Ordinary Differential Equations

Ordinary

Differential Equations: Methods and Applications

W. T. Ang and Y. S. Park

Universal Publishers Boca Raton

Ordinary Differential Equations: Methods and Applications

Copyright © 2008 W. T. Ang and Y. S. Park All rights reserved.

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Universal Publishers Boca Raton, Florida • USA

2008

ISBN-10: 1-59942-975-6/ISBN-13: 978-1-59942-975-5 (paper) ISBN-10: 1-59942-974-8/ISBN-13: 978-1-59942-974-8 (ebook)

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Library of Congress Cataloging-in-Publication Data Ang, W. T., 1961- Ordinary differential equations : methods and applications / W.T. Ang and Y.S. Park. p. cm. Includes bibliographical references and index. ISBN 978-1-59942-975-5 (pbk. : alk. paper) 1. Differential equations. I. Park, Y. S., 1964- II. Title. QA372.A598 2008 515'.352--dc22 2008026023

To our parents“Everywhere, we learn from those whomwe love” Johann Wolfgang von Goethe

Preface

This introductory course in ordinary differential equationsis intended for junior undergraduate students in applied mathe-matics, science and engineering. It focuses on methods of solu-tions and applications rather than theoretical analyses. Appli-cations drawn mainly from dynamics, population biology andelectric circuit theory are used to show how ordinary differen-tial equations appear in the formulation of problems in scienceand engineering.

The calculus required to comprehend this course is rather el-ementary, involving differentiation, integration and power seriesrepresentation of only real functions of one variable. A basicknowledge of complex numbers and their arithmetic is also as-sumed, so that elementary complex functions which can be usedfor working out easily the general solutions of certain ordinarydifferential equations can be introduced. The pre-requisites justmentioned aside, the course is mainly self-contained.

The course comprises six chapters.Chapter 1 gives the basic concepts of ordinary differential

equations, explaining what an ordinary differential equation isand what is involved in solving such an equation. It also illus-trates how ordinary differential equations can be derived fromphysical laws or basic principles for two specific examples ofproblems.

In Chapter 2, methods of solution are given for some firstorder ordinary differential equations. The equations studiedinclude those which can be written in separable form, thosewhich are linear, and the nonlinear Bernoulli differential equa-tion. Mathematical models which describe population growthare given as examples of applications involving first order ordi-nary differential equations.

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In Chapter 3, the mathematical theory for constructing gen-eral solutions of second order linear ordinary differential equa-tions is studied. It is applied to obtain general solutions of sec-ond order linear ordinary differential equations with constantcoefficients and the Euler-Cauchy equations. Also discussedis the extension of the theory to higher order linear ordinarydifferential equations.

Chapter 4 shows how linear ordinary differential equationswith constant coefficients arise in the formulation of problemsinvolving electric circuits and spring-mass systems. Specificexamples of problems are solved.

Chapter 5 introduces the power series method and the Frobe-nius method for deriving series solutions of rather general ho-mogeneous second order linear ordinary differential equations.The methods studied can be applied to solve some well knownordinary differential equations in mathematical physics, such asthe Legendre’s equation and the Bessel’s equation, giving riseto particular special functions, but those equations and the as-sociated special functions are not examined in this course.

Chapter 6 describes some simple numerical methods forsolving first and second order ordinary differential equations.For a particular example of applications, the second order non-linear ordinary differential equation which governs the motionof a swinging pendulum is solved numerically.

Exercises are set not only to test the understanding of stu-dents but sometimes also to impart additional insights into thematerials studied. Suggested solutions to all the exercises aregiven at the end of the chapters. To promote the use of thiscourse for self-study, the solutions provided are by and largecomplete with details.

