersystemequivalentto Eq.(19)is

dxdt

==Y,

dy 2 .-==-w sInx-Cydt ')

(34))

and again the critical points are of the form (nn,0)where n is an integer. In Prob-lems9 through 11we askyou to verify that). If n is odd, then (nn,0) is an unstable saddlepoint of (34),just as in the

undampedcase;but. If n is even and c2 >4w2, then (nn,0) is a nodal sink; whereas. If n is even and c2 < 4w2, then (nn,0) is a spiral sink.)

Figure 7.5.10illustrates the phaseplane trajectories for the more interesting

underdamped case,c2 < 4w2. Other than the physically unattainable separatrix)

4)

2)

\037 0)

-2)

-4)

-31t) o) 21t) 31t)-21t) -1t) 1t)

x)

FIGURE7.5.10.Position-velocityphaseplane portrait for the damped pendulumsystem x'= y, y' = - sin x -

\037y. The(black)separatricesareemphasized.)))

trajectoriesthat enter unstablesaddlepoints,everytrajectoryeventually is \"trapped\"

by one of the stablespiral points (nn,0)with n an even integer. What this means isthat even if the pendulum starts with enoughenergy to goover the top, after a certain(finite) number of revolutions it has lost enough energy that thereafter it undergoesdampedoscillationsaround its stable(lower)equilibrium position.)

lID.\037roE!\037.\037S........)In Problems1through 4, show that the given system is almostlinear with (0,0) as a criticalpoint, and classify this criti-calpoint as to type and stability. Usea computer system orgraphing calculatorto construct a phaseplaneportrait thatillustrates your conclusion.

dx dy1.-= 1- eX + 2y, -= -x-4 sin ydt dt

2.dx = 2 sin x + sin y.dy = sin x + 2 sin y (Fig.7.5.11)dt dt)

2It)

It)

\037 0)

-It)

-2It)

-2It -It 0 It 2ItX)

FIGURE7.5.11.Trajectoriesof the

system in Problem2.dx dy3.-= eX + 2y -1,-= 8x + eY - 1dt dtdx. dy4. -= sin x cosy- 2y,-= 4x-3cosxsinydt dt)

Find and classifyeachof the criticalpoints of the almost lin-ear systems in Problems5 through 8. Usea computer systemorgraphing calculatorto construct aphaseplaneportrait thatillustrates your findings.

dx . dy5.-= -x+ sin y, -= 2xdt dtdx dy .6.-= y, -= SIn n x -

ydt dtdx dy7.-= 1-eX-Y ,-= 2sinxdt dtdx . dy.8.-d = 3 sin x + y, -= sin x + 2y

t dt

Problems9 through 11dealwith the dampedpendulum systemx'= y, y' = -w2 sinx- cy.)

9.Show that if n is an odd integer, then the critical point(n1f, 0) is a saddlepoint for the damped pendulum sys-tem.)

10.Show that if n is an even integer and c2 > 4w 2, then the

critical point (nn, 0) is a nodal sink for the damped pen-dulum system.11.Show that if n is an even integer and c2 < 4w 2, then the

critical point (nn, 0) is a spiral sink for the damped pen-dulum system.)

In eachofProblems 12through 16,a second-orderequation

of the form x\" + f (x,x') = 0, corresponding to a certain

mass-and-springsystem, is given. Find and classify the criti-

calpoints of the equivalent first-ordersystem.

12.x\" +20x-5x3 = 0:Verify that the critical points resemblethose shown in Fig. 7.5.4.

13.x\" + 2x'+ 20x-5x3 = 0:Verify that the critical pointsresemblethose shown in Fig. 7.5.6.

14.x\" -8x + 2x3 = 0:Herethe linear part of the forceis re-pulsive rather than attractive (as for an ordinary spring).

Verify that the critical points resemblethose shown in

Fig. 7.5.12.Thus there are two stable equilibrium pointsand three types ofperiodicoscillations.

15.x\" + 4x-x2 = 0:Herethe forcefunction is nonsymmet-rico Verify that the criticalpoints resemblethose shown in

Fig. 7.5.13.16.x\" +4x-5x3+x5 = 0:Theideahereis that terms through

the fifth degreein an odd force function have beenre-tained. Verify that the criticalpoints resemblethoseshown

in Fig. 7.5.14.)

In Problems17through 20, analyze the criticalpoints of the

indicated system, usea computer system to construct an illus-

trative position-velocityphaseplaneportrait, and describethe

oscillationsthat occur.)

17.Example 2 in this sectionillustrates the caseof dampedvibrations of a soft mass-springsystem. Investigate an

exampleof damped vibrations of a hard mass-spring sys-tem by using the sameparameters as in Example 2,exceptnow with f3 =

-\037< o.

18.Example 2 illustrates the caseof damped vibrations of asoft mass-springsystem with the resistanceproportionalto the velocity. Investigate an exampleof resistancepro-portional to the squareof the velocity by using the same

parameters as in Example 2, but with resistanceterm

-cx'lx'linstead of -ex'in Eq.(12).19.Now repeatExample 2 with both the alterations corre-

sponding to Problems 17and 18.That is,take f3 = -\037

<oand replacethe resistanceterm in Eq.(12)with -cx'lx'l.)))

y)

y)

x)

FIGURE7.5.12.Thephaseportrait for Problem 14.)


x) x)



20.The equations x' = y, y' = - sin x -iYIYI model a

damped pendulum system as in Eqs.(34)and Fig. 7.5.10.But now the resistanceis proportional to the squareof the

angular velocity of the pendulum. Comparethe oscilla-tions that occurwith those that occurwhen the resistanceis proportional to the angular velocity itself.)

Problems21through 26 outline an investigation of the periodT ofoscillationofa mass on a nonlinear spring with equationofmotion)

d2xdt 2 + 4> (x) = o. (35)

If 4> (x) = kx with k > 0, then the spring actually is linearwith periodTo = 2nj-Jk.21.Integrate once(asin Eq.(6\302\273

to derive the energy equation

4y 2+V(x)=E, (36)where y = dxjdt and

Vex) =lX

</J(u) duo (37)

22.If the mass is releasedfrom rest with initial conditionsx (0) = Xo, Y(0)= 0 and periodicoscillationsensue,con-cludefrom Eq.(36) that E = V (xo) and that the time Trequired for onecompleteoscillationis

4 lxO duT (38)=

\037 0 ,JV(xo) - V(u)

23.If 4> (x) = kx - fJx3 as in the text, deducefrom Eqs.(37)

and (38)that

T = 4\037 (XOdx

(39)10 J(xJ - u 2)(2k -fJxJ - fJu

2))

24. Substitute u = Xo cos4> in (39)to show that)

2To lTC(2 d4>T= ,n ,J1- E 0 J1- 11- sin 2

4>)

(40))

where To = 2nj-Jkis the linear period,)

fJ 2E =

k xo,)1 E

and 11-= -- .

2 l-E) (41))

25.Finally, use the binomial seriesin (31)and the integralformula in (32)to evaluate the elliptic integral in (40)and

thereby show that the periodT ofoscillation is given by)

To

(1 9 2 25 3

)T= 1+-11-+-11-+-11-+....,Jl - E 4 64 256)

(42))

26.If E = fJx5jk is sufficiently small that E2 is negligible, de-

ducefrom Eqs.(41)and (42)that)

T \037 To (1+\037E)

= To (1+\037\037 x\037)

. (43))

It follows that. If fJ > 0, so the spring is soft, then T > To, and

increasing Xo increasesT, so the larger ovals in

Fig. 7.5.4correspondto smaller frequencies.. If fJ < 0, so the spring is hard, then T < To,and increasing Xo decreasesT, so the larger ovalsin Fig. 7.5.2correspondto larger frequencies.)))


7.5Application)

C L)

FIGURE7.5.17.A simplecircuit with an active element.)

\"!he\" \037,\037Y!v\037.!.g\037,..,\037\037,\037\".y\037,!!\"der P

\037\037nnH\037H9.\037,\037,!.!,\037,.\037..\037,\" H)

The British mathematicalphysicistLordRayleigh(John William Strutt, 1842-1919)introduced an equation of the form)

mx\" +kx ==ax'-b(x')3) (1))

to modelthe oscillationsof a clarinet reed.With y == x' we get the autonomoussystem) ,x ==

y,)

,y ==)

-kx+ay -by3) (2))

m)

whosephaseplane portrait is shown in Fig.7.5.15(for the casem ==k ==a ==b =1).The outward and inward spiral trajectories converge to a \"limit cycle\"solutionthat correspondsto periodicoscillationsof the reed.The periodT (and hencethe

frequency) of theseoscillationscan be measuredon a tx-solutioncurve plotted asin Fig.7.5.16.This periodof oscillation dependsonly on the parameters m, k, Q,and b in Eq. (1)and is independentof the initial conditions (why?).)

3) 2.01.51.00.5

\037 0.0-0.5-1.0-1.5-2.00 5 10 15 20 25 30 35 40

t)

H)2)

1)

\037O)

-1)

-2)

o 1 2 3x)

FIGURE7.5.15.Phaseplaneportrait for the Rayleigh system in

(2)with m = k = a = b = 1.)

FIGURE7.5.16.Thelx-solutioncurve with initial conditions

x(O)= 0.01,x'(0)= O.)

Chooseyour own parameters m, k, a, and b (perhapsthe least four nonzerodigits in your student ID number), and use an available ODE plotting utility to

plot trajectories and solution curves as in Figs.7.5.15and 7.5.16.Change one ofyour parameters to seehow the amplitude and frequency of the resulting periodicoscillationsare altered.)

Van derPol'sEquationFigure 7.5.17showsa simpleRLCcircuit in which the usual (passive)resistanceRhas beenreplacedwith an activeelement (suchasa vacuum tube or semiconductor)acrosswhich the voltagedropV isgiven by a known function f (I) of the current I.Of course,V ==f (I) ==IR for a resistor.If we substitute f (I) for IR in the familiarRLC-circuitequationLI'+RI+Q/C ==0 of Section2.7,then differentiation givesthe second-orderequation)

ILI\"+f'(/)/'+-==o.

C)(3))))


In a 1924study of oscillatorcircuits in early commercialradios,Balthasarvan derPol (1889-1959)assumedthe voltage drop to be given by a nonlinear function ofthe form f(I)= bI3 -aI,which with Eq.(3)becomes)

ILI\"+ (3b12-a)I'+-= o.

C)(4))

This equation is closelyrelated to Rayleigh'sequation and has phaseportraits re-semblingFig.7.5.15.Indeed,differentiation of the secondequation in (2)and the

resubstitution x'= y yield the equation)

my\" + (3by2-a)y' + ky = 0,) (5))

which has the sameform asEq.(4).If we denote by T the time variable in Eq. (4) and make the substitutions

I = px,t = T/,JLC, the result is)

d2x(

2 2)Iidx-+ 3bp x -a --+x = o.

dt2 L dt)

With p = ,Ja/(3b) and JL = a ,JC/L, this gives the standardform)

x\" + JL(x2- l)x'+x = 0) (6))

of van derPol'sequation.For every nonnegativevalue of the parameter JL, the solutionof van der Pol's

equation with x(0)= 2,x'(0)= 0 is periodic,and the correspondingphaseplanetrajectory is a limit cycleto which the other trajectoriesconverge(asin Fig.7.5.15).It will be instructive for you to solve van der Pol'sequation numerically and to

plot this periodictrajectory for a selectionof values from JL = 0 to JL = 1000or more.With JL = 0 it is a circleof radius 2 (why?). Figure 7.5.18shows the

periodictrajectory with JL = 1,and Fig.7.5.19showsthe correspondingx(t) andyet) solutioncurves.When JL is large, van derPol'sequation isquite \"stiff\" and the

periodictrajectoryismoreeccentricas in Fig.7.5.20,which wasplottedusing MAT-LAB'S stiff ODEsolver ode15s.The correspondingx(t) and yet) solution curvesin Figs.7.5.21and 7.5.22reveal surprising behaviorof thesecomponentfunctions.)

4321

\037 0-1-2-3-4-4 -3-2-1 0 1 2 3 4

x)

4321

\037 0\037

-1-2-3 y(t)

-40 5 10 15 20

t)

20001500 f.t

= 10001000500

\037 0-500

-1000-1500-2000-4 -3-2-1 0 1 2 3 4

x)

FIGURE7.5.18.Thephaseplanetrajectory ofa periodicsolution ofvan derPol'sequation with M = 1,aswell as sometrajectoriesspiraling inand out.)

FIGURE7.5.19.x(t)and yet)solution curvesdefining the

periodicsolution of van der Pol'sequation with M = 1.)

FIGURE7.5.20.Thephaseplanetrajectory of the periodicsolution ofvan der Pol'sequation with

J-t = 1000.)))

2) x(t))

15001000500)

I I I I I

- --

yet)-

- -

- -- T -- -

I I I I I)

15001000500

\037 0-500

-1000-1500)

4)

-2)

\037 0-500

-1000-1500)

\037 0)

I\ T) -I)

-4o 500 100015002000 25003000

t)

o 500 10001500200025003000t)

1614.28) 1614.285t)

1614.29)

FIGURE7.5.21.Graph ofx(t)with tl = 1000.)

FIGURE7.5.22.Graph of yet) with

tl = 1000.)FIGURE7.5.23.Theupper spike in

the graph of yet).)

Each alternates long intervals of very slow change with periodsof abrupt changeduring very short time intervals that correspondto the \"quasi-discontinuities\"that

are visiblein Figs.7.5.21and 7.5.22.For instance,Fig.7.5.23showsthat, betweent = 1614.28and t = 1614.29,the value of yet) zoomsfrom near zero to over 1300and backagain. Perhapsyou can measure the distancebetweenx-or y-interceptsto show that the periodof circuit around the cyclein Fig.7.5.20is approximatelyT = 1614.Indeed,this calculationand the constructionof figureslikethose shown

here may serve as a good test of the robustnessof your computer system'sODEsolver.

You might alsoplot other trajectories for JL = 10,100or 1000that (likethe

trajectories in Fig. 7.5.18)are \"attracted\" from within and without by the limit

cycle.The origin lookslikea spiral point in Fig.7.5.18.Indeed,show that (0,0)is a spiralsourcefor van derPol'sequation if 0 < JL < 2 but is a nodal sourceifJL >2.)

lIDChaosin Dynamical\037ystems)

In precedingsectionswe have lookedat populationgrowth and mechanicalsystemsfrom a deterministic point of view-withthe expectation that the initial state of a

physicalsystem fully determinesits future evolution. But many common systemsexhibit behavior that sometimesappearschaotic, in the sensethat future states maynot seemreliably predictablefrom a knowledgeof initial conditions.This sectionincludesprojectmaterial illustrating the phenomenonof chaos,which is a topic ofmuch current interest in scienceand engineering.)

PopulationGrowthandPeriodDoubling)In Section1.7we introduced the logisticdifferentialequation)

dP 2-=aP-bPdt)

(a,b > 0)) (1))

that modelsa bounded(rather than exponentiallygrowing) population. Indeed,ifthe population P(t) satisfiesEq.(1),then as t -+ +00,P(t) approachesthe (finite)

limiting population M = ajb.We discusshere a \"discrete\"version of the logistic)))

7.6Chaosin DynamicalSystems 543)

equation in the form of a type of \"differenceequation\" that has beenstudiedexten-sively in the past,but only recently has beendiscoveredto predict quite exoticand

unexpectedpatters of behavior for certainpopulations.In orderto solveEq. (1)numericallyas in Section6.1,we first choosea fixed

stepsizeh >0 and considerthe sequenceof discretetimes)

to, t1, t2, ..., tn, tn+1,...,) (2))

where tn+1= tn + h for each n. Beginning with the initial value Po = P(to),wethen calculate approximations)

PI,P2, ..., Pn, Pn+1 , ...) (3))

to the true values P(t1),P(t2),P(t3),...of the actual population pet).For in-stance,Euler'smethod for the logisticequation in (1)consistsof calculatingthe

approximationsin (3)iteratively by means of the formula)

Pn+1 = Pn + (aPn -bP;).h.) (4))

Now supposethat the population is one for which the stepsizeh can becho-senso that the approximationscalculatedusing Eq.(4)agreeto acceptableaccuracywith the actual population values. This might be the case,for instance,for an ani-mal or insectpopulation in which all reproductiontakesplacewithin short-duration

breedingseasonsthat recur at regular intervals. If h is the interval between suc-cessivebreedingseasons,then the population Pn during one breedingseasonmaydependonly on the population Pn-1during the previous season,and Pn may com-pletely determine the population Pn+1 during the next breedingseason.

Solet us assumefor the sakeof discussionthat the successivevalues Pn =P(tn ) of the populationare given by the equation)

Pn+1 = Pn + (aPn -bP;).h.) (4))

Thus we replacethe originaldifferentialequation in (1)with a \"discrete\"differenceequation)

\037Pn = (aPn -bP;)\037t) (5))

that gives the populationdifference \037Pn = Pn+1-Pn in terms of the time differenceh = \037t and the precedingpopulation Pn.

Equation (4)can be rewritten as the logisticdifferenceequation)

Pn+1 = rPn -sp;,) (6))

where)r = 1 +ah and s = bh.) (7))

The substitution)r

Pn = -XnS)

(8))

in Eq.(6)simplifiesit still further to)

Xn+1= rxn (1-xn ).) (9))

At this point we focus our attention on the final iterative formula in Eq. (9).Beginning with given values of Xo and r, this formula generatesa sequenceXl, X2,

X3, ...of values correspondingto the successivetimes t1, t2, t3, .... We may)))

think of Xn , the value at time tn, as the fraction of the maximum population that theenvironment can support.Assuming that the limiting fractionalpopulation)

Xoo = lim Xnn\037oo)

(10))

exists,we want to investigatethe way in which Xoo dependson the growth parameterr in Eq.(9).That is, if we regard r as the input to the processand Xoo as the output,we askhow the output dependson the input.

The iteration in Eq. (9) is readily implemented in any available calculatoror computer language.Figure 7.6.1showsillustrative Maple,Mathematica, andMATLAB codefor a simpleprogram that beginswith Xl = 0.5and calculates andassemblesa list of the first coupleof hundred (k = 200)iterates with r = 1.5.)

,.......,.M1.l(1/l\037r...i) :1:t:il,i:::ii\037\037:\037!:!!_i(ill!.liii\037)i\037TiAB'):::..::,::\"::::_>:::':\":---:,-;

,-'-\":\"_:-::(-::-,;:::-::---')

/.;<>:::::;:.::::::,\302\273v:::;:t?y:: ::::::::,- \", ,.-,---.;- ,\037 ',-,'-,,\",:.: ;,;.:\" -..,

> :>:: :\037:..-\037::..

'--:;:;'::: ,',

\037-:::;::::-<';:/::/\"<-\",,,::-..;', -,;- .-\";

,.:\037:;:i':: ::::.::\037 :-\037:>:;-;-\037

- '-,

,-':\037. ::,,\". ,\

r = 1.5;x = Table[n,{n,1,200});x[[l))= 0.5;For[n=2,n<=200,

n=n+1,z = x[ [n-1)) ;x[[n))= r*z*(l-z\302\273);)

r := 1.5:)x = array(1..200):x[l) := 0.5:for n from 2 to 200do)

Z .-.-)x[n-1)::= r*z*(l-z):)x[n))

od:)

r = 1.5;x = 1:200;x(l) = 0.5;for n = 2:200

z = x(n-1);x(n) = r*z*(l-z);end)

FIGURE7.6.1.Maple,Mathematica, and MATLAB versionsofa simple iteration program.)

Becauser = 1 + ah in (7),only values of r greater than 1 are pertinent toour idealizedmodelof discretepopulation growth. It turns out that, for a typicalsuch value of the growth parameter r entered at the first line, the results do not

dependmaterially on the initial valueXl. After a reasonablenumber of iterations-the number requireddependson the value of r-thevalue of Xn generally appearsto \"stabilize\" on a limiting value Xoo as in Eq.(10).For example,Fig.7.6.2showsresultsof runs of our simpleiteration program with the values r = 1.5,2.0,and 2.5of the growth rate parameter,yielding limiting (fractional)populations)

Xoo = 0.333333,) 0.500000,) and) 0.6000000,)

respectively.Thus it appears(sofar) that Xoo existsand that its value grows moder-ately as r increases.)

,:::\037::_- ::: C:'.:>':_\037:,-:--'-,- ':';';:

lli.jJi[,:\"ii';),--,,'-:: :.>::\037: :-'-':' ,-'- - -,-;-,-,

\"\"-':-'::':.\"\"-,)

,...,.,...,.......,

.i\037\"\"i\037'ij'.\037\037ij.1.;'..:.!...;.,..'..'...!.'.,........... ;.......,i.....;i:....'\037.5)

\", -,--,-;-::; :,,-::, ,:\037'.:; ::.;::\\ -::' :'----. ---'- -, --,. ,--'

..

'...'..>\"\"..:..:..'it:....j:'.:c;u.u\037.;::;:;;;.iJ;\\\037.\037:')

':\".;:,,:',::.,;--:-::::-,-:;-::,.-::,-':.'-\"'---\";:'\"::''.'';'''--'--''' \"'-\"- -,)

xlx2x3)

0.50.37500.3516)

x197x198x199x200)

0.33330.33330.33330.3333)

0.50.50000.5000)

0.50000.50000.50000.5000)

0.50.62500.5859)

0.60000.60000.60000.6000)

FIGURE7.6.2.Iterates with growth parameters r = 1.5,2.0,and 2.5.)))

EXERCISE1:Try severalother values of the growth rate parameterin the range1 < r < 3. Do your resultssupport the conjecture that the limiting population

alwaysexistsand is an increasing function of r?The resultsin Fig.7.6.3show that the conjecturestated in Exercise1 is false!

With growth rate parameters r = 3.1and r = 3.25,the (fractional)population failsto stabilize on a singlelimiting population. (We calculatedover a thousand iteratesto make sure.)Instead,it oscillatesbetween two different populations in alternatemonths (thinking of a month as our unit of time).For instance, with r = 3.25weseethat)

XIOOI = XI003 = XI005 = ... \037 0.4953,)whereas)

XI002 = XI004 = XI006 = ... \037 0.8124.Thus we have not a singlelimiting population,but rather a \"limiting cycle\"consist-ing of two distinct populations(asillustrated graphicallyin Fig.7.6.4).Furthermore,when the growth rate is increasedto r = 3.5,the periodof the cycledoubles,andnow we have a limiting cyclewith a periodof 4-thepopulationcyclesrepeatedlythrough the four distinct values0.5009,0.8750,0.3828,and 0.8269(Fig.7.6.5).)

::.....:,.:x\037.'.\037I.:.;I.\037

.:\037fi.;; .....;.'.'.1....:.:.\\....'.;....,'.: :.;../..t:;'\037.\037.;.\037.\037..t......;!.,...........

....:

r.......::;.....3...25;;,''::;: :;':_:;::;::';\037::.\037::::\037::::,:_::..\037: :\037:;:\037:,;\";;';

-,;; .;.'.;.' .,',)

r = 3.5)

Xlx2x3x4)

0.50000.77500.54060.7699)

0.50000.81250.49510.8124)

0.50000.87500.38280.8269)

X1001 0.5580 0.4953 0.5009x1002 0.7646 0.8124 0.8750x1003 0.5580 0.4953 0.3828x1004 0.7646 0.8124 0.8269x1005 0.5580 0.4953 0.5009x1006 0.7646 0.8124 0.8750x1007 0.5580 0.4953 0.3828x1008 0.7646 0.8124 0.8269)

FIGURE7.6.3.Cycleswith period 2 with r = 3.1and r = 3.25;acyclewith period4 with r = 3.5.)

1.0 1.0x =0.8124 x =0.8750-

0.8 0.8 -0.6 0.6

,,-.....

,,-.....s.: s.:'-\" '-\"\037 \0370.4 0.4

x =0.38280.2 0.2

0.0 0.0988 990 992 994 996 998 1000 988 990 992 994 996 998 1000n n)

FIGURE7.6.4.Thegraph ofx(n)= xn , showingthe period2 cycleof iterates obtained with r = 3.25.)

FIGURE7.6.5.Thegraph ofx(n) =xn , showingthe period4 cycleof iterates obtained with r = 3.5.)))

n) x\

10011002100310041005100610071008)

0.50600.88740.35480.81270.54050.88170.37030.8278)

10091010101110121013101410151016)

0.50600.88740.35480.81270.54050.88170.37030.8278)

FIGURE7.6.6.Theperiod8cycleobtained with r = 3.55.)

EXERCISE2: Try valuesof the growth rate parameter in the range2.9< r <3.1to determine as closelyas possiblejust where the singlelimiting population splits(as r increases)into a cycle of period2. Doesthis appearto happen just as rexceeds3?

The resultsshown in Fig.7.6.6indicate that a cyclewith period8 is obtainedwith the growth rate parameter value r = 3.55.Events are now changing quiterapidly.)

EXERCISE3: Verify that a cyclewith period16is obtained with the growth rateparameter value r = 3.565.)

EXERCISE4: Seeif you can find a cycleof period32 somewherebetween r =3.565and r = 3.570.)

This is the phenomenonof perioddoublingfor which the innocuous-lookingiteration Xn+l = rxn (1-xn ) has becomefamous in recent years. As the growthrate parameter is increasedbeyond r = 3.56,perioddoubling occursso rapidlythat utter chaosappearsto breakout somewherenear r = 3.57.Thus the graphshown in Fig.7.6.7indicatesthat, with r = 3.57,the earlierperiodicity seemstohave disappeared.No periodiccycleis evident, and the population appearsto bechanging (from one month to the next) in someessentiallyrandom fashion. Indeed,the deterministicpopulation growth that is observedwith smallerparametervaluesseemsnow to have degeneratedinto a nondeterministicprocessof apparently ran-dom change.That is, although the entire sequenceof populationvalues is certainlydeterminedby the valuesXl = 0.5and r = 3.57,successivepopulationvaluesfor n

large do not now appearto be \"predicted\"or determined in any systematicfashionby the immediatelyprecedingvalues.)

1.0I I I I I I

0.8\037

\037\037 \037

0.6\037

\037

\037\037

\037.-..\037 -'-\"\037 0.4--

0.2--)

I)

-)

\037l\037

\037)

-)-)-)

0.0 I I I I I I I

1000 1005 101010151020 1025 1030 1035 1040n)

FIGURE7.6.7.With r = 3.57:Chaos!)

The solutionsand applicationsmanuals accompanyingthis text includeMAT-

LAB, Mathematica, and other versions of a program calledPICHFORK.This pro-gram producesa visual presentationof the way in which the behaviorof our iteration

dependson the value of the growth parameter r. For each value of r in the inputinterval a < r < b (the horizontal axis in the resulting diagram), 1000iterationsare first carriedout to achieve \"stability.\" Then the next 250values of X generatedby the iteration are plotted on the vertical axis-thatis, the screenpixelat (r,x) is\"turned on.\" The descriptivelynamed \"pitchfork diagram\" that resultsthen showsat a glancewhether a given value of r correspondsto a cycle(with finite period)orto chaos.If the resolution in the picture sufficesto make it clearthat only finitely

many values of x are placedabove a given value of r, then we seethat the iterationis \"eventually periodic\"for that specificvalue of the growth rate parameter.)))

1.0

0.8

0.6\037

0.4

0.2)

0.02.8 3.0 3.2 3.4 3.6 3.8 4.0

r)

FIGURE7.6.8.Thepitchforkdiagram with 2.8< r < 4.0,O< x < 1.)

\037)

1.0)

0.8)

0.6)

0.4)

0.2)

0.03.8 3.82 3.84 3.86 3.88 3.9

r)

FIGURE7.6.9.Thepitchforkdiagram with 3.8< r < 3.9,O< x < 1.- -)


Figure 7.6.8showsthe pitchfork diagram for the range 2.8< r < 4.0.Scan-ning from left to right, we seea singlelimiting populationuntil r \037 3, then a cyclewith period2 until r \037 3.45,then a cycleof period4, then one of period8, andso forth, rapidly approachingthe darknessof chaos.But note the verticalbandsof\"white space\"that appearin the diagram between r == 3.6and r == 3.7,betweenr == 3.7and r == 3.8,and again between r == 3.8and r == 3.9.Theserepresentregionswhere (periodic)orderreturns from the precedingchaos.

For instance, Fig.7.6.9showsthe interval 3.8< r < 3.9,where we observea cycleof period3 that emergessuddenly from the chaosnear r == 3.83,and then

splitssuccessivelyinto cyclesof periods6,12,24,...(Figs.7.6.10and 7.6.11).This perioddoublingbeginning with a cycleof period3 is especiallysignificant-afundamental article by JamesYorke and T.-Y.Li in the 1975American Mathemat-icalMonthly was entitled \"PeriodThree ImpliesChaos.\"Accordingto this article,the existenceof a cycleof period3 (for an appropriate iteration) implies the exis-tenceof cyclesof every other (finite) period,as well as chaotic \"cycles\"with noperiodat all.)

1.0 x =0.9594 1.0 x =0.9612x =0.9582

0.8 0.8

0.6 0.6.-.. .-.. x =0.5009\037 x =0.4880 \037'-'\" '-'\" x =0.4718

\037 \0370.4 0.4

0.2x =0.1494 0.2 x =0.1540- x=0.1432

0.0100210041006100810101012 0.0100210041006100810101012n n)

FIGURE7.6.10.Thegraph ofx(n) = xn ,showing the period 3 cycleof iteratesobtained with r = 3.84.)

FIGURE7.6.11.Thegraph ofx(n) = Xn ,showing the period6 cycleof iteratesobtained with r = 3.845.)

PROJECT1:UseProgramPICHFORKto searchfor other interesting cycles,andverify their apparent periodsby appropriate iterative computations. For instance,you should find a cyclewith period10between r == 3.60and r == 3.61,and onewith period14betweenr ==3.59and r ==3.60.Can you find cycleswith period5and 7? If so,lookfor subsequentperioddoubling.A run of PICHFORKrequiresseveral hundred thousand iterations, so it will help if you have a fast computer(orone you can leave running overnight).

As we scanthe pitchfork diagram (Fig.7.6.8)from left to right, we spot the

successivevaluesrl,r2, r3, ...of the growth rate parameterat which a bifurcationor qualitative change in the iteration Xn+l == r Xn (1-xn ) occursas the value of r isincreasedfurther. Theseare the discretevalues of r at which any sufficiently smallincreasein the growth parameterdoublesthe periodof the iteration. In the 1970sthe

LosAlamosphysicistMitchellFeigenbaumdiscoveredthat a certainorderunderliesthis perioddoubling toward chaos:)

. rk - rk-lhm ==4.66920160981....

k\037oo rk+l - rk)(11))

The fraction on the left in Eq.(11)is the ratio of the lengths of successiveconstant-period\"windows\" in the pitchfork diagram. It is the fact that this ratio approaches)))

J(t))

(a)) (b)) (c))

FIGURE7.6.12.Equilibriumpositions ofa mass on a filament:(a) stableequilibrium with x < 0;(b) unstable equilibrium at x = 0;(c) stableequilibrium with x > o.)

a limit as k --* +00,rather than the specificvalue of this limit, that demonstratesa sort of orderunderlying the perioddoublingobservedwith the particular iterationXn+1 = rxn (1-xn ). On the other hand, it is now known that preciselythe sameFeigenbaum constant 4.66920160981...plays exactly the samerolefor a widevariety of period-doublingphenomenaarising in many different areas of science.)

PROJECT 2: Feigenbaum useda (now long obsolete)HP-65pocketcalculator(rather than a powerful computer) to carry out the computationsleading to the dis-coveryof his famousconstant. Perhapsyou would liketo useiterative computationsand/orPICHFORKto isolatethe first few bifurcationvaluesr1,r2, r3, ...with suffi-cient accuracy to verify that the limit in (11)is approximately4.67.You can consultpages124-126of T. Gray and J.Glynn, ExploringMathematicswith Mathematica(New York: Addison-Wesley,1991)for a fancier approach.)

PeriodDoublinginMechanicalSystemsIn Section7.5we introduced the second-orderdifferentialequation)

mx\" +ex'+kx + fJx3 = 0) (12))

to modelthe free velocity-dampedvibrations of a masson a nonlinearspring.Recallthat the term kx in Eq. (12)representsthe force exertedon the massby a linearspring,whereasthe term fJx

3 representsthe nonlinearity of an actual spring.We want now to discussthe forcedvibrations that result when an external

force F(t) = Fa coswt acts on the mass.With such a force adjoined to the systemin Eq.(12),we obtain the forcedDuffing equation)

\037) mx\" +ex'+kx + fJx3 = Fa coswt) (13))

for the displacementx(t)of the massfrom its equilibrium position.For most valuesof the parameters it is impossibleto solveEq.(13)explicitlyfor x(t).Nevertheless,its solutionscan beportrayedqualitatively by means of [numericallyapproximated]phaseplane trajectories likethose we usedin Section7.5to describefree vibrationsof nonlinear mechanicalsystems.

