differential equations

Chapter 1

The Wonderful World ofDifferential Equations

The essential fact is simply that all the pictures which science now draws of nature,and which alone seem capable of according with observational fact, are mathemat-ical pictures. . . .

Sir James Jeans1

1.1 What are Differential Equations?Just what are differential equations? Following the wisdom of the old Chinese proverbthat one picture is worth more than a thousand words, we defer our answer until wehave provided a picture of sorts. Table 1.1 is, so to speak, a collage of various typesof differential equations. With one exception, these are well-known equations drawnfrom different scientific and technical disciplines. A sense of their importance maybe realized from their ability to mathematically describe, or model, real-life situations.The equations come from the diverse disciplines of demography, ecology, chemicalkinetics, architecture, physics, mechanical engineering, quantum mechanics, electricalengineering, civil engineering, meteorology, and a relatively new science called chaos.The same differential equation may be important to several disciplines, although fordifferent reasons. For example, demographers, ecologists, and mathematical biologistswould immediately recognize

dp

dtD rp;

the first equation in Table 1.1, as the Malthusian law of population growth . It is used topredict populations of certain kinds of organisms reproducing under ideal conditions

1See [?, ch. 5]. Sir James Jeans (18771946)was a British mathematical physicist, Cambridge Universitylecturer, Princeton University professor of applied mathematics, and author of a number of popular worksof science, of which The Mysterious Universe [?] was one of his most famous. His treatise, Problemsof Cosmogony and Stellar Dynamics (1917), on the behavior of fluids in space contributed to a greaterunderstanding of the origin and evolution of the universe.

1

2 CHAPTER 1. THE WONDERFUL WORLD OF DIFFERENTIAL EQUATIONS

whereas physicists, chemists, and nuclear engineers would be more inclined to re-gard the equation as a mathematical portrayal, or model, of radioactive decay. Evenmany economists and mathematically minded investors would recognize this differ-ential equation, but in a totally different context: it also models future balances ofinvestments earning interest at rates compounded continuously.

Another example is the van der Pol equation

d2x

dt2 ".1 x2/dx

dtC x D 0;

which came from modeling oscillations of currents in the nonlinear electrical circuits ofthe first commercial radios. For many years, it was the subject of research by electricalengineers and mathematicians alike.

Table 1.1: Differential Equations Modeling Real-Life Situations

Differential Equation Situation

dp

dtD rp

The Malthusian law of population growth isused to model the populations of certain kindsof organisms living in ideal environments forlimited lengths of time t . It gives the rate atwhich a population p changes with respect tot . The value of the constant r depends on theorganism.

dx

dtD k.A x/2

This second-order reaction rate law gives therate at which a single chemical species com-bines to produce a new species, such as methylradicals combining in a gas to form ethanemolecules. See Atkins [?, p. 134].

d2y

dx2D C

L

sAC

L

2Cdy

dx

2

The graph of the solution models the shape ofthe Gateway Arch in St. Louis, where y is itsheight at a distance x from one end of its base.The constants A, C , and L relate the lengthsof the base, top, and centroid. The GatewayArch has the shape of an inverted catenary.A catenary is a curve that has the shape of achain suspended from two points at the samelevel. The equation used to design the cate-nary curve shape of Arch can be found at thewebsite: www.nps.gov/jeff.

1.1. WHAT ARE DIFFERENTIAL EQUATIONS? 3

Differential Equation Situation

md2x

dt2C b dx

dtC kx D F.t/

This equation models the motion of a dampedmass-spring system subjected to a time-dependent force F.t/.

22m

d2

dx2C E 1

2kx2

D 0

This equation from quantum mechanics is thetime-independent Schrodingers equation forthe one-dimensional simple harmonic oscilla-tor. The constant is defined in terms ofPlancks constant h by D h=2 .

.1 x2/d2y

dx2 2x dy

dxC .C 1/y D 0

This equation is known as Legendres differ-ential equation and is one of several equationsused for calculating the energy levels of thehydrogen atom.

EId4y

dx4D w.x/

This differential equation models the verticaldisplacement y.x/ of a point located a dis-tance x from the fixed end of a beam of uni-form cross section, where w.x/ represents theload at x; E and I are constants.

x00 ".1 x2/x0 C x D 0The van der Pol equation models the currentat time t in an electrical circuit with nonlinearresistance.

4xy2y.4/

3 3x4y5y006 D cos9.x10/ This is just one mean-looking equation con-cocted by the author.

dx

dtD .y x/

dy

dtD rx y xz

dz

dtD xy bz

This set of three differential equations, calledthe Lorenz system, is a overly simplified ver-sion of a complicated system of twelve equa-tions used to model convection in the atmo-sphere. The Lorenz system models the chaoticrotational motion of a wheel with leaking com-partments of water symmetrically positionedaround its rim. See Appendix A for more in-formation.