W. T. Ang and Y. S. Park, Singapore, 2008, 2014

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Contents

1 Basic concepts 101.1 What is an ODE? . . . . . . . . . . . . . . . 101.2 Solving an ODE . . . . . . . . . . . . . . . . . . 11

1.2.1 General solution . . . . . . . . . . . . . . 121.2.2 Particular solution . . . . . . . . . . . . . 121.2.3 Exact closed form solution . . . . . . . . . 12

1.3 Exercise I . . . . . . . . . . . . . . . . . . . . . . 131.4 Why study ODEs? . . . . . . . . . . . . . . . . . 14

1.4.1 ODE for a body in motion . . . . . . . . 141.4.2 ODE for a pursuit problem . . . . . . . . 18

1.5 Exercise II . . . . . . . . . . . . . . . . . . . . . . 201.6 Solutions to Exercise I . . . . . . . . . . . . . . . 221.7 Solutions to Exercise II . . . . . . . . . . . . . . 24

2 First order ODEs 272.1 Preamble . . . . . . . . . . . . . . . . . . . . . . 272.2 First order ODEs in separable form . . . . . . . . 272.3 Linear 1st order ODEs . . . . . . . . . . . . . . . 33

2.3.1 Homogeneous linear 1st order ODEs . . . 342.3.2 Nonhomogeneous linear 1st order ODEs . 34

2.4 Bernoulli differential equation . . . . . . . . . . . 362.5 Population dynamics . . . . . . . . . . . . . . . . 38

2.5.1 Malthus theory of unlimited growth . . . 382.5.2 Verhulst theory of limited growth . . . . . 39

2.6 Exercise III . . . . . . . . . . . . . . . . . . . . . 412.7 Solutions to Exercise III . . . . . . . . . . . . . . 43

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3 Second order linear ODEs 493.1 Preamble . . . . . . . . . . . . . . . . . . . . . . 493.2 General solution of homogeneous 2nd order lin-

ear ODE . . . . . . . . . . . . . . . . . . . . . . . 503.2.1 Linearly independent functions . . . . . . 503.2.2 Construction of general solution . . . . . 52

3.3 Homogeneous 2nd order linear ODEs with con-stant coefficients . . . . . . . . . . . . . . . . . . 55

3.4 Euler-Cauchy equations . . . . . . . . . . . . . . 643.5 Exercise IV . . . . . . . . . . . . . . . . . . . . . 703.6 Solving nonhomogeneous ODEs . . . . . . . . . . 71

3.6.1 Finding a particular solution by guesswork 723.6.2 Method of variation of parameters . . . . 78

3.7 Extension to higher order linear ODEs . . . . . . 813.7.1 General N -th order linear ODEs . . . . . 813.7.2 General solution of a homogeneous ODE . 813.7.3 General solution of a nonhomogeneous lin-

ear ODE . . . . . . . . . . . . . . . . . . 823.8 Exercise V . . . . . . . . . . . . . . . . . . . . . . 833.9 Solutions to Exercise IV . . . . . . . . . . . . . . 853.10 Solutions to Exercise V . . . . . . . . . . . . . . 89

4 Circuits and springs 964.1 Preamble . . . . . . . . . . . . . . . . . . . . . . 964.2 Electric circuits . . . . . . . . . . . . . . . . . . . 96

4.2.1 Basic electrical components . . . . . . . . 964.2.2 Voltage across an electric component . . . 984.2.3 ODEs in electric circuits . . . . . . . . . . 99

4.3 Exercise VI . . . . . . . . . . . . . . . . . . . . . 1094.4 Spring-mass systems . . . . . . . . . . . . . . . . 109

4.4.1 A simple spring-mass system . . . . . . . 1094.4.2 A more complicated spring-mass system . 115

4.5 Exercise VII . . . . . . . . . . . . . . . . . . . . . 1184.6 Solutions to Exercise VI . . . . . . . . . . . . . . 1194.7 Solutions to Exercise VII . . . . . . . . . . . . . 124

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5 Series solutions 1285.1 Preamble . . . . . . . . . . . . . . . . . . . . . . 1285.2 Review of power series . . . . . . . . . . . . . . 1285.3 Power series method for ODEs . . . . . . . . . . 1315.4 Exercise VIII . . . . . . . . . . . . . . . . . . . . 1465.5 Frobenius method . . . . . . . . . . . . . . . . . 1475.6 Exercise IX . . . . . . . . . . . . . . . . . . . . . 1615.7 Solutions to Exercise VIII . . . . . . . . . . . . . 1625.8 Solutions to Exercise IX . . . . . . . . . . . . . . 167