The Hooke'sconstantk ispositivefor a typical spring that resistsdisplacementfrom equilibrium. But there do exist simplemechanical systemsthat emulate aspring having a negativeHooke'sconstant. For example,Fig.7.6.12showsa massm atop a verticalmetal filament. We assumethat the thin metal filament can oscillateonly in a vertical plane,and behaves likea flexiblecolumn that \"buckles\" or bendswhen the massis displacedto either sideof the verticalposition.Then there is onestableequilibrium point to the left (x < 0) and another to the right (x > 0), but

the vertical equilibrium position (x = 0) is unstable.When the massis displacedslightly from this unstable equilibrium position,the internal force exertedon it isrepulsiverather than attractive;this correspondsto a negativevalueof k in Eq.(13).If a periodicforce is exertedon the massby (say)an oscillating electromagneticfield, and air resistancedampsits oscillations,then Eq. (13)with k < 0 but with

c >0 and fJ >0 isa reasonablemathematicalmodelfor its horizontaldisplacementfunction x(t).

In the absenceof both damping and the external force, the phaseplane trajec-toriesof the free oscillationsof the masswould resemblethose shown in Fig.7.5.12(with Problem 14in Section7.5).The massbehavesas though it is repelledby the

unstable critical point at x = 0 but is attracted by eachof the two stablecriticalpoints symmetricallylocatedon either sideof the origin.)))

;;:..,. 0.0)

-0.5)

-1.0)


We saw in Section2.6that in the linearcasea periodicexternalforce F(t) =Focoswt causesa steady periodicresponsex(t) = Ccos(wt-a) with the samefrequency w. The amplitude C of the steady periodicresponseis proportionaltothe amplitude Fo of the externalforce.For instance, if the periodicexternalforceisdoubledin amplitude, then the only change in the responseis that its amplitude isdoubledas well.

To illustrate the quite different behaviorof a nonlinearsystem,we take k = -1and m = c = f3 = w = 1 in Eq.(13),so the differentialequation is)

1/ I 3 DX +x -x+x = r0cost .) (14))

As an exerciseyou may verify that the two stable critical points are (-1,0) and

(1,0).We want to examine the dependenceof the (presumably steady peri-odic)responsex(t) on the amplitude Fo of the periodicexternal force of period2n/w = 2n.

Figures7.6.13through 7.6.16show the solutionsofEq.(14)obtainedwith thesuccessivevalues Fo = 0.60,0.70,0.75,and 0.80of the amplitude of the externalforce.In eachcasethe systemwas solvednumerically with initial conditionsx(O)=1,x'(0)= 0and the resulting solutionplottedfor the range 100< t <200 (to showthe steady periodicresponseremaining after the initial transient responsehas died)

1.5)

Forced Duffing equation (14)with F0 = 0.60)

1.5)

Forced Duffing equation (14)with Fo = 0.60)

1.0) 1.0)

0.5)

-0.5)

0.5)

\037 0.0)

-1.0)-1.5-1.5-1.0-0.5 0.0 0.5 1.0 1.5

x)

-1.5100 120 140 160 180 200

t)

FIGURE7.6.13(a).Period2Jr responsewith Fo = 0.60:phaseplane trajectory.)

-1.5-1.5-1.0-0.50.0 0.5 1.0 1.5x)

0.5)

;;:..,. 0.0)

-0.5)

-1.0)

FIGURE7.6.13(b).Period2Jr responsewith Fo = 0.60:solution x(t).)

1.5)


1.5)I I

\037 \037

\\I I I 1\\1 \037 I \037

\037 -- -

I I I I)


1.0) 1.0)

0.5)

\037 0.0)

-0.5)

-1.0)

-1.5100 120 140 160 180 200

t)

FIGURE7.6.14(a).Period4Jr responsewith Fo = 0.70:phaseplane trajectory.)

FIGURE7.6.14(b).Period4Jr responsewith Fo = 0.70:solution x{t).)))

1.5) 1.5)

1.0) 1.0)

0.5) 0.5)

;;:..,. 0.0) \037 0.0)

-0.5) -0.5)

-1.0) -1.0)

-1.5-1.5-1.0-0.5 0.0 0.5 1.0 1.5x)

-1.5100 120 140 160 180 200

t)

FIGURE7.6.15(a).Period8Jl' responsewith Fo = 0.75:phaseplane trajectory.)

FIGURE7.6.15(b).Period8Jl' responsewith Fo = 0.75:solution x(t).)

Forced Duffing equation (14)with F0 = 0.80)


1.5) 1.5)

1.0) 1.0)

0.5) 0.5)

;;:..,. 0.0) \037 0.0)

-0.5) -0.5)

-1.0) -1.0)

-1.5-1.5-1.0-0.50.0 0.5 1.01.5x)

-1.5100 150 200 250 300

t)

FIGURE7.6.16(a).Chaoticresponsewith

Fo = 0.80:phaseplane trajectory.)

FIGURE7.6.16(b).Chaoticresponsewith

Fo = 0.80:solutionx(t).)

out). Part (a) of eachfigure showsthe phaseplane trajectoryx = x(t), y = x'(t),and part (b) showsthe actual solution curve x = x(t) in the tx-plane.Part (a)exhibitsthe qualitative characterof the solutionmore vividly, but part (b)is requiredto determine the periodand frequencyof the solution.

Figure 7.6.13showsa simpleoscillation of period2]'( of the massaroundthe right-hand critical point. In the ensuing sequenceof figures we seesuccessiveperioddoubling and finally chaosas the amplitude of the externalforce is increasedin the range from Fa = 0.6to Fa = 0.8.This perioddoubling toward chaosisa common characteristic of the behavior of a nonlinear mechanical system as an

appropriatephysicalparameter (suchasm, c,k, f3, Fa, or (J)in Eq. (13\302\273is increased

or decreased.No such phenomenonoccursin linear systems.)

PROJECT 3: Usean ODE plotting utility to seewhether you can reproduceFigs.7.6.13-7.6.16.Then investigate the parameter range 1.00< Fa < 1.10forthe force constant in Eq.(14).With Fa = 1.00you should seea period6]'(phaseplane trajectory that encirclesboth stablecritical points (as well as the unstableone).The perioddoublesaround Fa = 1.07and chaossets in around Fa = 1.10.Seewhether you can spota secondperioddoubling somewherebetween Fa = 1.07and Fa = 1.10.Produceboth phaseplane trajectories and tx-solution curves onwhich you can measure the periods.)))

Warning: You should not expectyour own hardware and ODEsoftwareto

replicatethe exactdetail of the \"chaotictangle\" shown in Fig.7.6.16.To explainwhy, let'sregard the forced Duffing equation (in (14\302\273

with Fa = 0.80as an input-output systemhaving the initial point (x(0),x'

(0\302\273as input and the corresponding

solution x(t) as output. This input-output system is chaotic in the sensethat verysmall changesin the input may causevery large changesin the output. For in-stance,the data shown in the table in Fig.7.6.17weregeneratedusing MATLAB's

sophisticatednumerical solver ode45with two nearby initial points and two dif-ferent error tolerance settings. Solving the same numerical initial value problemrepeatedly with different error tolerancescan provide someindicationof the relia-bility of the results;significant discrepanciescertainlysuggestlittle reliability. With

initial conditionsx(O)= 1,x'(0)= 0it looksplausible(though hardly certain)that

x(100) \037 -1.1and x(200)\037 -0.6,but the value of x(300)remainsquite uncer-tain. By contrast, with initial conditionsx(0)= 1.000001,x'(0)= 0 it appearsthat

perhapsx(200)\037 -0.3instead.If so,then a visually significant change in the solu-tion resultsfrom a changeof initial conditionsno larger than might beexpectedfromaccumulatedroundofferror in the courseof a numericalapproximationprocessthat

relieson machine arithmetic. In this event, any numerically computed solution islikely to divergeappreciablyfrom the true solutionover a long time interval. Hencewe cannot be confident of the fine structure in a numerically generated trajectorysuch as that shown in Fig.7.6.16.Investigations like this suggestonly that theactual long-intervalsolution with initial conditionsx(O)= 1,x'(0)= 0 is not peri-odic,but instead \"wanders\" backand forth in a seeminglyunpredictableor chaoticfashion. Thus the qualitative character of the solution indicated in Fig.7.6.16mayapproximatereality without necessarilypresentingan accuratepictureof the precisedetail of the trajectory.This behaviorof solutionsof the forced Duffing equation isnot yet fully understood and remains a subjectof current research.An interesting

expositionwith further referencescan be found in Chapter 15of Dan SchwalbeandStan Wagon, VisualDSolve(New York: Springer-Verlag,1997).)

;\\ :'-:':;:::\037::::\302\253'>,:::\\,

/1)..i\037i\037\037\037\037\037I;[tj.II!!\037'\037']ii;:it]ii\037\037!;i!;i;\037,\037\037iltiti..

/!j.;\037;Jt\037rsE\342\202\254tli;\037iI1D,I{,,/t: '\037:\037If.T(J)li#;:lij\037i2r)

x(t)With>ErrTol::::10--8)

x(t)with

ErrTol:::10---12);.\"

;:::>:;',' :SY-C':\037::,:;:':;:'\037\037''.,' .,)

','\"

\"\"->,;'>;'.;.:\037\037.;:<'/'::v;) ';\037;P:\037:;::\037;: :;:,:\037);;\037::':;:{\\:;-C):;:\037;,:::::;::).::-::';::::\"

'- ,.,.'.-.)

o100200300)

1-1.1125-0.5823-1.2850)

1-1.1125-0.5828-0.1357)

1.000001-1.1125-0.2925-0.0723)

1.000001-1.1125-0.2816-0.1633)

FIGURE7.6.17.MATLAB attempts to approximate the solution of the forcedDuffing

equation x\" + x'-x + x3 = (0.80)cost on 0 < t < 300with x'(0)= 0 and two differentvalues ofx(O)and two different error tolerances(ErrTol denoting the value used for both theabsoluteand the relative error tolerancein ode45).)

The LorenzStrangeAttractorThe substitution of Xl = x,X2 = x' in the forced Duffing equation in (13)yieldsatwo-dimensionalnonlinear system of first-orderdifferentialequations,and period-doubling phenomena are characteristicof such systems.But in higher dimensionseven more exoticphenomena occurand are currently the subjectof much activeinvestigation. All this work stemsultimately from the original investigation of an

extraordinarythree-dimensionalnonlinear system by the mathematical meteorolo-gistE.N.Lorenz,who later describedits discoveryas follows.)))

50)

N)

25)

o 10x)

FIGURE7.6.18.TheLorenztrajectory in spacewith initialvalues x(0)= -8,y (0)= 8,z(O)= 27,and 0 < t < 40.)

By the middle1950s\"numerical weatherprediction,\"i.e.,forecastingbynumerically integrating such approximations to the atmospheric equa-tions as couldfeasibly be handled, was very much in vogue,despitetherather mediocreresultswhich it was then yielding. A smaller but de-termined group favored statistical prediction....I was skeptical,anddecidedto test the idea by applying the statistical method to a set ofartificial data, generatedby solving a system of equations numerically.... The first task was to find a suitablesystemof equations to solve....The systemwouldhave to besimpleenough...and the generalsolutionwould have to be aperiodic,sincethe statisticalpredictionof a periodicserieswould be a trivial matter, oncethe periodicity had beendetected...[In the courseof talks with Dr.Barry Saltzman]he showedme somework on thermal convections, in which he useda system of sevenordi-nary differentialequations.Mostof his solutionssoonacquiredperiodicbehavior, but one solution refusedto settle down.Moreover, in this so-lution four of the variables appearedto approach zero.Presumably theequations governing the remaining three variables, with the terms con-taining the four variables eliminated, would alsopossessaperiodicso-lutions. Uponmy return Iput the three equations on our computer,andconfirmed the aperiodicity which Saltzman had noted.We werefinallyin business.[Quotedin E.Hairer,S.P.Norsett, and G.Wanner, SolvingOrdinary DifferentialEquations I (New York: Springer-Verlag,1987).])

The famousLorenzsystemof differentialequations is given by)

dx-= -ax+ay,dt

dydt

= px- y-xz,

dzdt

= -f3z+xy.)

(15))

Figure 7.6.18showsa plot in spaceof a trajectoryobtainedby numericalinte-gration of the Lorenz systemwith parametervalues f3 =

\037,

a = 10,and p = 28.Asthis trajectory is traced in \"real time,\" the moving solutionpoint P(x(t),yet), z(t\302\273

appearsto undergo a random number of oscillationson the right followedby a ran-dom number of oscillationson the left, then a random number on the right followedby a random number on the left, and soon.Given the meteorologicalorigin of theLorenz system,one naturally thinks of a random number of cleardays followedbya random number of rainy days,then a random number of cleardays followedby arandom number of rainy days,and soon.

Further investigation of this Lorenz trajectory showsthat it is not simply os-cillating backand forth around a pair of critical points (asFig.7.6.18may initially

suggest).Instead,as t \037 +00,the solution point pet) on the trajectory wandersbackand forth in spaceapproachingcloserand closerto a certain complicatedsetof points whosedetailedstructure is not yet fully understood.This elusive set that

appearssomehow to \"attract\" the solution point is the famous Lorenz strange at-tractor.)

PROJECT4: Sometimesthe behaviorof a trajectory is clarifiedby examining its

projectionsinto one or more coordinateplanes.First usean ODEplotting utility toproducethe xz-projectionof the Lorenz trajectory shown in Fig.7.6.19,using the)))

40)

30)

N)

20)

10)

-10) ox)

10)

FIGURE7.6.19.Thexz-projectionof the Lorenztrajectory with initial point(-8,8,27)and 0 < t < 60.)

6)

4)

N)

2)

o)

FIGURE7.6.20.TheRosslerband illustrated with a trajectoryplotted with x(O)= 2, y(O)= 0,z(O)= 3, and 0 < t < 400.)


sameparametervaluesas listedfollowing(15).Plot alsothe xy- and yz-projectionsof this samesolution. Next, experimentwith different parametervaluesand initial

conditions.For instance, seeif you can find a periodicsolution with p = 70 (andf3 =

\037,a = 10as before)and initial values Xo = -4and zo = 64. To get a

trajectory that almost repeatsitself, you will needto try different valuesof Yo in the

range 0 <Yo < 10and lookat xz-projectionsas in Fig.7.6.19.)

PROJECT 5: Another much-studied nonlinear three-dimensionalsystem is the

Rosslersystem)

dx-= -y -z,dt

dy-=x+ay,dt

dz-=f3- yz +xz.

dt)

(16))

Figure 7.6.20showsa plot in spaceof a trajectory by numerical integration of theRosslersystemwith parametervaluesa = 0.398,f3 = 2, and y = 4.This trajectoryspiralsaround and around as it approachessomesort of \"chaoticattractor\"-theso-calledRosslerband that lookstwisted, somewhat likea Mobiusstrip in space.Investigatethe perioddoubling toward chaosthat occurswith the Rosslersystem asthe parameter a is increased,beginning with a = 0.3,a = 0.35,and a = 0.375(take f3 = 2 and y = 4 in all cases).)

In this sectionwe have given just a taste of the ideas that are the focus ofcontemporaryapplicationsof nonlinear systems.To seehow these ideascome full

circle,consult the discussionof the Lorenz systemon pages117-123of the bookbyHaireret al.referencedpreviously.There you will seea certainaspectof the Lorenztrajectory describedvisually by means of a picture that looksvery much like the

pitchfork diagram shown in Fig.7.6.8,together with the very same Feigenbaumconstant4.6692...!

For an engaging account of the historical background to this final sectionofChapter 6,seeJamesGleick,Chaos:Making a New Science(New York: Viking

Press,1987).For more detaileddiscussionsof the forced Duffing, Lorenz, andRosslerequations, seeJ.M.T. Thompson and H.B.Stewart,NonlinearDynamicsand Chaos(New York: John Wiley, 1986).)))

R E.,.,

F.\"...,

E..,....

R..'

E N C E....,.

S F 0 R......\037.

.. .. ... .\037..'........\"

.. .. .., ,-

FURTHER STUDY)

The literature of the theory and applicationsof differentialequations is vast. Thefollowing list includesa selectionof booksthat might beuseful to readerswho wish

to pursue further the topicsintroduced in this book.)

1.ABRAMOWITZ, M. and I. A. STEGUN,Handbookof Mathematical Func-tions.New York: Dover, 1965.A comprehensivecollectionof tables to which

frequent referenceis made in the text.

2. BIRKHOFF,G.and G.-C.ROTA, Ordinary DifferentialEquations (2nded.).New York: John Wiley, 1969.An intermediate-leveltext that includesmorecompletetreatment of existenceand uniquenesstheorems, Sturm-Liouvilleproblems,and eigenfunctionexpansions.

3.BRAUN, M.,Differential Equations and Their Applications (3rd ed.).NewYork: Springer-Verlag,1983.An introductory text at a slightly higher levelthan this book;it includesseveralinteresting \"casestudy\" applications.

4. CHURCHILL,R. V., OperationalMathematics, 3rd edition. New York:

McGraw-Hill,1972.The standard reference for theory and applicationsofLaplacetransforms, starting at about the same levelasChapter7 of this book.

5.CHURCHILL,R. V. and J.W. BROWN, Fourier Seriesand Boundary Value

Problems,3rd edition. New York: McGraw-Hill,1978.A text at about the

samelevel asChapters9 and 10of this book.6.CODDINGTON,E. A., An Introduction to Ordinary DifferentialEquations.

EnglewoodCliffs, N.J.:Prentice Hall,1961.An intermediate-levelintroduc-tion; Chapters 3 and 4 include proofs of the theorems on power seriesand

Frobenius seriessolutions stated in Chapter 8 of this book.

7. CODDINGTON,E. A. and N. LEVINSON, Theory of Ordinary DifferentialEquations. New York: McGraw-Hill,1955.An advanced theoretical text;Chapter 5 discussessolutionsnear an irregular singular point.

8. DORMAND, J.R.,Numerical MethodsforDifferentialEquations.BocaRa-ton: CRCPress,1996.Morecompletecoverage of modem computationalmethodsfor approximatesolutionof differentialequations.

9.HABERMAN, R.,ElementaryAppliedPartial DifferentialEquations,3rd edi-tion. UpperSaddleRiver, N.J.:PrenticeHall,1998.A next stepbeyondChap-ters 9 and 10of this book,but still quite accessible.)

555)))

556 Referencesfor Further Study)

10.HUBBARD, J.H.and B.H.WEST, DifferentialEquations:A DynamicalSys-tems Approach. New York: Springer-Verlag,1992(part I) and 1995(Higher-DimensionalSystems).Detailedtreatment of qualitative phenomena, with abalancedcombinationof computationaland theoreticalviewpoints.

11.INCE,E.L.,Ordinary DifferentialEquations.New York: Dover, 1956.First

publishedin 1926,this is the classicolderreferencework on the subject.

12.LEBEDEV, N.N.,SpecialFunctionsandTheirApplications.NewYork: Dover,1972.A comprehensiveaccountof Besselfunctions and the other specialfunc-tions of mathematicalphysics.

13.LEBEDEV, N. N.,I.P. SKALSKAYA, and Y. S.UFLYAND, Worked Problemsin AppliedMathematics. NewYork: Dover,1979.A largecollectionof appliedexamplesand problemssimilar to thosediscussedin Chapter 10of this book.

14.McLACHLAN,N. W., BesselFunctionsforEngineers,2nd edition.London:Oxford University Press,1955.Includesnumerous physical applicationsofBesselfunctions.

15.McLACHLAN,N.W., Ordinary Non-Linear DifferentialEquations in Engi-neeringand PhysicalSciences.London:Oxford University Press,1956.Aconcreteintroduction to the effects of nonlinearity in physical systems.

16.POLKING,J. C.and D. ARNOLD, Ordinary Differential Equations UsingMATLAB (2nd ed.).UpperSaddleRiver, N.J.:Prentice Hall,1999.A manualfor using MATLAB in an elementary differential equations course;basedonthe MATLAB programs dfieldand pplanethat are usedand referencedinthis text.

17.PRESS,W. H.,B.P. FLANNERY, S.A. TEUKOLSKY,and W. T.VETTER-LING,NumericalRecipes:TheArt of ScientificComputing.Cambridge:Cam-bridgeUniversity Press,1986.Chapter 15discussesmodem techniquesfor thenumerical solution of differentialequations.This edition includesFORTRAN

programs;Pascaland C editions have alsobeenpublished.

18.RAINVILLE, E.,Intermediate DifferentialEquations, 2nd edition.New York:

Macmillan, 1964.Chapters3 and 4 include proofs of the theorems on powerseriesand Frobeniusseriessolutionsstated in Chapter 8 of this book.

19.SAGAN, H.,Boundary and Eigenvalue Problemsin Mathematical Physics.New York: John Wiley, 1961.Discussesthe classicalboundary valueproblemsand the variational approach to Sturm-Liouville problems,eigenvalues,andeigenfunctions.

20.SIMMONS,G.F.,DifferentialEquations.New York: McGraw-Hill,1972.An

introductory text with interestinghistorical notes and fascinatingapplicationsand with the most eloquentpreface in any mathematicsbookcurrently in print.

21.TOLSTOV,G.P.,FourierSeries.NewYork: Dover,1976.An introductory text

including detaileddiscussionof both convergenceand applicationsof Fouriersenes.)

22.THOMPSON,J.M. T.and H.B.STEWART, Nonlinear Dynamicsand Chaos.New York: John Wiley, 1986.Includesmoredetaileddiscussionsof the forcedDuffing, Lorenz,and Rosslersystems(among others that exhibit nonlinearchaosphenomena).)))

Referencesfor Further Study 557)

23.WEINBERGER, H.F.,A First Coursein Partial DifferentialEquations. NewYork: Blaisdell,1965.Includesseparationof variables,Sturm-Liouvillemeth-ods,and applicationsof Laplacetransform methodsto partial differential equa-tions.

24.WEINSTOCK,R.,Calculusof Variations. New York: Dover, 1974.Includesvariational derivationsof the partial differentialequationsof vibrating strings,membranes,rods,and bars.)))

APPENDIX)Existenceand Uniquenessof Solutions)

In Chapter 1 we saw that an initial value problem of the form)

dydx

= f(x,y), y(a) = b) (1))

can fail (on a given interval containing the point x = a) to have a unique solution.For instance, the answer to Problem 33 in Section1.3says that the initial valueproblem)

dyx2dx+i= 0, y(O)= b) (2))

has no solutions at all unlessb = 0,in which casethere are infinitely many solu-tions.According to Problem31of Section1.3,the initial value problem)

dy-= -V I - y2, y(O)= 1dx)

(3))

has the two distinct solutions YI (x)= 1 and Y2 (x)= cosx on the interval 0 < x <JT. In this appendixwe investigateconditions on the function f (x,y) that sufficeto guarantee that the initial valueproblemin (1)has one and only one solution, andthen proceedto establishappropriateversionsof the existence-uniquenesstheoremsthat werestated without proof in Sections1.3,2.1,2.2,and 5.1.)

_ Existenceof Solutions)The approachwe employ is the methodof successiveapproximations,which wasdevelopedby the French mathematician EmilePicard (1856-1941).This method isbasedon the fact that the function y(x) satisfiesthe initial value problem in (1)onthe openinterval I containingx = a if and only if it satisfiesthe integral equation)

y(x) = b +lx

f(t,y(t\302\273dt) (4))

for all x in I. In particular, if y(x) satisfiesEq. (4),then clearly y(a) = b, anddifferentiationof both sidesin (4)-usingthe fundamental theorem of calculus-yieldsthe differentialequationy'(x)= f(x,y(x\302\273.)

559)))

560 Appendix)

Example1)

To attempt to solve Eq.(4),we begin with the initial function

Yo(x) = b,) (5))

and then define iteratively a sequenceYl, Y2, Y3, ...of functions that we hopewill

convergeto the solution. Specifically,we let

Yi(x)=b+LX

f(t,Yo(t\302\273dt and Y2(x)=b+LX

f(t,Yi(t\302\273dt. (6)

In general,Yn+l is obtained by substitution of Yn for Y in the right-hand side in

Eq. (4):)

Yn+i (x)= b +LX

f(t,Yn(t\302\273 dt. (7)

Supposewe know that eachof thesefunctions {Yn(x)}\037 is defined on someopeninterval (the samefor each n) containingx = a, and that the limit

y(x) = lim Yn(x) (8)n-+oo)

existsat eachpoint of this interval. Then it will follow that

y(x) = lim Yn+l (x)= lim [b + r f(t,Yn(t\302\273 dt ]n-+oo n-+oo Ja

=b+ lim rf(t,Yn(t\302\273dt (9)

n-+oo Ja)

= b + r I (t, limYn(t\302\273 ) dt (10)

Ja n-+oo)

and hencethat)

y(x) = b +LX

f(t,y(t\302\273 dt,

provided that we can validate the interchangeof limit operationsinvolved in passingfrom (9)to (10).It is thereforereasonableto expectthat, under favorableconditions,the sequence{Yn (x)}defined iteratively in Eqs. (5)and (7)will converge to a so-lution Y (x) of the integral equation in (4),and henceto a solution of the originalinitial value problemin (1).)

\037 \"\"\037 \037\037..., \"\"'\" \"'\" \"\"... ........ .....\",..... .....,v.... \"\"........ ...,\"\".... \".... \"...\".............v\037\037\"\"'....,'\".. \"\"''' \"'\" \" O..-.............'O;.'N.\"'...\"'....;.>\".\"'\" .....\"\"'..................,\"\"'\" \"..... '\"\"\"'\" '\" \"'''' \"'\" ,\"\".',,,.

To apply the method of successiveapproximationsto the initial valueproblem

dydx

= Y, y(O)= 1, (11)we write Eqs.(5)and (7),thereby obtaining

Yo (x)= 1, Yn+i(X) = 1 +lx

Yn(t)dt. (12)

The iteration formula in (12)yields

Yi (x)= 1 +lx

1dt = 1 +x,

Y2(X) = 1 +lx

(1+t)dt= 1 +x+ 1X2,

Y3(X) = 1 +lx

(1+ t + 1t2) dt = 1 +x + 1X2 +\037x3,)))

Example2)

Appendix 561)

and)

Y4(X) = 1 +lx

(1+ t + !t2 + it3) dt)

= 1 +x+\037x2

+\037x3

+2\037

x4.

It is clearthat we are generating the sequenceof partial sums of a power seriessolution; indeed,we immediatelyrecognizethe seriesas that of y(x) = eX. Thereis no difficulty in demonstratingthat the exponentialfunction is indeedthe solution

of the initial value problemin (11);moreover,a diligent student can verify (usinga proof by induction on n) that Yn (x),obtained in the aforementionedmanner, isindeedthe nth partial sum for the Taylorserieswith center zero for y(x) = eX. .)

To apply the method of successiveapproximationsto the initial valueproblem)

dy-= 4xy,dx

we write Eqs.(5)and (7)as in Example1.Now we obtain)

(13))y(O)= 3,)

Yo(x) = 3, Yn+l(X) = 3 +lx

4tYn(t)dt.) (14))

The iteration formula in (14)yields)

Yl (x)= 3 +lx

(4t)(3)dt = 3 +6x2,

Y2(X) = 3 +lx

(4t)(3+6t2) dt = 3 +6x2+6x4,

Y3(X) = 3 +lx

(4t)(3+6t2 +6t4) dt = 3 +6x2 +6x4 +4x6,)

and)

Y4(X) = 3 +lx

(4t)(3+6t2 +6t4 +4t6) dt

= 3 +6x2 +6x4 +4x6 +2x8.)

It is again clearthat we are generatingpartial sums of a powerseriessolution.It isnot quite so obvious what function has such a powerseriesrepresentation,but theinitial value problemin (13)is readily solvedby separationof variables:

00 (2x2)ny(x) = 3exp(2x

2) = 3L ,

On.n=

= 3 +6x2 +6x4 +4x6 +2x8 + \037xl0 + ....) .)

In somecasesit may be necessaryto compute a much large number of terms,either in orderto identify the solutionor to usea partial sum of its serieswith largesubscriptto approximate the solution accuratelyfor x near its initial value. Fortu-

nately, computer algebra systemssuch asMapleand Mathematicacan perform the)))

562 Appendix)

Example3)

symbolicintegrations(asopposedto numericalintegrations)of the sort in Examples1 and 2.If necessary,you couldgenerate the first hundred terms in Example2 in amatter of minutes.

In general,of course,we apply Picard'smethod becausewe cannot find asolutionby elementarymethods.Supposethat we haveproduceda large number ofterms of what we believeto be the correctpowerseriesexpansionof the solution.We must haveconditionsunder which the sequence{Yn (x)}providedby the methodof successiveapproximationsis guaranteed in advance to convergeto a solution. Itisjust asconvenientto discussthe initial value problem)

dx-= f(x, t), x(a)= bdt)

(15))

for a system of m first-orderequations,where)

x=)

XI

X2

X3)

II b l

12 b2

f= 13 and b= b3

1m bm)xm)

It turns out that with the aid of this vector notation (which we introduced in Sec-tion 5.3),most resultsconcerning a single[scalar] equation x' = I(x,t) can begeneralizedreadily to analogous resultsfor a system of m first-orderequations,asabbreviated in (15).Consequently,the effort of using vector notation is amply jus-tified by the generality it provides.

The method of successiveapproximationsfor the system in (15)callsfor usto compute the sequence{xn (t)}\037 of vector-valuedfunctions of t,)

Xl n (t)X2n(t )

( ) X3n(t)Xn t =)

xmn (t))

defined iteratively by)

xo(a)= b, Xn+l (t) = b +it f(xn(s),s)ds.) (16))

Recall that vector-valuedfunctions are integratedcomponentwise.)

Considerthe m-dimensional initial value problem)

dx-= Ax, x(O)= bdt)

(17))

for a homogeneouslinear system with m x m constant coefficientmatrix A. Theequations in (16)take the form)

Xo(t) = b, Xn+l = b +lx

Axn(s)ds.) (18))))

Example4)

Example5)

Appendix 563)

Thus)

Xl (t) = b +1t

Abds = b +Abt = (I+At)b;

X2(t) = b +1t

A(b+Abs)ds= b +Abt + 1A2bt 2 = (I+At + 1A

2t 2)b)

and)

X3(t) = b +1t

A(b+Abs + 1A2bs2

) ds= (I+At + 1A2t 2 + iA3t3)b.)

We have therefore obtained the first several partial sums of the exponentialseriessolution)

X(t) = eAtb =(f:

(Atr )

bO

n.n=)

(19))

of (17),which was derivedearlier in Section5.7.) .)The key to establishingconvergencein the method of successiveapproxima-

tions is an appropriatecondition on the rate at which f(x, t) changeswhen x variesbut t is held fixed. If R is a region in (m + 1)-dimensional(x,t )-space,then the

function f(x,t) is said to be Lipschitzcontinuouson R if there existsa constantk >0 such that)

If(XI, t) - f(X2, t)1 <klxl - x21) (20))

if (Xl, t) and (X2, t) are points of R. Recall that the normof an m-dimensionalpointor vector X is defined to be)

Ixi = jxi+xi+x\037+\",+x\037.

Then IXI-x21is simply the Euclideandistancebetween the points Xl and X2.)

(21))

..... .....

Let f (x,t) = x2exp(-t 2

) sin t and let R be the strip 0 < x < 2 in the xy-plane.If(Xl, t) and (X2, t) are both points of R, then

If(XI,t) - f(X2,t)1 =I exp(-t2) sintl.lxl+x21.lxI

-x21< 41xI-x21,)

becauselexp(-t2) sin tl< 1 for all t and IXI +x21<4 if Xl and X2 are both in the

interval [0,2].Thus f satisfies the Lipschitz condition in (20)with k = 4 and istherefore Lipschitzcontinuous in the strip R. .)

...\037. \037 ...........H.... ...... \"\"

Letf(x,t) = t,JX on the rectangle R consistingof the points (x,t) in the xt-planefor which 0 <x < 1 and 0 < t < 1.Then, taking Xl = X, X2 = 0,and t = 1,wefind that)

1II(x,0-1(0,01= ,JX=

,JXlx-01.

Becausex-I /2 -* +00as X -* 0+, we seethat the Lipschitz condition in (20)cannot be satisfied by any (finite) constant k > O. Thus the function f, though

obviouslycontinuouson R, is not Lipschitzcontinuouson R. .)))

564 Appendix)

Example6)

x)

FIGUREA.I. An infinite slab in

(m + 1)-space.)

Suppose,however,that the function I(x,t) has a continuouspartial derivativeIx(x,t) on the closedrectangle R in the xt-plane,and denote by k the maximumvalueof

\037Ix(x,t) Ion R. Then the meanvaluetheoremof differentialcalculusyields)

I/(XI,t) - I(X2,t)1 = I/x(x, t).(Xl -x2)1)

for somex in (Xl, X2), so it follows that)

I/(XI,t) - I(X2,t)1 <klxl -x21)

becauseI/x(x, t)1 <k.Thus a continuouslydifferentiablefunction I(x,t) definedon a closedrectangle is Lipschitzcontinuous there.Moregenerally, the multivari-ablemean value theorem of advancedcalculuscan beusedsimilarly to prove that avector-valuedfunction f(x, t) with continuously differentiablecomponentfunctionson a closedrectangular region R in (x,t)-spaceisLipschitzcontinuouson R.)

The function I(x,t) = x2 is Lipschitzcontinuouson any closed[bounded]regionin the xt-plane.But considerthis function on the infinite strip R consistingof the

points (x,t) for which 0 < t < 1 and x is arbitrary. Then

II(Xl, t) - I(X2,t)1 = Ixi- x\0371= IXI +x21. IXI

-x21.)