Even though the equations in Table 1.1 come from diverse fields, they do have somecommon features. The foremost feature shared by all of them is that they have at leastone derivative, which is precisely what makes them differential equations in the firstplace! We make special note of this by formally defining what is meant by a differentialequation.


Differential Equation

Definition 1.1. A differential equation is an equation that involves one or morederivatives of some unknown function or functions.

To complicate matters, there are various types of differential equations: chief amongthem are ordinary differential equations , partial differential equations , and integro-differential equations . The equations in Table 1.1 are all examples of ordinary differ-ential equations because they only involve ordinary derivatives. Ordinary derivativesare the derivatives that we study in a first (single-variable) calculus course. Partial dif-ferential equations are equations involving derivatives called partial derivativeshowa partial derivative differs from an ordinary derivative is discussed later on, after thereview of ordinary derivatives in the next section. To give us an inkling of what partialdifferential equations look like, here is a classic example:

@u

@tD k @

2u

@x2:

It is used to model the conduction of heat through an extremely thin metal bar, whereu.x; t/ is the temperature at the point x in the bar at time t .

Integro-differential equations involve not only derivatives of unknown functionsbut also their integrals. For example, in Chapter ?? we will solve integro-differentialequations that look like

x0.t/ D f .t/ CZ t0

k.t u/x.u/ du:

This book is devoted to a study of ordinary differential equations. Even so, therewill be brief forays at times into topics involving very simple partial differential equa-tions, integral equations, and integro-differential equations.

Before we formally define what is meant by an ordinary differential equation, letspoint out some features that the equations in Table 1.1 have in common. First weobserve that each equation in Table 1.1 contains a single independent variable and oneor more dependent variables. It is a relatively simple matter to tell these two types ofvariables apart from the derivatives themselves, since differentiation always takes placewith respect to the independent variable. Obviously then, the other variable, the onebeing differentiated, is the dependent variable.

Example 1.1. The first entry in Table 1.1 is the Malthusian law of population growth:

dp

dtD rp:

Translated into words, the equation says that the rate at which the current populationp of an organism changes with respect to the time t is equal to the product of theconstant r and the current population p. The time t is the independent variable and thepopulation p is the dependent variable.


Example 1.2. The equation

EId4y

dx4D w.x/

in Table 1.1 models the vertical displacement of a beam. Since y is differentiated withrespect to x, the independent variable is x and the dependent one is y.

The name dependent variable is befitting because it describes the type of variable itis: it depends in some functional way on the independent variable as prescribed by thedifferential equation, although this dependence is not always possible. For example,

y2 C .y0/2 D 1is a differential equation; even so, no real-valued function 2 can fulfill the prescript thatthe sum of its square and the square of its derivative is equal to a negative number.

Space on a page can be saved by replacing Leibniz notation , which uses the Latind for denoting derivatives, such as

dp

dt;

d2x

dt2;

d4y

dx4;

with a shorthand notation that uses primes ( 0 ) or overdots (P / for differentiation. Inprime notation , the derivatives

dy

dxand

dp

dt

are writteny0 and p0;

respectively. A shortcoming of this notation is that the independent variable is notexplicitly stated.

The overdot notation is reserved for derivatives that are taken with respect to thetime t . For example, Pp means dp=dt . Thus, the Malthusian population law

dp

dtD rp

in the overdot notation becomesPp D rp:

We also have to be aware of the orders of the derivatives appearing in equations.The derivatives

dy

dx; Pp; z0

are first-order derivatives, whereas the derivativesd2x

dt2; y00; Rp

2A function is real-valued when every evaluation of it results in a real number. Even though the functioni sinx, where i2 D 1, satisfies the differential equation, it is a complex-valued solution, not a real-valuedsolution.


are second-order derivatives . Of course, some differential equations have derivativesof even higher order: third-order derivatives such as

y000;d3x

dt3

or fourth-order derivatives such asd4x

dt4

or even higher. It is easy to lose track of the number of primes or overdots when theorder is more than three. In such a case, it is customary to use either Leibniz notationor to use superscripts enclosed in parentheses to denote such derivatives: for example,d4y=dx4 or y.4/ is preferred over y0000. The nth derivative of y with respect to x iswritten as dny=dxn or as y.n/ .