6 Numerical methods 1766.1 Preamble . . . . . . . . . . . . . . . . . . . . . . 1766.2 Euler’s method for 1st order ODEs . . . . . . . . 1766.3 Second order ODEs . . . . . . . . . . . . . . . . . 1836.4 Oscillation of a pendulum . . . . . . . . . . . . . 187

6.4.1 Nonlinear ODE . . . . . . . . . . . . . . . 1876.4.2 ODE for ‘very small’ oscillation . . . . . . 1886.4.3 Numerical solution for ‘larger’ oscillation 189

6.5 Numerical prudence . . . . . . . . . . . . . . . . 1916.6 Exercise X . . . . . . . . . . . . . . . . . . . . . . 1916.7 Solutions to Exercise X . . . . . . . . . . . . . . 194

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Chapter 1

Basic concepts

1.1 What is an ODE?

An equation which contains the derivative(s) of a yet to bedetermined function y(x) (a function of one variable) is calledan ordinary differential equation (ODE) in y(x).

Below are some examples of ODEs in y(x):

(1)dy

dx− 2x = 0

(2)d2y

dx2+ 3

dy

dx− 2y(x) = x5

(3) x3d3y

dx3+ 6x2

d2y

dx2− 3xy(x) = sin(2x)

(4) 2x2(y(x))10d4y

dx4+ 3x

d2y

dx2= xy(x)

An ODE in y(x) is said to be of order N if dNy/dxN is thehighest order derivative of y(x) present in the ODE.

In the examples above, (1) is an ODE of order 1 (or 1storder ODE); (2) is of order 2; (3) is of order 3; and (4) is oforder 4.

It may be sometimes convenient to use the notation

y0(x) =dy

dx, y00(x) =

d2y

dx2, y000(x) =

d3y

dx3, · · · , y(N)(x) = dNy

dxN.

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Thus, we may write the 4th order ODE in (4) above as

2x2(y(x))10y0000(x) + 3xy00(x) = xy(x),

or even more simply as

2x2y10y0000 + 3xy00 = xy

if it is already understood that y is a function of x. Note thaty10 above refers to y raised to the power 10, not to be mistakenas the 10th order derivative y(10) = d10y/dx10.

More generally, we may regard an ODE in y(x) as an equa-tion of the general form

F (x, y, y0, y00, y000, · · · , y(N−1), y(N)) = 0.Here F denotes a mathematical expression involving x, y, y0,y00, y000, · · · , y(N−1) and y(N).

1.2 Solving an ODE

Given an ODE in y(x), we are interested in finding functionsy(x) that satisfy the equation, that is, we are interested insolving the ODE.

For the purpose of illustration, let us now consider solvingthe ODE

y00 − 6x = 0which may be rewritten as

d2y

dx2= 6x.

Wemay solve the ODE above by directly integrating it twicewith respect to x, that is,

dy

dx=

Z6xdx = 3x2 + C

⇒ y =

Z(3x2 + C)dx = x3 + Cx+D.

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Here C and D are constants which may have any values (arbi-trary constants).

Thus, we find that y(x) = x3 + Cx + D (where C and Dare arbitrary constants) satisfies the ODE y00 − 6x = 0.

Not all ODEs may be easily solved by direct integration likein the example above. In subsequent chapters, we will look atvarious methods for solving ODEs.

1.2.1 General solution

From the example above, it appears that solving an ODE isreally “undoing the derivatives” in the ODE.

Roughly speaking, if an ODE is of order N, we are requiredto integrate it N times in order to solve it. Consequently, Narbitrary constants appear in the solution obtained.

For our purpose, we regard any function y(x) with N ar-bitrary constants in it as a general solution of an N -th orderODE, if the function satisfies the ODE.

Thus, we may regard y(x) = x3 + Cx + D as a generalsolution of the ODE y00 − 6x = 0.

1.2.2 Particular solution

If some or all of the arbitrary constants in a general solution ofan ODE assume specific values, we obtain a particular solutionof the ODE.

Examples of particular solutions of the ODE y00 − 6x = 0are y(x) = x3+x+D, y(x) = x3+Cx−2 and y(x) = x3−3x.

1.2.3 Exact closed form solution

We regard a solution of an ODE to be in exact closed form ifthe solution can be directly expressed in terms of elementaryfunctions1.