BecauseIXI + x21can be made arbitrarily large, it follows that I is not Lipschitzcontinuouson the infinite strip R. .

If I is an interval on the t-axis,then the setof all points (x,t) with t in I is an

infinite strip or slabin (m + I)-space(as indicated in Fig.A.l). Example6 showst that Lipschitzcontinuity of f(x, t) on such an infinite slabis a very strong condition.

Nevertheless,the existenceof a solutionof the initial valueproblem

dx-= f(x, t), x(a)= bdt)

(15))

under the hypothesis of Lipschitzcontinuity of f in such a slabis of considerableimportance.)

THEOREM1 GlobalExistenceof Solutions)Let f be a vector-valuedfunction (with m components)of m + 1 real variables,and let I be a [boundedor unbounded]openinterval containing t = a.If f(x, t)is continuous and satisfies the Lipschitzcondition in (20)for all t in I and forall Xl and X2, then the initial value problemin (15)has a solutionon the [entire]interval I.)

Proof:We want to show that the sequence{xn (t)}\037 of successiveapproxi-mations determined iteratively by

xo(a)= b, Xn+l (t) = b +it f(xn(s),s)ds) (16))

converges to a solution x(t) of (15).We seethat each of thesefunctions in turn iscontinuous on I,aseachis an [indefinite]integral of a continuousfunction.

We may assumethat a = 0,becausethe transformation t -* t + a converts(15)into an equivalentproblemwith initial point t = O.Also,we will consideronlythe portion t >0 of the interval I;the details for the caset <0 are very similar.)))

Appendix 565)

The main part of the proof consistsin showing that if [0,T] is a closed(andbounded)interval contained in I,then the sequence{xn (t)}convergesuniformly on[0,T] to a limit function x(t).This means that, given E >0,there existsan integerN such that)

IXn (t)-x(t ) I

< E) (22))

for all n > N and all t in [0,T].For ordinary (perhapsnonuniform) convergencethe integer N, for which (22)holdsfor all n > N, may dependon t, with no singlevalue of N working for all t in I. Oncethis uniform convergenceof the sequence{xn (t)}has beenestablished,the following conclusionswill follow from standard

theoremsof advancedcalculus(seepages620-622of A. E.Taylorand W. R. Mann,AdvancedCalculus,3rd ed.(New York: John Wiley, 1983\302\273:)

1.The limit function x(t) is continuouson [0,T].2.If N is sochosenthat the inequality in (22)holdsfor n >N, then the Lipschitz

continuity of f impliesthat)

If(xn(t), t) - f(x(t),t)1 <klxn(t) -x(t)1<kE

for all t in [0,T] and n > N, so it follows that the sequence{f(xn (t),t)}\037

convergesuniformly to f(x(t),t) on [0,T].3.But a uniformly convergentsequenceor seriescan be integratedtermwise,so

it follows that, on taking limits in the iterative formula in (16),

x(t)= lim Xn+l (t) = b + lim t f(xn(s),s)dsn-+oo n-+oo Jo)

= b + t lim f(xn(s),s)ds;Jo n-+oo)

thus)

X(t) = b +it f(x(s),s)ds.

4.Becausethe function x(t)iscontinuouson [0,T],the integral equationin (23)(analogous to the one-dimensionalcasein

(4\302\273 impliesthat x'(t) = f(x(t),t)on [0,T].But if this is true on everyclosedsubinterval of the open interval I,then it is true on the entire interval I as well.)

(23))

It therefore remains only to prove that the sequence{xn (t)}\037 convergesuni-

formly on the closedinterval [0,T].Let M be the maximum value of If(b, t)1 for t

in [0,T].Then

IXl(t) -xo(t)1= it f(xo(s),s)ds<it If(b,s)lds< Mt. (24))

Next,

IX2(t) -Xl (t)1 = it[f(Xl (s),s)-f(xo(s),s)]ds <k it IXI (s)-Xo(s)I ds,)

and hence)

IX2(t) -xl(t)1<k itMsds= 1kMt2.) (25))))

566 Appendix)

We now proceedby induction. Assume that

< M (kt)nIXn (t)-Xn-1(t)I ==

-k

., .

n.)(26))

It then follows that

IXn+l (t)-Xn (t)I= it f(xn (s),s)- f(Xn-l (s),s)]ds

<k it Ixn(s)-Xn-l(s)1ds;)

consequently,)

it M (ks)nIXn+I (t)-xn(t)1 < k -. ds.

o k n!It follows upon evaluating this integral that)

M (kt)n+IIx I(t ) -x (t )I <-.n+ n ==

k (n + I)!Thus (26)holdson [0,T] for all n > 1.

Hencethe terms of the infinite series)00

xo(t)+L [Xn (t)-Xn-I (t)]n=I)

(27))

are dominated (in magnitude on the interval [0,T])by the terms of the convergentserIes)

00 M (kT)n+I M kT

\037 T .(n + 1)!

= T(e- 1),) (28))

which is a seriesof positive constants.It therefore follows (from the WeierstrassM-teston pages618-619of Taylor and Mann) that the seriesin (27)convergesuniformly on [0,T].But the sequenceof partial sums of this seriesis simply our

original sequence{xn (t)}\037of successiveapproximations,so the proof of Theorem

1 is finally complete. ..)

\037LinearSystems\037\037 ,\"u,,\"\" _u'\" \"\",,,,,\"'''',, \"\"_\"\",,_,, _ \"\"___\037 \037\"\"\"\"\",,)

An important applicationof the globalexistencetheoremjust given is to the initial

value problem)

dxdt

= A(t)x+g(t), x(a)= b) (29))

for a linear system,where the m x m matrix-valuedfunction A(t) and the vector-valued function g(t) are continuous on a [boundedor unbounded] open interval Icontaining the point t = a. In orderto apply Theorem 1 to the linear system in

(29),we note first that the proof of Theorem 1 requiresonly that, for each closedand boundedsubinterval J of I, there existsa Lipschitzconstant k such that)

If(XI, t) - f(X2, t)1 <klxI - x21) (20))

for all t in J (and all Xl and X2). Thus we do not needa singleLipschitz constantfor the entire openinterval I.)))

Appendix 567)

In (29)wehave f(x, t) = A(t)x+g,so

f(XI, t) - f(X2, t) = A(t)(XI -X2).) (30))

It therefore suffices to show that, if A(t) is continuous on the closedand boundedinterval J,then there is a constant k such that)

IA(t)xl < klxl) (31))

for all t in J.But this follows from the fact (Problem 17)that)

IAxl <IIAII

. lxi,) (32))

where the norm IIAII of the matrix A is defined to be)

(m

)1/2

IIA II=

ij;/aij)2.) (33))

BecauseA(t) is continuouson the closedand boundedinterval J,the norm IIAII isboundedon J, soEq. (31)follows, as desired.Thus we have the following globalexistencetheorem for the linear initial valueproblem in (29).)

THEOREM2 Existencefor LinearSystemsLet the m xm matrix-valuedfunction A(t) and the vector-valuedfunction g(t)becontinuous on the [boundedor unbounded]open interval I containing the pointt = a.Then the initial value problem

dxdt

= A(t)x+g(t), x(a)= b) (29))

has a solutionon the [entire] interval I.)

As we saw in Section5.1,the mth-order initial valueproblem

x(m) +al(t)x(m-l)+ ...+am-l (t)x'+am (t)x= p(t),x(a) = bo, x'(a) = b I , ...,x(m-l)(a) = bm-l)

(34))

is readily transformedinto an equivalentm x m systemof the form in (29).It there-fore follows from Theorem 2 that if the functions al (t),a2(t), ..., am (t) and p(t)in (34)are all continuouson the [boundedor unbounded]open interval I containingt = a, then the initial valueproblem in (34)has a solutionon the [entire]interval I.)

_ LocalExistence)

In the caseof a nonlinear initial valueproblem)

dx-= f(x, t), x(a)= b,dt)

(35))

the hypothesis in Theorem 1 that f satisfies a Lipschitz condition on a slab(x,t)(t in I,all x) is unrealistic and rarely satisfied.This is illustrated by the followingsimpleexample.)))

568 Appendix)

Example1) Considerthe initial value problemdy 2dx

= x, x(0)= b >O. (36)

As we saw in Example6,the equation x' = x2 doesnot satisfy a \"strip Lipschitzcondition.\"When we solve (36)by separationof variables,we get)

bx(t) = .

1- bt)(37))

Becausethe denominatorvanishes for t = lib,Eq. (37)provides a solutionof theinitial value problemin (36)only for t < 1Ib, despitethe fact that the differentialequationx' = x2 \"looksnice\" on the entire real line-certainlythe function appear-ing on the right-hand sideof the equation is continuouseverywhere.In particular, ifb is large,then we have a solutiononly on a very small interval to the right of t = O..

Although Theorem 2 assuresus that linearequations have globalsolutions,Example7 showsthat, in general,even a \"nice\" nonlineardifferentialequationcanbe expectedto have a solution only on a small interval around the initial point t =a, and that the length of this interval of existencecan dependon the initial valuex(a) = b, as well as on the differentialequation itself. The reasonis this: If f(x, t)is continuously differentiable in a neighborhood of the point (b,a) in (m + 1)-dimensionalspace,then-asindicated in the discussionprecedingExample6-wecan concludethat f(x, t) satisfiesa Lipschitzcondition on somerectangularregionR centeredat (b,a), of the form)

It -aI< A, IXi -bil< B

i) (38))

(i = 1,2,...,m). In the proof of Theorem 1,we need to apply the Lipschitzcondition on the function f in analyzing the iterative formula

Xn+l (t) = b +1t

f(xn(s),s)ds. (39)

The potential difficulty is that unlessthe values of t are suitably restricted,then thepoint (xn (t), t) appearing in the integrand in (39)may not liein the region R wherefis known to satisfy a Lipschitzcondition. On the other hand, it can be shown that-on a sufficiently smallopeninterval J containing the point t = a-thegraphsof thefunctions {xn (t)}given iteratively by the formula in (39)remain within the regionR, so the proof of convergencecan then be carriedout as in the proof of Theorem1.A proof of the following localexistencetheorem can be found in Chapter 6 ofG.Birkhoff and G.-C.Rota, Ordinary DifferentialEquations, 2nd ed.(New York:John Wiley, 1969).)

THEOREM3 LocalExistenceof Solutions)Letfbea vector-valuedfunction (with m components)of the m +1 real variablesXl, X2, ...,Xm , and t. If the first-orderpartial derivativesof f all existand arecontinuous in someneighborhoodof the point x= b, t = a, then the initial valueproblem)

dx-= f(x, t), x(a)= b,dt)

(35))

has a solution on someopeninterval containing the point t = a.)))

Appendix 569)

BI Uniquenessof Solutions)It is possibleto establishthe existenceof solutions of the initial value problem in

(35)under the much weakerhypothesisthat f(x, t) ismerelycontinuous;techniquesother than those used in this sectionare required.By contrast, the Lipschitzcon-dition that we usedin proving Theorem 1 is the key to uniquenessof solutions. In

particular, the solutionprovidedby Theorem 3 is unique near the point t == a.)

THEOREM4 Uniquenessof SolutionsSupposethat on someregion R in (m + 1)-space,the function fin (35)iscontin-uous and satisfiesthe Lipschitzcondition)

If(XI, t) - f(X2, t)1 <k .IXI

- x21.) (20))

If Xl (t) and X2(t) are two solutions of the initial problem in (35)on someopeninterval I containingx ==a, such that the solutioncurves (Xl (t), t) and (X2(t), t)both liein R for all t in I,then Xl (t) ==X2 (t) for all t in I.)

We will outline the proof of Theorem4 for the one-dimensionalcasein which

x is a real variable. A generalizationof this proof to the multivariable casecan befound in Chapter 6 of Birkhoff and Rota.

Let us considerthe function)

cfJ(t) == [XI(t)-X2(t)]2) (40))

for which cfJ(a) ==0, becauseXl (a) ==x2(a)==b.We want to show that cfJ(t) = 0,so that Xl (t) = X2(t).We will consideronly the caset >a; the details are similarfor the caset <a.

If we differentiateeach sidein Eg.(40),we find that

IcfJ

'(t)

I

== 12[XI (t) -X2(t)] .[x\037 (t) - x\037(t)] I

==1

2[XI (t) -X2(t)] .[f(XI(t), t) - f(X2(t),t)] I

<2klxI(t)-x2(t)12 ==2kcfJ(t),)

using the Lipschitzconditionon f. Hence)

cfJ

I(t) <2k cfJ (t ) .) (41))

Now let us temporarily ignore the fact that cfJ(a) == 0 and compare cfJ(t) with the

solution of the differentialequation)

<p'(t) ==2ka,) (44))

and this is easilyproved (Problem 18).Hence

o < [Xl (t)-X2(t)]2< [Xl (a)-X2(a)]2e2k(t-a).)))

570 Appendix)

Example1)

On taking squareroots,we get)

o <IXI (t) -X2(t) I

<IXI (a)-X2(a)lek(t-a).) (45))

But Xl (a)-X2(a)==0,so (45)impliesthat Xl (t) = X2(t).)

....The initial value problem)

dx==3x2/3 X (0) ==0

dt ') (46))

has both the obvious solution Xl (t) = 0 and the solution X2 (t) == t 3 that is readilyfound by separation of variables. Hencethe function f (x,t) must fail to satisfy aLipschitzcondition near (0,0).Indeed,the mean value theorem yields)

If(x,O)- f(O,0)1== Ifx(x, 0)1.Ix - 01)

for someX between0 and x. But fx (x,0) ==2x-I /3 is unboundedas X ----+ 0,sono

Lipschitzcondition can be satisfied. .)

_ Well-PosedProblemsand MathematicalModels)

In addition to uniqueness,another consequenceof the inequality in (45)is the factthat solutionsof the differentialequation)

dxdt

== f(x,t)) (47))

dependcontinuously on the initial value x(a); that is, if Xl (t) and X2(t) are two

solutions of (47)on the interval a < t < T such that the initial values Xl (a) andx2(a)are sufficiently closeto one another, then the valuesof Xl (t) and X2(t) remaincloseto one another. In particular, if IXI (a)-X2 (a)I

<8, then (45)impliesthat)

IXI (t)-X2(t) I<8ek(T-a) == E) (48))

for all t with a < t <T.Obviously,we can make E assmallaswe wish by choosing8 sufficiently closeto zero.

This continuity of solutions of (47)with respectto initial values is importantin practical applicationswhere we are unlikely to know the initial value Xo ==x(a)with absoluteprecision.For example,supposethat the initial value problem)

dxdi ==f(x,t), x(a) ==Xo) (49))

modelsa population for which we know only that the initial population is within

8 > 0 of the assumedvalue Xo. Then even if the function f(x,t) is accurate,the solution x(t) of (49) will be only an approximation to the actual population.But (45)impliesthat the actual population at time t will be within 8ek(T-a)of the

approximate populationx(t). Thus, on a given closedinterval [a,T],x(t) will bea closeapproximation to the actual population provided that 8 > 0 is sufficientlysmall.

An initial value problemis usually consideredwellposedas a mathematicalmodelfor a real-worldsituation only if the differential equationhas unique solutions)))

Appendix 571)

that are continuous with respectto initial values. Otherwiseit is unlikely that theinitial valueproblemadequatelymirrors the real-worldsituation.

An even stronger \"continuous dependence\"of solutions is often desirable.Inaddition to possibleinaccuracy in the initial value, the function f (x,t) may not

modelpreciselythe physical situation. For instance, it may involve physicalpa-rameters (suchas resistancecoefficients)whosevalues cannot be measuredwith

absoluteprecision.Birkhoff and Rota generalizethe proof of Theorem4 to estab-lish the following result.)

THEOREM5 ContinuousDependenceof SolutionsLet x(t)and yet) be solutionsof the equations)

dx-= f(x, t) anddt)

dy-= g(y, t)dt)

(50))

on the closedinterval [a,T].Let f and gbe continuousfor a < t < T and for xand y in a commonregion D of n-spaceand assumethat f satisfiesthe Lipschitzcondition in (20)on the region D.If)

If(z, t) -g(z,t)! <JL) (51))

for all t in the interval [a,T] and all z in D,it then follows that)

Ix(t)-y(t)1< Ix(a)-y(a)1\302\267 ek(t-a)+\037

[ek(t-a)-1] (52))

on the interval [a,T].)

If JL >0 is small, then (51)impliesthat the functions f and gappearingin thetwo differential equations, though different, are \"close\"to each other. If E > 0 is

given, then it is apparent from (52)that)

Ix(t) -y(t)!< E) (53))

for all t in [a,T] if both Ix(a)-y(a)!and JL are sufficiently small.Thus Theorem5 says(roughly) that if both the two initial valuesand the two differential equationsin (50)are closeto eachother, then the two solutionsremain closeto eachother fora < t <T.- -

For example,supposethat a falling body is subjectboth to constantgravita-tional acceleration g and to resistanceproportional to somepower of its velocity,so (with the positiveaxisdirecteddownward)its velocity v satisfiesthe differential

equation)

dv-= g-cvp .dt)

(54))

Assume,however, that only an approximationc to the actual resistancec and an

approximationp to the actual exponentp are known. Then our mathematical modelis basedon the differentialequation

du --= g-cuPdt)

(55))

instead of the actual equation in (54).Thus if we solve Eq. (55),we obtain onlyan approximationu(t) to the actual velocity vet). But if the parameterscand p are)))

572 Appendix)

sufficiently closeto the actual values c and p,then the right-hand sidesin (54)and(55)will be closeto each other. If so,then Theorem 5 impliesthat the actual and

approximatevelocity functions vet) and u(t)are closeto each other. In this casethe

approximation in (55)will be a goodmodel of the actual physical situation.)

_._\037\037\037.!>l\037!!!\037....)

In Problems1through 8, apply the successiveapproximationformula to compute Yn (x) for n < 4. Then write the expo-nential seriesfor which theseapproximations arepartial sums

(perhaps with the first term or two missing; for example,

eX - 1 = x + 1.x2 + 1.x3 + ..l.x4 + ...)2 6 24 .)

dy dy1.dx

= y, y(O)= 3 2.dx

= -2y,y(O)= 4

dy dy3.-= -2xy,y(O)= 1 4. -= 3x2y, y(O)= 2

dx dxdy dy5.-d

= 2y + 2, y(O)= 0 6.-= x + y, y(O)= 0x dx

dy7.dx

= 2x(1+ y), y(O)= 0

dy8.dx

= 4x(y+ 2X2), y(O)= 0)

In Problems9 through 12,compute the successiveapproxima-tions Yn (x)for n < 3; then comparethem with the appropriatepartial sums of the Taylor seriesof the exactsolution.

dy dy9.-= x + y, y(O)= 1 10.-= y + eX, y(O)= 0dx dxdy 2 dy I 311.-= y , y(O) = 1 12.-= 2Y , y(O)= 1dx dx

13.Apply the iterative formula in (16)to compute the firstthree successiveapproximations to the solution of the ini-tial value problem

dxdi=2x-y , x(O)= 1;

dy

dt= 3x - 2y,) y (0)= -1.)

14.Apply the matrix exponential seriesin (19)to solve(inclosedform) the initial value problem)

I

[1

x (t) =0) \037 ]x,) xeD)= U l)

(Suggestion: Show first that)

[1 1

]n

[In

]o 1--

0 1)

for eachpositive integer n.)15.For the initial value problem dyjdx = 1 + y3, y(l) = 1,

show that the secondPicard approximation is)

Y2(X) = 1+ 2(x-1)+ 3(x-1)2+ 4(x- 1)3+ 2(x- 1)4.)

Then compute Y2 (1.1)and Y2(1.2).The fourth-order

Runge-Kutta method with step size h = 0.005yields

y(1.1)\037 1.2391and y(1.2)\037 1.6269.16.For the initial value problem dyjdx = x2 + y2, y(O) = 0,

show that the third Picard approximation is)

1 1 2 1y (x) = _x3 + _x7 + _Xll+ x I5 .3 3 63 2079 59535)

Compute Y3(1). The fourth-order Runge-Kutta method

yields y(l) \037 0.350232,both with step sizeh = 0.05and

with step sizeh = 0.025.17.Proveas follows the inequality IAxl <

IIAII.lxi, where A

is an m x m matrix with row vectorsaI, a2,..., am, and xis an m-dimensional vector.First note that the componentsof the vectorAx area I \302\267 x, a2 \302\267 x, ..., am \302\267 X, SO)

[ ]1/2

IAxl =\037?

ai.X)2)

Then use the Cauchy-Schwarzinequality (a.X)2 0))

for t > a. Multiply both sidesby e-kt , then transposeto

show that)ddt [</J(t)e-

kt

] < 0

for t > a.Then apply the mean value theorem to concludethat)

</J(t) < </J(a)ek(t-a))

for t > a.)))

ANSWERSSELECTED)

TOPROBLEMS)

Chapter1) 19.C= 6)10)

5)

Section1.1)

11.If y = YI = x-2, then y'(x) = -2x-3 and yl/(x) = 6x-4, sox2

yl/ + 5xy' + 4y = x2(6x-4) + 5x(-2x-3) +4(x-2) =6x-2-10x-2 +4x-2 = O.If y = Y2 = x-21nx, then

y'(x) = x-3-2x-31nx and yl/(x) = -5x-4 + 6x-41nx, sox2

yl/ + 5xy' + 4y =x2(-5x-4 + 6x-41nx) + 5x(x-3-2x-31nx) +4(x-21nx) = o.) -5)

13.r =\037)

14.r = :f:\037

15.r = -2, 1)

5)

-10-5 0 5x

20. C= 1120)

16.r = i(-3:f:\037)17.C= 2)

\037o) \037 0)

-5 -20-5 0 5 -10 -5 0 5 10

x x

18.C= 3 21.C= 710

5

5)

-5)

\037 0)\037o)

-5-5) o

x)

5)

-10-2) -1) o

x)

1) 2)

573)))

574 AnswerstoSelectedProblems)

22. C= 1 26. C= -1f5)

\037 0)

-5-20 -10

23. C= -5630

20

10

\037 0

-10-20

-300 1

24. C= 17)

10)

5)

\037 0)

-5)

0 10 20 -100 5 10x x

27. y' = x + y 28. y' = 2y /x

29. y' = x/(1-y) 31.y' = (y -x)/(y + x)32. dP/dt = k\037 33.dv /dt = kv 2

35.dN /dt = k(P -N) 37. y = 1or y = x

39.y = x2 41.y =\037

eX

42. y = cosx or y = sin x

43. (b) Theidentically zerofunction x (0)= 044. (a) Thegraphs (figure below) of typical solutions with k =

\037

suggest that (for each) the value x(t) increases without bound ast increases.

52 3

x)

oo 1 234

t

(b) Thegraphs (figure below) of typical solutions with k = -\037

suggest that now the value x (t) approaches 0 as t increaseswithout bound.

6)

30)

20)

10)

\037 0)

-10)

-20)

-30o 0.5 1 1.52 2.5 3 3.54 4.5 5

x)

25. C= 1f /4)

4)

2)

\037 0)

-2)

4)

3)

\037)

2)

1)

5

4

\037 3

2

1

00 1 2 3 4

t)

-4-2) 2)

45. P(t) = 100/(50-t); P = 100when t = 49,and P = 1000when t = 49.9.Thus it appears that P(t) grows without boundas t approaches 50.)

-1) ox)

1)))

AnswerstoSelectedProblems575)

46. vet) = 50/(5+ 2t); v = 1 when t = 22.5,and v =1\037

whent = 247.5.Thus it appears that v(t) approaches 0 as t increaseswithout bound.)

{

-t if 0 :St :S5,21.x(t) = :0:_\037t2

-25 if 5 :St :S10.)

47. (a) C= 10.1;(b) No such C,but the constant function y(x) = 0satisfies the conditions y' = y2 and y(O)= o.)

40)

;:::. 2023. v(t) = -(9.8)t+49,so the ball reaches its maximum height

10 (v = 0)after t = 5 seconds.Its maximum height then is

y(5) = 122.5(m).24. v(5) = -160ft/s

0 25. The carstops when t \037 2.78(s),sothe distance traveled before0 2 4 6 8 10

stopping is approximately x(2.78)\037 38.58(m).t26. (a) y \037 530m (b) t \037 20.41s (c)t \037 20.61s

20.{

lt 2 if 0 :St :S5, 27. Yo \037 178.57(m)x (t) = 2255t -- if5:St:S10. 28. v(4.77) \037 -192.64ft/s2

29. After 10secondsthe carhas traveled 200ft and is traveling at 7040 ft/s.

30.a = 22ft/s 2; it skids for 4 seconds.

30 31.Vo = 10,J30(m/s), about 197.18kmlh

32. 60m 33. 20\037 \037 63.25(ft/s)

;:::. 20 34. 460.8ft 36. About 13.6ft37. 25(mi) 38. 1:10pm

10 39. 6 mph 40. 2.4mi

41.5\037

\037 181.33ft/s 42. 25mi

0 43. Time:6.12245x 109 S \037 194years;

0 2 4 6 8 10 Distance: 1.8367x 1017 m \037 19.4light-yearst 44. About 54mi/h)

Section1.2)1.y(x) = x2 + x + 3

3.y(x) =\037(2X3/2

-16)

5.y(x)=2,Jx+2-5)7. y (x) = 10tan -Ix

9.y(x) = sin-I x

11.x(t) = 25t 2 + lOt + 20

13.x(t) = 4t3 +5t)

15.x(t) =\037(t

+ 3)4-37t -26

16.x(t) = \037 (t + 4)3/2-5t _ 29

3 3)

17.x(t)=4[(t+l)-I+t -l])19.x(t) =

{\037\037t_ lt 2 _ 25

2 2)

40)

30)

30)

2.y (x) =\037

(x -2)2 + 1

4. y (x) = -1Ix + 6

6.y(x) = ![(x2 +9)3/2-125]3)

;:::. 20)

10)

8.y (x) = 4 sin 2x+ 1

10.y(x) = -(x+ l)e-X + 2

12.x(t) = -10t2- 15t + 5

14.x(t) =\037t3

+ 4t2 -7t+4)

oo) 2) 4) 8) 10)6)

t)

22. x(t) =11:2-1;

i(-5t2 + lOOt -290))

if 0 :St :S3,if 3 :St :S7,if7:St:s10.)

40)

30)

if 0 :St :S5,if5:St< 10.) ;:::. 20)

10)

oo) 6) 8) 10)2) 4)

t)))

576 AnswerstoSelectedProblems

Section1.3 5. 3

2

1. 3 1

2 \037 0

1 -1\037 0 -2

-1 -3-3 -2 -1 0 1 2 3

-2 x

-3-3 -2 -1 0 1 2 3

x 6. 3

2. 3 2

2 1

1\037 0

0 -1\037

-1 -2

-2 -3-3 -2 -1 0 1 2 3

x-3-3 -2 -1 0 1 2 3

x

7. 33. 3I 2I

2 I/

1/

1

\037 0\037 0 -1

-1-2

-2-3

2 3-3 -3 -2 -1 0 1

-3 -2 -1 0 1 2 3 xx)

4. 3 8. 3

2 2

11

\037 0\037 0

-1 -1-2 -2-3

0 2 3 -3-3 -2 -1 1 -3 -2 -1 0 1 2 3x

x)))

2 .,.I ./. 1 .1. ./.} ./. ./ I .\" .z./. ./

I / / / / / / / / / / / /i / / I / / / / / / I / // / / / / ,/ /' /' ,/ / / / / (2,?)

1 / l{./-.\".-- --,/0,(./ /( //-:'--..........----:/// / ,/ ,,--,//

(0,0)I \"'\037/'\037 0

i // /, / ///.--..... /

1 2 3 -1 . . I ./ {/-.\".----./0;i'.I -/ I1 / / / / / ,/ /' /' ,/ / / / / // / / / ( / / / / / / 1 / / // / / / / / / / / / / / / / /-2 .,., ./ / 1- ./.}./ ./ I / 1 /. ./.I-2 -1 0 1 2

x)

9. 3

2

1

\037

0

-1-2

-3-3 -2 -1 0

x

10. 3

2

1

\037 0

-1-2-3

-3 -2 -1 0x)

2) 3)1)

11.12.13.14.)

A unique solution exists in someneighborhood ofx = 1.A unique solution exists in someneighborhood ofx = 1.A unique solution exists in someneighborhood ofx = o.Existence but not uniqueness isguaranteed in someneighborhood ofx = o.Neither existence nor uniqueness isguaranteed in anyneighborhood ofx = 2.A unique solution exists in someneighborhood ofx = 2.A unique solution exists in someneighborhood ofx = o.Neither existence nor uniqueness isguaranteed.

A unique solution exists in someneighborhood ofx = o.A unique solution exists in someneighborhood ofx = o.Your figure should suggest that y(-4) \037 3; an exact solution ofthe differential equation gives y(-4)= 3+e-4 \037 3.0183.)

15.)

16.17.18.19.20.21.)

\037)

54321

o-1-2-3-4-5-5-4 -3-2-1 0) 1 2 345)

.I . . . ../ .,. .. . . ..././-..,..J.. . . .. ./.I-

. ./.. I . .. . . .. ./J.I I.1-.)

\\ \\ \\

. ... . ....\\. . . .. .

\\.. .

\\.

.\\..

\\. ... .

\\.. .

\\..

.\\.. \\. .

. .\\-

. .\\

..\\,

, \\. .)

x)

22. y(-4) \037 -3)


23. Your figure should suggest that y(2) \037 1;the actual value iscloserto 1.004.)

24. y(2) \037 1.525. Your figure should suggest that the limiting velocity is about 20

ft/sec (quite survivable) and that the time required to reach19ft/sec isa little lessthan 2 seconds.An exact solution givesv(t) = 19when t =

\037

In 20\037 1.8723.)

40353025

:::.2015105o

o)

\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\

.\\.\\ \\.\\.\\ \\ \\ \\ \\,.\\ '\\ \\..\\ \\

\\ \\\037

\\ \\ \\ \\ \\\037

\\ \\ \\ \\ \\

\\\" ,.'\\\" , , '\\.,.\"\\' , , '\\. \\. ,..,, , , , , , , , , , , , , ,.,',\037

\" \"',,, \" .\037 '- ,'\", ........)

-----..:.------------). .\037/...;--r',/ ,/ ,/ ,//..//.// / / /,f../../II / / i)

\037\037

,/ ,//, // /I..// /)

....-:.//.'.--:'./,/ ,/ ,/ ,/ ,/.I / /' / // / / / // . /. . I 1..// / I / /)

3) 4) 5)1) 2)

t)

26. A figure suggests that there are 40deerafter about 60months; amore accurate value is t \037 61.61.The limiting population is 75deer.

27. The initial value problem y' = 2,.jY, y(O)= b has no solution if

b < 0;a unique solution if b > 0; infinitely many solutions if

b = O.)

\037)

(0,0))

x)

28.) Theinitial value problem xy' = y, y(a) = b has a uniquesolution if a

=1= 0; infinitely many solutions if a = b = 0;nosolution if a = 0but b =1=

o.Theinitial value problem y' = 3y 2/3,y(a) = b always has

infinitely many solutions defined for all x.However, if b =1= 0)

29.)))

then it has a unique solution near x = a.)

\037)

x)

30. The initial value problem y' = -)1-y2, y(a) = b has aunique solution if Ibl < 1;no solution if Ibl > 1,and infinitelymany solutions (defined for all x ) if b = :1::1.

31.The initial value problem y' = )1-y2, y(a) = b has a uniquesolution if Ibl < 1;no solution if Ibl > 1,and infinitely manysolutions (defined for all x) if b = :I::1.)

1)

\037)

-1)

-n12 nl2x)

32. The initial value problem y' = 4x,JY, y(a) = b has infinitelymany solutions (defined for all x) if b

\037 0;no solutions if b < O.However, if b > 0 then it has a unique solution near x = a.

33. The initial value problem x2y' + y2 = 0,y(a) = b has a unique

solution with initial point (a, b) if a=1= 0, no solution if a = 0

but b =1= 0, infinitely many solutions if a = b = O.)

6)

4)

2)

\037 0)

-2)

-4)

-6-6 -4 -2 0 2 4 6x)

34. (a) If y(-I)= -1.2then y(l) \037 -0.48.If y(-I)= -0.8then

y(l) \037 2.48.(b) If y(-3)= -3.01then y(3) \037 -1.0343.If y(-3)= -2.99then y(3) \037 7.0343.

35. (a) If y(-3)= -0.2then y(2) \037 2.019.If y(-3)= +0.2then

y(2) \037 2.022.In either case,y(2) \037 2.02.(b) If y(-3) \037 -0.5then y(2) \037 2.017.If y(-3) \037 +0.5then

y(2) \037 2.024.In either case,y(2) \037 2.02.)