All of the previous derivatives are known as ordinary derivatives. When we take theordinary derivative of a function, the term ordinary indicates that we are dealing witha function of a single variable. In other words, an ordinary derivative is a derivativeof a function of a single independent variable with respect to that variable. The wordordinary qualifies the word derivative, distinguishing between the derivatives of single-variable calculus from the ones of multivariable calculus. Multivariable calculus dealswith functions of two or more variables; their derivatives are called partial derivatives.They will be introduced shortly; but for now, lets review the definition and meaning ofthe ordinary derivative of a function.

1.1.1 Ordinary DerivativesLets review the meaning of an ordinary derivative with an example. Imagine stretch-ing a filament-like copper wire of length 25 centimeters tautly along a straight line. Letthe line serve as the x-axis and the left end of the wire designate the location of theorigin. Suppose that the wire is heated unevenly in such a way that each of its pointseventually reaches a constant temperature but that the temperature generally variesfrom point to point. Even though in reality the wire is a three-dimensional object, itsvery thinness suggests that variations in temperature along the y- and z-directions arenegligible. Consequently, the wire may be regarded ideally as a one-dimensional math-ematical object: the line segment extending from x D 0 cm to x D 25 cm. Now sup-pose that Table 1.2 gives temperature measurements, accurately to the ten-thousandthplace, at 5-centimeter intervals along the wire.

Table 1.2: Temperatures at Points of an Unevenly Heated Wire

x (cm) 0 5 10 15 20 25T (C) 100.0000 99.8740 99.4980 98.8720 97.9960 96.8700

The average rate of change of the temperature with respect to x, as x changes from10 cm to 15 cm, is given by the difference quotient T=x, where T is the changein the temperature corresponding to x, the change in the x-coordinate. Thus,


T

xD T .15/ T .10/

15 10 D98:8720 99:4980

5D 0:1252 C=cm:

Since the negative sign comes from T , the quantity 0:1252 C is interpreted as thetemperature decrease per centimeter (roughly) as x increases from 10 cm to 15 cm.Likewise, the difference quotient when x decreases from 10 cm to 5 cm is

T

xD T .5/ T .10/

5 10 D99:8740 99:4980

5 D 0:0752C=cm:

The negative sign results from the decrease in x accompanied by the increase in T .Consequently, the quantity 0:0752 C roughly estimates the temperature increase percentimeter as x decreases from 10 cm to 5 cm. Equivalently, we can view the temper-ature as decreasing roughly 0:0752 C per centimeter as x increases from 5 cm to 10cm.

If no other temperature measurements are available to us, we could use one ofthe two previously calculated values as a rough estimate of the rate of change of thetemperature near x D 10 cm. Better yet, we could use their average. Even so, a changeof 5 centimeters in x is a big jump when it comes to estimating the rate of change in thetemperature near x D 10 cm. Better estimates could be obtained with smaller jumpsin x. For example, suppose that we are also able to measure the temperature at x D 11cm. Then an estimate of the rate at which the temperature decreases starting at x D 10cm is given by the difference quotient

T

xD T .11/ T .10/

11 10 :Suppose the measured temperature at x D 11 cm is 99:3928 C; then,

T

xD 99:3928 99:498

1D 0:1052 C=cm:

This provides us with a new estimate of the rate at which the temperature decreaseswhen x D 10 cm; namely, 0:1052 C per centimeter. This is an improvement over theprevious estimates since x is now smaller by a factor of 5.

Of course, even better estimates than the previous ones could be obtained by hav-ing data for even smaller values of x. The ideal situation would be to know thetemperature at every point of the wire. Then the instantaneous rate of change of thetemperature at x D 10 cm would be given precisely by the limit that the differencequotient T=x approaches as x approaches 0, where x D x 10 and T isthe corresponding change in the temperature from 10 cm to x cm. In mathematicalnotation, we express this by writing

limx!0

T

xD lim

x!0T .10Cx/ T .10/

x:

This limit is known as the derivative of the temperature at x D 10 cm and is symbolizedby T 0.10/. For example, lets suppose that the temperature at every point of the wire isgiven by the function

T .x/ D 100 0:0002x 0:005x2:


In fact, the values in Table 1.2 were constructed using this function. Thus, the temper-ature change T from x D 10 cm to x D 10Cx cm is

T D T .10Cx/ T .10/D 100 0:0002.10Cx/ 0:005.10Cx/2 99:4980D 0:1002x 0:005.x/2:

It follows that the corresponding difference quotient is

T

xD 0:1002 0:005x:

Hence, T=x approaches 0:1002 C=cm as x approaches 0. We conclude thatT 0.10/ D 0:1002 C=cm, which means that the temperature decreases at an instan-taneous rate of 0:1002 C per centimeter when x D 10 cm. Of course, the derivativeT 0.10/ can be obtained more easily from the derivative rules of calculus. At any valueof x,