1For our purpose, a function is regarded as elementary if it can becalculated directly using the function keys of an ordinary scientific handcalculator. Thus, cos(x), sin(x), exp(x), g(x) = x3−1 and f(x) = cos3(x2+1)− exp(−x2) are elementary.

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Thus, the general solution y(x) = x3+Cx+D of the ODEy00 − 6x = 0 is in exact closed form. If the values of C and Dare given, we may readily evaluate y(x) = x3+Cx+D for anyvalue of x by using an ordinary hand calculator.

We may not be able to find exact closed form solutions ofsome ODEs. For example, take the ODE

dy

dx=sin(x)

x

whose solution is theoretically given by

y =

Zsin(x)

xdx.

It is, however, not possible to express the integral (on the righthand side) in terms of elementary functions. Thus, the ODEdoes not have an exact closed form solution.

1.3 Exercise I

1. Check by direct substitution whether y(x) = ex+2e−x isa solution of each of the following ODEs in y(x) or not.(Substitute y(x) = ex + 2e−x into the left hand side of agiven ODE and simplify to see if it is possible to obtainthe right hand side.)

(a) y00 − y = 0(b) y000 + 3y00 − 2y0 = 2ex − 4e−x(c) 2y00 + 2y0 − 3y = e2x + 5e−x(d) y000 + 3y00 − y0 − 3y = 0

2. Find out whether each of the following functions is a so-lution of the ODE p0000(x)− 5p00(x)− 36p(x) = 0 or not.

(a) p(x) = 2 sinh(3x)

(b) p(x) = cosh(2x)

(c) p(x) = cos(3x)− sin(3x)(d) p(x) = 2 sin(2x) + 3 cos(2x)

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3. If y(x) = α sin(2x) + β cos(2x) is a solution of the ODEy00 + 4y0 + 3y = 3 sin(2x), find the constants α and β.(Hint. Substitute y(x) = α sin(2x) + β cos(2x) into theODE and choose the constants α and β in such a waythat the equation is satisfied for all x.)

4. If y (x) = α+ βx+ γx2 is a solution of the ODE x3y000 +x2y00 − y = x2 + 2x+ 1, find the constants α, β and γ.

5. Each of the following ODEs in y(x), if rewritten in an ap-propriate form, can be solved by direct integration. Findgeneral solutions of these ODEs.

(a)√xy0 = x6 + 5x

(b) y0ex = 5(c) y00 + 6x2 − 2 = 0(d) xy0 + y = 4x3 + 2x (Hint.

d

dx(xy) = x

dy

dx+ y)

6. Verify that

y(x) =

s5

2 cos(x) + 4 sin(x) + 10e2x

is a solution of the ODE y0 + y − y3 sin(x) = 0. (Hint.Differentiate the solution with respect to x to obtainy0 = (sin(x) − 2 cos(x) − 10e2x)y3/5. You may do thisby squaring both sides of the solution first.)

1.4 Why study ODEs?

Some problems in science and engineering may be formulatedin terms of ODEs. Below are two examples of such problems.More examples will be given in later chapters.

1.4.1 ODE for a body in motion

Consider the motion of a body which is dropped at some heightabove the ground.

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We assume that the body, which is pulled by gravity to-wards the earth, moves along a straight line which is perpen-dicular to the surface of the earth (that is, it moves along avertical path). In addition to gravity, the body is also actedupon by a force due to air resistance.

A sketch of the situation is given in Figure 1.1. Let s(t)be the vertical downward displacement (in meter) of the bodyfrom a fixed point P (which may lie on the path of motion)2.We are interested in finding s(t), that is, the position of thebody at time t.

Figure 1.1

All motion obeys Newton’s law which may be stated asfollows.

If the mass of a moving body is constant, then thetotal force (in newton) acting on the body is equalto the product of the mass (in kilogram) and the

2We define s(t) to be the downward displacement of the object from Pat time t. This means that s(t) > 0 if the object is below P at time t, ands(t) < 0 if it is above P. It is, of course, possible to choose P in such a waythat s(t) > 0 at all time t during which the object is falling.

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acceleration (in meter per second per second) of thebody.

“Force” and “acceleration” are vectors. In simple terms, avector is a quantity which has a magnitude and a direction.Thus, Newton’s law as stated above implies that the force andthe acceleration are in the same direction.