Section1.4)

1.y(x) = Cexp(-x2)

3.y(x) = Cexp(-cosx)

5. y(x) = sin (C+ -fi)7. y(x) = (2X

4/3 + C)3/2)

2.y(x) = 1/(x2+ C)4. y(x) = C(1+ X)4

6.y(x) = (X3/2 + C)2

8.y(x) = sin-I (x2 + C))

9.y(x) = C(1+x)/(1-x)

10.y(x) = (1+x)/[1+ C(1+x)]-1

11.y(x) = (C-X2)-1/2)

212.y2 + 1 = CeX

13.In(y4 + 1)= C+4sinx

14.3y + 2y 3/2 = 3x + 2X3/2 + C

15.1/(3y 3)-2/y = l/x+ In Ixl + C

16.y(x) = sec-l (C.Jl + x2)

17.In 11 + y I= x + 4x2 + C

18.y(x) = tan (c-

\037

-x)

19.y(x) = 2exp(eX ) 20. y(x) = tan(x 3+ 1f /4)

21.y2 = 1+ .Jx2-16 22. y(x) = -3exp(x4-x)

23. y(x) = 4(1+ e2x-2) 24. y(x) = f sin x

25. y(x) = x exp(x2-1) 26. y(x) = 1/(1-x2-x3)

27. y = In(3e2X -2) 28. y(x) = tan-I (-fi-1)

29. (a) General solution y(x) = -1/(x-C);(b) The singularsolution y(x) = o.(c)In the following figure we seethat there isa unique solution through every point of the xy- plane.)

6

4

2

\037 0

-2-4

-6-6 -4 -2 0 2 4 6x)

30.General solution y(x) = (x -C)2;singular solution y(x) = o.(a) No solution if b < 0; (b) Infinitely many solutions (for all

x) if b\037 0; (c)Two solutions near (a, b) if b > o.

31.Separation ofvariables gives the samegeneral solution

y = (x -C)2as in Problem 30,but the restriction that

y' = 2,JY \0370 implies that only the right halves of the parabolas

qualify as solution curves. In the figure below we seethat

through the point (a,b) there passes(a) No solution curve if

b < 0, (b) a unique solution curve if b > 0, (c)Infinitely many)))

solution curves if b = O.)

75)

50)

\037)

25)

o)

-15-10-5 0x)

5 10 15)

32.General solution y (x) = ::I::sec(x-C);singular solutionsy(x) = ::1::1.(a)Nosolutioniflbl < 1; (b) A unique solution iflbl > 1;(c)Infinitely many solutions if b = ::I::1.

33.About 51840persons 34. t \037 3.87hr

35.About 14735years 36.Age about 686years37.$21103.48 38. $44.5239.2585mg 40. About 35years41.About 4.86x 109 years ago42. About 1.25billion years43. After a total ofabout 63min have elapsed44. About 2.41minutes

45. (a) 0.495m; (b (8.32x 10-7)/0; (c) 3.29m46. (a) About 9.60inches; (b) About 18,200ft47. After about 46days48. About 6 billion years49. After about 66min 40s50. (a) A(t) = 10.32t/15; (b) About 20.80pu; (c)About 15.72

years51.(a) A(t) = 15.(\037r/5; (b) approximately 7.84su; (c)After about

33.4months52.About 120thousand years ago53. About 74 thousand years ago54. 3hours 55.972s56. At time t = 2048/1562\037 1.31(in hours)58.1:20P.M.59.(a) y(t) = (8-7t)2/3; (b) at 1:08:34P.M.;

(c) r =\037o [[;\037 0.15(in.)

60. About 6 min 3 sec61.Approximately 14min 29s62. The tank is empty about 14secondsafter 2:00P.M.63. (a) 1:53:34P.M.; (b) r \037 0.04442ft \037 0.53in.

64. r =7\0370

.J3ft, about3\037

in.

65. At approximately 10:29A.M.)

Section1.51.y(x) = 2(1-e-X )3.y(x) = e-3x(x2 + C)5. y(x) = x + 4x-27. y(x) = 5X I/2 + CX-I/29.y(x) = x(7 + In x)11.y(x) = 0)

2.y(x) = (3x + C)e2x

24. y(x) = (x + C)eX

6.y(x) = x2 + 32/x58.y(x) = 3x + CX-I/310.y(x) = 3x3 + CX 3/212.y(x) = i X5 -56x-3)


13.y(x) = (eX +e-X )/2 14.y(x) = x31nx + 10x3

15.y(x) = [1-5exp(-x2)]/216.y(x) = 1 +e-sinx

17.y(x) = (1+ sinx)/(1+x)18.y(x) = x2(sin x + C)19.y(x) =

\037

sin x + Ccscx20. y(x) = -1+ exp (x + \037X2)21.y(x) =x3sinx

22. y (x) = (x3 + 5)eX2

23.y(x) = x3(2+ Ce-2x)24. y(x) =

\037

[1+ 16(x2 + 4)-3/2]25.y (x) = [exp(-\037 X2)] [3(x2+ 1)3/2-2]26. x(y) = 1/2y2 + C/y427. x(y) = eY (C+

\037y2)

28. x(y) =\037

[y + (1+ y2)(tan-1

y + C)]30. y(x) = XI/2 Itt- I /2cost dt

29. y(x) = [exp(x2)][C+ \037,Jn erf(x)]

32. (a) y(x) = sinx -cosx; (b) y(x) = Ce-X + sinx -cosx;(c)y(x) = 2e-x + sin x -cosx

33. After about 7 min 41s34. About 22.2days35. About 5.5452years36. (a) x(t) = (60-t) -(60-t)3/3600;(b) About 23.09Ib37. 393.751b38. (a) x(t) = 50e-t/20; (b) y(t) = 150e-t/4o-100e-t/2o ;

(c)56.25lb39. (b) Ymax

= 100e-1 \037 36.79(gal)41.(b) Approximately $1,308,28343. -50.0529,-28.0265,-6.0000,16.0265,38.052944. 3.99982,4.00005,4.00027,4.00050,4.0007345. x(t) = 20(1-e-t/IO );x = 10after t = 10In 2 \037 6.93months.46. x(t) =

12

0\302\2601

(101-102e-t/1O +cost+ 10sint);x = 10after

t = 6.47months.)

Section1.6)2.y2 = x2(1nx + C))1.x2-2xy -y2 = C

3.y(x) = x (C+ In Ixl)24. 2tan- I

(y Ix) -In(y2/x2 + 1)= 2lnx + C5.1nlxyl=C+xy-1 6.2ylny=x+Cy7. y3 = 3x3(C+ In Ix I) 8.y = -xIn(C-In x)9.y(x) = x/(C-In Ixl) 10.x2 + 2y2 = Cx6

11.y = C(x2 + y2)12.4x2 + y2 = x2(1nx + C)213.y+ JX2 +y2=CX2

15.x2(2xy + y2) = C

16.x=2\037 x+y+I-21n(I+\037 x+y+l)+C17.y(x) = -4x+ 2tan(2x + C)18.y = In(x + y + 1)+ C19.y2 = x/(2+ Cx5)21.y2 = 1/(Ce-2X -1)23.y(x) = (x + CX 2)-325.2x3

y3 = 3\037 1+x4 + C27.y(x) = (x4 + CX)I/329.sin2

y = 4x2 + Cx31.x2+ 3xy + y2 = C33.x3 + 2xy

2 + 2y3 = C

35.3x4 + 4y3 + 12y In x = C

36.x + eXY + y2 = C37. sin x +x In y + eY = C)

14.x -Jx2+ y2 = C)

20.y3 = 3+ Ce-3x2

22.y3 = 7x/(7Cx7 + 15)24. y2 = e2x /(C+ Inx)26.y3 = e-X(x + C)28. y = In(Cx2 +x2e2x )30.x2-2xeY -e2y = C32.2X2 -xy + 3y2 = C34.x3

+X2y2 + y4 = C)))


38. x2+ 2x tan- I

y + In(1 + y2) = C39. 5x3

y3 + 5xy4 + y5 = C

40. eX sin y +x tan y = C41.x2y-1 + y

2x-3 + 2y l/2 = C42. xy-2/3 + X-3/2y = C43. y(x) = AX2 + B44. x(y) = Ay

2+ B45. y(x) = A cos2x+ Bsin 2x46. y (x) = x2 + A In x + B47. y(x)=A-lnlx+BI48. y(x) = In x + AX-2 + B49. y(x) = ::I::(A+ BeX )1/2

50. y(x) = In I sec(x+ A)I - \037X2+ B

51.x(y) = -k(y3+ Ay + B)52. Ay

2- (Ax + B)2= 153. y(x) = A tan(Ax + B)54. Ay 2(B-x) = 158. y = exp(x

2 +c/x2)59. x2-2xy -y2

-2x-6y = C60. (x + 3y + 3)5= C(y-x -5)61.x = tan(x -y) + sec(x -y) + c64. y(x) = x +e-x2 [C+ \037,Jn erf(x)]-165. y(x) = x + (C-X)-I69. Approximately 3.68mi

Section1.721.x(t) =

2-e-t)

102.x(t) =1+ ge-lOt)

2+ e-2t

3.x(t) = 22-e-t)

4

3

2

\037 1

0

-1-2

1 2 3t)

3(1-e-12t )4. x(t) =2(1+ e-12t

))

3

2

1

\037 0

-1-2-3

1t)

3)

405.x(t) =3 158- e- t)

2)

\037 1)

o)

-1o) 2 3

t)

4) 5)1)

15)10)

105

\037 5 \037

00

-5 -50 1

0 0.25 0.5t t)))


106.x(t) ==152+ 3e 1)

5)

22. About 34.66days

23.(a) lim x(t) == 200grams (b)\037

In 3 \037 1.37secondst\037oo)

10)

-5o) 0.25

t)

0.5)

24. About 9.24days

25. (a) M == 100and k == 0.0002;(b) In the year 203526. 50In

\037

\037 5.89months

27.(a) 100ln\037

\037 58.78months; (b) 100ln2\037 69.31months

28. (a) The alligators eventually die out. (b) Doomsday occurs after

about 9 years 2 months.

29.(a) P(140)\037 127.008million; (b) About 210.544million;

(c)In 2000we get P \037 196.169,whereas the actual 2000population was about 281.422million.

31.a \037 0.3915;2.15x 106 cells37.k \037 0.0000668717,M \037 338.02738. k \037 0.000146679,M \037 208.250

39.P(t) = Poexp(kt

+ 2:sin27ft);

the colored curve in the

figure below shows the graph with Po == 100,k == 0.03,andb == 0.06.It oscillates about the black curve which representsnatural growth with Po == 100and k == 0.03.We seethat the two

agree at the end ofeach full year.)

\037)

o)

7. x(t) == 11-4e-281)

77)

15)

o)

P120)

10)

\037 5)

115)

-50 0.1 105t

8.x(t) ==221 t

17-4e9It 1 2 3 4 5

30 Section1.8)

20)

-10o) 0.01

t)

0.02)

1.Approximately 31.5s3.400/(ln2)\037 577 ft

5.400In 7 \037 778 ft

7. (a) 100fUsec; (b) about 23secand 1403ft to reach 90fUsec

8.(a) 100fUsec; (b) about 14.7seeand 830ft to reach 90fUsee

9.50ft/s10.About 5 min 47 s11.Timeof fall: about 12.5s12.Approximately 648ft

19.Approximately 30.46ft/s; exactly 40ft/s20. Approximately 277.26ft22. Approximately 20.67ft/s; about 484.57s23. Approximately 259.304s24. (a) About 0.88em; (b) about 2.91km

25.(b) About 1.389kmlsee;(c)rmax == 1OOR I19\037 5.26R26. Yes

28. (b) After about 8\037

minutes it hits the surface at about 4.116kmlsec.

29. About 51.427km

30. Approximately 11.11kmlsec(ascompared with the earth's

escapevelocity ofabout 11.18kmlsec).)

\037 10)

o)

9.484 10.20weeks11.(b) P(t) ==(\037t

+ 10)2

24012.pet) ==20-t

18013.P(t) ==30_ t

Po14.P(t)-1+ k Pot16.About 27.69months 17.About 44.22months19.About 24.41months20. About 42.12months

200 .21. 6/5\037 153.7milhon

1 + e-)))


Chapter1ReviewProblems1.Linear: y(x) = x3(C+ lnx)2.Separable: y(x) = x/(3-Cx-x In x)3.Homogeneous: y(x) = x/(C-lnx)4. Exact: x2

y3 + eX -cosy = C5.Separable: y(x) = Cexp(x-3-x-2)6.Separable: y(x) = x/(1+ Cx + 2x Inx)7. Linear: y(x) = x-2(C+ Inx)8.Homogeneous: y(x) = 3Cx/(C-x3) = 3x/(1+ Kx 3)9.Bernoulli: y(x) = (x2 + CX-1)210.Separable: y(x) = tan (C+x +

\037X3)11.Homogeneous: y(x) = x/(C-31nx)12.Exact: 3x2y3 + 2xy4 = C

13.Separable: y(x) = 1/(C+ 2X2 -x5)14.Homogeneous: y2 = x2/(C+ 21n x)15.Linear: y(x) = (x3 + C)e-3x

16.Substitution: v = y -x; solution: y -x-I= Ce2x(y -x + 1)

17.Exact: eX + eY + eXY = C18.Homogeneous: y2 = CX 2

(X2 _

y2)19.Separable: y(x) = X2/(X

5 + Cx2 + 1)20. Linear: y(x) = 2X-3/2 + Cx-321.Linear: y(x) = [C+ In(x -1)]/(x+ 1)22. Bernoulli: y(x) = (2x4 + CX 2)323. Exact: xeY + y sin x = C24. Separable: y(x) = X 1 /2/(6x2 + CX 1 /2 + 2)25. Linear: y(x) = (x + 1)-2(x3 + 3x2 + 3x + C)

= x + 1+ K (x + 1)-226. Exact: 3X3/2

y4/3-5X6/5y3/2 = C

27. Bernoulli: y(x) = X-I (C+ Inx)-1/328. Linear: y(x) = X-I (C+ e2x)29. Linear: y(x) = (x2 +x + C)(2x+ 1)-1/230. Substitution: v = x + y; solution:

x = 2(x+ y)I/2 -21n[1+ (x + y)I/2] + C31.Separable and linear

32. Separable and Bernoulli

33. Exact and homogeneous34. Exact and homogeneous35. Separable and linear

36. Separable and Bernoulli)

Chapter2Section2.1)1.y(x) =

\037ex

-\037e-x2.y (x) = 2e3x -3e-3x

3.y(x) = 3cos2x + 4sin 2x4. y(x) = lOcos5x-2sin 5x5.y(x) = 2eX -e2x

6.y(x) = 4e2x + 3e-3x

7. y(x) = 6-8e-X8. y(x) =

\037

(14-2e3x )9.y(x) = 2e-X +xe-X10.y(x) = 3e5x -2xe5x

11.y(x) = 5eX sinx12.y(x) = e-3x (2cos2x+ 3sin 2x)13.y(x) = 5x-2X2

14.y(x) = 3x2-16/x315.y(x)=7x-5xlnx16.y(x) = 2cos(1nx)+ 3sin(lnx)21.Linearly independent)

22.Linearly independent


24. Linearly dependent



28. y(x) = 1-2cosx-sinx

29.There is no contradiction becauseif the given differential

equation isdivided by x2 to get the form in Eq. (8),then the

resulting coefficient functions p(x) = -4/xand q(x) = 6/x2

are not continuous at x = O.)

33.y(x) = clex +C2e2x35.y(x) = CI + C2e-5x

37.y(x) = Cle-x/2+ C2ex

39.y(x) = (CI+ c2x)e-x/241.y(x) = Cl e-4x/3 + C2e5x/2

43. y\" + 10y'= 045. y\" + 20y'+ 100y = 047. y\" = 049. The high point is

(In\037, \0376 ).50. (-In2,-2)53. y(x) = CIX-4+ C2X3

55. y(x) = CI + c2lnx)

34.y(x) = cle-5x + C2e3x

36. y(x) = CI + C2e-3x/2

38.y(x) = cle-x/2 + C2e-3x/2

40. y(x) = (CI+ c2x)e2x/3

42.y(x) = CIe-4x/7 + C2e3x/5

44.y\"- 100y = 0

46.y\"-110y'+ 1000y = 0

48. y\"-2y' -y = 0)

52. y(x) = CIX + C2/X

54. y(x) = CIX-3/2 + C2XI/2

56. y(x) = X2(CI+ c2lnx))

Section2.2)

1.2.3.4.5.6.

13.14.15.16.17.18.19.20.21.22.23.24.

38.39.40.41.42.)

15. (2x)-16. (3x2)-6 . (5x -8x2) = 0

(-4)(5)+ (5)(2-3x2) + (1)(10+ 15x2) = 01.0+ 0 . sin x + 0 .eX = 0

(6)(17)+ (-51)(2sin2x) + (-34)(3cos2x) = 01. 17-34.cos2x + 17 .cos2x = 0

(-I)(eX ) + (l)(coshx) + (1)(sinhx) = 0

y(x) =\037ex

-\037e-2x

y(x) =\037(3eX

-6e2x + 3e3x)

y(x) = (2-2x+ x2)ex

y(x) = -12ex + 13e2x -10xe2x

y(x) =\037(29

-2cos3x-3sin3x)

y(x) = eX (2-cosx-sin x)y(x) = x + 2X2 + 3x3

y(x) = 2x-x-2 +x-2lnx

y(x) = 2cosx-5sin x + 3x

y(x) = 4e2x -e-2x -3

y(x) = e-X +4e3x -2

y(x) = eX (3cosx+4sinx)+x + 1

1Y2(X) =

3\"xY2 (x) = xex/2

Y2(X) = xeX

Y2 (x) = x + 2

Y2(X) = 1+X2)))


Section2.81.y(x) = Cl e2x + C2e-2x2.y(x) = Cl + C2e3x/2

3.y(x) = Cl e2x + C2e-5x4. y(x) = c}ex/2 + C2e3x

5. y(x) = cle-3x +C2xe-3x6.y(x) = e-5x12

[c,exp (\037x.J5)+ C2exp (-\037 x.J5)]

7. y(x) = c}e3x/2 + c2xe3x/28.y(x) = e3x(Clcos2x + C2 sin 2x)9.y(x) = e-4x(Clcos3x + C2sin 3x)10.y (x) = c}+ C2X + C3 X2 + C4e-3x/511.y(x) = Cl + C2X + C3e4x + C4xe4x12.y(x) = Cl + C2ex+ C3xex+ C4x2ex13.y(x) = c}+ C2e-2x /3 + c3xe-2x/314.y(x) = clex + C2e-x+ C3 cos2x+ C4 sin 2x

15.y(x) = Cl e2x + C2xe2x+ C3e-2x+ C4xe-2x16.y(x) = (Cl+ C2X) COS 3x + (C3+ C4X) sin 3x

17.y(x) = c,cos(x/.J2)+ C2sin (x/.J2)+ C3 cos(2x/.J3)+C4sin (2xIJ3)18.y(x) = Cte2x + C2e-2x+ C3 COS 2x+ C4 sin 2x

19.y(x) = Cl eX + C2e-x+ C3 xe-x

20. y(x) =e-xl2

[(C,+ C2X)cosOx.J3)+ (C3+ C3 X) sin Ox.J3)]21.y(x) = 5eX + 2e3x

22.y (x) = e-x13

[3cos(x/.J3)+5.J3sin (x/.J3)]23.y(x) = e3x(3cos4x-2sin 4x)24. y(x) =

\037

(-7+ e2x + 8e-x/2)25. y(x) = i (-13+ 6x+ ge-2x /3)26. y(x) =

k (24-ge-5x -25xe-5x )27.y(x) = C} eX + C2e-2x + C3 xe-2x

28.y(x) = cle2x + C2 e-x + C3e-x/2

29.y(x) = c,e-3x +e3xI2[C2

cosOx.J3)+ C3 sin(\037x.J3)]

30.y(x) = cle-x + C2e2x + C3 cos(xJ3)+ C4 sin(xJ3)31.y(x) = clex + e-2x(C2cos2x+ C3 sin 2x)32.y(x) = Cte2x + (C2+ C3X + c4x2)e-x33.y(x) = Cl e3x + e-3x(C2cos3x+ C3 sin 3x)34. y(x) = Cl e2x/3 + C2 COS 2x+ C3 sin 2x35. y(x) = Cl e-x/2 + C2e-x/3+ C3 COS 2x+ C4sin 2x36.y(x) = cle7x/9 +e-X (c2cosx+ C3 sin x)37.y (x) = 11+ 5x+ 3x2 + 7eX

38.y(x) = 2e5x -2coslOx

39.y(3)-

6y\" + 12y'-8y = 040. y(3)

-2y\" + 4y'-8y = 0

41.y(4)- 16y = 0

42. y(6) + 12y(4)+ 48y\" + 64y = 044. (a) x = i, -2i (b) x = -i,3i45. y(x) = C} e-ix + C2e3iX

46. y(x) = cle3ix + C2e-21X

47. y (x) =c,exp ([1+ i.J3]x) + C2 exp (-[1+ j.J3] x)48. y(x) =

t (ex+ exp[\037 (-1+ i.J3)x] + exp

[\037 (-1-i.J3)x])49. y(x) = 2e2x -5e-X + 3cosx-9sin x52. y(x) = C} cos(3Inx)+ C2 sin(3Inx)53. y(x) = X-3

[Cl cos(41nx)+ C2sin(41nx)]54. y(x) = Cl + c21nx + C3X-3)

55. y(x) = c}+ X2(C2+ c31nx)56. y(x) = Cl + c21nx+ c3(lnx)2

57. y(x) = Cl + x3(C2X-J3+ C3X+J3)

58. y(x) = X-1[Cl + c2lnx+ c3(1nx)2])

Section2.4)

1.Frequency: 2 radls (1/nHz); period: n s2.Frequency: 8 radlsec (41nHz);period: n 14sec3.Amplitude: 2 m; frequency: 5rad/s;

period: 2n 15s4. (a) x(t) =

\037\037

cos(12t-a) with

a = 2n -tan- 1 (5/12)\037 5.8884;(b) Amplitude:

\037\037

m; period: n 16sec6.About 7.33mi

7. About 10450ft8.29.59in.

10.Amplitude: 100cm; period: about 2.01sec11.About 3.8in.

13.(a) x(t) = 50(e-2t /5-e-r/2); (b) 4.096exactly

14.(a) x(t) = 25e-t/5cos(3t-a) with a = tan- 1 (3/4)\037 0.6435;(b) envelope curves x = ::I::25e-r/5; pseudoperiod 2n 13

15.x(t) = 4e-2t -2e-4t , overdamped; u (t) = 2cos(2J2t))

2)

t)

-2)

16.x(t) = 4e-3t -2e-7t , overdamped;

u(t) \037 2/\302\245I

cos(.J21t -0.2149))

2)

t)

-2)

17.x(t) = 5e-4t (2t + 1),critically damped;u(t) \037 \037,J5cos(4t-5.8195))

5)

t)

-5)))


18.x(t) = 2e-3t cos(4t - 3;),underdamped;u (t) =

\037

cos(5t _ 3;))

t)

-1)

19.x(t) \037

\037.J3T3e-5t /2cos(6t-0.8254),underdamped;

u(t) \037

I\037,J233cos( Ii t -0.5517))

4)

t)

-4)

20. x(t) \037 13e-4t cos(2t-1.1760),underdamped;

u(t) \037

I\302\245cos(2,J5t -0.1770)

5)

-5)

t)

21.x(t) \037 10e-5t cos(10t-0.9273),underdamped;u(t) \037 2.J14cos(5,J5t -0.6405))

6)

-6)

22. (b) The time-varying amplitude is\037,J3,

the frequency is4,J3rad/ s,and the phase angle isn /6.

23. (a) k \037 7018Ib/ft; (b) After about 2.47s34. Damping constant: C \037 11.511b/ft/s;spring constant:

k \037 189.681b/ftSection2.5)1.Yp(x)

= 21

5e3x

2.y p (x) = -\037

(5+ 6x)3.Yp(x) =

3\037(cos3x -5sin 3x)

4. Yp(x) =\037(-4eX + 3xeX )

5. Yp(x) =2\037

(13+ 3cos2x-2sin 2x)6.Yp(x)

=3\0373

(4-56x+ 49x2)7. Yp(x)

= -i(eX -e-X ) =-\037

sinh x8.Yp(x) =

\037xsinh 2x

9.Yp(x) = -\037

+I\037

(2x2-x)eX)

10.11.12.13.14.15.16.17.18.19.20.21.22.23.24.25.26.27.28.29.30.31.32.33.34.35.36.37.38.39.40.41.42.)

Yp(x) = i (2x sin 3x -3x cos3x)yp(x) = k(3x

2-2x)yp(x) = 2x+ 4x sin xyp(x) =

\0375eX(7 sinx -4cosx)

yp(x) =2\037

(-3x2ex + x3ex)yp(x) = -17yp(x) =

8\\ (45+ e3x -6xe3x + 9x2e3x)yp(x) =

\037

(x2sin x -x cosx)yp(x) = -

I\037(24xeX -19xe2x + 6x2e2x)

yp(x) = k(10x2-4x3 +x4)

yp(x) = -7+\037xex

yp(x) = xeX(A cosx+ Bsinx)yp(x) = Ax3 + Bx4 + Cx5 + Dxex

yp(x) = Ax cos2x+ Bx sin 2x + Cx2cos2x + DX2 sin 2xyp(x) = Ax + Bx2 + (Cx+ Dx2)e-3x

yp(x) = Axe-x + Bx2e-x + Cxe-2x + Dx2e-2x

yp(x) = (Ax + Bx2)e3x cos2x+ (Cx+ Dx2)e3x sin 2xyp(x) = Ax cosx+ Bx sinx + Cxcos2x+ Dxsin 2x

yp(x) = (Ax + BX2 + Cx3)cos3x+ (Dx+ Ex2 + Fx3) sin3x

yp(x) = Ax3ex + Bx4ex + Cxe2x + Dxe-2x

yp(x) = (A + Bx + Cx2)cosx+ (D+ Ex + FX2) sinx

y (x) = cos2x+\037

sin 2x+ 4xy(x) = i (15e-X -16e-2x + eX)

y(x) = cos3x -1

25 sin 3x + k

sin 2xy(x) = cosx-sinx + 4xsin xy(x) = eX(2cosx-

\037

sin x) + 4x+ 1y(x) =

1\0372(234+ 240x-ge-2x -33e2x - 12x2-4x4)

y(x) = 4-4ex + 3xex +x - 4x2ex + ix3exy(x) =

815[e-X (176cosx + 197sin x) -(6cos3x + 7 sin 3x)]

y(x) = -3+ 3x - 4X2 + iX3 +4e-x +xe-Xy(x) =

\037(5e-X+ 5ex + lOcosx -20)

yp(x) = 255-450x+ 30x2 + 20x3 + 10x4-4x5

y(x) = 10e-x + 35e2x + 210cosx+ 390sinx+ yp(x) where

y p (x) is the particular solution ofProblem 41.(b) y(x) = CI COS 2x+ C2sin 2x

I I 3+ - cosx --cos X4 20

y(x) = e-x/2[Cl

cos(4xv'3)+ Czsin (4xv'3)] +2\037

(-3cos2x+ 2sin2x)+4\0372

(15cos4x+4sin4x)y(x) = CI cos3x+ C2 sin3x +

2\037

-I\037

cos2x-5\037

cos4xy(x) = CI cosx+ C2sinx + /6(3xcosx+ 3x2sinx) +

1\0378(3sin 3x -4xcos3x)

yp(x) =\037ex

yp(x) = -I\037

(6x + l)e-2X

yp(x) = x2e2x

yp(x) =I\037

(4xcosh2x -sinh 2x)yp(x) = -

\037

(cos2xcosx -sin 2xsin x) + 21

0(cos5xcos2x +sin 5xsin 2x)= -

kcos3x (!)

yp(x) = -ixcos3x

y p (x) =\037

x sin 3x +\037

(cos3x)In I cos3x I

y p (x) = -1-(cosx) In I cscx -cotx I

yp(x) =k (1-x sin 2x)

yp(x) = -\037ex

(3x + 2)yp(x) = x3(1nx -1)yp(x) =

\037X4

yp(x) = -752X4/3

yp(x) = Inx

1

1+xI

1yp(x) = -x2+xln-+ -(1+x2)Inll-x21I-x 2)

43.)

44.)

45.46.)

47.48.49.50.51.)

52.53.54.55.56.58.59.60.61.62.)))

Section2.6)1.x(t) = 2cos2t -2cos3t)

t)

-3)

2.x(t) =\037

sin 2t -sin 3t)

I-)

21t)-I)

t)

3.x(t) = .J138388cos(10t-a) + 5cos(5t-fJ) witha = 2n -tan- 1 (1/186)\037 6.2778and

fJ = tan- 1(4/3)\037 0.9273.)

375)

-375)

4. x(t) = 2,JT06cos(5t-a) + 10cos4t witha = n -tan- 1(915)\037 2.0779)

30) I_21t

_,)

t)

-30)

5. x(t) = (xo-C)coswot + Ccoswt,where C= Fo/(k -m(2)7. xsp(t) ==

:\037cos(3t-a) with a == n -tan- 1 (12/5)\037 1.9656)

3)

t)

-3)

8.xsp(t) ==2\037

cos(5t-a) with a = 2n -tan- 1(3/4)\037 5.6397)


-1)

9.Xsp (t) ==

v'4\037OOIcos(lOt -a) with

a = n + tan- 1(199/20)\037 4.6122)

0.1)

XI\037,. \037\037,..,. \037\037,..\"'\037 t\037 \037 \037 \037 \037 \037 \037 \037 \037

\0371\0371t)

\037) ,)

-0.1\"\\ \\

FI

10.xsp(t) =719\037

-J6Tcos(10t-a) with

a = n + tan- 1 (171/478)\037 3.4851)

\\)

1)

x)

ft

r\037 ft ft

h \" Ir\\ \" \" \" \" \" \" \"\037

v v V i'\" v v v v Vl\037

n

\037

Fl\037

\037 \037 ,)

t)

-1)

11.xsp(t) =\037 cos(3t-a) with a == n -tan- 1 (3) \037 1.8925

xtr(t) =\037.J2

e-2t cos(t-fJ) with fJ = 2n -tan- 1 (7) \037 4.8543)

0.5)

t)

-0.5)

12.xsp(t) =3\037

cos(3t-a) with a = n + tan- 1 (2/5)\037 3.5221xtr(t) =

6\037e-3t cos(2t-fJ) with fJ = tan- 1 (512)\037 1.1903

0.5)

t)

-0.5)

13.xsp(t) ==v'31\037\0379

cos(10t-a) with

a = n -tan- 1 (10/37)\037 2.9320xtr(t) = 2 I

\037\037\037\0374

e-t cos(5t-fJ) with

fJ = 2n -tan- 1(421/12895)\037 6.2505)))


10)

t)

-10)

14.xsp(t) = .J485cos(t-a) with a = tan- 1(22) \037 1.5254xtr(t) = .J3665e-4t cos(3t-fJ) with

fJ = n + tan- 1 (52/31)\037 4.1748)

t)

-30)

15.C(w) = 2/.J4 + w4; there is no practical resonance frequency.)

c)

1)

w10)5)

16.C(w) = 10/.J25+ 6w2 + w4; there is no practical resonancefrequency.)

c)

2)

1)

5)

w10)

17.C(w) = 50/.J2025-54w2 + w4; there ispractical resonance at

frequency w = 3,J3.)

C)

w)

10) 20)

18.C(w) = 100/.J422500- 1200w2 + w4; there is practicalresonance at frequency w = 10,J6.)

C0.4)

w)

25 50

19.w = .J384rad/sec (approximately 3.12Hz)

20. w \037 44.27rad/sec (app roximately 7.05Hz)21.Wo = .J(g/L)+ (k/m )22.Wo = Jk/(m + I/a2)23.(a) Natural frequency: .JIOrad/s (approximately 0.50Hz);

(b) amplitude: approximately 10.625in.)

Section2.7)

1.I(t) = 4e-5t

2.I(t) = 4(1-e-5t )3.I(t) = 1:5(cos60t + 12sin 60t -e-5t )4. I(t) = 5(e-lOt -e-20t ); Imax = I( /0 In 2) = 5/4.5. I(t) =

\037e-lOtsin 60t

6.Isp(t) =3\037(-21cos60t + 22sin 60t) = (5/../37)cos(60t-a),

where a = n -tan- 1(22/21)\037 2.3329.7. (a) Q(t) = EoC(1-e-t/RC );I(t) = (Eo/R)e-t/RC

8.(a) Q(t) = 10te-5t ; I(t) = 10(1-5t)e-5t ;(b) Qmax = Q(I/5)= 2e-1.

9.(a) Q(t) = (cos120t+ 6sin 120t-e-20t )/1480I(t) = (36cos120t-6sin120t+e-20t )/74(b) Isp

=7\037

(6cos120t-sin 120t)= k cos(120t-a) with

a = 2n -tan -1.!..611.Isp(t) = a sin(2t -8)with 8 = 2n -tan- 1(1/6)\037 6.118012.Isp(t) = lD sin(10t -8)with 8 = 2n -tan- 1 (1/4)\037 6.038213.Isp(t) =

\037sin(5t -8)with 8 = 2n -tan- 1(2/3)\037 5.6952

14.Isp(t) \037 0.9990sin(100t -0.8272)15.Isp(t) \037 0.1591sin(60nt -4.8576)16.Isp(t) \037 1.6125sin(377t -1.2282)17.I(t)=-25e-4t sin3t

18.I(t) =15

7\302\2601

(1ge-lOt -18e-20t -e-t)19.I(t) = 10e-20t -10e-10t -50te-lOt

20.I(t) = -37ffi

e-3t /2(v'TIcostv'TI/2 + 27 sin t v'TI/2)+\037\037

(cos2t + 6sin 2t))

IIIsp)

-5)

t21t)1t)

21.I(t) = -\037\037

e-t(12cos3t+47sin3t)+\037\037

(2cos5t+3sin5t))))

3.I

5 4.

t 5.-5)

22.I (t) \037 _e-25t (0.1574cos25t,JT59+ 0.0262sin 25t,JT59)+(0.1574cos60nt+ 0.0230sin60nt))

I)

0.2)

t)

-0.2)

Section2.8)1.Only positive eigenvalues {n

2n 2/4}and associated

eigenfunctions {cos(nnx12)}for n == 1,3,5, . . . .2.Eigenvalue Ao == 0 with eigenfunction Yo (x) = 1,and positive

eigenval ues {n2

} with associated eigenfunctions {cosnx} forn == 1,2,3, . ...