T 0.x/ D ddx

100 0:0002x 0:005x2 D 0:0002 0:01x:

In particular, at x D 10,T 0.10/ D 0:0002 0:01.10/ D 0:1002 C=cm:

Now that we have reviewed the meaning of a derivative with an example, letsconsider derivatives in general. It is often the case when dealing with a function, call itf .x/, that the instantaneous rate3 with which it changes with respect to x needs to bedetermined. This rate is computed from the difference quotient for f :

f .x Cx/ f .x/x

:

This gives the average rate of change of f from x to xCx. The instantaneous rateof change of f at x is the limit of the difference quotient as x approaches 0, whichis expressed by writing

limx!0

f .x Cx/ f .x/x

; (1.1.1)

provided that this limit actually exists. This limit is what is meant by the ordinaryderivative of f at x. It is symbolized by f 0.x/ or by dy=dx if y is the dependentvariable. The result of (1.1.1), if the limit exists, is an expression in x that representsthe ordinary derivative f 0. It is as much a function of x as is f ; however, its domainmay differ from f s. The domain of f 0 consists of all x-values for which the limit(1.1.1) exists. For example, the domain of the real-valued function

f .x/ D px3Usually the term instantaneous is omitted.


consists of all nonnegative numbers (x 0/. For these values of x, with the exceptionof x D 0, the difference quotient (1.1.1) converges to 1=.2px/. Therefore,

f 0.x/ D 12px

and its domain consists of all positive numbers (x > 0/. Of course, the simplest wayto find derivatives is not to apply (1.1.1) directly but to use the rules of differentiationlearned in elementary calculus. When x is assigned a particular value, say x D a, thenf 0.a/ is a number that represents the (instantaneous) rate of change of f at x D a.

1.1.2 Ordinary Differential EquationsNow that we have reviewed the definition of the ordinary derivative, we can state whatis meant by the type of differential equation known as an ordinary differential equation.

Ordinary Differential Equation

Definition 1.2. An ordinary differential equation is an equation involving oneindependent variable; one or more dependent variables, each of which is a func-tion of the independent variable; and ordinary derivatives of one or more of thedependent variables.

Each of the equations in Table 1.1 fits the description in Definition 1.2; accordingly,they are all ordinary differential equations. Still, a few words need to be said about thesituation described in the last entry of the table, which unlike the others requires morethan one differential equation to model it, namely, the system of three equations:

Px D .y x/Py D rx y xz (1.1.2)Pz D xy bz:

When a steady, uniformly distributed shower of water falls over a wheel with leakycompartments symmetrically positioned around its rim, it can be shown that togetherthese three equations model the rotational motion of the wheel. 4 Observe that the equa-tions are coupled or linked together by their sharing of the three dependent variables:x, y, and z. As a result, we say that they make up a system of ordinary differentialequations. This system, known as the Lorenz system or the Lorenz equations, is leg-endary in being one of the catalysts in initiating a branch of mathematics and scienceknown as chaos. The equation Py D rxyxz is still considered an ordinary differen-tial equation, even though it contains all three dependent variables. The reason for thisis that Py is an ordinary derivative and all three dependent variables depend only on thesingle independent variable t . This is implied from the form of the three equations in

4For more information and a derivation of the equations, see the section entitled Chaotic Motion inWater Wheels and the Lorenz Equations in the Appendix.


the system. Likewise, the other two equations are also ordinary differential equations.In other words, the Lorenz system consists of three ordinary differential equations.

One characteristic of a differential equation is the order of the highest derivativeappearing in the equation. For example, if a differential equation contains a derivativeof second order (a second derivative) but none of higher order, then we say that the dif-ferential equation is second order or of order 2. Legendres equation listed in Table 1.1is a second-order equation. The Lorenz system consists of three first-order equations.Other examples are:

(a) dxdt

D k.a x2/ is an equation of order 1 (or 1st order);

(b) 2xyy0 C .yy0/2 D y2 is a 1st order equation;

(c) EI d4y

dx4D w.x/ is an equation of order 4;

(d) 4xy2y.4/3 3x4y5y006 D cos9.x10/ is a 4th order equation.Order of an Ordinary Differential Equation

Definition 1.3. The order of an ordinary differential equation is said to be n if theorder of the highest derivative appearing in the equation is n.