For the situation depicted in Figure 1.1, Newton’s law issimply:

(total downward force acting on body)

= (mass of body)× (downward acceleration of body).

As mentioned above, we assume that there are only twotypes of forces acting on the body, namely gravity and air re-sistance.

The gravitational force has a magnitude given bymg, wherem is the mass of the body and g is the acceleration due togravity (that is, g ' 9.81 meter per second per second nearplanet earth). The gravitational force acts downward.

The magnitude of the force due to air resistance is takento be given by k|s0(t)|. Here k is a positive coefficient whichdepends on, among other things, the shape of the body. Thederivative s0(t) is the downward velocity3 of the body. In thiscase, s0(t) is always greater than zero at any time t during themotion as the body is always moving towards the ground. Theforce due to air resistance acts upward against the motion ofthe body.

It is obvious that

(total downward force on body) = mg − k|s0(t)|= mg − ks0(t)

(since s0(t) > 0).

The downward acceleration of the body is given by s00(t).Newton’s law now becomes

mg − ks0(t) = ms00(t).3“Downward velocity” implies that s0(t) > 0 if the object is moving

downward, and s0(t) < 0 if it is moving upward.

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The above equation is an ODE in s(t) which may be rewrit-ten as

s00(t) +k

ms0(t)− g = 0.

If we can solve the ODE for s(t), we can predict the positionof the body at any time t during which it is moving towardsthe earth. For the case in which k is a non-zero constant, theODE may be solved using methods of solution in later chapters(see Problem 3 in Exercise V on page 84).

For now, let us consider the special case in which the effectof air resistance on the motion is so small that it may be ig-nored, that is, we take k/m in the ODE to be zero. For such acase, the ODE which describes Newton’s law of motion reducesto

s00(t)− g = 0 or s00(t) = g.

This simpler ODE may be solved by directly integrating ittwice with respect to t. Bearing in mind that g is a constant,we obtain

ds

dt=

Zgdt = gt+ C

⇒ s(t) =

Z(gt+ C)dt =

1

2gt2 +Ct+D.

Here C and D are arbitrary constants.To calculate the position of the body, we need to know the

values of the constants C and D. The values of these constantsmay be determined if the displacement and velocity of the bodyare known at a certain time.

For example, if we are told that s0(t) = 3 (meter per second)when t = 0 (second) then

3 = g · 0 +C ⇒ C = 3.

Furthermore, if s(t) = 2 (meter) at t = 0, we obtain

2 =1

2g · 02 + 3 · 0 +D⇒ D = 2.

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Thus, if there is no air resistance, and if the displacementand velocity at t = 0 are 2 and 3 (in the appropriate units)respectively, then the displacement of the body is given by

s(t) =1

2gt2 + 3t+ 2

at any time t during which the body is falling towards theground.

1.4.2 ODE for a pursuit problem

With reference to a Cartesian coordinate system denoted byOxy, consider the following pursuit problem. At time t = 0, anavy ship is at (a, 0), where a > 0, and it spots a small boatat (0, 0). The boat moves with a constant speed u along thepositive y axis. The navy ship decides to chase after the boatby moving at a constant speed v and in a direction that isalways directed at the boat. We are interested in finding thepath of the navy ship.

Figure 1.2

A geometrical sketch of the problem is given in Figure 1.2.Let the path be given by the curve y = p(x), where p(x) is an

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unknown function yet to be determined. The position of theboat at time t is B (0, ut) (along the y axis). At time t, thenavy ship is located at N (x, p(x)). Now the line NB (dottedline in Figure 1.2) is tangential to the path y = p(x) at Nbecause the navy ship is always directed at the boat. Thus, thegradient of NB is equal to the gradient of the tangent to thecurve y = p(x) at the point N (x, p(x)), that is,

p0(x) =p(x)− ut

x⇒ xp0(x) = p(x)− ut.

Since the ship is moving, x and y are functions of t. Let usdifferentiate the above equation once with respect to t. Usingthe chain rule for differentiation, we obtain

xp00(x)dx

dt+ p0(x)

dx

dt= p0(x)

dx

dt− u

which simplifies to

xp00(x)dx

dt= −u.