3.Only positive eigenvalues {n2/4}for n == 1,2,3,.. . . The nth

eigenfunction Yn (x) is cos(nx12)if n isodd, sin(nx 12)if n iseven.

4. Eigenvalue Ao == 0 with eigenfunction Yo(x) = 1,and positiveeigenvalues {n

2/4}for n == 1,2,3, .... The nth eigenfunctionYn (x) is sin(nx 12)if n is odd, cos(nx12)if n iseven.

5. Only positive eigenvalues {n2n 2/64}for n == 1,2,3, . . . . The

nth eigenfunction Yn(x) iscos(nnx/8)+ sin(nnx/8) if n isodd, cos(nnx18)-sin(nn x18)if n iseven.

7-8.In the figure below, points of intersection of the curve Y == tan zwith the lines Y == xz are labeled with their z-coordinates. Weseethat an liesjust to the right of the vertical line

z == (2n - l)n12,while fJn liesjust to the left of the line

z == (2n + 1)n 12.)

Y15.

10 16.17.

z 18.19.

-1020.

Chapter3Section3.1 21.)

1.Y(X)==co (I+X+ x2 + X3 +...) ==coex;p==+oo2 3!

(4x 42x2 43x3 44x4

)2.y(x) == Co 1+-+-+-+-+...== coe4x .I! 2! 3! 4! '

p==oo)

AnswerstoSelectedProblems 587)

(3x (3X)2 (3X)3 (3X)4

)Y (x) == Co 1-2+ 2!22-

3!23+ 4!24-...==

coe-3xj2 ; p == +00

(X2 x4 x6

)?

y(x) == Co 1- IT+ 2!-

3T+ . . . == coe-x-;p = 00

y(x) == Co(1+ x3+ \037 + \037 +...) == coexp(.!.x

3).3 2!32 3!33 3 'P == +00

(X x2 x3 x4

)2co6.y(x) == Co 1 +

\"2+ 4+

8\"+ 16+ . . . ==

2 _ x ;p = 2

2 3 Co I7. Y (x) == Co(1+ 2x + 4x + 8x +...)==2 ; p = 2

1 - x

(X x2 x3 5x4

)8.y(x) == Co 1+ 2-

8\"+ 16- 128+... == co\037;

p==1Co9. Y (x) == Co(1+ 2x+ 3x2+ 4x3 + . . . ) ==

(1_ x)2; P == 1

(3x 3x2 x3 3x4

)y(x) == Co 1 -2+8+ 16+ 128+... = co(1-x)3/2;

p==1

y(x) = Co(1+\037:

+::+ \037\037 +...)+cI (X + x3

+ x5+ x7 +...)3! 5! 7!

== Cocosh x + CI sinh x; p == +00

((2X)2 (2X)4 (2X)6

)y(x) == Co 1+2!+4!+6!+... +CI

((2x)3 (2x)5 (2x)7

)\"2(2x) +3!+51+7!+ . . . ==

Cocosh2x+ \037 sinh 2x;p == 002

((3X)2 (3X)4 (3X)6

)y(x) ==Co 1-2!+4!-6!+...+\037

(3X _ (3X)3 + (3X)5 _ (3X)7 +...)3 3! 5! 7!== Cocos3x+ kc, sin 3x;p == +00

y(x) = x + Co(1-

\037:

+::- \037\037 +...)+ (CI_ 1)(X _ x3

+ x5_ x7 +...)3! 5! 7!

== x + Cocosx+ (CI-1)sin x; p == 00

(n + l)cn == 0 for all n\037 0,soCn == 0 for all n

\037o.

2ncn == Cn for all n\037 0, soCn == 0 for all n

\037o.

Co== CI == 0 and Cn+ I== -ncn for n

\037 1,thus Cn == 0 for all

n\037

O.Cn == 0 for all n

\0370

(n + 1)(n + 2)Cn+2 == -4cn ;3[

(2x)3 (2x)5 (2x)7

]3.

y(x) == - (2x) --+---+... == -sln2x2 3! 5! 7! 2

(n + 1)(n + 2)Cn+2 == 4clI ;

[(2X)2 (2X)4 (2X)6

]y(x) == 2 1+-+-+-+...= 2cosh2x2! 4! 6!

n(n + 1 )Cn+1 == 2ncn -Cn-I;x3 x4 x5

y(x) == x +x2 +-+-+-+...== xeX

2! 3! 4!n(n + l)cn+1 == -ncn + 2Cn-l;y == e-2x

As Co== CI == 0 and (n2-n + l)cn + (n - l)cn-1== 0 for n

\037 2,Cn == 0 for all n

\0370)

10.)

11.)

12.)

13.)

14.)

22.23.)))

Section3.2)00 00

1 ( ) L 2n L 2n+ 1 Co+ Cl x. Cn+2 == Cn ; Y x == Co x + CI x == ; p == 11-x2

n=O n=O)

2.Cn+2 == -\037cn; P == 2;

00 (_I)nx2n 00 (-I)nx2n+1y(x) ==

Co'\"\"' + CI

'\"\"'\037 2n \037 2nn=O n=O)

3.(n + 2)Cn+2 == -cn ;00

(_I)n x2n 00(_I)nx2n+l

Y(X) = Co\037 n!2\"

+ Cl\037 (2n + I)!!; P = +00

4. (n + 2)Cn+2 == -en+ 4)cn ; P == 1;00 00

y(x) == CoL(-I)n(n+ l)x2n + kCI L (-It(2n+ 3)X2n + 1

n=O n=O)

00X2n+ 1

5. 3(n + 2)Cn+2 == ncn ; P == ,J3;y(x) == Co+ Cl Ln=O (2n + 1)3n

6.(n + 1)(n+ 2)Cn+2 == (n -3)(n -4)cn ; P == 00;y(x) == co(1+ 6x2 +x4) + CI (x + x3)

7. 3(n + 1)(n + 2)Cn+2 == -en-4)2Cn ;

Y(x) = Co(1_

8\0372+

\0374 ) +

(X3 x5 00 (-I)n(2n-5)!!X2n+l

)Cl x -2+ 120+ 9

\037 (2n + 1)!3\"

8.2(n + 1)(n + 2)Cn+2 == (n -4)(n + 4)cn ; y(x) == co(1-4x2 +4

(5X3 7x5 00 (2n-5)!!(2n+3)!!X2n +l

)2x ) + CI x --+-+ '\"\"'

4 32 \037 (2n + 1)!2n

9.(n + 1)(n+ 2)Cn+2 == (n + 3)(n + 4)cn ; P == 1;00 C 00

y(x) == CoL(n+ 1)(2n+ l)x2n + -iL(n+ 1)(2n+ 3)X2n+ 1

n=O n=O)

10.3(n + 1)(n+ 2)Cn+2 == -en-4)cn ;

Y(x) = Co(1+2\0372

+\037; ) +

(X3 x5 00 (-I)n(2n_ 5)!!X2n + 1

)Cl x +

6\"+ 360+ 3

\037 (2n + 1)!3\"

11.5(n + 1)(n+ 2)Cn+2 == 2(n -5)cn ;

Y(X)=Cl(X-\037x;+

;;\037 )+

(2 x4 X6 00

(2n -7)!!2nX2n

)Co 1-x +-+-+ 15'\"\"'

10 750 \037 (2n)!5n)

12.C2 == 0;(n + 2)Cn+3 == Cn ;

y(x) == Co(1 +f X3\"

)+ CI

\037X3n +1

n=12.5...(3n -1) f1n!3n)

13.C2 == 0; (n + 3)Cn+3 == -Cn ;00 (-1)nx3n 00 (-1)n x3n+ 1

y(x) == CoL + CI'\"\"'

n=O n!3n f11.4...(3n + 1)

14.C2== 0; (n + 2)(n + 3)Cn+3 == -Cn ;

(00 (-1)nx3n

)y(x) == Co 1+L +

n=13n . n!.2. 5 ... (3n -1)

00 (-I)nx3n+1Cl

\037 3\" . n!.1.4...(3n + 1))

15.C2== C3 == 0; (n + 3)(n + 4)Cn+4 == -Cn ;

(00

(_I)n x4n

)y(x) == Co 1 +L +

n= 1

4n . n! . 3 . 7 ...(4n -1)00 (-I)nx4n+1clL

n=O4n .n! . 5 .9. . . (4n + 1)

16.y(x) == x

17.y(x) == 1+X200 (-I)n(x _

1)2n18.y(x) == 2 '\"\"' ; converges for all x\037 n!2nn=O)

00

19.y(x) ==k L (2n + 3)(x- 1)2n+l;converges if 0 < x < 2

n=O)

20.21.22.23.)

y(x) == 2-6(x-3)2;converges for all x

y(x) == 1+ 4(x+ 2)2;converges for all x

y (x) == 2x+ 6

2C2+ Co== 0; (n + 1)(n + 2)Cn+2 + Cn + Cn-I == 0 for n\037 1;

x2 x3 x3 x4YI(x) == 1-2-

6\"+...;Y2(X) == x -

6\" -12+...)24.)

x3 x5 x6YI (x) == 1+-+-+-+ . . ..

3 5 45 'x3 x4 x5

Y2(X) == x +-+-+-+ . . .365C2 ==

C3 == 0, (n + 3)(n +4)Cn+4 + (n + l)cn+1 + Cn == 0 forx4 x7

n \037 O.YI (x) == 1--+-+ ....- , 12 126 'x4 x5

Y2(X) == x ----+ . ..12 20

(X6 x9

) (X7 X10

)y(x) == Co 1--+-+ ... + CI x --+-+ . . .30 72 42 90

x2 x3 x4 x5 29x6 13x7 143x8y(x) == l-x-2+3-24+ 30+ 720 + 630

-40320+...;

y(0.5) \037 0.4156)

25.)

26.)

27.)

28.) (X2 x3

) (X3 x4

)y(x) == Co 1--+-+ . .. + CI x --+-+ . . .2 6 6 12

1 2 1 6YI(x)==I--x+-x+....2 720 '1 3 1 5

Y2(X) == x --x --x +...6 60

(X2 x3

) (X2 x4

)y(x) == Co 1--+-+ . . . + CI x --+-+ . . .2 6 2 18

The following figure shows the interlaced zerosof the 4th and5th Hermite polynomials.)

29.)

30.)

33.)

Y)

-3) 3x)

34. The figure below results when we use n == 40terms in eachsummation. But with n == 50we get the samepicture asFig. 3.2.3in the text.)))


Y)

(00 (-1)nx2n

)24. YI(X) =X I/3 1+\" ,

\037 2n . n!.7.13...(6n + 1)00

(_I)n x2n

Y2(X) = 1+L 2n .n!.5.11...(6n -1)n=1)

x)

25.)00 (-I)nxn

Y I (x) = X1/2 \" = X1/2 e-x/2,\037 n!.2nn=O00 (-1)nxn

yz(x) = 1+\037 (2n _ I)!!)Section3.3)

1.Ordinary point 2.Ordinary point

3.Irregular singular point 4. Irregular singular point

5. Regular singular point; rl = 0, r2 = -16.Regular singular point; rl = 1,r2 = -27. Regular singular point; r = -3,-3

8.Regular singular point; r =\037,

-3

9.Regular singular point x = 1

10.Regular singular point x = 1

11.Regular singular points x = 1,-112.Irregular singular point x = 2

13.Regular singular points x = 2,-214.Irregular singular points x = 3,-315.Regular singular point x = 2

16.Irregular singular point x = 0, regular singular point x = 1

17.YI (x) = cos,JX,Y2(X) = sin,JX)

26.)00 x2n

YI (x) = XI/2 \" = XI/2 exp (\037X2),\037 n!.2nn=O00

2n x2nyz(x) = 1+

\037 3 .7...(4n - 1))

1 1 .27. YI (x) = -cos3x,Y2(X) = - sin 3xx x)

Y)

1 Y2)

x)

-1YI)

00 xn00 xn

18 (x) -\"Y (x) -X- I/2\"

\302\267 YI -\037 n!(2n + I)!!'2 -

\037 n!(2n - I)!!)11.228. YI (x) = -cosh2x,Y2(X) = - sinh xx x)

19.)(

00 n

)_ 3/2

XYl (x) -x 1+ 3

\037 n!(2n + 3)!!'00 xn

Yz (x) = 1-x -L I (2 _ 3)II2 n. n ..n=)

8)

Y)

20.)00 (-I)n2nxn

YI (x) = XI/3\" ,\037 n!.4.7...(3n + 1)

00 (-I)n2nxn

Y2(X) = \"\037 n!.2 .5...(3n -1)

(00 x2n

)YI (x) = x 1+ ,

\037 n!.7 .11...(4n + 3)

(00 x 2n

)Y2(X) = X- I/2 1+

\037 n!.1.5...(4n -3)

(00 (-1)nx2n

)Yl (x) = x3/Z 1+

\037 n!.9.13...(4n + 5),

-I(

00(_I)n-Ix2n

)Y2(X) = x 1+

\037 n! . 3 .7 ...(4n -1)

(00 2n

)1/2

X

YI (x) = x 1+ ,\037 2\" .n!.19.31...(12n+ 7)

(00

x2n

)Y2(X) = X-2/3 1+

\037 2\" .n!.5 .17...(l2n -7))

6)

4)

2)Y2)

x)

21.) 1) 2)

1 x 1 x29.YI (x) = -cos-,Y2(X) = - sin -x 2 x 2)

22.)Y)

1)

41t)

0.5)

23.)x)

-0.5)))


Y)

1 00 (_1)n-13nxn7. Yl (x) = 2(2-6x+ 9x2), Y2(X) = L ( 2)'x n=l n + .)

30. YI(X) = COSX2, Y2(X) = sinx2)

1)

x48.YI(x)=3+2x+X2,Y2(X)= (l-x)2)

x)

x2 x4 x69.YI (x) = 1+ 22 + 22 .42 + x2 .42 . 62 + .. . ,

(X2 5x4 23x6

)Y2(X) = YI (x). lnx -\"4

+ 128-

3456+...)

31.YI (x) = XI/2 coshx,Y2(X) = XI/2 sinhx)

10.Yl (x) = x (I-

\037:

+ 2t42-

22 .=:.62 +...).(

X2 5x4 23x6

)Y2(X) = YI (x). lnx +\"4

+ 128+ 3456+...)

Y)

1)

11.Yl (x) = x2

(I_ 2x+

3\0372

_\0373 +...).

(l1x2 49x3

)Y2(X) = YI (x). lnx + 3x+4+18+ ...)

x)

(X 3x2 x3 x4

)12.YI (x) = x2 1-

2\"

+ 20-

30+ 168-...,

Y2 (x) = YI (x) .(-3\0373+

2\037X+

7\037+ ...

))

34.)

x2YI (x) = x +-,5

(5x 15x2 5x3

)Y2(X) = X- I/2 1-2-8-48 +...

1

(2 10x3

)YI (x) = x- 1+ lOx + 5x +9+ ...,

1 2

(llx l1x2 671x3

)Y2(X) =x/I+---+ + ...20 224 24192

(X2 x4

)YI (x) = x 1--+-+ ...42 1320 '

(7X2 19x4

)Y2(X) = X- I/2 1--+-+...24 3200)

3

(2 4x3

)13.YI (x) = x 1-2x+ 2x -3+ ...,

Y2(X) = YI (x) .(2lnx _ \037 _ \037 + 4x + ...

)2X2 X 3)

32.)

33.) (2x x2 2x3 x4

)14.YI (x) = x2 1-5+ 10

-105+ 336

-...,

(1 1 1 13

)Y2(X) = YI(X). -4x4

+15x3

+100x2

-1750x

+...;Y2 (x) contains no logarithmic term.)

Section3.4)

(00 (-1)nx2n

)16.YI (x) = X3/2 1+ \" ,

\037 2n .n!.5.7...(2n + 3)

(00 (-1)nx2n

)Y2(X) = X-3/2 1+ \"

\037 2n .n! . (-1).1. 3. ..(2n -3))

1+x 00 xn1.Yl (x) = -:;Z'Y2(X) = 1+ 2\037 (n + 2)!

1

(X2 x3

)00 xn2.Yl(X) =

x2 l+x+2+(j 'Y2(X)=1+24\037(n+4)!3.Yl (x) =

\0374 (I_ 3x+

9\0372

_9\0373 ).00 (-I)n3nxn

Y2(X) = 1+ 24\037 (n +4)!

1(

3x 9x2 9x3 27x4

)4. YI (x) =

x5 1-5+ 50-

250+ 5000 '00 (-I)n3nxn

Y2(X) = 1+ 120\037 (n5)!.5n

3x x2 x35. YI (x) = 1+ 4 +

\"4+ 24'

Y2(X) = x5

(I + 120f:(n + I

\037\037n )n=l (n + ).4

(8

\037 (2n +5)!!Xn

)6.Yl(X)=X

1+5\037n!(n+4)!2n)

Section3.52.

0.5

x

-D.5 J-I12(x)

4. J312(x)0.5

x

-0.5)

1 85. J4(x) =\"2 (x2-24)Jo(x) +

\"3\"(6-X2)J1(x)

X X)))

12.Thefollowing figure corroborates the calculated value y(O)= 3.)

y)

x)

x2JI(x) +xJt(x)-f Jo(x)dx+ C

(x3-4x)Jl (x) + 2X2 Jo(x) + C

(x4-9x2)JI (x) + (3x3-9x)Jo(x) +9f Jo(x)dx + C

16.-XJI(x) + f Jo(x)dx + C

2xJl (x) -x2Jo(x) + C3x2JI(x) + (3x-x3)JO (x) -3f Jo(x)dx + C

(4x3-16x)Jl (x) + (8x2-x4)JO (x) + C

-21,(x) + f Jo(x)dx + C4

Jo(x) - -Jl(x)+ Cx)

13.14.15.)

17.18.19.20.)

21.26.)

0.3)

x)

-0.3)

Section3.6)1.2.3.4.5.6.7.8.9.

10.11.12.)

y(x) = X [Cl Jo(x) + C2YO(X)]1

y(x) = -[Cl Jl (x) + C2Yl (x)]x

y(x) = X [ClJl/2(3x2) + C2J_l/2(3x2)]y(x) = x3

[ClJ2(2xl/2) + C2Y2(2xl/2)]y(x) = X- I/3 [Cl Jl/3 (\037X3/2)

+ C2J-l/3(\037X3/2)]

y(x) = X- l/4[ClJO (2X

3/2) + C2YO (2X3/2)]

y(x) = X-I [Cl Jo(x) + C2YO(X)]

y(x) = x2[ClJl (4X

l /2) + C2Yl (4Xl/2)]

y(x) = Xl/2 [ClJl/2(2x3/2) + C2J_l/2(2x3/2)]

y(x) = X- l/4 [ClJ3/2 (\037X5/2)+ C2J-3/2 (\037X5/2)]

y(x) = Xl/2 [ClJl/6 (\037X3)+ C2 J-l/6(\037X3)]

y(x) = Xl/2 [ClJl/S (\037XS/2)+ C2J-l/5(\037X5/2)])

Chapter4Section4.1)1.lis2, s > 0 2.21s3, S > 03.e1(s-3), s > 3 4. s1(s2 + 1),s > 05. 11(S2-1),s > 16.

\037

[lis-s1(s2 +4)], s > 07. (1-e-S)ls,s > 08.(e-S -e-2S )ls,s > 0 9.(1-e-S-se-s)ls2, s > 0

10.(s-1+ e-s)ls2, s > 011.

\037,JJrS-3/2+ 3s-2, S > 0

12.(45n-192s3/2)/(8s7/2), s > 0)


13.S-2-2(s-3)-1,S > 3

14.3,JJr/(4s5/2) + I/(s+ 10),s > 0

15.S-l+ S(S2-25)-1,S > 5 16.(s+ 2)/(S2+4), s > 0

17.cos22t =\037(1

+cos4t);\037 [S-l+ SI(S2+ 16)],s > 0

18.3/(S2+ 36),s > 0

19.S-l+ 3s-2 + 6s-3 + 6s-4, s > 020. I/(s-1)2,s > 1 21.(S2-4)/(S2+4)2, S > 0

22.\037 [SI(S2-36)-S-l] 23.

\037t3

24. 2,Jtln 25.1- \037t3/2n-l/2

26.e-5t 27.3e4t

28.3cos2t +\037

sin 2t 29.\037

sin 3t -3cos3t

30. -cosh2t -\037

sinh 2t 31.\037

sinh 5t -10cosh 5t

32.2u(t -3)37. f(t) = 1-u(t -a).Your figure should indicate that the graph

of f contains the point (a, 0),but not the point (a, 1).38.f(t) = u(t -a)-u(t -b). Your figure should indicate that the

graph of f contains the points (a, 1)and (b, 0),but not the

points (a, 0) and (b, 1).39.Figure 4.2.8shows the graph of the unit staircase function.)

Section4.2)

1.x(t) = 5cos2t) 2.x(t) = 3cos3t +\037

sin 3t

4. x(t) =\037

(7e-3t -3e-St )

6.x(t) =\037

(cost -cos2t)

8.x(t) =\037

(1-cos3t))

3.x(t) =\037

(e2t -e-t)5.x(t) =

\037

(2sin t -sin 2t)

7. x(t) =k (9cost -cos3t)

9.x(t) = i (2-3e-t +e-3t )

10.x(t) =\037

(2t -3+ 12e-t -ge-2t)11.x(t) = 1,y(t) = -212.x(t) =

\037

(e2t -e-t -3te-t), y(t) =\037

(e2t -e-t +6te-t)

13.x(t) = -(2/\037)

sinh(t/\037),

y(t) = cosh(t/\037)

+ (1/\037)sinh

(t/\037)

14.x(t) =:i (2t -3sin 2t), y(t) = -

k (2t + 3sin 2t)

15.x(t) =\037 (2+e-3t /2[cos(rt12)+ r sin(rt 12)]),

y(t) =2\\ (28-get + 2e-3t /2[cos(rt12)+4r sin(rt 12)])where

r=\037)

16.x(t) = cost + sin t, Y (t) = et -cost, Z(t) = -2sin t

17.f(t) =\037

(e3t -1)

18.f(t) =\037(1

-e-St )

19.f(t) =\037

(1-cos2t) =\037

sin2t

20.f(t) =\037

(6sin 3t -cos3t + 1)21.f (t) = t -sin t

22. f(t) = i(-1+ cosh3t)23. f(t) = -t+ sinh t

24. f(t) =\037(e-2t

-2e-t + 1))))


Section4.31.24/(s-n)s 2.\037,JJr (s +4)-S/23.3n I[(s+ 2)2+ 9n2] 4. ,J2(2s+ 5)1(4s2+4s+ 17)5.

\037e2t6.(t -t2)e-t

7. te-2t 8.e-2t cost9.e3t (3cos4t+ \037 sin4t) 10.

316e2t/3 (8cos\037t

-5sin\037t)11.

\037

sinh 2t 12.2 + 3e3t

13.3e-2t -5e-St 14.2+ e2t -3e-t15.21S(eSt-1-5t) 16.

I\037S[e2t (5t-2)+e-3t (5t+2)]

17./6(sinh 2t -sin 2t) 18.e4t(1+ 12t+ 24t 2 +

33

2t3)

19.k(2cos2t+ 2sin2t -2cost-sint)20. 3

12[e2t (2t -1)+e-2t (2t + 1)]

21.\037e-t (5sin t -3t cost -2t sin t)

22.\037

et/2[(4t+ 8)cost + (4-3t) sin t]27. \037e-3t(8cos4t+ 9sin4t)28.

\037

(1-2e2t + e4t) 29.k (-6t+ 3sinh 2t)

30.1\037

[2e-t -e-2t (2cos2t + sin 2t)]31./s(6e2t -5-e-3t ) 32.

\037

(cosht + cost)33. x(t) = r(coshrt sin rt -sinh rt cosrt) where r = 1/,J234.

\037

sin 2t + ksin 3t 35./6(sin 2t -2t cos2t)

36.S10

[2e2t + (lOt -2)cost - (5t + 14)sin t]37. ;0[(5t - l)e-t + e-2t (cos3t + 32sin 3t)]38.

s\037O

e-3t (489cos3t + 307sin 3t) +1\0370

(7cos2t + 6sin 2t)39.

20)

-20)

Section4.4)1.

\037t22.(eat

-at -1)la23. \037(sint-tcost) 4.2(t-sint)5. teat 6.(eat -ebt)/(a -b)7.

k (e3t -1) 8.\037

(1-cos2t)9.

S14(sin 3t -3t cos3t) 10.(kt -sin kt)1k3

11.\037

(sin 2t + 2t cos2t) 12.\037

[1-e-2t (cost + 2sin t)]13./0 (3e3t -3cost + sin t) 14.

k (cost -cos2t)15.6sI(S2+ 9)2,s > 0 16.(2s3-24s)/(S2+4)3,S > 017.(S2-4s-5)I(s2-4s+ 13)2, s > 0

2(3s2 + 6s+7)18. s > 0(s+ 1)2(S2+ 2s+ 5)2

'19.

\037

n -arctan s = arctan (1Is),s > 020.

\037

In(s2 +4)-In s,s > 0 21.In s -In(s -3),s > 322. In (s+ 1)-In (s-1),s > 123. -(2sinh 2t)It 24. 2(cos2t -cost)It

25. e-2t + e3t -2cost)It 26.(e-2t sin 3t)It

27. 2(1-cost)lt 28. k(t sint -t2cost)29. (s+ I)X'(s)+ 4X(s) = 0;x(t) = Ct3e-t, C i= 030. X (s) = AI(s + 3)3;x(t) = Ct2e-3t, C i= 031.(s-2)X'(s)+ 3X (s) = 0;x(t) = Ct2e2t , C i= 0)

32.(S2+ 2s)X'(s)+ (4s+ 4)X(s)= 0;x(t) = C(1-t -e-2t -te-2t ), C i= 0

33.(S2+ I)X'(s)+4sX(s) = 0;x(t) = C(sin t -t cost), C i= 0

34. x(t) = Ce-2t (sin 3t -3t cos3t), C i= 0)

Section4.5)

1.f(t) = u(t -3) . (t -3))

fit))

t)

3)

2.f(t) = (t - l)u(t -1)-(t -3)u(t -3))

f(t))

2)

3.f(t) = u(t -1).e-2(t-l))

fit))

1)

t)

1) 3)

t)

1)

4. f(t) = et-1u(t -1)-e2et-2u(t -2))

-5)

-10)

fit)

1)

t)

1)))


5. f(t) = u(t -n) .sin(t -n) = -u(t-n) sin t) 10.f(t) = 2u(t -n) cos2(t -n) -2u(t -2n)cos2(t -2n)= 2[u(t -n) -u(t -2n)]cos2t)

fit)

1)fit)

2)

t)

t1t)

-1)-2)

6.f (t) = u (t -1).cosn (t -1)= -u (t -1)cosn t)

fit)

1)

t)

11.f (t) = 2[1-U 3(t)];F(s) = 2(1-e-3s)/s12.F(s)= (e-S -e-4S )/s13.F(s)= (1-e-2ns)/(s2+ 1)14.F(s)= s(1-e-2s)/(s2+ n2)15.F(s)= (1+e-3ns)/(s2+ 1)16.F(s)= 2(e-nS -e-2ns)/(s2+4)17.F(s)= n(e-2s+e-3s)/(S2+ n2)18.F(s)= 2n(e-3s+e-5S )/(4s2 + n2)19.F(s)= e-S (S-l+S-2)20. F(s)= (1-e-S )/s221.F(s)= (1-2e-S +e-2s)/s228. F(s)= (1-e-as -ase-aS )/[s2(1-e-2aS )]31.x(t) =

\037

[1-u(t -n)]sin2t)

-1)

7. f(t) = sint -u(t -2n) sin(t -2n) = [1-u(t -2n)]sint)

fit)

1) x(t))

- t)

12)

t)

1t)

-1)

1t)

8.f(t) = cosnt-u(t -2)cosn(t-2) = [1-u(t -2)]cosnt)32.x(t) = g(t) -u(t -2)g(t -2) where g(t) =

/2 (3-4e-t +e-4t ).)fit)

1 x(t)

0.2

t2 0.1

-1t)

9.f(t) = cosnt + u(t -3)cosn(t -3) = [1-u(t -3)]cosnt)

fit)

1)

33.x(t) =k [1-u(t -2n)](sint

-\037 sin3t)

x(t))

0.1)

t)t)

21t)

-0.1)-1)))


34. x(t) = g(t) -u(t - 1)[g(t-1)+ h(t -1)]where

g(t) = t -sint and h(t) = 1-cost.)

x(t)

0.5)

-0.5)

35. x(t) =\037 {-I+ t + (t + l)e-2t +

u(t -2)[1-t + (3t -5)e-2(t-2)]})

x(t))

14)

t)

2) 4)

36. i(t) = e-lOt -u(t - l)e-lO(t-l)

37. i(t) = [1-u(t -2n)]sin lOOt

38. i (t) =\037\037

[1- (u -n)](coslOt -coslOOt)

39. i(t) =5\037 [(1-e-SOt )2-u(t -1)(1+ 98e-50(t-l) -9ge-'OO (t-')]

40. i(t) =5\037 [(1-e-SOt -50te-SOt ) -u(t -1)(1-e-SO(t-l)

+2450te-50(t-I))]

41.x(t) = 21sin t1sin t)

x(t)

2)

-2)

11:)

t)

00

42. x(t) = g(t) + 2L(-I)nu (t -nn)g(t -nn) wheren=1

g(t) = 1-ke-t (3cos3t + sin 3t))

x(t)2)

2) 4)t)

-2)

Section4.6)

1.x(t) =\037

sin 2t)

x)

1)

12)

1-2)-1)

2.x(t) =\037

[1+ u(t -n)]sin 2t)

x)

1)

12)

-1)

3.x(t) =\037[1

-e-2t -2te-2t ] + u(t -2)(t -2)e-2(t-2))

x)

0.5)

0.25)

1) 2) 3)

4. x(t) = -2+ t + 2e-t + 3te-t)

x)

1.5)

1)

0.5)

1) 2)

5.x(t) = 2u(t -n)e-(t-n) sin(t -n))

x)

0.5)

t)

1t) 21t) 31t)))

6.x(t) = i (t -2u(t -3n)) sin 3t)

x)

2)

t)

-2)

7. x(t) = [2-e27r u(t -n) + e47r u(t -2n)]e-2t sin t)

x)

0.3)

t)

1t) 21t) 31t)

8.x(t) = (2+ 5t)e-t -u(t -2)(t -2)e-(t-2))

x3)

2)

t)

1)

2) 6)4)

9.x(t) =[\037(sin2-r)f(t--r)dT

10.x(t) =l'u-3'f(t --r) d-r

11.x(t) =[(e-3' sinh -r)f(t --r)d-r

I 1t12.x(t) = - (e-2r sin 2r:)f(t-r:)dr:

2 013.(a) mxE(t) = p[t2-uE(t)(t -E)2]/(2E);

(b) If t > E, then mxE(t) = p(2Et -E2)/(2E),and hencemxE(t) \037 pt aSE \037 0;(c) mv = (mx)' = (pt)' = o.

15.Thetransform ofeachof the two given initial value problems is(ms2 + k)X (s) = mvo = Po.

17.(b) i(t) = e-lOO(t-l)UI(t)-e-lOO(t-2)U2(t).1ft>2, theni(t) = _(eIOO _1)elOOO -t) < o.