Besides an independent variable and a dependent variable (or variables), most ofthe ordinary differential equations listed in Table 1.1 contain quantities known as pa-rameters. A parameter does not change in value with changes in the value of theindependent variable; however, its value may change when the situation or experimentis modified. For example, consider the simple differential equation

dp

dtD rp;

which is known as the Malthusian law when it is used to predict the populations ofcertain types of species. The quantity r is a constant for a given species; that is, itsvalue does not change with time. Yet its value will most likely change if it is applied toa different species. Another example is the damped mass-spring system

md2x

dt2C b dx

dtC kx D F.t/;

which has three parameters: m, b, and k. The parameter m is the mass of a body towhich a spring is attached. The parameter k measures the stiffness of the spring, andb measures the retardation in the motion of the body due to damping forces, such asfriction. None of these parameters depends on time, yet their values would change ifthe body and spring were replaced by some other body and spring.


1.1.3 Solutions of Ordinary Differential EquationsGenerally speaking, when dealing with a differential equation, the goal is to find outas much as possible about its solutions. However, the meaning of the word solutionin the context of differential equations is sometimes misunderstood by students. Thereason for this goes back to its meaning in algebra, trigonometry, and calculusinthese worlds, solutions are numbers. But in the world of differential equations solutionsare not numbers . We illustrate by comparing a solution of an algebraic equation to thatof a differential equation. The solution of

2x 3 D x C 7

is 10, a number! Why is it a solution? The answer, of course, is that 10 satisfies thisequation. Substitution of 10 for the unknown x results in the left- and right-handsides of the equation being equal: the left-hand side (LHS) becomes

LHS D 2x 3 D 2.10/ 3 D 17;

which equals the right-hand side (RHS)

RHS D x C 7 D 10C 7 D 17:

By contrast, the solutions of differential equations are functions, not numbers. A simpleexample is provided by the differential equation

dy

dxD 2x:

A solution is y D x2. In fact, every function of the form y D x2 C C , where C is aconstant, is a solution. The reason for this is that these functions satisfy the equation.When we substitute x2 C C for the unknown y and differentiate, we obtain 2x,which is precisely the right-hand side of the equation:

LHS D ddx

y D ddx

.x2 C C / D 2x equals RHS D 2x:

No other functions have derivatives equal to 2x, aside from those of the form x2CC .Consequently, these are the only solutions of the differential equation. Lets take a lookat some more examples.

Example 1.3. Suppose someone alleges that y D lnx is a solution of

xy00 D y0: (1.1.3)

Determine whether it really is a solution.

Solution. We have not yet learned any methods for solving differential equations. Butwe do not need any to answer this question. All that is required is for us to substitute


the first and secondderivatives of ln x into the equation to see if a true statement results.With these substitutions, the left-hand side of the equation is

LHS D xy00 D x d2

dx2.ln x/ D x d

dx

1

x

D x

1x2

D 1

x;

while the right-hand side is

RHS D ddx

.ln x/ D 1x:

Since both the left- and right-hand sides turn out to be equal to x1, the functiony D lnx is a solution.

There is another matter to consider. Since a solution is a function, we need to beaware of the domain of its definition and where it solves the differential equation. Asfor this example, we know from calculus that the domain of ln x is the interval .0;1/.Since ln x also satisfies (1.1.3) on this interval, we say that the maximal interval forwhich y D lnx is a solution of (1.1.3) is .0;1/.Example 1.4. The function y D x2 is also alleged to be a solution of (1.1.3). Is itreally?

Solution. Substituting, we find that

LHS D x d2

dx2.x2/ D x d

dt.2x/ D 2x

whereasRHS D d

dx.x2/ D 2x:

Since the LHS RHS, we conclude that y D x2 is not a solution of the differentialequation.

Example 1.5. Is y D 1 a solution of (1.1.3)?Solution. This might be interpreted as: Is 1 a solution? But that would be incorrect.The question really asks: Does the constant function y.x/ D 1 satisfy the equation?Now this may seem like quibbling over semantics but there is a point to be made:

Solutions of algebraic equations are numbers. Solutions of differentialequations are functions.

Now the answer: Since both the first and second derivatives of this function are equalto 0 at all values of x, it satisfies the equation for all 1 < x


Solution of an Ordinary Differential Equation

Definition 1.4. A solution of an ordinary differential equation with one dependentvariable is a differentiable function of the independent variable that satisfies theequation on some interval. In other words, if we substitute the function for thedependent variable, we obtain a result that is valid on the interval.

Example 1.6. Is y D sin x a solution of the equation y2 C .y0/2 D 1?Solution. When we substitute sin x for y, we obtain y2C .y0/2 D sin2 xC cos2 x D 1by the Pythagorean identity of trigonometry. Since this is true for all real numbers,y D sinx is a solution on the interval .1;1/.