Note that dx/dt and dy/dt are the x and y components ofthe velocity of the navy ship. Since its speed is v, we may writer

(dx

dt)2 + (

dy

dt)2 = v.

From y = p(x) and the chain rule for differentiation, we findthat

dy

dt= p0(x)

dx

dt.

Thus, r(dx

dt)2 (1 + (p0(x))2) = v

⇒ dx

dt= ± vp

1 + (p0(x))2.

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From Figure 1.2, it is clear that dx/dt < 0 as the x-coordinateof the ship decreases during the pursuit of the boat. Thus, wetake

dx

dt= − vp

1 + (p0(x))2.

Substituting the above expression for dx/dt into the equa-tion xp00(x)dx/dt = −u, we obtain the 2nd order ODE in p(x)given by

xp00(x)− uv

p1 + (p0(x))2 = 0

If we can solve this ODE for p(x), we obtain a formula for thepath taken by the navy ship when it chases after the small boat.The two arbitrary constants in the general solution of the ODEmay be determined from the conditions p(a) = 0 and p0(a) = 0.(Can you see how the conditions come about?)

The ODEmay be converted into a 1st order ODE and solvedas explained in Chapter 2 (on page 30).

1.5 Exercise II

1. An extremely thin wire lies along the x-axis from x = 0to x = ` (` is the length of the wire). The temperaturein the wire varies from one point to the next along thewire, that us, it is a function of x. If it is denoted byT (x), then that under certain conditions T (x) satisfiesthe simple 2nd order ODE

d2T

dx2= 0.

If temperature is given by T0 and T` at x = 0 and x = `respectively, find the temperature throughout the wholewire.

2. The tangent to a curve at the point (x, y) has gradientgiven by x3+2x+1. If the equation of the curve is givenby y = p(x), write down an ODE in p(x). Find the curvegiven that (1, 2) is a point on it.

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3. The tangent to a curve at the point (x, y) has gradientgiven by (x − y)/x. If the equation of the curve is givenby y = p(x), write down an ODE in p(x). Find the curvegiven that (1, 1) is a point on it. (Hint. Use the formula

d

dx(xp(x)) = x

dp

dx+ p(x) to help you to solve the ODE.)

4. A body is moving to the right along a horizontal line. Itsspeed at time t ≥ 0 is given by 8t3 + 3t2 + 1. If w(t) isthe distance of the body from a fixed point P on the line,write down an ODE in w(t) for the motion of the body.Where is the body at time t = 2 given that it is a unitdistance away from P at time t = 0?

5. A body of 1 kilogramme is moving to the right along ahorizontal line. It is acted upon by a rightward forcegiven by F (t) = 1 + e−2t (in newton) at time t (in sec-ond). If s(t) is the rightward displacement (in meter) ofthe body from a fixed point P on the line, use Newton’slaw of motion to write down an ODE in s(t). Find thedisplacement of the body at time t = 1 given that it is atrest at the point P at time t = 0.

6. The uniform temperature of a body at time t is given byT (t). The rate of change of the body temperature per unittime is given by c(T (t) − Tambient), where Tambient is theconstant temperature of the surrounding atmosphere andc is a constant. Express the statement in the last sentenceby writing down an ODE in T (t). What can you sayabout the constant c? (Hint. To find out something aboutc, ask whether the body should be gaining or losing heatto its surrounding when T (t) > Tambient. What aboutwhen T (t) < Tambient?)

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1.6 Solutions to Exercise I

1. From y = ex + 2e−x, we have y0 = ex − 2e−x, y00 =ex + 2e−x = y and y000 = ex − 2e−x = y0. Thus:

(a) y00 − y = 0 (from y00 = y).Yes. (y = ex + 2e−x is a solution of the ODE.)

(b) y000 + 3y00 − 2y0 = 2ex + 8e−x 6= 2ex − 4e−x.No. (y = ex + 2e−x is not a solution of the ODE.)

(c) 2y00 + 2y0 − 3y = ex − 6e−x6= e2x + 5e−x. No.

(d) y000 + 3y00 − y0 − 3y= y0 + 3y − y0 − 3y = 0. Yes.