18.i(t) = [1-u(t -n)]sin lOt)

0019.i (t) = Lu(t -nn /10)sin lOt

n=O)

0020. i(t) = L(-I)nu (t -nn/5) sin lOt

n=O)


0021.i(t) = Lu(t-nn/l0)e3n7re-30t sint

n=O)

i(t))

0.1)

t)

1t)

-0.1)

00

22. x(t) = Lu(t -2nn) sin tn=O)

X(t)

5)

-5)

t1t)

Chapter5Section5.1)1.

x\037

= X2, x\037

= -7Xl -3X2 + t2

2.x\037

= X2, x\037

= X3, x\037

= X4, x\037

= -Xl + 3X2 -6X3 +cos3t3.

x\037

= X2, t2x\037= (1-t2

)XI-tX2

4.x\037

= X2, x\037

= X3, t3x\037= -5XI-3tX2 + 2t2x3 + lnt

5 ' , , 2 +. Xl =X2, X2 =X3, X3 =X2 COS XI

6.x\037

= X2, x\037

= 5XI -4YI, Y\037

= Y2, Y\037

= -4X I + 5YI

7.x\037

= X2, y\037

= Y2, x\037

= -kXI .(x\037

+ y\037)-3/2,

Y\037

= -kYI .(x\037 + y\037)-3/2

8.x\037

= X2, x\037

= -4XI+ 2YI-3X2

y\037

= Y2, Y\037

= 3XI -YI-

2Y2 + COS t

9 ' , , , 3 +2\302\267 Xl = X2, YI= Y2, Zl = Z2, X2 = Xl -YI Zl,

Y\037

= Xl + YI-4zh z; = 5XI -YI

-Zl

10.x\037

= X2, x\037

= xI(1-YI)

Y\037

= Y2, Y\037

= YI(1 -Xl)

11.x(t) = Acost + Bsint, y(t) = Bcost-Asint)

54321

\037 0-1-2-3-4-5

-5 -4-3-2-10 1 2 3 4 5)

X)))


12.x(t) = Aet + Be-t , y(t) = Aet -Be-t) 15.x(t) = A cos2t + B sin 2t, y(t) = 4Bcos2t -4A sin 2t)

54/ 1'1'3/ 1'1'2/ 1'1'l1'il'

\037 0-1)

t t tt t tt t ti \037

J, \037 tJ J tJ J \037

J J \037)

2 3 4 5) -5-4-3-2-10 1 2 3 4 5)x)

x)

13.x(t) = A cos2t + B sin 2t, y(t) = -Bcos2t + A sin 2t;x(t) = cos2t, y(t) = sin 2t) 16.x(t) = A cos4t+ B sin4t, y(t) = 4Bcos4t- 4A sin4t)

54321

\037 0-1-2-3-4-5

-5-4-3-2-10 1 2 3 4 5x)

54321

\037 0-1-2-3-4-5 \037\037\037\037

-5-4-3-2-10 1 2 3x)

14.x(t) = A coslOt + Bsin lOt, y(t) = BcoslOt -A sin lOt;x(t) = 3coslOt +4sin lOt, y(t) = 4coslOt -3sin lOt)

17.x(t) = Ae-3t + Be2t , y(t) = -3Ae-3t + 2Be2t ;x(t) = e2t ,y(t) = 2e2t)

54321

\037 0-1-2-3-4-5

-5-4-3-2-10 1 2 3 4 5) 1 2 345)x) x)))


18.x(t) = Ae-2t + Be-St , y(t) = 2Ae-2t + 5Be-St ; A =13

7 and

B = -13

1 in the particular solution.)

Section5.2)

5J J \037

\037JJ,J,J2J\037JJIJJJJ

\037 0-1-2-3-4-5

-5-4-3-2-10x)

1.x(t) = al e-t + a2e2t, y(t) = a2e2t)

l' l' l'l' l' l' l'l' l' l' l'l' l' l' f)

54321

\037 0-1-2 ,l,

-3i-4-5

-5 -4-3-2-10 1)

1 2 3 4 5)

2 345)19.x(t) = _e-2t sin 3t, y(t) = e-2t (3cos3t + 2sin 3t)) x)

2.x(t) = (CI+ c2t)e-t,y(t) = (CI- \037C2+ c2t)e-t)

1 2 3 4 5)

54321

\037 0-1-2-3 \037

-4-5

-5-4 -3 -2-10 1 2 3 4 5)

x)

20. x(t) = (A + Bt)e3t, y(t) = (3A + B + 3Bt)e3t)

x)

3.x(t) =\037

(e3t -e-2t ), y(t) =\037

(6e3t -e-2t))

t i \037i \037 JJ J J JJ J J J)

54321

\037 0-1-2-3-4-5

:\\a

-5-4-3-2-10 1 2 3 4 5)

1 2 3 4 5)

x)

27.2(/;- I\037) + 50/1 = 100sin 60t,2(/\037-I;)+ 25/2 = 0)

28. I;= -20(/1-h), I\037

= 40(11 -12)) x)))

598) AnswerstoSelectedProblems)

4. x(t) =\037

(3e2t -e-2t ), y(t) =\037

(3e2t -5e-2t)

54321

\037 0-1-2-3-4-5

-5-4-3-2-10 1 2 3 4 5x)

5.x(t) = e-t (al cos2t + a2 sin 2t),y(t) = -

\037e-t [(al+ a2)cos2t + (a2-al) sin 2t]

5\037

4321

\037 0-1-2-3-4-5

-5-4-3-2-101 2 3 4 5x)

6.x(t) = e-2t (3cos3t + 9sin 3t), y(t) = e-2t (2cos3t -4sin 3t)

54321

\037 0-1-2-3-4-5

-5-4-3-2-101 2 3 4 5)

x)

7. x(t) = ale2t +a2e3t - It +...!..y(t) = -2ale2t -a2e3t - 'J.:t - \0373 18' 3 98.x(t) = CIet + C2e3t + e2t , y(t) = -CIet + C2e3t9.x(t) = 3alet +a2e-t- k (7COS 2t +4sin 2t),y(t) = alet + a2e-t- k (2COS 2t +4sin 2t)10.x(t) = et, y(t) = -et

11.x(t) = al cos3t + a2 sin 3t -;\037

et -\037e-t,

y(t) =\037

[(aI -a2)cos3t + (aI + a2)sin 3t]+ /o et

12.x(t) = CIe2t + C2e-2t+ C3 e3t + C4e-3t,y(t) = -CIe2t -C2e-2t+ \037C3e3t +

\037C4e-3t13.x(t) = al cos2t + a2 sin 2t + b l cos3t + b2sin 3t,y(t) =

\037

(al cos2t + a2 sin 2t) -2(bl COS 3t + b2sin 3t)14.x(t) = CI COS 2t + C2sin 2t +

\037

sin t,

y(t) = CI COS 2t + C2sin 2t + C3 COS 2tv'2 +C4sin 2t v'2+ 2

41

sin t)

15.x(t) = al cost+ a2 sint + b l cos2t+ b2sin2t,y (t) = a2 cost -a I sin t + b2cos2t -b I sin 2t

17.x(t) = al cost + a2 sin t + b le2t + b2e-2t ,y(t) = 3a2 cost -3al sin t + b le2t -b2e-2t

18.x(t) =\037

(4cIe3t -3c2e-4t),y(t) = CI e3t + C2e-4t,z(t) =

\037

(-4cIe3t + 3c2e-4t)

19.x(t) = al + a2e4t + a3e8t, y(t) = 2al-2a3e8t,z(t) = 2al-2a2e4t + 2a3e8t

20.x(t) = ale2t +a2e-t+ \037te-t, y(t) = ale2t + b2e-t -\037te-t,

z(t) = ale2t - (a2+b2 +\037)e-t

-\037te-t)

23.Infinitely many solutions

24. No solution

25. Infinitely many solutions

26. Two arbitrary constants

27.No arbitrary constants

28. No solution

29. Four arbitrary constants)

31.II (t) = 2+ e-5t

[-2cos(lOt/,)6)+4,)6sin (lOt/,)6)),I2(t) = (20/,)6)e-5t sin (10t/,)6)

32. II(t) =13

121 (120e-25t /3 -120cos60t+ 1778sin60t),

h(t) =13

121 (-240e-25t /3+ 240cos60t+ 1728sin60t)

33.II (t) =\037

(2+ e-60t ), I2(t) =\037

(1-e-60t )

37.(a) x(t) = al cos5t+ a2 sin5t + b l cos5t.j3+ b2sin5t.j3,y(t) = 2alcos5t + 2a2 sin 5t - 2bl COS 5t.j3-2b2sin 5t.j3;(b) In the natural mode with frequency WI = 5, the massesmovein the samedirection, whereas in the natural modewith

frequency W2 = 5.j3they move in opposite directions. In eachcasethe amplitude of the motion ofm2 is twice that ofmi.

39.x(t) = aI cost + a2 sin t + b I COS 2t + b2sin 2t,y(t) = 2alcost + 2a2 sin t -b l COS 2t -b2sin 2t.In the natural mode with frequency WI = 1the massesmoveinthe same direction, with the amplitude ofmotion of the secondmass twice that of the first mass. In the natural modewith

frequency W2 = 2 they move in opposite directions with thesame amplitude of motion.)

3

2

1\037

\037 0

-1-2-3

0 1t)

/YI=2cos(t))

21t 31t

t)))

3

2

1

?-.\037 0

-1\\X2 =cos(2t)-2

-30 1t 21t 31t

t)

40. x(t) = al cos5t + a2 sin 5t + b l coslOt + b2sin lOt,yet) = 2alcos5t + 2a2 sin 5t -b l COS lOt -b2sin lOt.)

41.x(t) = al cost + a2 sin t + b l cos3t + b2sin 3t,yet) = al cost + a2 sin t -b l COS 3t -b2sin 3t.In the natural mode with frequency WI = 1 the massesmove inthe samedirection, while in the natural mode with frequencyW2 = 3 they move in opposite directions. In eachcasethe

amplitudes ofmotion of the two massesareequal.)

2)

1)

Xl =YI =cos(t))

?-.\037

0)

-144.

-20 1t 21t 45.t

2Y2 =-cos(3t)

1

?-. 0\037

-1)

-2o) 1t) 21t)

t)

42. x(t) = al cost + a2 sin t + b l cos2t + b2sin 2t,y(t) = al cost + a2 sin t -

\037blCOS 2t -

\037b2sin 2t.)

43. x(t) = al cost + a2 sin t + b l cost,J5+ b2sin t,J5,yet) = al cost + a2 sin t -b l cost,J5-b2sin t,J5.)


In the natural mode with frequency WI = 1 the massesmove inthe samedirection, while in the natural modewith frequency

W2 = ,J5they move in opposite directions. In each casethe

amplitudes ofmotion of the two massesare equal.)

2)

1)

Xl =YI =cos(t))

?-.\037

0)

-1)

-2o) 1t) 21t)

t)

2)

1)

?-.\037

0)

-1)'\\

X2 =cos(t{5))-2

o) 1t) 21t)

t)

x(t) = al cost,J\"i + a2 sin t,J\"i + b l cos2t + b2sin 2t,yet) = al cost,J\"i + a2 sin t,J\"i -b l COS 2t -b2sin 2t.)

x(t) = al cost,J\"i + a2 sin t,J\"i + b l cost,J8+ b2sin t,J8,yet) = al cost,J\"i + a2 sin t,J\"i -

\037blcost,J8-

\037b2sin t,J8.

In the natural mode with frequency WI = ,J\"i the two massesmove in the samedirection with equal amplitudes ofoscillation.In the natural modewith frequency W2 = ,J8= 2,J'2the twomassesmove in opposite directions with the amplitude ofm2

being half that ofmI.)

2)

1)

?-.\037

0)

-1)'\\

Xl =YI

=cos(t-Y2))

-2o) 1t) 21t)

t)))


2)

1)

?-.\037

0)

-1)'\\

X2 =Y2 = COS(t -{8))-2

o) 1t) 21t)

t

46. x(t) = al cos2t + a2 sin 2t + b l cos4t + b2sin 4t,yet) = al cos2t + a2 sin 2t -b l COS 4t -b2sin 4t.)

Section5.31.(a)

[i\037

-\037\037 l (b)

[\037\037\037 J.

(c)[\037\037

-\037\037 l (d)[-\037\037

-3\037]

2.(AB)C=A(BC) =[::::\037\037

1O\037 lA(B + C)= AB + AC = [

-\037\037 :::::]

3.AB=[ \037\037 _\037 ]; BA=[

_\037\037 -1\037 l\037

]o 8 -134. Ay = [3 2

24t2.-co

+s5t

], Bx =[

2t\0371\037;-t

]t - sIn t cost 6 2 -tt - e

5. (a)[

2\037 4; \037

]; (b)

[-

\037\037

\037

-i\037

];

-27 34 45 -25 -19 26

(c)[

1\037 3\037 -1\037

];(d)

[-\037\037 \037\037 -l\037

];

16 58 -23 11 22 6

(e)[

30

t4

2t -;1

]-5 2 7 -t

7. det(A) = det(B) = 0 8.det(AB) = det(BA) = 144

9 (AB)' _ [1 -8t + 18t2 1 + 2t - 12t2 + 32t 3

]\302\267- 3+ 3t2-4t 3 8t + 3t2 + 4t 3

11.x = [\037 J.pet) =[\037

-\037 J.r(t) = [\037 ]13.x = [\037 J.pet) = [; -1J.r(t) = [ 3\037

\037 ]15.x =

[\037 ].pet) =[f f 1].r(t} =

[ \037 ]17.x =

[\037

], pet) =

[\037 -6 -\037

], f(t) =

[t\037

]Z 0 6 -7 t3

19.x =[\037: ]

, P(t) =[ \037 g \037 \037 ]

, f(t) =[ \037 ]X4 4 0 0 0 0)

21.) [2 t + 2t

]3t cle C2 e

W (t) = e ; x(t) = 3 t 2t- Cl e -C2e

[Ce3t + 2ce-2t

]Wet) = -5et ; x(t) =3

13t +

2-2tcle C2 e

[Ce2t + Ce-2t

]W (t) = 4;x(t) = 12t + 5

2 -2tCle C2e

[Ce3t + Ce2t

]W (t) = e5t ; x(t) = 13t 2

22t-cle- C2 e

[3ce2t + Ce-5t

]Wet) = 7e-3t ;x(t) = 21

2t + 32

-5tCle C2e

[2Clet -2C2e3t+ 2C3e5t

]W (t) = 16e9t ; x(t) = 2Clet -2C3e5t

Cl et + C2e3t + C3 e5t

[Cl e2t + C2 e-t

]Wet) = 3;x(t) = cle2t + C3e-t

2t ( + ) -tCl e -

C2 C3 e

[Cl + 2C2e3t-C3 e4t

]W (t) = -84e7t ;x(t) = 6CI + 3C2e3t+ 2C3e4t-13cI -2C2e3t+ C3 e4t

[3Cle-2t + C2et+ C3e3t

]Wet) = e2t ;x(t) = -2Cle-2t-C2et-C3 e3t

2Cle-2t + C2et

[

Cle-t+ C4et

]C3 et

Wet) = l;x(t)= -t +3 tC2e C4eCIe-t -2c3et

x = 2XI-

X2

X = 7Xl -2X2

X = 15xI -4X2

X =\037(3Xl

-2X2)

x = Xl + 2X2 + X3

x = 7XI + 3X2 + 5X3

x = 3XI -3X2 -5X3

X = -2XI+ 15x2-4X3

X = 3XI + 7X2 + X3-2X4

X = 13xI + 41x2 + 3X3- 12x4(a) X2 = tXI, soneither isa constant multiple of the other.(b) W (Xl, X2) = 0, whereas Theorem 2 would imply that W 1= 0if Xl and X2 were independent solutions ofa system of theindicated form.)

22.)

23.)

24.)

25.)

26.)

27.)

28.)

29.)

30.)

31.32.33.34.35.36.37.38.39.40.41.)

Section5.4)1.Xl (t) = CIe-t + C2e3t,X2 (t) = -CIe-t + C2e3t)

543

\03721 J

N 0\037

-1-2-3-4-5

-5-4-3-2-10) 5)

Xl)

2.Xl (t) = CIe-t + 3C2e4t,X2 (t) = -CIe-t + 2c2e4t3.General solution Xl (t) = Cle-t + 4C2e6t,X2(t) = -Cle-t + 3c2e6t

Particular solution Xl (t) = t (-e-t + 8e6t ),X2(t) = t(e-t + 6e6t

).)))

4. Xl (t) = Cle-2t + C2e5t, X2(t) = -6Cle-2t+ C2e5t)

54321

\037 0-1-2-3-4-5

-5 -4-3-2-10 1 2 3 4 5)

Xl)

5.Xl (t) = CIe-t + 7C2e5t, X2 (t) = CIe-t + C2e5t)

54321

\037 0-1-2-3-4 '\\

-5-5-4-3-2-10 1 2 3 4 5)

Xl)

6.General solution Xl (t) = 5CIe3t + C2e4t,X2 (t) = -6CIe3t -C2e4tParticular solution Xl (t) = -5e3t + 6e4t , X2(t) = 6e3t -6e4t

.)

54321o

-1-2-3-4-5 \037 \037 \037 \037

-5-4-3-2-10)

N\037)

12345)Xl)


7. Xl (t) = Clet + 2C2e-9t, X2(t) = Clet -3C2e-9t)

54321

\037 0-1-2-3-4-5

-5-4-3-2-10 1 2 3 4 5)

Xl)

8.Xl (t) = 5CIcos2t + 5C2sin 2t,X2(t) = (CI-2C2)cos2t + (2CI+ C2)sin 2t)

N\037)

54321

o-1-2-3-4-5 -7-7--7--7

-5-4-3-2-10) 1 2)

Xl)

9.General solution Xl (t) = 5CIcos4t + 5C2sin 4t,X2(t) = cI(2cos4t+4sin4t)+c2(2sin4t-4cos4t).Particular solution Xl (t) = 2cos4t -

\037l

sin 4t,

X2(t) = 3cos4t +\037

sin 4t10.Xl (t) = -2CIcos3t -2C2sin 3t,

X2(t) = (3CI+ 3C2)cos3t + (3C2-3CI)sin 3t11.General solution Xl (t) = et (CIcos2t -C2 sin 2t),

X2(t) = et (CIsin 2t + C2 COS 2t)Particular solution Xl (t) = -4et sin 2t, X2(t) = 4et cos2t)

N\037)

54321o

-1-2-3-4-5 \037

-5 -4 -3 -2-10 1 2 3 4 5)

Xl)

12.)Xl (t) = e2t (-5CI COS 2t -5C2sin 2t),X2(t) = e2t[(cl+ 2C2)cos2t + (-2CI+ C2)sin 2t]Xl (t) = 3e2t (CIcos3t -C2sin 3t),X2(t) = e2t [(cl+ C2)cos3t + (CI-C2)sin3t]Xl (t) = e3t (CIcos4t + C2 sin 4t),X2 (t) = e3t (CIsin 4t -C2COS 4t))

13.)

14.)))


15.XI (t) = 5e5t (CIcos4t -C2 sin 4t),X2(t) = e5t[(2cl+ 4C2)cos4t+ (4CI-2C2)sin4t]16.XI(t) = cle-IOt + 2c2e-IOOt, X2(t) = 2cle-lOt -5c2e-IOOt)

28. XI (t) = 15e-0.4t,X2(t) = 40(_e-O.4t +e-O.25t ).The maximum amount ever in tank 2 is about 6.85lb.)

54321

N 0\037

-1-2-3-4-5

\037

-5-4-3-2-10 1 2 3 4 5)

15)

5)

10)

\037)

Xl)

oo) 5) 10

t)

15) 20)

17.Xl (t) = Cle9t + C2 e6t + C3,X2 (t) = CIe9t -2C2e6t,X3(t) = Cle9t + C2 e6t-C318.Xl (t) = CIe9t + 4C3,X2(t) = 2Cle9t + C2 e6t-C3,X3(t) = 2Cle9t -C2e6t-C319.XI (t) = CIe6t + C2e3t + C3 e3t ,X2(t) = CIe6t -2c2e3t,X3(t) = cle6t + C2e3t -C3 e3t

20. Xl(t) = CIe9t + C2e6t+ C3 e2t,X2 (t) = CIe9t -2c2e6t,X3 (t) = CIe9t + C2e6t-C3 e2t

21.XI (t) = 6CI+ 3c2et + 2c3e-t,X2(t) = 2CI+ C2et+ C3 e-t ,X3(t) = 5CI+ 2c2et + 2c3e-t

22. Xl (t) = C2et+ C3 e3t ,X2(t) = Cle-2t -C2et-C3 e3t ,X3(t) = -Cle-2t +C3e3t

23. XI (t) = CIe2t + C3 e3t,X2(t) = -CIe2t + C2e-2t-C3e3t,X3 (t) = -C2e-2t+ C3 e3t

24. Xl (t) = clet+ c2(2cos2t -sin 2t) + C3(COS 2t + 2sin 2t)X2(t) = -clet-C2(3cos2t + sin 2t) + C3(COS 2t -3sin 2t)X3(t) = c2(3COS 2t + sin 2t) + c3(3sin 2t -COS 2t)

25. Xl (t) = CI + e2t [(C2+ C3)cos3t + (-C2+ C3)sin 3t],X2(t) = -CI+ 2e2t (-C2COS 3t -C3 sin 3t),X3(t) = 2e2t (C2cos3t + C3 sin 3t)

26. XI (t) = 4e3t -e-t(4cost -sin t),X2(t) = ge3t -e-t (9cost + 2sin t),X3(t) = 17e-t cost

27. XI (t) = 15e-o.2t , X2(t) = 15(e-o.2t -e-O.4t).The maximum amount ever in tank 2 isx2(51n 2) = 3.75lb.

15)

29.XI (t) = 10+ 5e-O.6t , X2(t) = 5-5e-O.6t)

15)

10)

\037)

5)

oo) 5) 10) 15)

t)

30.Xl (t) = 5CI+ C2e-O.65t,X2(t) = 8CI -c2e-O.65t.)

15)

5)

10)

\037)

\037)

oo) 5) 10) 15)

5)t)

oo) 5) 10

t)

15) 20)

31.Xl (t) = 27e-t,X2(t) = 27e-t -27e-2t,X3(t) = 27e-t -54e-2t + 27e-3t .The maximum amount of salt ever in tank 3 is x3(ln 3) = 4)))


pounds.

25

20

15\037

10

5

00 5

t)

X3(\037

In 5) \037 21.4663pounds.)

4035

30

25 Xl

\037 20

15

10

50

0) 4)t)

32. Xl (t) = 45e-3t ,X2(t) = -135e-3t + 135e-2t ,X3(t) = 135e-3t -270e-2t + 135e-t .Themaximum amount of salt ever in tank 3 is X3 (In 3) = 20pounds.)

35.Xl (t) = 10-\037 (55e-I8t -216e-Ilt

),X2(t) = 3-

\037 (165e-I8t -144e-llt ),X3(t) = 20+

\037 (220e-I8t -360e-llt).The limiting amounts of salt in tanks 1,2, and 3 are 10lb, 3 lb,and 20lb.)

4540 30

35 253025 Xl 20

\037\037

20 15

15 10105

5

0 00 5 0

t)

X3)

Xl)

X2)

t)

1)

33.Xl (t) = 45e-4t ,X2(t) = 90e-4t -90e-6t,X3(t) = -270e-4t + 135e-6t+ 135e-2t .Themaximum amount of salt ever in tank 3 is X3

(\037

In 3) = 20pounds.)

36.XI(t) = 4+e-3t/5[14cos(3t/l0)-2sin(3t/l0)],X2(t) = 10-e-3t/5[10cos(3t/l0)-10sin(3t/l0)],X3(t) = 4-e-3t/5[4cos(3t/l0)+ 8sin(3t/l0)].The limiting amounts of salt in tanks 1,2,and 3 are 4 lb, 10lb,and 4 lb.)

5)

X2)

4540353025

\037

2015105o

o)

15)

10)\037)

oo) 5

t)

10)

t)

2)

34. Xl (t) = 40e-3t ,X2(t) = 60e-3t -60e-5t ,X3(t) = -150e-3t + 75e-5t + 75e-t.Themaximum amount of salt ever in tank 3 is)

37. Xl (t) = 30+e-3t [25cos(t-V2)+ 10-V2sin(t-V2)],X2(t) = 10-e-3t[10cos(t-V2)-

2\037J2sin(tJ2)],

X3(t) = 15-e-3t[15cos(tJ2)+\037

J2sin(tJ2)].The limiting amounts of salt in tanks 1,2,and 3 are 30lb, 10lb,)))


and 15lb.)

555045403530

\037

252015105o

o)

Xl)

X3)

X2)

t)

2)

38. xI(t) = CIet,X2 (t) = -2cIet + C2e2t ,X3(t) = 3clet -3c2e2t+ C3e3t,X4(t) = -4cIet + 6c2e2t-4c3e3t+ C4e4t

39. XI (t) = 3cIet + C4e-2t ,X2 (t) = -2cIet + C3e2t -C4e-2t,X3(t) = 4cIet + C2 e-t ,X4(t) = CIet

40. XI (t) = CIe2t ,X2(t) = -3cle2t +3c2e-2t-C4e-St,X3 (t) = C3 eSt ,X4 (t) = -C2e-2t-3c3eSt

41.XI (t) = 2elO t + elSt = X4(t), X2(t) = _elOt + 2elSt = X3(t)

42.x(t) = Cl

[-

\037 ]+ Cz

U ]eZt + C3

[-

\037 ]e5t

43.x(t) = Cl

[-

\037 ]e-Zt + Cz

U ]e4t + C3

[-i]

eS'

44. x(t) = Cl

[-\037]

e-3t + Cz

[r]e6t+ C3

[ -\037]

elZ'

45. x(t) = Cl [j]e-3t+cz

[-t]+C3[1]e3t+C4

[=1]e&

46. x(t) =

CI

[_;]e-\037 +Cz

[j]eZt + C3

[_

i ]e4t +C4

[=!]eSt

47. x(t) =

Cl

[j]e-3t + Cz

[_

t ]e3t + C3

[1]e6t+ C4

[=1]e9t

48. x(t) =

Cl

[_i]

el& +cz[J]

e3Zt +C3

[-1]e\037t +C4

[j]eM')

1 0 10 3 7

49. x(t) = CI 3 e-3t + C2 0 +C3 1 e3t +1 -1 11 1 1)

0 21 0

C4 0 e6t+ Cs 5 e9t

1 21 1

0 1 01 0 1

50.x(t) = CI1 e-7t + C2

0 e-4t + C3

0 e3t +1 0 1

0 1 01 1 1

0 1 00 1 01

eSt + Cs0 e9t + C6

1 elltC4 0 0 -11 0 -10 1 0)

Section5.5)1.Thenatural frequencies are Wo = 0and WI = 2. In the

degenerate natural mode with \"frequency\" Wo = 0 the twomassesmove linearly with XI (t) = X2(t) = ao + bot. At

frequency WI = 2 they oscillate in opposite directions with equalamplitudes.

2.The natural frequencies are WI = 1 and W2 = 3.In the naturalmode with frequency WI, the two massesm I and m2 move in thesamedirection with equal amplitudes ofoscillation. At

frequency W2 they move in opposite directions with equalamplitudes.

3.Thenatural frequencies are WI = 1and W2 = 2.In the naturalmode with frequency w., the two massesm I and m2 move in thesamedirection with equal amplitudes ofoscillation. In thenatural mode with frequency W2 they move in opposite directionswith the amplitude ofoscillation ofm I twice that ofm2.

4. Thenatural frequencies are WI = 1 and W2 = ,J5.In the natural

mode with frequency WI, the two massesm I and m2 move in thesamedirection with equal amplitudes ofoscillation. At


5. The natural frequencies are WI = ,J2and W2 = 2. In the naturalmode with frequency WI, the two massesm I and m2 movein thesamedirection with equal amplitudes ofoscillation. At

frequency W2 they move in opposite directions with equal

amplitudes.

6.Thenatural frequencies are WI = ,J2and W2 = ,J8.In thenatural mode with frequency w., the two massesml and m2

move in the samedirection with equal amplitudes ofoscillation.In the natural mode with frequency W2 they move in oppositedirections with the amplitude ofoscillation ofm I twice that ofm2.

7. Thenatural frequencies are WI = 2 and W2 = 4. In the naturalmode with frequency WI, the two massesm I and m2 move in the

samedirection with equal amplitudes ofoscillation. At


8.XI(t) = 2cost+ 3cos3t-5cos5t,X2(t) = 2cost-3cos3t+ cos5t.We have a superposition of three oscillations, in which the twomassesmove(1)in the samedirection with frequency WI = 1

and equal amplitudes; (2)in opposite directions with frequencyW2 = 3 and equal amplitudes; (3)in opposite directions with

frequency W3 = 5 and with the amplitude ofmotion ofm I being5 times that ofm

2.)))

9.XI (t) = 5cost -8cos2t + 3cos3t ,X2 (t) = 5cost + 4cos2t -9cos3t.We have a superposition of three oscillations, in which the twomassesmove(1)in the samedirection with frequency uh = 1

and equal amplitudes; (2)in opposite directions with frequencyuh = 2 and with the amplitude ofmotion ofm I being twice thatofm2; (3)in opposite directions with frequency W3 = 3 and withthe amplitude ofmotion ofm2 being 3 times that ofmI.

10.XI(t) = cos2t-15cos4t+ 14cost,X2 (t) = cos2t + 15cos4t + 16cost.We have a superposition of three oscillations, in which the twomassesmove(1)in the same direction with frequency WI = 1and with the amplitude ofmotion ofm2 being 8/7 times that ofm I; (2)in the same direction with frequency W2 = 2 and equalamplitudes; (3)in opposite directions with frequency W3 = 4 and

equal amplitudes.11.(a) Thenatural frequencies areWI = 6 and W2 = 8.In mode 1

the two massesoscillate in the same direction with frequencyWI = 6 and with the amplitude ofmotion ofm I being twice thatofm2. In mode2 the two massesoscillate in opposite directionswith frequency W2 = 8 and with the amplitude ofmotion ofm2

being 3 times that ofmI.(b)x(t) = 2sin 6t + 19cos7t, Y (t) = sin 6t + 3cos7tWe have a superposition of (only two) oscillations, in which thetwo massesmove(1)in the samedirection with frequencyWI = 6 and with the amplitude ofmotion ofmi being twice thatofm2; (2)in the samedirection with frequency W3 = 7 and withthe amplitude ofmotion ofm I being 1913times that ofm2.

12.The system's three natural modesofoscillation have (1)natural

frequency WI = ,J2with amplitude ratios 1 :0:-1;(2)natural

frequency W2 = J2+ ,J\"i wi th ampli tude ratios 1 :-,J2:1;(3)natural frequency W3 = J2-,J2with amplitude ratios

1:,J2:1.13.Thesystem's three natural modesofoscillation have (1)natural

frequency WI = 2 with amp litude ratios 1:0:-1;(2)natural

frequency W2 = J4 + 2,J\"i with amplitude ratios 1:-,J2:1;(3)natural frequency W3 = J4-2,J\"i with amplitude ratios 1:,J2:1.

15.XI (t) =\037

cos5t -2cos5-J3t +\037

coslOt,

X2(t) =\037

cos5t + 4cos5-J3t +13

6 coslOt.We have a superposition of two oscillations with the natural

frequencies WI = 5 and W2 = 5-J3and a forced oscillation with

frequency W = 10.In eachof the two natural oscillations the

amplitude ofmotion ofm2 is twice that ofm I, while in theforced oscillation the amplitude ofmotion ofm2 is four timesthat ofmI.

20.x\037 (t) = -Vo, x\037(t)

= 0,x\037 (t) = Vo for t > rr: 1221.

x\037 (t) = -Vo, x\037(t)= 0,

x\037 (t) = 2vo for t > rr: 1222.

x\037 (t) = -2vo, x\037 (t) = Vo, x\037 (t) = Vo for t > rr: 1223.

x\037 (t) = 2vo, x\037 (t) = 2vo, x\037 (t) = 3vo for t > rr:1224. (a) WI \037 1.0293Hz; W2 \037 1.7971Hz.