An algebraic equation may not have real-valued solutions, such as x2 D 1. Thesame may be true of a differential equation. Consider, for instance,

y2 C .y0/2 D 1:Whatever real-valued, differentiable function is substituted for y, the left-hand side ofthe equation is always nonnegative. Consequently, there is no real-valued function thatsolves this equation.

1.1.4 Partial DerivativesWe introduce the concept of partial derivatives much in the same way as our review ofordinary derivatives by again considering temperature variations in an unevenly heatedcopper object. This time, however, instead of a filament-like copper wire, we imagineheating a thin, rectangular copper plate with a length of 25 cm and a width of 5 cm.The thinness of the plate allows us to ignore its thickness in the ensuing discussion; ineffect, we model the real, three-dimensional plate with an idealized two-dimensionalrectangle. Lets orient the plate so that two of its adjoining edges are along the x- andy-axes as shown in Figure 1.1.

Fig. 1.1: Unevenly heated copper plate


As with the previous example of the copper wire, lets imagine that the plate isheated unevenly in such a way that the temperature at each of its points stays constantbut that it generally varies from point to point. One possibility is given in Table 1.3,which gives temperatures accurately measured to the ten-thousandth place at the points.x; y/. Lets use the notation T .x; y/ to designate the temperature at .x; y/. In otherwords, T is the name that we give to the temperature function. Note that it dependson both x and yin other words, it is a function of two variables rather than just onevariable.

Table 1.3: Temperatures (C) at Points of an Unevenly Heated Plate

x (cm)0 5 10 15 20 25

0 99.0000 98.8750 98.5000 97.8750 97.0000 95.87501 100.0000 99.8740 99.4980 98.8720 97.9960 96.8700

y(cm) 2 101.0000 100.8710 100.4920 99.8630 98.9840 97.85503 102.0000 101.8660 101.4820 100.8480 99.9640 98.83004 103.0000 102.8590 102.4680 101.8270 100.9360 99.79505 104.0000 103.8500 103.4500 102.8000 101.9000 100.7500

Lets select a specific point on the plate, say .10; 1/, in order to investigate thetemperature changes near it. In Table 1.3, the temperatures in the row and columncontaining the temperature at .10; 1/, namely 99.4980 C, are boldfaced. Consider thetemperature changes along this row and column. The common feature shared by thetemperatures in the row is that all of them are at points with their y-coordinates equalto 1. If we restrict our attention to the y D 1 row, the temperature function may beviewed as a function of the single variable x. Let r .x/ denote its value at x. Note thatr .x/ is the same as T .x; 1/. The derivative

r 0.10/ D limx!0

r .10Cx/ r .10/x

(1.1.4)

gives the exact (instantaneous) rate of change of the temperature T at the point .10; 1/when x is varied but the y-coordinate is held at the constant value 1. One of the waysto denote this derivative is by writing Tx.10; 1/. The limit of the difference quotient(1.1.4) written entirely in terms of T is

Tx.10; 1/ D limx!0

T .10Cx; 1/ T .10; 1/x

: (1.1.5)

This particular limit is called the partial derivative of T with respect to x at the point.10; 1/.

In the same way, just as we found that (1.1.5) gives the rate of change of T at thepoint .10; 1/ when the value of y is kept fixed at 1, we can find another rate of change


of T at the point .10; 1/ by varying y but keeping x D 10. This means that this timethe measurements in the boldfaced column (the one with the heading 10) must be usedto compute the difference quotients. Let c.y/ denote the temperatures in this column.One possible estimate of the rate of change of T at the point .10; 1/, albeit a rough one,is given by the following difference quotient obtained by changing y from y D 1 toy D 2:c

yD c.1Cy/ c.1/

yD c.2/ c.1/

2 1 D100:4920 99:4980

1D 0:9940 C=cm:

Since c.y/ D T .10; y/, this translates toT

yD T .10; 1Cy/ T .10; 1/

y(1.1.6)

D T .10; 2/ T .10; 1/2 1 D

100:4920 99:49801

D 0:9940 C=cm:

The quantity 0:9940 C is roughly the temperature increase per centimeter as y in-creases from 1 cm to 2 cm when x is held fixed at 10. The exact rate of change at thepoint .10; 1/ keeping x D 10 is defined by the limit

limy!0

T

yD lim

y!0T .10; 1Cy/ T .10; 1/

y: (1.1.7)

However, there is a fly in the ointment: if we do not know the temperatures at all of thepoints in the plate that have their x-coordinates equal to 10, then the limit (1.1.7) canonly be estimated with a difference quotient, such as the one in (1.1.6). Nevertheless,the limit does exist because there is a temperature T .x; y/, whether known or not,associated with every point .x; y/ on the plate. The limit given by (1.1.7) is calledpartial derivative of T with respect to y at the point .10; 1/ and is denoted by thesymbol Ty.10; 1/.