2. (a) p = 2 sinh(3x), p0 = 6cosh(3x), p00 = 18 sinh(3x) =9p, p000 = 9p0 and p0000 = 9p00 = 81p. Thus,

p0000 − 5p00 − 36p = 81p− 45p− 36p = 0.

Yes, p = 2 sinh(3x) is a solution of the given ODE.

(b) p = cosh(2x), p0 = 2 sinh(2x), p00 = 4cosh(2x) = 4p,p000 = 4p0 and p0000 = 16p. Thus,

p0000 − 5p00 − 36p = 16p− 20p− 36p 6= 0.

No, p = 2 cosh(2x) is not a solution of the given ODE.(c) p(x) = cos(3x)− sin(3x) is not a solution.(d) p(x) = 2 sin(2x) + 3 cos(2x) is a solution.

3. Substitute y = α sin(2x)+β cos(2x) into ODE, we obtain

(−α− 8β) sin(2x) + (−β + 8α) cos(2x) = 3 sin(2x).

Thus, for the ODE to be satisfied for all x, we requireα+ 8β = −3 and 8α− β = 0. This gives α = −3/65 andβ = −24/65.

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4. Substitute y = α+ βx+ γx2 into ODE to obtain

−α− βx+ γx2 = x2 + 2x+ 1.

This gives α = −1, β = −2 and γ = 1.

5. (a) Rewrite ODE as

dy

dx= x11/2 + 5x1/2

and integrate directly once to obtain

y =2

13x13/2 +

10

3x3/2 +C.

(b) Rewrite ODE as

y0 = 5e−x

and integrate to obtain

y = −5e−x + C.

(c) Rewrite ODE as

d2y

dx2= −6x2 + 2

and integrate twice to obtain

y = −12x4 + x2 + Cx+D.

(d) Rewrite ODE as

d

dx(xy) = 4x3 + 2x.

Integrate once to obtain

xy = x4 + x2 +C

⇒ y = x3 + x+C

x.

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6. It may be very tedious to proceed by directly differenti-ating

y(x) =

s5

2 cos(x) + 4 sin(x) + 10e2x.

Instead, rewrite the above as

y2 =5

(2 cos(x) + 4 sin(x) + 10e2x)

⇒ y2(2 cos(x) + 4 sin(x) + 10e2x) = 5

and differentiate both sides with respect to x to obtain

2yy0(5

y2) + y2(−2 sin(x) + 4 cos(x) + 20e2x) = 0

⇒ 10y0 = −y3(−2 sin(x) + 4 cos(x) + 20e2x)⇒ 10y0 + 10y − 10y3 sin(x)= y[

5(2 sin(x)− 4 cos(x)− 20e2x)2 cos(x) + 4 sin(x) + 10e2x

+ 10

+−50 sin(x)

2 cos(x) + 4 sin(x) + 10e2x]

= 0 ⇒ y0 + y − y3 sin(x) = 0 (as required).

1.7 Solutions to Exercise II

1. The conditions are T (0) = T0 and T (`) = T`. Solving theODE, we find that

d2T

dx2= 0⇒ dT

dx= C ⇒ T = Cx+D.

Applying the conditions, we obtain

T (0) = T0 = D

T (`) = T` = C`+ T0 ⇒ C =T` − T0`

.

Thus, the required temperature is

T (x) =T` − T0`

x+ T0.

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2. The ODE giving the tangent of the curve is

dp

dx= x3 + 2x+ 1.

Integrating, we obtain

p(x) =1

4x4 + x2 + x+C.

From the point (1, 2), we know p(1) = 2. Thus,

1

4+ 1 + 1 + C = 2⇒ C = −1

4.

The required curve is

y =1

4x4 + x2 + x− 1

4.

3. The ODE is

dp

dx=

x− px

⇒ xp0 + p = x⇒ d

dx(xp) = x.

Integrating, we obtain

xp =1

2x2 +C ⇒ p =

1

2x+

C

x.

From the point (1, 1), we have p(1) = 1, that is, C =−1/2. Thus, the required curve is

y =1

2x− 1

2x(for x > 0).

4. The ODE is

dw

dt= 8t3 + 3t2 + 1

which can be integrated to obtain

w = 2t4 + t3 + t+ C.

From w(0) = 1, we find that C = 1. Thus, w(2) = 42+1 =43.

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