(b) VI \037 28mi/h; V2 \037 49mi/h27.WI = 2,JIO,VI \037 40.26(ft/s (about 27 mi/h),

W2 = 5-V5, V2 \037 71.18ft/s (about 49mi/h)28.WI \037 6.1311,VI \037 39.03ftls (about 27 mi/h)

W2 \037 10.3155,V2 \037 65.67ftls (about 45milh)29.WI \037 5.0424,VI \037 32.10ft/s (about 22mi/h),

W2 \037 9.9158,V2 \037 63.13ftls (about 43mi/h)

Section5.6)1.Repeated eigenvalue A = -3,eigenvector v = [1 -1]T;

XI (t) = (CI+ C2 + c2t)e-3t,X2(t) = (-CI-c2t)e-3t)


54321

\0370

-1-2-3-4-5

-5-4-3-2-10 1 2 3 4 5)

Xl)

2.Repeated eigenvalue A = 2, single eigenvector v = [1 1]T;Xl (t) = (CI+ C2+ C2t)e2t,X2(t) = (CI+ C2 t)e2t)

54321

N 0\037 -1

-2-3-4-5

-5-4-3-2-10 1 2 3 4 5)

Xl)

3.Repeated eigenvalue A = 3,eigenvector v = [-2 2]T;XI(t) = (-2CI+C2-2c2t)e3t,x2(t)= (2CI+2c2t)e3t)

4. Repeated eigenvalue A = 4, single eigenvector v = [-1 1]T;XI(t) = (-CI+C2-c2t)e4t,x2(t)= (CI+c2t)e4t)

5. Repeated eigenvalue A = 5,eigenvector v = [ 2 _4]T;Xl (t) = (2CI+ C2 + 2c2t)e5t, X2(t) = (-4CI-4c2t)e5t)

54321

N 0\037 -1

-2-3-4-5 \037

-5-4-3-2-10 1 2 3 4 5)

Xl)

6.Repeated eigenvalue A = 5, single eigenvector v = [-4 4]T;Xl (t) = (-4CI+ C2

-4c2t)e5t, X2(t) = (4CI+ 4c2t)e5t)))


54321

N 0\037 -1

-2-3-4-5

-5-4-3-2-1012 3 4 5)

Xl)

7. Eigenvalues A = 2,2,9 with three linearly independenteigenvectors; XI (t) = CI e2t + C2e2t , X2 (t) = CIe2t + C3e9t ,X3 (t) = C2 e2t

8.Eigenvalues A = 7, 13,13with three linearly independenteigenvectors; XI (t) = 2cIe7t -C3e13t,X2 (t) = -3cIe7t + C3e13t,X3 (t) = CIe7t + C2 el3t

9.Eigenvalues A = 5,5,9 with three linearly independenteigenvectors; XI (t) = CI eSt + 7c2eSt+ 3c3e9t, X2(t) = 2cIeSt,X3(t) = 2c2eSt+ C3e9t

10.Eigenvalues A = 3,3,7 with three linearly independenteigenvectors; XI (t) = 5cIe3t -3c2e3t+ 2c3e7t,X2(t) = 2cle3t + C3e7t,X3(t) = C2e3t

11.Triple eigenvalue A = -1ofdefect 2;XI (t) = (-2C2+ C3

-2c3t)e-t,X2(t) = (CI-C2+ C2t-'C3t+ \037c3t2)e-t,X3(t) = (C2+ c3t)e-t

12.Triple eigenvalue A = -1ofdefect 2;XI (t) = e-t (CI+ C3 + C2t+ \037C3t2)

X2(t) = e-t(ci+ C2t+ tC3t2),X3(t) = e-t (C2+ C3t)

13.Triple eigenvalue A = -1ofdefect 2;XI (t) = (CI+ C2t+ tC3t2)e-t,X2(t) = (2C2+ C3 + 2c3t)e-t,X3(t) = (C2+ c2t)e-t

14.Triple eigenvalue A = -1ofdefect 2;XI (t) = e-t(5cI+ C2+ C3 + 5C2t+ C3t+ \037C3t2),

X2(t) = e-t(-25cI-5C2-25c2t -5C3t- 2;C3 t2),X3 (t) = e-t(-5CI+ 4C2-5C2t+ 4C3t- \037C3t2)15.Triple eigenvalue A = 1 ofdefect 1;XI (t) = (3cI+ C3

-3c3t)et,X2(t) = (-CI+c3t)et,x3(t)= (C2+c3t)et

16.Triple eigenvalue A = 1 ofdefect 1;XI (t) = et(3cI+ 3C2+ C3)X2(t) = et(-2cI-2C3t),X3(t) = et(-2C2 + 2C3t)

17.Triple eigenvalue A = 1 ofdefect 1;XI(t) = (2cI+ c2)et ,X2(t) = (-3C2+C3+6c3t)et,X3(t) = -9(cl+ c3t)et

18.Triple eigenvalue A = 1 ofdefect 1;XI (t) = et(-CI-2C2+ C3),X2(t) = et (C2+ C3t),X3(t) = et (CI-2C3t)

19.Double eigenvalues A = -1and A = 1,eachwith defect 1;XI (t) = CIe-t + C4et ,X2 (t) = C3et,X3(t) = C2e-t+ 3c4et,X4(t) = CIe-t -2c3et)

20. Eigenvalue A = 2 with multiplicity 4 and defect 3;XI (t) = (CI+ C3 + C2t+ C4t+ tC3t2 +

\037c4t3)e2t,

X2(t) = (C2+ C3t+ tC4t2)e2t,X3(t) = (C3+ c4t)e2t, X4(t) = C4e2t

21.Eigenvalue A = 1with multiplicity 4 and defect 2;XI (t) = (-2C2 + C3

-2c3t)et,X2 (t) = (C2+ c3t)et ,X3(t) = (C2+C4+c3t)et,x4(t)= (CI+C2t+ t C3t2)et

22.Eigenvalue A = 1 with multiplicity 4 and defect 2;XI (t) = (CI+ 3C2+ C4+ C2t+ 3C3t+ tC3t2)et,X2(t) = -(2C2 -C3+ 2c3t)et,X3(t) = (C2+c3t)et ,X4(t) = -(2cI+ 6C2+ 2C2t+ 6C3t+ c3t2)et

23. x(t) = clvle-t + (C2V2 +c3v3)e3t with VI = [1 -1 2 ]T,V2 = [4 0 9]T,V3 = [0 2 I]T

24. x(t) = clvle-t + (C2V2 + c3v3)e3t with VI = [ 5 3 -3]T,v2=[4 0 -1]T,v3=[2-1 O]T

25. x(t) =[CIVI + C2(Vlt + V2) + C3 (tVlt2 + V2t + V3)] e2t with

VI = [-1 0 -1]T, V2 = [-4 -1 0]T, and

V3 = [1 0 O]T26.x(t) =[CIVI + C2(Vlt + V2) + C3 (tVlt2 + V2t + V3)] e3t with

VI = [0 2 2]T,V3 = [ 2 1 _3]T,and

V3 = [1 0 O]T27.x(t) = [CI VI + C2(VI t + V2) + C3 v3]e2t with

VI = [-5 3 8]T,V2 = [1 0 O]T,V3 = [1 1 0 ]T28.x(t) =[CIVI + C2(Vlt + V2) + C3 (tVlt2 + V2t + V3)]e2t with

VI = [119-289O]T,V2 = [-1734 17]T,and

V3 = [1 0 O]T29. x(t) = [CI V I + C2(vIt + V2)]e-t + [C3V3 + C4(V3t + V4)]e2t with

VI = [1 -3 -1 -2]T,V2 = [ 0 1 0 O]T,V3 = [ 0 -1 1 0 ]T,V4 = [0 0 2 I]T30.x(t) = [CIVI + C2(Vlt + v2)]e-t + [C3V3 + C4(V3t + v4)]e2t , with

VI = [ 0 1 -1 -3]T,V2 = [0 0 1 2]T,V3 = [-1 0 0 O]T,V4 = [ 0 0 3 5]T31.x(t) = [CIVI +C2(Vlt +V2)+ C3 (tVlt2 + V2t + V3) + C4V4] et with

V I = [42 7 -21 -42]T,

V2 = [34 22 -10 -27 ]T, V3 = [-1 0 0 0]T,V4 = [ 0 1 3 0 ]T32.x(t) = (CIVI +c2v2)e2t+ (C3V3 +C4V4+csvs)e3t with

VI = [8 0 -3 1 0]T,V2 = [I 0 0 0 3]T,v3=[3 -2 -1 0 O]T,V4 = [ 2 -2 0 -3 0 ]T,Vs = [1 -1 0 0 3]T

33. x I (t) = [ cos4t sin 4t 0 0]Te3t ,X2 (t) = [-sin 4t cos4t 0 0]Te3t,X3 (t) = [ t cos4t t sin 4t cos4t sin 4t ]

Te3t,X4(t) = [ -tsin 4t t cos4t -sin 4t cos4t ]Te3t

( ) _

[3cos;/\037\037 sin 3t

]2t34. x I t - 0 e ,

sin 3t

( ) _

[3sin 3t c\037si\037os 3t

]2t

X2 t - 0 e ,-cos3t

[

3cos3t + t sin 3t

]( ) _ (3t -10)cos3t -(3t + 9)sin 3t 2t

X3 t -sin 3t

e ,t sin 3t)))


35.36.)

[

-tcos3t + 3sin 3t

]X4(t) = (3t + 9)cos3t + (3t -10)sin 3t 2t-COS 3t

e-tcos3t

XI (t) = X2(t) = vo(1-e-t); lim Xl (t) = lim X2(t) = Vot\037oo t\037oo

XI (t) = vo(2-2e-t -te-t),X2(t) = vo(2-2e-t -te-t - l.t2e-t).lim Xl (t) = lim X2 (t) = 2vo

2 't\037oo t\037oo)

21.eAt = [1it I

-tt ]

22.eAt = [1

\0379\037t 14t

6t ]23.eAt =

[1

rt

1

ott \037

t

1/]24. eAt =

[5/:13;t2 \037 7t .\037?: 8t2

]3t 0 1-3t

25.eAt =[e\037t

5\037\037t2t lx(t) = eAt [i]26.eAt = [1:::7t e\037t lx(t) = eAt [-lg]27.eAt =

[\0372?t

(3t

i\037r)et 1x(t) = eAt

[n28.eAt =

[1;::5t e\037t \037

], x(t) = eAt

[\037g

](20t + 150t2)e5t 30te5t e5t 60

[

1 2t 3t + 6t 2 4t + 6t 2+ 4t 3

]29 A 0 1 6t 3t + 6t2

t.et = 0 0 e1 2t 'o 0 0 1

x(t) = eAt

U ])

Section5.7)

1.4>(t)= [_:::::lx(t) =\037 [_;::\037 :::]

2.4>(t)=[\037

-2:::lx(t) =\037 [:\037 15;;\037 ]

3.c)(t) = [5cos4t . -5sin 4t

]2cos4t + 4SIn 4t 4cos4t -2sin 4t 'x(t) = 1.

[-5sin 4t

]4 4cos4t -2sin 4t

4. 4>(t)= e2t

[\037

1it lx(t) = e2t

[1it]

5. c)(t) = [ 2cos3t. -2Sin3t]-3cos3t + 3SIn 3t 3cos3t + 3sin 3t '

x(t) = 1. [3cos3t -sin 3t

]3 -3cos3t + 6sin 3t

6.c)(t) = e5t

[cos4t-2sin4t 2cos4t+ 2sin4t

]2cos4t 2sin 4t 'x(t) = 2e5t

[cos4t. + sin 4t

]SIn 4t

[6 3et 2e-t

] [-12+12et +2e-t

]7. c)(t) = 2 et e-t , x(t) = -4+4et +e-t

5 2et 2e-t -10+ Set + 2e-t

[0 et e3t

] [et

]8.c)(t) = e-2t -et _e3t , x(t) = -et +e-2t

_e-2t 0 e3t _e-2t

9.eAt = [2e::-et -2e33t + 2et

]e -et -et + 2et

A

[-2+ 3e2t 3-3e2t

]10.e t -- -2 + 2e2t 3-2e2t

11.eAt = [3e3t-2e2t -3e3t + 3e2t

]2e3t -2e2t -2e3t + 3e2t

A

[-3et +4e2t 4et -4e2t

]12.e t -- -3et + 3e2t 4et -3e2t

13.eAt = [4e:t-3et -4e3t + 4et

]3et -3et -3e3t +4et

1 At

[-Set+ ge2t 6et -6e2t

]4. e = -12et+ 12e2t get _ Se2t

15.eAt = [5e\037t

-4et -10e2t + 10et

]2et -2et -4e2t + 5et

16.eAt = [-get + 1

0\0372t

15et -15e2t

]-6et + 6et 10et -ge2t

[e4t + e2t e4t _ e2t

]17 eAt - I\302\267

- 2 e4t _ e2t e4t + e2t

18.eAt = \037

[e2t + e6t _e2t + e6t

]2 _e2t + e6t e2t +e6t

[4eIOt + e5t 2elOt -2e5t

]19eAt - 1.\302\267

- 5 2elO t -2e5t elOt +4e5t

eAt _ \037 [e5t +4el5t -2e5t + 2eI5t

]20. -5 -2e5t + 2eI5t 4e5t + el5t)

30.)

[

1

eAt = e3t 6t9t + ISt 2

12t + 54t 2 + 36t 3

x(t) = eAt

U ])

o1

6t9t + 18t2)

oo1

6t) \037l)

33. x(t) = [CI c?sht + C2 sinh t

]CI sInh t + C2 cosht

[1 2t 3t + 4t 2

]36.eAt = et

g \037 it

[e2t 3e2t -3et 13e2t -(13+ 9t)et

]37. eAt = 0 et 3tet

o 0 et

[e5t 4elO t -4e5t 16elOt -(16+ 50t)e5t

]38.eAt = 0 elOt 4eIOt -4e5t

o 0 e5t

39. eAt =

[\037 3\037\037t

o 0o 0)

[e3t35. eAt =

0)

4te3t

]e3t)

12e2t -(12+ 9t)et (51+ 18t)et -(51-36t)e2t

]3e2t -3et 6et -(6-9t)e2t

e2t 4e3t -4e2t

o e\037)

40. eAt =

[e\037t

4t e2t (4t + St2)e2t l00e3t -(100+ 96t + 32t2)e2

]e2t 4t e2t 20e3t -(20+ 16t)e2t

0 e2t 4e3t -4e2t

0 0 e3t)))


Section5.8)1.x(t) =\037, y(t) =-\037

2.x(t) =k (1+ 12t),y(t) = -

\037

(5+ 4t)

3.x(t) =7\0376

(864e-t +4e6t-868+ 840t -504t 2),y(t) =

7\0376(-864e-t + 3e6t+ 861-882t + 378t 2)

4. x(t) =8\037

(9ge5t -8e-2t -7et), y(t) =8\037

(9ge5t +48e-2t -63et)5.x(t) =

\037

(-12-e-t -7te-t), y(t) =\037

(-6-7te-t)6.x(t) = -

2\0376(91+ 16t)et , y(t) = 3

12(25+ 16t)et

7. x(t) =4\037O

(36get + 166e-9t -125cost -105sin t),y(t) =

4\037o(36get -24ge-9t -120cost-150sint)

8.x(t) = \037(17cost+2sint),y(t) = \037(3cost+5sint)9.x(t) =

\037

(sin 2t + 2t cos2t + t sin 2t), y(t) =\037t

sin 2t

10.x(t) = /3et(4cost -6sin t), y(t) =/3et(3cost + 2sin t)

11.x(t) =\037

(1-4t +e4t), y(t) =\037

(-5+ 4t + e4t)12.x(t) = t2, y(t) = -t2

13.x(t) = t(1+ 5t)et , y(t) =-\037tet

14.x(t) =k (-2+ 4t -e4t + 2te4t), y(t) = tt (-2+ e4t)

15.(a) Xl(t) = 200(1-e-tjlO ), X2(t) = 400(1+e-tjlO -2e-tj20 )(b) Xl (t) -+ 200and X2(t) -+ 400as t -+ +00(c) Tank 1:about 6 min 56s; tank 2:about 24min 34s

16.(a) Xl (t) = 600(1-e-tj20 ), X2(t) = 300(1+e-tjlO -2e-tj20 )(b) Xl (t) -+ 600and X2(t) -+300as t -+00(c) Tank 1:about 8 min 7 sec;tank 2:about 17min 13sec

17.Xl (t) = 102-95e-t -7e5t, X2(t) = 96-95e-t -e5t

18.Xl (t) = 68-110t-75e-t +7e5t, X2(t) = 74-80t -75e-t +e5t

19.Xl(t) = -70-60t + 16e-3t + 54e2t ,X2(t) = 5-60t -32e-3t + 27e2t

20. Xl (t) = 3e2t + 60te2t -3e-3t , X2(t) = -6e2t + 30te2t +6e-3t

21.Xl (t) = -e-t -14e2t + 15e3t, X2 (t) = -5e-t -10e2t + 15e3t

22. Xl (t) = -10e-t -7te-t + 10e3t -5te3t,X2(t) = -15e-t -35te-t + 15e3t -5te3t

23. XI (t) = 3+ 11t + 8t2, X2 (t) = 5+ 17t + 24t 2

124. Xl(t) = 2+t+lnt,x2(t)= 5+3t--+3lntt)

25. Xl(t) = -1+8t +cost-8sint,X2 (t) = -2+ 4t + 2cost -3sin t

26. Xl (t) = 3cost -32sin t + 17t cost + 4t sin t,X2 (t) = 5cost -13sin t + 6t cost + 5t sin t

27. Xl (t) = 8t3 + 6t4, X2(t) = 3t 2-2t 3 + 3t4

28. Xl (t) = -7+ 14t -6t 2 + 4t 2 ln t,X2 (t) = -7+ 9t -3t 2 + In t -2t In t + 2t 2 In t

29. Xl (t) = t cost - (In cost)(sin t),X2(t) = t sin t + (In cost)(cost)

30. Xl(t) = tt2cos2t,X2(t) = tt

2sin2t

31.Xl (t) = (9t2 + 4t 3)et, X2 (t) = 6t 2et , X3 (t) = 6tet

32. Xl (t) = (44+ 18t)et + (-44+ 26t)e2t ,X2(t) = 6et + (-6+ 6t)e2t , X3(t) = 2te2t

33. Xl (t) = 15t2 + 60t 3 + 95t 4 + 12t5, X2(t) = 15t2 + 55t 3 + 15t4,X3(t) = 15t2 + 20t 3, X4(t) = 15t2

34. Xl (t) = 4t 3 + (4+ 16t+ 8t2)e2t,X2(t) = 3t2 + (2+ 4t)e2t ,X3(t) = (2+ 4t + 2t 2)e2t , X4(t) = (1+ t)e2t)

Chapter6Section6.1)

In Problems 1 through 10we round off the indicated values to 3decimal places.1.Approximate values 1.125and 1.181;true value 1.2132.Approximate values 1.125and 1.244;true value 1.3593.Approximate values 2.125and 2.221;true value 2.2974. Approximate values 0.625and 0.681;true value 0.7135. Approximate values 0.938and 0.889;true value 0.8516.Approximate values 1.750and 1.627;true value 1.5587. Approximate values 2.859and 2.737;true value 2.6478.Approximate values 0.445and 0.420;true value 0.4059.Approximate values 1.267and 1.278;true value 1.287

10.Approximate values 1.125and 1.231;true value 1.333)

Problems 11through 24call for tables ofvalues that would occupytoo much spacefor inclusion here. In Problems 11through 16we givefirst the final x-value, next the corresponding approximate y-valuesobtained with step sizesh = 0.01and h = 0.005,and then the final

true y-value. (All y-values rounded off accurate to 4 decimal places.)11.1.0,-0.7048,-0.7115,-0.718312.1.0,2.9864,2.9931,3.000013.2.0,4.8890,4.8940,4.899014.2.0,3.2031,3.2304,3.258915.3.0,3.4422,3.4433,3.444416.3.0,8.8440,8.8445,8.8451)

In Problems 17 through 24we give first the final x-value and then the

corresponding approximate y-values obtained with step sizesh = 0.1,h = 0.02,h = 0.004,and h = 0.0008respectively. (Ally-values rounded off accurate to 4 decimal places.))

17.1.0,0.2925,0.3379,0.3477,0.349718.2.0, 1.6680,1.6771,1.6790,1.679419.2.0,6.1831,6.3653,6.4022,6.409620. 2.0,-1.3792,-1.2843,-1.2649,-1.261021.2.0,2.8508,2.8681,2.8716,2.872322.2.0,6.9879,7.2601,7.3154,7.326423. 1.0,1.2262,1.2300,1.2306,1.230724. 1.0,0.9585,0.9918,0.9984,0.999725.With both step sizesh = 0.01and h = 0.005,the approximate

velocity after 1secondis 16.0ftlsec (80%of the limitingvelocity of20ftlsec); after 2 secondsit is 19.2ftlsec (96%of the

limiting velocity).26.With both step sizesh = 1and h = 0.5,the approximate

population after 5 years is 49deer(65%of the limitingpopulation of75 deer);after 10years it is 66deer(88%of the

limiting population).27.With successivestep sizesh = 0.1,0.01,0.001,...the first four

approximations to y(2) we obtain are 0.7772,0.9777, 1.0017,and 1.0042.It therefore seemslikely that y(2) \037 1.00.

28.With successivestep sizesh = 0.1,0.01,0.001,...the first four

approximations to y(2) we obtain are1.2900,1.4435,1.4613,and 1.4631.It therefore seems likely that y(2) \037 1.46.)))

29.)

-1.0-0.7-0.4-0.1

0.20.5)

30.)

1.81.92.0)

31.)

0.70.80.9)

Section6.21.)

0.10.20.30.40.5)

1.81001.63811.48241.34161.2142)

1.80971.63751.48161.34061.2131)

Note: In Problems 2 through 10,we give the value ofx, the

corresponding improved Euler value ofy, and the true value ofy.

2.0.5, 1.3514,1.31913.0.5,2.2949,2.29744. 0.5,0.7142,0.71315.0.5,0.8526,0.85136.0.5, 1.5575,1.55767. 0.5,2.6405,2.64758.0.5,0.4053,0.40559.0.5, 1.2873,1.2874

10.0.5, 1.3309,1.3333)

In Problems 11through 16we give the final value ofx, the

corresponding values ofy with h = 0.01and with h = 0.005,and thetrue value of y.)

11.1.0,-0.71824,-0.71827,-0.71828)


12.1.0,2.99995,2.99999,3.0000013.2.0,4.89901,4.89899,4.8989814.2.0,3.25847,3.25878,3.2588915.3.0,3.44445,3.44445,3.4444416.3.0,8.84511,8.84509,8.84509)

In Problems 17through 24we give the final value ofx and the

corresponding values ofy for h = 0.1,0.02,0.004,and 0.0008.

17.1.0,0.35183,0.35030,0.35023,0.3502318.2.0, 1.68043,1.67949,1.67946,1.6794619.2.0,6.40834,6.41134,6.41147,6.4114720. 2.0,-1.26092,-1.26003,-1.25999,-1.2599921.2.0,2.87204,2.87245,2.87247,2.8724722. 2.0,7.31578,7.32841,7.32916,7.3292023. 1.0,1.22967,1.23069,1.23073,1.2307324. 1.0,1.00006,1.00000,1.00000,1.0000025. With both step sizesh = 0.01and h = 0.005the approximate

velocity after 1second is 15.962ft/sec (80%of the limiting

velocity of20ft/sec); after 2 secondsit is 19.185ft/sec (96%ofthe limiting velocity).

26. With both step sizesh = 1and h = 0.5the approximatepopulation after 5 years is49.391deer(65%of the limiting

population of75 deer);after 10secondsit is66.113deer(88%ofthe limiting population).

27. With successivestep sizesh = 0.1,0.01,0.001,...the first

three approximations to y(2) we obtain are 1.0109,1.0045,and

1.0045.It therefore seemslikely that y(2) \037 1.0045.28. With successivestep sizesh = 0.1,0.01,0.001,...the first four

approximations to y(2) we obtain are 1.4662,1.4634,1.4633,and 1.4633.It therefore seemslikely that y(2) \037 1.4633.

29. Impact speedapproximately 43.22m1s30. Impact speedapproximately 43.48m1s

Section6.81.y(0.25)\037 1.55762;y(0.25)= 1.55760.

y(0.5) \037 1.21309;y(0.5)= 1.21306.Solution: y = 2e-x)

In Problems 2 through 10we give the approximation to y(0.5),itstrue value, and the solution.)

2.1.35867,1.35914;y =\037e2x3.2.29740,2.29744;y = 2ex -1

4. 0.71309,0.71306;y = 2e-x +x-I5.0.85130,0.85128;y = -eX+ x + 26.1.55759,1.55760;u = 2exp (-x2

)7. 2.64745,2.64749;y = 3exp (-x3

)8.0.40547,0.40547;y = In(x + 1)9.1.28743,1.28743;y = tan

\037

(x +]f)10.1.33337,1.33333;y = (1-X2)-111.Solution: y(x) = 2-eX.)

0.0 1.00000 1.00000 1.000000.2 0.77860 0.77860 0.778600.4 0.50818 0.50818 0.508180.6 0.17789 0.17788 0.177880.8 -0.22552 -0.22554 -0.225541.0 -0.71825 -0.71828 -0.71828)))


In Problems 12through 16we give the final value ofx, the

corresponding Runge-Kutta approximations with h = 0.2and withh = 0.1,the exact value of y, and the solution.)

12.1.0,2.99996,3.00000,3.00000;y = 1+2/(2-x)13.2.0,4.89900, 4.89898,4.89898;y = .v8+x4

14.2.0,3.25795,3.25882,3.25889;y = 11(1-In x)

15.3.0,3.44445,3.44444,3.44444;y = x +4x-2

16.3.0,8.84515,8.84509,8.84509;y = (x6-

37)1/3)

In Problems 17through 24we give the final value ofx and the

corresponding values ofy with h = 0.2,0.1,0.05,and 0.025.

17.1.0,0.350258,0.350234,0.350232,0.35023218.2.0, 1.679513,1.679461,1.679459,1.67945919.2.0,6.411464,6.411474,6.411474,6.41147420. 2.0,-1.259990,-1.259992,-1.259993,

-1.25999321.2.0,2.872467,2.872468,2.872468,2.87246822.2.0,7.326761,7.328452,7.328971,7.32913423. 1.0,1.230735,1.230731,1.230731,1.23073124. 1.0,1.000000,1.000000,1.000000,1.00000025. With both step sizesh = 0.1and h = 0.05,the approximate

velocity after 1secondis 15.962ft/sec (800/0of the limitingvelocity of20ft/sec);after 2 secondsit is 19.185ft/sec (960/0ofthe limiting velocity).

26. With both step sizesh = 6 and h = 3,the approximatepopulation after 5 years is49.3915deer(650/0of the limitingpopulation of75 deer);after 10years it is66.1136deer(880/0 ofthe limiting population).

27. With successivestep sizesh = 1,0.1,0.01,...the first four

approximations to y(2) we obtain are 1.05722,1.00447,1.00445and 1.00445.Thus it seemslikely that y(2) \037 1.00445accurate to 5 decimal places.

28. With successivestep sizesh = 1,0.1,0.01,...the first fourapproximations to y(2) we obtain are 1.48990,1.46332,1.46331,and 1.46331.Thus it seemslikely that y(2) \037 1.4633accurate to 5 decimal places.

29. Timealoft: approximately 9.41seconds30. Timealoft: approximately 9.41seconds

Section6.4)

The format for the first eight answers is this: (x(t), y(t\302\273at t = 0.2

by the Euler method, by the improved Euler method, by the

Runge- Kutta method, and finally the actual values.)

1.(0.8800,2.5000),(0.9600,2.6000),(1.0027,2.6401),(1.0034,2.6408)

2.(0.8100,-0.8100),(0.8200,-0.8200),(0.8187,-0.8187),(0.8187,-0.8187)

3.(2.8100,2.3100),(3.2200,2.6200),(3.6481,2.9407),(3.6775,2.9628)

4. (3.3100,-1.6200),(3.8200,-2.0400),(4.2274,-2.4060),(4.2427,-2.4205)

5. (-0.5200,2.9200),(-0.8400,2.4400),(-0.5712,2.4485),(-0.5793,2.4488)

6.(-1.7600,4.6800),(-1.9200,4.5600),(-1.9029,4.4995),(-1.9025,4.4999)

7. (3.1200,1.6800),(3.2400,1.7600),(3.2816,1.7899),(3.2820,1.7902))

8.(2.1600,-0.6300),(2.5200,-0.4600),(2.5320,-0.3867),(2.5270,-0.3889)

9.At t = 1we obtain (x, y) = (3.99261,6.21770)(h = 0.1)and

(3.99234,6.21768)(h = 0.05);the actual value is(3.99232,6.21768).

10.At t = 1we obtain (x, y) = (1.31498,1.02537)(h = 0.1)and

(1.31501,1.02538)(h = 0.05);the actual value is(1.31501,1.02538).

11.At t = 1we obtain (x, y) = (-0.05832,0.56664)(h = 0.1)and

(-0.05832,0.56665)(h = 0.05);the actual value is(-0.05832,0.56665).)

12.We solved x'= y, y' = -x+ sin t, x(O)= y(O)= O.With

h = 0.1and alsowith h = 0.05we obtain the actual value

x(1.0)\037 0.15058.)

13.Runge-Kutta, h = 0.1:about 1050ft in about 7.7s

14.Runge- Kutta, h = 0.1:about 1044ft in about 7.8s

15.Runge- Kutta, h = 1.0:about 83.83mi in about 168s

16.At 40\302\260: 5.0s,352.9ft; at 45\302\260: 5.4s,347.2ft; at 50\302\260: 5.8s,334.2ft (all values approximate)

17.At 39.0\302\260 the range is about 352.7ft. At 39.5\302\260 it is 352.8;at 40\302\260,

352.9;at 40.5\302\260, 352.6;at 41.0\302\260, 352.1.)

18.Just under 57.5\302\260)

19.Approximately 253ft/s

20. Maximum height: about 1005ft, attained in about 5.6s;range:about 1880ft; time aloft: about 11.6s)

21.Runge-Kutta with h = 0.1yields these results:

(a) 21400ft, 46s,518ft/s; (b) 8970ft, 17.5s; (c) 368ft/s (att \037 23).)

Chapter7)

Section7.1)

1.Unstable critical point: x = 4;)

x(t) = 4+ (xo-4)et)

8)

\037 4)

oo) 4) 5)1) 2 3

t)))


2.Stable critical point: x = 3;)

x(t) = 3+ (xo-3)e-t)5. Stable critical point: x = -2;unstable critical point: x = 2;

2[xo + 2+ (xo-2)e4t]x(t) =

Xo + 2- (xo-2)e4t)

6)

2)

\037 3)

\037 0)

-2)

oo) 1) 2 3

t)

4) 5)

o 0.5 1 1.52 2.53 3.54 4.55t)

3.Stable critical point: x = 0; unstable critical point: x = 4;) 6.Stable critical point: x = 3;unstable critical point: x = -3;)4xo

x(t) =Xo + (4-xo)e4t) 3[xo-3+ (xo + 3)e6t ]

x(t) =3-Xo + (xo + 3)e6t)

8)

3)4)

\037)

\0370)

o)

-3)

-4o) 1) 2) 3) 4) 5)

o) 1) 2) 3) 4) 5)t)

t)

4. Stable critical point: x = 3;unstable critical point: x = 0;

3xox(t) =

Xo + (3-xo)e-3t)

7. Semi-stable (seeProblem 18)critical point: x = 2;)

(2t - l)xo-4tx(t) =

tXo-2t - 1)

6)

4)

3

\037 \037 2

0

0

-3 0 1 2 3 4 50 1 2 3 4 5t t)))


8.Semi-stable critical point: x = 3;) 11.Unstable critical point: x = 1;)

x(t) = (3t + l)xo-9t

txo -3t + 1)1 1= -2t

(x(t) -1)2 (XO-

1)2)

6)

4)

2)

\037 3)

\037)

o) o)

2) 3) 4) 5)

t)

( )4(1-xo) + (xo-4)e3t

xt =1-Xo + (xo-4)e3t)

-20 1 2 3 4 5

t

12.Stable critical point: x = 2;

1 1(2-X(t\302\2732

- +2t(2-xoF)

9.Stable critical point: x = 1;unstable critical point: x = 4;)

7)

1)

6)

4)

4)

\037)

\037 2)

0-20 1 2 3 4 5

t -20 1 2 3 4 5

t)

( )2(5-xo) + 5(xo-2)e3t

x t =5-Xo + (xo-2)e3t)

For eachofProblems 13through 18we show a plot ofslopefield and

typical solution curves. The equilibrium solutions of the givendifferential equation are labeled, and the stability or instability ofeachshould beclearfrom the picture.

13.) x'= (x+2)(x-2)2)

10.Stable critical point: x = 5; unstable critical point: x = 2;)

8)

-1o) 1) 2) 3) 4) 5)

4

2

\037 0

-2

-4

0 1 2 3 4t)

5)

\037)

2)

t)))

14.) x'=x(x2 -4))18.) x'=X3 (X

2-4))

\037 0)

2)

4)

2)

\037 0)

-2-4

0 1 2 3 4 0 1 2 3t t

19.There are two critical points if h < 2\037,

one critical point if15. x'= (x2-4)2 h = 2

\037,

and no critical points if h > 2\037.

The bifurcation

, , I I I Idiagram is the parabola (e-5)2= 25-10h in the he-plane.

4 , , I I , , 20. There are two critical points if s <l\037'one critical point if

, , , I , , , s =l\037'and no critical points if s >

l\037

.The bifurcation diagram2 is the parabola (2e-5)2= 25(1-16s)in the se-plane.

Section7.2\037 0 1.7.1.13 2. 7.1.15 3. 7.1.18 4. 7.1.12-2 5.7.1.11 6. 7.1.17 7. 7.1.14 8. 7.1.16

I I I A I I I I I 9.Equilibrium solutions x(t) = 0, ::l:2.The critical point (0,0) in, , , I I I I I

, , I , , I , I the phase plane looks like a center, whereas the points (::l:2,0)-4 , , , , , , , , look like saddle points., I , I , , I I

0 1 2 3 4 5t

16. x'= (x2-4)3

4, I

, ,, I ;>.., 0, I

2 I I

\037 0-5

-2 -5 0 5XI=I_Q I I

, , , , , x, I , I I-4 , I , I I 10.Equilibrium solution x(t) = O.The critical point (0,0) in the, I , I I

0 1 2 3 4 phase plane looks like a spiral sink.t

517. x'=x2 (x2-4)

4

2 ;>.., 0

\037 0

-2 -5-3 0 3

-4 x

0 1 2 3 4 11.Equilibrium solutions x(t) = ... , -2Jr,-Jr,0,Jr, 2Jr, ....The

t phase portrait shown in the solutions manual suggests that the)))


critical point (nJr, 0) in the phase plane isa spiral sink if n iseven, but isa saddle point if n isodd.)

12.Equilibrium solution x(t) = O.The critical point (0,0)in the

phase plane looks like a spiral source, with the solution curvesemanating from this source spiraling outward toward a closedcurve trajectory.)

4)

2)

\037 0)

-2)

-4) -2 -1) o) 1) 2)

x)

13.Solution x(t) = xoe-2t , y(t) = yoe-2t.The origin is a stableproper node similar to the oneillustrated in Fig. 6.1.4.)

14.Solution x(t) = xoe2t , y(t) = yoe-2t.The origin is an unstablesaddle point.)

5)

\037 0 18.

19.-5

-5 0 5 20.x)

15.Solution x(t) = xoe-2t , y(t) = yoe-t . Theorigin isa stable node.)