Suppose that the temperature at every point of the plate is given by the function 6

T .x; y/ D 99C y 0:0002xy2 0:005x2: (1.1.8)Lets compute the partial derivative Ty.10; 1/ by taking the limit of the difference quo-tient in (1.1.7) as follows:

limy!0

T .10; 1Cy/ T .10; 1/y

D limy!0

y 0:004y 0:002.y/2y

D limy!0

.0:996 0:002y/ D 0:996 C=cm:

Therefore, Ty.10; 1/ D 0:996 C=cm.The previous example serves as an introduction to the definition of a partial deriva-

tive. The rate of change of a function of two variables with respect to one of its vari-ables, as the other one is held constant, is known as a partial derivative of the function.

6In fact, the values in Table 1.3 were actually computed with this function.


Since there are two variables, there are two first-order partial derivatives.7 A partialderivative of a function of two variables is obtained by taking the limit of a differencequotient for that function, just as an ordinary derivative of a function of one variableis the result of taking the limit of its difference quotient. We just have to state pre-cisely what is meant by the differences quotients for a function of two variables andhow to take their limits. Using the example of the temperature function as a guide, thefollowing definition should come as no surprise.

First-Order Partial Derivatives

Definition 1.5. Let F be a function of x and y. The partial derivative of F withrespect to x, denoted by Fx , is defined by

Fx.x; y/ D limx!0

F.x Cx; y/ F.x; y/x

; (1.1.9)

provided the limit exists. Likewise, the partial derivative of F with respect to y,denoted by Fy , is defined by

Fy.x; y/ D limy!0

F.x; y Cy/ F.x; y/y

; (1.1.10)

provided this limit exists.

Example 1.7. Find the partial derivative Fx of the function F.x; y/ D 5x2y.Solution. By (1.1.9),

Fx.x; y/ D limx!0

5.x Cx/2y 5x2yx

D limx!0

5x2y C 10xxy C 5.x/2y 5x2yx

D limx!0

.10xy C 5xy/ D 10xy:

We should point out that there is another way to denote partial derivatives. Fx.x; y/and Fy.x; y/ are also expressed by writing

@

@xF.x; y/ and

@

@yF.x; y/;

respectively. The curly d distinguishes partial derivatives, expressed by symbols like@=@x and @=@y, from ordinary derivatives, which are expressed with the Latin d , suchas d=dx or d=dy.

7There are higher-order partial derivatives too, just as there are higher-order ordinary derivatives. How-ever, we only discuss first-order partial derivatives in this chapter. In a future chapter, we will need to talkabout second-order partial derivatives but that can wait for now.


When we examine definition (1.1.9) and look at the result of the example, we seethat taking the partial derivative of F with respect to x is really the same as taking theordinary derivative of F with respect to x, if at the same time we regard y as beingheld at a constant value. Symbolically, we could write

Fx.x; y/ D ddx

F.x; y D constant/:

Similarly, the process of taking the partial derivative of F with respect to y can beremembered symbolically as

Fy.x; y/ D ddy

F.x D constant; y/:

With this viewpoint, it becomes a relatively simple matter to take the partial derivativesof two-variable functions, such as F.x; y/ D 5x2y. The partial derivative of F withrespect to x is

Fx.x; y/ D @@x

5x2y

D ddx

F.x; y D constant/ D ddx

.5x2y/yDconstant D 10xy;

as we already determined in Example 1.7. Similarly, the partial derivative of F withrespect to y is

Fy.x; y/ D @@y

5x2y

D ddy

F.x D constant; y/ (1.1.11)

D ddy

.5x2y/yDconstant D 5x

2:

In practice,

d

dxF.x; y D constant/ and d

dyF.x D constant; y/

are mental steps but are not written down. We employ this merely as a pedagogic aidfor newcomers to this subject. Once you become adept at finding partial derivatives,you may think (1.1.11) but should write

Fy.x; y/ D @@y

5x2y

D 5x2:Example 1.8. Evaluate the first-order partial derivatives of the temperature function(1.1.8)

T .x; y/ D 99C y 0:0002xy2 0:005x2 Cat the point .10; 1/.