5)

\037 0)

-5-5) o

x)

5)

16.Solution x(t) = xoet , y(t) = yoe3t .The origin is an unstable)

improper node.)

5)

\037 0)

-5-5) o

x)

5)

17.Solution x(t) = A cost + Bsin t, y(t) = Bcost -A sin t. The

origin isa stable center.)

\037)

5)

ox)

5)

Solution x(t) = A cos2t + Bsin 2t,y(t) = -2Bcos2t + 2A sin 2t. The origin isa stable center.

Solution x(t) = A cos2t + Bsin 2t, y(t) = Bcos2t -A sin 2t.The origin isa stable center.

Solution x(t) = e-2t (A cost + Bsin t),y(t) = e-2t [(-2A + B)cost - (A + 2B)sin t].The origin isastable spiral point.)

\037 0)

5 /')

-5-5) o

x)

II:

5)

23. The origin and the circlesx2 + y2 = C > 0; the origin isa stablecenter.)

24. The origin and the hyperbolas y2-x2 = C;the origin is an)))


unstable saddle point.) Section7.8)

1.Asymptotically stable node)

5)

5)

\037 0)

\037 0)

-5-5) o

x)

5) -5-5) o

x)

5)

25.The origin and the ellipsesx2 + 4y2 = C > 0; the origin isastable center.)

2.Unstable improper node)

3.Unstable saddle point)

\037 0)

5)

5)

\037 0)

-5-5) o

x)

5)

ox)

4. Unstable saddle point)

26. The origin and the ovals of the form x4 + y4 = C > 0;the originisa stable center.)

5. Asymptotically stable node)

5)

-4)

-5-5) o) 5)

4

\037 0

\037 0)

x)

-4) ox) 6.Unstable node)))


7. Unstable spiral point) 13.Unstable saddle point: (2,2))

5) 5)

\037 0) \037 0)

-5-5) o

x)

5)

-5-5) o

x)

5)

8.Asymptotically stable spiral point) 14.Unstable saddle point: (3,4))

9.Stable, but not asymptotically stable, center) 15.Asymptotically stable spiral point: (1,1))

\037 0 \037 0JJJ\037

-50 5 -5 0 5x x)

5) 5)

10.Stable, but not asymptotically stable, center) 16.Unstable spiral point: (3,2))

11.Asymptotically stable node:(2, 1)) 17.Stable center:(\037,

-\037 ))

5) 5)

\037 0) \037 0)

-5-5) o

x)

5)

-5-5) o

x)

5)

12.Unstable improper node:(2,-3)) 18.Stable, but not asymptotically stable, center: (-2,-1))))


19.(0,0) isa stable node. Also, there isa saddle point at

(0.67,0.40).)(::1::0.82,::1::5.06)and nodal sourcesat (::1::3.65,=r=0.59).)

2)

6

4

2

;>.., 0

-2-4

-6-6 -4 -2 0 2 4 6

x)

;>.., 0)

-2)

-2) ox)

2)

20.(0,0) is an unstable node. Also, there isa saddle point at

(-1,-1)and a spiral sink at (-2.30,-1.70).)23. (0,0) isa spiral sink. Also, there isa saddle point at

(-1.08,-0.68).)3)

3'\"\037 '\\)

\037)

;>.., 0;>.., 0

-1\037 \"-

-2\0371-3 -3 '\\ \037

-3 0 3x -3 -2 -1 0 1 2 3

x)

21.(0,0) is an unstable saddle point. Also, there is a spiral sink at

(-0.51,-2.12).)

5)

24. (0,0) is an spiral source.No other critical points are visible.)

5)

;>.., 0)

;>.., 0)

-5-5) o

x)

5)

-5-5) o

x)

5)

22.(0,0) is an unstable saddle point. Also, there are nodal sinks at)))

25. Theorem 2 implies only that (0,0) isa stable sink-either a nodeor a spiral point. The phase portrait for -5:Sx, y :S5 showsalso a saddle point at (0.74,-3.28)and spiral sink at

(2.47,-0.46).The origin looks like nodal sink in a secondphase portrait for -0.2<x, y <0.2,which alsoreveals asecond saddle point at (0.12,0.07).)

27. Theorem 2 implies only that (0,0) isa center or a spiral point,but doesnot establish its stability. The phase portrait for-2 :Sx, y < 2 shows alsosaddle points at (-0.25,-0.51)and

(-1.56,1.64),plus a nodal sink at (-1.07,-1.20).The originlooks like a likely center in a secondphase portrait for-0.6<x, y < 0.6.)

5)2)

\037 0)\037 0)

-2-5

-5 0 -2 0 2x x

0.2 0.6)

\037 0)\037)

-0.2-0.2)

-0.6)

ox)

0.2) -0.6) ox)

0.6)

26. Theorem 2 implies only that (0,0) is an unstable source.Thephase portrait for -3:Sx, y < 3 shows alsosaddle points at

(0.20,0.25)and (-0.23,-1.50),aswell asa nodal sink at

(2.36,0.58).)

28. Theorem 2 implies only that (0,0) isa center or a spiral point,but doesnot establish its stability (though in the phase portrait it

looks like a likely center). The phase portrait for-0.25:Sx :S0.25,-1:Sy :S1 alsoshows saddle points at

(0.13,0.63)and (-0.12,-0.47).)

3) 1)

\037 0\037 0

tl' l' l'l' l' i -1l' i t-3

-3 0 3 -0.25 0 0.25x

x)))


29.There is a saddle point at (0,0).The other critical point (1,1)isindeterminate, but looks like a center in the phase portrait.)

37. Note that the differential equation ishomogeneous.)

2) Section7.4)

\037 0)

1.Linearization at (0,0):x'= 200x,y' = -150y;phase planeportrai t:)

5)

-2)

-2) ox)

2)

\037 0)

30.There isa saddle point at (1,1)and a spiral sink at (-1,1).)

3

-5-5 0 5

x\037 0)

Linearization at (75,50):u' = -300v,v' = 100u; phase planeportrai t:)

-3-3)

5)

ou)

ox)

3)

31.There isa saddle point at (1,1)and a spiral sink at (-1,-1).)3)

\037 0)

\037 0) -5-5)

-3) 5. The characteristic equation is)..,2 + 45)..,+ 126= O.)

-3) ox)

3) 7. The characteristic equation is (-24-)..,)2-2 . (18)2= O.Phase

plane portrait:)

32.There isa saddle point at (2, 1)and a spiral sink at (-2,-1).)3)

5)

\037 0)\037

0)

-3)

-3) ox)

3)

-5-5) o

u)

5)))


Phase plane portrait for the nonlinear system in Problems 4-7:) portrai t:)

5)

\037) \0370)

(15,0))

o) 5) 10 15 20x)

-5-5) o

u)

5)

9.The characteristic equation is A2 + 58A-120= O.) 15.The characteristic equation is A

2 + 2A-24= O.)

10.The characteristic equation is (A + 36)(A + 18)-576= O.Phase plane portrait:)

17.The characteristic equation is A2-4A + 6 = O.Phase plane

portrait:)

5) 5)

\037 0) \0370)

-5-5) o

u)

5)

-5-5) o

u)

5)

Phase plane portrait for the nonlinear system in Problems 8-10:) 19.The characteristic equation isA2 + 10= O.Phase plane portrait:)

5

20(0,14)

15

\037 \037 010

5 (12,6)(20,0)

0 -50 5 10 15 20 -5

x u)

12.The characteristic equation is A2 + 2A-15= O.) 21.The characteristic equation is A

2-A -6 = O.)

13.The characteristic equation isA2+ 2A + 6 = O.Phase plane) 22. The characteristic equation is A

2-5A+ 10= 0 .Phase plane)))

portrait:)

5)

\037 0)

-5-5) o)

u)

24. The characteristic equation is A2 + 5A-14= o.

25. The characteristic equation is A2 + 5A + 10= O.Phase plane

portrait:)

5)

\0370)

-5-5) o

u)

5)

26.Naturally growing populations in competitionCritical points: nodal source (0,0) and saddle point (3,2)Nonzero coexisting populations x(t) = 3,yet) = 2

27.Naturally declining populations in cooperationCritical points: nodal sink (0,0) and saddle point (3,2)Nonzero coexisting populations x(t) = 3,yet) = 2)

5)

(3,2))\037)

(0,0)o)

o) 5)x)

28. Naturally declining predator, naturally growing prey populationCritical points: saddle point (0,0) and apparent stable center(4,8))


Nonzero coexisting populations x(t) = 4, yet) = 8)

15)

\037)

5)

o)

(0,0)o 5 10

x)

15)

29.Logistic and naturally growing populations in competitionCritical points: nodal source (0,0),nodal sink (3,0),and saddle

point (2,2)Nonzero coexisting populations x(t) = 2,yet) = 2)

5)

\037)

(2,2))

o)

(3,0))(0,0)o) 5)

x)

30.) Logistic and naturally declining populations in cooperationCritical points: saddle point (0,0),nodal sink (3,0),and saddle

point (5,4)Nonzero coexisting populations x(t) = 5,yet) = 4Logistic prey, naturally declining predator populationCritical points: saddle points (0,0) and (3,0),spiral sink (2,4)Nonzero coexisting populations x(t) = 2,yet) = 4)

31.)

5

(2,4))

\037)

o)

(0,0)o) 5)

x)

32.) Logistic populations in cooperationCritical points: nodal source (0,0),saddle points (10,0)and

(0,20),nodal sink (30,60)Nonzero coexisting populations x(t) = 30,yet) = 60Logistic prey and predator populationsCritical points: nodal source (0,0),saddle points (30,0) and

(0,20),nodal sink (4,22))

33.)))


Nonzero coexisting populations x(t) = 4, y(t) = 22) 4. Eigenvalues: -1::i:2i; stable spiral point)

4)

t tt 1\\; \\)

\037

'\\ \037

\037 \037

t l'l' l'l' l'

-7 J0 4x)

40)

\037 0)(4,22))

\037

20)

(0,20))

o)

(0,0)o) 10

x)

(15,0)20)

-4-4)

34. Logistic prey and predator populationsCritical points: nodal source (0,0), saddle points (15,0) and

(0,5),spiral sink (10,10)Nonzero coexisting populations x(t) = 10,y(t) = 10)

5. Critical points: (0,nn) where n is an integer; an unstable saddle

point if n iseven, a stable spiral point if n isodd)

31t)

21t)

5)

\037 0

-1t

-21t

-31t\037

\037 \037 \037

-5 0 5x)

Section7.5)

1.Eigenvalues: -2,-3;stable node)

\037 0)

6.Critical points: (n, 0) where n is an integer; an unstable saddle

point if n iseven, a stable spiral point if n isodd7. Critical points: (nn, nn) where n is an integer; an unstable

saddle point if n iseven, a stable spiral point if n isodd)

-5-10) o) 10)

31t

21t

1t

\037 0

-1t

-21t

-31t)

x)

2.Eigenvalues: 1,3;unstable node)

3.Eigenvalues: -3,5;unstable saddle point)

3) -31t -21t -1t 0 1t 21t 31t)

x)

-3-3) o) 3)

8.Critical points: (nn, 0) where n is an integer; an unstable node if

n iseven, an unstable saddle point if n isodd9.If n isodd then (nn, 0) is an unstable saddle point.10.If n isodd then (nn, 0) isa stable node.11.(nn, 0) isa stable spiral point.12.Unstable saddle points at (2,0) and (-2,0),a stable center at

(0,0)13.Unstable saddle points at (2,0) and (-2,0),a stable spiral point

at (0,0)14.Stable centers at (2,0) and (-2,0),an unstable saddle point at

(0,0))

\037 0)

x)))


15.A stable center at (0,0) and an unstable saddle point at (4,0)16.Stable centers at (2,0), (0,0),and (-2,0),unstable saddlepoints at (1,0)and (-1,0)

17.(0,0) isa spiral sink.

10)

20. (nn, 0) isa spiral sink if n iseven, a saddle point if n is odd.)

Appendix1.Yo = 3,YI = 3+ 3x,Y2 = 3+ 3x +

\037

x2,Y3 = 3+ 3x +

\037X2+ \037X3,

Y4 = 3+ 3x +\037

x2 +\037

x3 + kx4; Y (x) = 3ex

Yo = I'YI = l-x2'Y2 = l-x2 +

\037X4,

Y3 = 1 -x2 +\037

x4- i x6,Y4 = 1 -x2 +

\037X4

-iX6 + d4x 8; y(x) = exp (-x2

)Yo = 0, YI = 2x, Y2 = 2x+ 2x2,Y3 = 2x + 2x2 +

\037

x3,

Y4 = 2x + 2x2 +\037

x3 +\037

x4; Y (x) = e2x -1O

2 2 + 14 2 + 14+ 16Yo = , YI = X , Y2 = x 2X , Y3 = x 2X 6X ,Y4 = x2 +

\037X4+ iX6 +

2\037

x 8; y(x) = exp(x2)- 1

Yo = 1,YI = (1+x)+ \037X2, Y2 = (1+x +X2) + ix3,Y3 = (1+x +X2 +

\037X3)+

2\037X4;

Y (x) = 2ex -1-x = 1+ x + x2 +\037

x3+ . . .

11.Yo = 1,YI = 1 +x,Y2 = (1+x+X2) +\037x3,

Y3 = (1+x+X2 +x3) +\037X4

+\037X5

+\037X6

+\0373

X7;

1

y(x) =-= 1+ x + x2 + x3 +x4 +x5+...I-x12.Yo = 1,YI = 1+

\037

x, Y2 = 1 +\037

x +\037

x3+ kx3+

\037

x4,Y3 = 1+

\037

x +\037

x2 +1

56x3 +

\037

x4 + . . . ; Y (x) = (1-x)-I/2

13.[\037\037]

- [-\037 l[\037: ]- [-\037 1\037\037 l)

5 3.\037 0

5.-5

7.

0 5 9.x)

18.(0,0) is a spiral sink; the points (::i:2,0) are saddle points.)

\037 0)

5\037

tt \037

\037J,

\037t

\037

\037

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INDEX)

Boldfacepagenumbers indicate where terms are defined.)

AAbel'sformula, 112,124Acceleration,12Addition (ofmatrices),348Adriatic Sea,514Air resistance,85

proportional to squareof velocity,88

proportional to velocity, 22,86Airy equation, 217,259Airy function, 218Alligator population, 82,83Almost linear system, 502,508

stability, 508Ampere, 173Amplification factor, 165Amplitude, 139Analog computers, 174Analytic function, 196Argument (ofcomplexnumber), 132Arnold, David,31Artin, Emil, 251Associatedhomogeneousequation,

101,121,149,151,356,362Asymptotic approximations, 257Asymptotic stability, 494,507,508Augmented coefficientmatrix, 360Automobile:

two-axle,392vibrations, 173

Autonomous differential equation, 481critical point, 481equilibrium solution, 481stable critical point, 482unstable critical point, 482

Autonomous system, 488linearized,502

Auxiliary equation, seeCharacteristicequation

Average error, 79)

BBatted baseball,469,474)

Beats,164Bernoulli, Daniel (1700-1782),248Bernoulli equation, 63Bessel,Friedrich W. (1784-1846),248Bessel'sequation, 124,194,228,245,

248,257,300modified, 263parametric, 255

Besselfunction:

asymptotic approximations, 257identities, 253modified, 263order1,first kind, 232,order1,secondkind, 246,247order4,232order

\037,

246order n, secondkind, 252orderp, first kind, 251orderzero,first kind, 228orderzero,secondkind, 245solutions in terms of, 258

Bifurcation, 485,511,525,537diagram, 486Hopf,512point, 486

Big bang, 44Binomial series,195,206,233Birth rate, 74Blackhole,94Boundary value problem, 181Brachistochroneproblem, 46Broughton Bridge, 166Buckledrod, 191Buckling ofvertical column, 259Buoy, 146Bus orbit (Moon-Earth),473)

CCantilever, 188Capacitor,173Carbon-14,37Carrying capacity, 77Cart with flywheel, 162)

Cascade,55Catenary, 46Cauchy-Schwarzinequality, 570Cello,178Center,494

ofpower series,196stable,494,506

Chain (ofgeneralizedeigenvectors),398,401

Chaos,542and period-doubling, 547

Characteristicequation, 109,125ofmatrix, 368complexroots, 131,133distinct real roots, 109,126repeatedroots, 111,129,133

Characteristicvalue, seeEigenvalueChurchill, Ruel V., 274,302,303Circular frequency, 139,142Clairaut equation, 73Clarinet reed,540Clarke,Arthur, 19Clepsydra,45Closedtrajectory, 495Coefficientmatrix, 355Column vector, 349Compartmental analysis, 371Competing species,516Competition and cooperation,522Competition, measureof, 517Competition system, 517Complementary function, 121,362Complexeigenvalue, 374Complex-valuedfunction, 130Complex-valuedsolution, 374Compound interest, 37Conservation ofmechanical energy,

137,166Constant acceleration,13Constant thrust, 95Continuous dependenceof solutions,

569Convergenceofpowerseries,195)

1-1)))

1-2 Index)

Convolution (offunctions), 297Cooperationand competition, 522Corrector(improved Euler), 446Coulomb, 173Criterion for exactness,68Critical buckling force,191Critical damping, 141Critical point (ofautonomous

equation), 481Critical point (ofsystem), 488

asymptotic stability, 494center,494classification,509isolated,500,503node,491ofpredator-prey system, 514saddlepoint, 492spiral point, 495spiral sink, 495spiral source,495stability, 492

Critical speed(ofwhirling string), 187Crossbow,85,87,89,450,460,474Cumulative error, 435,443Cycloid,346)

DDampedmotion, 136

nonlinear, 530Dampedpendulum oscillations,537Damping constant, 136Death rate, 74Decayconstant, 38Defect(ofeigenvalue), 396Defectiveeigenvalue, 396Degeneratesystem, 343deLaplace,PierreSimon

(1749-1827),275Deltafunction, 317

inputs, 318and step functions, 320

Density offorce,188Deflectionofbeam, 187

deflection curve, 187Dependenceon parameters, 485Dependentvariable missing, 70Derivative:

ofcomplex-valuedfunction, 130ofmatrix function, 354

Determinant, 352Differenceequation, 543Differential equation, 1

autonomous, 481Bernoulli, 63Clairaut, 73dependent variable missing, 70)

differential form, 67exact,67first-order, 7general solution, 10,36,106homogeneous,60,101independent variable missing, 71linear, 47,100normal form, 7order,6ordern, 113ordinary, 7partial, 7particular solution, 10reduciblesecond-order,70Riccati,73separable,32singular solution, 36solution, 2,6

Differential equations and

determinism, 148Differential form, 67Dirac,P.A. M.(1902-1984),317Diracdelta function, 317Direction field, 19,331,489Displacementvector,381Distinct real eigenvalues,369Doomsdaysituation, 82Doomsdayversus extinction, 81Downward motion, 89Drag coefficient,87Drug elimination, 38Duffing equation, 548Duhamel'sprinciple, 322Duplication, caseof, 154Dynamic damper, 391)

Endpoint conditions, 188Endpoint problem, 181,189Equidimensional equation, 222Equilibrium position, 135Equilibrium solution, 22,50,481

stableor unstable, 493of system, 488

Error function, 54Error:

in the Euler method, 442in the improved Euler method, 446in the Runge-Kutta method, 454

Escapevelocity, 92Euler, Leonhard(1707-1783),248,

431Euler buckling force,191Euler equation, 113,135Euler'sformula, 130Euler'smethod, 432

cumulative error,435,443improved, 445localerror,435roundoff error,436for systems, 464

Exact equation, 67Existence,uniqueness of solutions, 23,

24,50,104,114,557,565,567for linear systems, 565global,562local,566of solutions of systems, 334,

Exponential growth, seeNatural

growth

Exponential matrix, 411,417Exponential order,273Exponential series,196Exponents (ofa differential equation),

223External force,136

vector,389)

EEarth-Moon satelliteorbits, 472Eigenfunction, 182Eigenvalue, 182

complete,393complex,374defective,396distinct real,369for matrix, 367multiplicity 2, 397multiplicity k, 393

Eigenvalue method, 367,368Eigenvalue problem, 182Eigenvector, 367

rank r generalized,398Elecuncalresonance,178Elementary row operations,360Elimination, method of, 339Elimination constant, 38Elliptic integral, 536)

FFamous numbers, 442,452,463Farad (unit ofcapacitance),173Feigenbaum, Mitchell, 547Feigenbaum constant, 547,553Fibonaccinumber, 207First-order equation, 7First-order system, 329,355Flagpole,261Flight trajectories,65Flywheel on cart, 162Folia ofDescartes,513ForcedDuffing equation, 548Forcedmotion, 136Forcedoscillations:)))

damped, 168and resonance,388undamped, 162

Forcedvibrations, 101Formal multiplication of series,197Fourier, Joseph,(1768-1830),248,546Freemotion, 136

damped, 141undamped, 138

Freeoscillations,345,384Freespace,97Freevibrations, 101Frequency, 139

resonance,178Frequency equationFrobenius, Georg(1848-1919),222Frobenius series,222

solutions, 225From the Earth to the Moon, 92,94Fundamental matrix, 408Fundamental matrix solutions, 408Fundamental theorem of algebra, 125,

368Funnel, 482)

Gg,14G,90Gamma function, 250,268Gauss'shypergeometric equation, 232Generalpopulation equation, 75Generalsolution, 10,36,106,107

ofhomogeneousequation, 120of nonhomogeneous equation, 121

Generalizedeigenvector,398Generalizedfunctions, 324Geometricseries,195,233Gleick,James,553Globalexistenceof solutions, 562Gzyx, 18)

HHailstone,56Half-life,40Halley'scomet,478Hard spring, 528

oscillation,528Harvesting a logisticpopulation, 453,

483Heaviside,Oliver(1850-1925),275Henry (unit of inductance), 173Hermite equation, 217Hermite polynomial, 217Hole-through-Earth problem, 146Homicidevictim, 45Homogeneousequation, 60,101)

Hooke'slaw, 135,526Hopf bifurcation, 512Hypergeometric equation, series,

232-233Hypocycloid,347)

I

Identity principle, 199Imaginary part, 130Impedance,175Implicit solution, 35Improper integral, 267Improper node,505Improved Euler method, 445

error in, 446for systems, 465

Impulse, 316Ince,E.L.,264,343Independent variable missing, 71Indicial equation, 223Inductor, 173Inhibition, measureof, 517Initial condition, 4,8Initial position, 12Initial velocity, 12Initial value problem, 8, 104,114,557

and elementary row operations,359for linear systems, 360ordern, 114

Integrating factor, 47InverseLaplacetransform, 271Inversematrix, 351Irregular singular point, 220Isolatedcritical point, 500)

JJacobianmatrix, 502Jump, 271)

KKansasCity (skywalk collapse),166Kepler,Johannes (1571-1630),336

laws ofplanetary motion, 336,477Kinetic energy, 166Kirchhoff's laws, 173,328Kutta, Wilhelm (1867-1944),453)

LLakesErie, Huron, and Ontario, 53Languagefamilies, 43Laplacetransform, 267

and convolution, 298ofderivative, 277derivatives of transforms, 299)

Index 1-3)

differentiation, 299existence,273for s large,274generalpropertiesof, 272of higher derivatives, 278and initial value problems, 278of integral, 284integrals of transforms, 301inverse, 271inverse transforms ofseries,304linearity of, 269and linear systems, 281notation, 271ofperiodicfunction, 310products of transforms, 297translation on the s-axis,289translation on the t -axis,305uniqueness of inverse, 274

Legendrepolynomial, 216Legendre'sequation, 194,358,218Limit cycle,513Limiting population, 23,77

Limiting solution, 484Limiting velocity, 22Lineardependence,independence,

105,107,116,117of vector-valued functions, 357

Lineardifferential equation, 47,100Linear system, 333

almost linear, 502associatedhomogeneous equation,

356eigenval ue method, 368first-order, 355general solution, 358homogeneous,334nonhomogeneous, 334,362,420solution, 334,355,362upper triangular form, 360

Linearity ofLaplacetransform, 269Linearization, 502Linearizedsystem, 502Lipschitzcontinuous, 561Localerror, 435Localexistenceof solutions, 566Logarithmic decrement,148Logisticdifferenceequation, 543Logisticequation, 23,76,452,542

competition situation, 80with harvesting, 453,483joint proportion situation, 80limited environment situation, 80

Logisticpopulations, interaction of,522

Lorenz,E.N.,551Lorenzstrange attractor, 551Lorenzsystem, 552)))

1-4 Index)

Lunar lander, 13,90,467)

MMaclaurin series,196Manchester (England) bridge collapse,

166Massmatrix, 381Mass-spring -dashpotsystem, 101Mathematical model,4Mathematical modeling, 4Matrix, 348

addition, 348augmented, 360coefficient,355columns, 348determinant, 352diagonal, 412elements,348elementary row operations,360equali ty, 348exponential, 411fundamental, 408identity, 351,359inverse, 352multiplication, 350nilpotent, 413nonsingular,352order,351principal diagonal, 352rows, 348singular, 352subtraction, 349transpose, 349zero,348

Matrix differential equation, 407Matrix exponential solutions, 414

general,416Matrix-valued function, 354

continuous, 354differentiable, 354

Mechanical-electricalanalogy, 174Mechanicalvibrations, 343Method of elimination, 339Method ofFrobenius, 222

logarithmic case,240nonlogarithmic case,233the exceptionalcases,242

Method ofsuccessiveapproximations,557

Method ofundetermined coefficients,149

for nonhomogeneous systems, 421MexicoCity earthquake, 166Mixture problems,52Modem JazzQuartet, 640Modulus (ofcomplexnumber), 132)

Multiplicity of eigenvalue, 393)

NNatural frequency, 163,345,384

ofbeam, 644Natural growth and decay,37Natural growth equation, 38Natural mode ofoscillation,345,384Newton, Sir Isaac(1642-1727),85,

336Newton's law ofcooling,2,40,57,480Newton's law of gravitation, 90,336,

472Newton's method, 125,185,261Newton's secondlaw of motion, 13,

15,85,95,136,162,166,186,326,381

Nilpotent matrix, 413Nodal sink, 504Nodal source,504Node,491

improper, 492,504,505proper,491,505

N onelementary function, 430Nonhomogeneous equation, 101,121,

122Nonhomogeneous system, 334,362Nonlinear pendulum, 533

periodofoscillation,535,539Nonlinear spring, 527,528Nonsingular matrix, 352Noonburg, Anne, 264Norm, 561)

oOhm (unit ofresistance),173Operationaldeterminant, 341Operator,polynomial differential, 127,

340Orderofdifferential equation, 6Ordinary differential equation, 7Ordinary point, 208

solution near, 209Oscillatingpopulations, 516Overdamping, 141)

pPainleve transcendant, 264Parachute, 87,94,440,450,457,460Parameters,variation of, 158,160Parametric Besselequation, 255Partial differential equation, 7partial fraction decomposition,289Particular solution, 10)

Peacefulcoexistenceof two species,520

Pendulum, 137,146,171,262nonlinear, 533variable length, 262

Period,139Perioddoubling, 546,550

in mechanical systems, 548Periodicfunction, 310Periodicharvesting and restocking,

453Phaseangle, 139Phasediagram, 481Phaseplane,489

position-velocity, 527Phaseplane portrait, 331,489Pheny lethy lamine, 44Physical units, 14Picard,Emile (1856-1941),557Pitchfork diagram, 486Piecewisecontinuous function, 271

jump, 271Piecewisesmooth function, 277Pit and the Pendulum, The,262Pitchfork diagram, 546Poe,Edgar Allan (1809-1849),262Polarform (ofa complexnumber), 132Polking, John, 31,499Polynomial differential operator, 127,

340Population equation, 75Population explosion,75,82Population growth, 37

and perioddoubling, 542Position function, 12Position-velocity phaseplane,527Potential energy, 166Powerseries,194Powerseriesmethod, 194Powerseriesrepresentation, 195Practicalresonance,169Predation, 522Predator, 513Predator-prey situation, 513Predator-prey system, 514Predictor (improved Euler), 446Predictor-correctormethods, 445Prey, 513Principleof superposition, 102,113

for nonhomogeneous equations, 155for systems, 356

Principia Mathematica, 85,336Product ofmatrices,350Propernode,505Pseudofrequency,142Pseudoperiod,142Pure resonance,165)))

R

Radio frequencies,178Radioactive decay,37Radius ofconvergence,201Railway cars,385,391,402Rank r generalizedeigenvector,398Rayleigh, Lord(John William Strott,

1842-1919),540Rayleigh's equation, 540Reactance,176Realpart, 130Recurrencerelation, 200

many-term, 213two-term, 213

Reduciblesecond-orderequation, 70Reduction oforder, 124,238

formula, 239Regular singular point, 220Repeatedquadratic factors,294Resistance:

proportional to velocity, 86proportional to squareof velocity,

88Resistancematrix, 403Resistor, 173Resonance,165,389

electrical,178practical, 169pure, 165and repeatedquadratic factors,294Riccati equation, 73,262

RLCcircuit, 173,308,320initial value problems, 176integrodifferential equation, 308

Rocketpropulsion, 95Rodrigues'formula, 217Rosslerband, 553Rosslersystem, 553Row vector,349Runge, Carl (1856-1927),453Runge-Kutta method, 453

error in, 454for systems, 466variable step sizemethods, 471)

SSaddlepoint, 492,504Saltzman, Barry, 552Sawtooth function, 288,314Schwartz, Laurent, 324Secondlaw ofmotion, 13,15,85,136,

162,166,186,326,381Second-ordersystem, 382,383Separableequation, 32Separatrix, 520,529Series:)

Index 1-5)

binomial, 195,206,233convergent, 195exponential, 196formal multiplication, 197geometric,195,233hypergeometric,233identity principle, 199Maclaurin, 196power,194radius ofconvergence,201shift of index, 200Taylor, 196termwise addition, 197termwise differentiation, 198trigonometric, 205

Shift of index of summation, 200Simpleharmonic motion, 139Simplependulum, 137Sineintegral function, 51Singular matrix, 352Singular point, 208

irregular, 220regular, 220

Singular solution, 36Sink, 492Skydiver, 457,463Skywalk, 166Slopefield, 19,489Snowplow problem, 45Sodium pentobarbitol, 43Soft spring, 528

oscillation,529Soft touchdown, 14,18Solution curve, 19,331Solution:

ofdifferential equation, 2,7of system, 326on an interval, 6linear first-order, 50existence,uniqueness, 23,24,50,

104,114,557general,10,36implicit, 35one-parameterfamily, 5singular, 36

Source,492Spacecraftlanding, 467Spiral point, 506Spiral sink, 495Spiral source,495Spout, 482Spring constant, 136Squarewave function, 276,288,311,

314,549Stability:

of almost linear systems, 508of linear systems, 507)

Stablecenter,506Stablecritical point, 492Staircasefunction, 315Star, seePropernodeStatic displacement, 165Static equilibrium position, 136Steadyperiodiccurrent, 175Steadyperiodicsolution, 390Stepsize,431,445,471Stiffness matrix, 382Stirling's approximation, 56Stokes'drag law, 148Stonehenge,40Strange attractor, 552Substitution methods, 59Superposition principle, 102,113

for nonhomogeneous equations, 155for systems, 356

Survival of a single species,517Swimmer's problem, 16Systems analysis, 322)

TTaylor series,196Temperature oscillations,indoor, 57Terminal speed,86Termwisedifferentiation ofseries,198Termwiseinverse Laplace

transformation, 304Thermal diffusivity, 7Thresholdpopulation, 81Thresholdsolution, 484Time lag, 139,176Time-varying amplitude, 142Torricelli'slaw, 2,41Trace-determinant plane,513Trajectory, 331,488

closed,495Transfer function, 322Transient current, 175Transient solution, 168,390Translated seriessolutions, 212Triangular wave function, 288,311Trigonometric series,205Two-dimensional systems, 330)

uUndamped forcedoscillations,162Undamped motion, 136Underdamping, 142Undetermined coefficients,149,151,

155,421Unicyclemodelofcar, 167Uniform convergence,563Uniquenessof solutions, seeExistenceUnit impulse response,323)))

1-6 Index)

Unit on-offfunction, 288Unit square wave function, 288Unit staircasefunction, 276,287Unit step function, 271,305Unit stepresponse,323Unstable critical point, 482,492Upward motion, 88U.S.population, 78)

Vvan der Pol,Balthasar (1889-1959),

541van der Pol'sequation, 541Vandermonde determinant, 123Variable gravitational acceleration,90)

Variation of parameters, 158,160for nonhomogeneous systems, 423,

425Variable step size,471Vector, 349

scalarproduct, 350Velocity, 12

limiting, 22Verhulst, Pierre-Fran\037ois (1804-1849),

78,84Verne, Jules (1828-1905),92,94Vertical motion with gravitational

acceleration,15with air resistance,85

Viscosity, 148Voltage drop, 173Volterra, Vito (1860-1940),514)

WWater clock,45Watson, G.N. (1886-1965),248Weight function, 322Well-posedproblemsand

mathematical models,568Whirling string, 185World population, 39Wronski, J.M. H.(1778-1853),117Wronskian, 106,107,117,119

of vector-valued functions, 357)

yYorke, James,547Young's modulus, 191,259)))

elementary differential equations 6th edition - edwards and penny_high

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