Solution. First we find the partial derivative of F with respect to x as follows:

Tx.x; y/ D @@x

T .x; y/ D @@x

.99Cy0:0002xy20:005x2/ D 0:0002y20:01x:Next we evaluate the result at the point .10; 1/:

Tx.10; 1/ D .0:0002y2 0:01x/.10;1/

D 0:0002.1/2 0:01.10/ D 0:1002 C=cm:Similarly, the partial derivative of F with respect to y is

Ty.x; y/ D @@y

T .x; y/ D @@y

.99C y 0:0002xy2 0:005x2/ D 1 0:0004xy:

Consequently,

Ty.10; 1/ D .1 0:0004xy/.10;1/

D 1 0:0004.10/.1/ D 0:996 C=cm:If you recall, the result Ty.10; 1/ D 0:996 C=cm was also obtained by directly apply-ing the difference quotient. Note the ease with which we can find partial derivatives byapplying the rules of differentiation that we already know from calculus.

1.1.5 Partial Differential EquationsUp to now we have explained what ordinary differential equations are and given exam-ples. They basically are equations containing ordinary derivatives. Likewise, equationscontaining partial derivatives (but not ordinary derivatives) are called partial differentialequations. In this book we are not concerned with partial differential equations per se;nevertheless, we will need to know a little about them. As it turns out, some methodsfor solving certain kinds of ordinary differential equations involve partial derivativesand some basic equations containing them. We will see this in Chapter 8. Lets beginwith the definition of a so-called partial differential equation.

Partial Differential Equation

Definition 1.6. A partial differential equation is an equation containing morethan one independent variable, one or more dependent variables, and partialderivatives of one or more of these dependent variables.

Our first example of a partial differential equation is

@u

@tD k @

2u

@x2:

It models the conduction of heat through an extremely thin metal bar, where u.x; t/ isthe temperature at the point x in the bar at time t . This equation is known as the one-dimensional heat equation. 8 There are two independent variables: the spatial variable

8For more information, see Churchill [?].

1.2. ORIGINS OF BASIC ORDINARY DIFFERENTIAL EQUATIONS 19

x and the temporal variable t . The variable u depends on both of them and so it isthe dependent variable. The parameter k is called the thermal conductivity of the bar.What makes this a partial differential equation is that it is an equation containing partialderivatives. Whereas @u=@t is a first-order partial derivative, @2u=@t2 is a second-orderpartial derivative, similar to second-order ordinary derivatives. Therefore, this is anexample of a second-order partial differential equation. We will say a little more abouthigher-order partial derivatives in a later chapter.

The only partial differential equations that we need to consider in this book arefirst-order equations of the form

@u

@xD f .x; y/ or @u

@yD g.x; y/; (1.1.12)

where the dependent variable u is a function of both x and y. An example of anequation of the first form is

@u

@xD 2xy sinx:

An example of one of the second form is

@u

@yD 5y4 C 2xy

x2 C y2 C 10:

In Section 1.2.3, we will discuss how to find solutions of partial differential equationsof the two forms shown in (1.1.12). For the sake of brevity, it is common to use the ab-breviations ode for ordinary differential equation and pde for partial differentialequation.

1.2 Origins of Basic Ordinary Differential EquationsAs Section 1.1 points out, ordinary differential equations arise when we attempt to usemathematics to model certain real-life situations. Equations that are judged good mod-els imitate reality closely, provide insight and understanding, and predict well. Findingjust the right equation, or equations as the case may be, could be quite complicatednot only because of the mathematics but because other disciplines are involved as well.In spite of this, we will ease our way into a study of differential equations by start-ing with some of the more elementary ones that arise from modeling relatively simplesituations. Throughout the rest of this chapter, we present examples of elementarydifferential equations that result when

(a) the rate of change of a quantity is known or can easily be determined;

(b) the rate of change of a quantity is known or conjectured to be proportional toanother quantity;

(c) Newtons second law of motion is used to model the motion of a body.


1.2.1 Rates of ChangeModeling real-life situations frequently involves the rates of change of quantities. Fromstudying calculus, we have learned that the ( instantaneous) rate of change of a quan-tity, dependent solely on a single variable, is given by its ordinary derivative with re-spect to the variable. As a result, when the rate of change of a quantity is given orcan be determined, that information can be expressed mathematically as an ordinarydifferential equation. Lets take a look at some examples of typical situations involvingrates of change.

Bathtub

Water flows out of a spout into a bathtub at the rate of 3 gallons per minute. Trans-lated in the succinct language of mathematics, this verbal statement becomes the dif-ferential equation

dN

dtD 3;

where N .t/ denotes the number of gallons of water that has flowed into the bathtubafter t minutes.

Temperature along a heated wire

In our review of the meaning of an ordinary derivative, we determined that thetemperature T along the copper wire changes at a rate of 0:0002 0:01x degreesCelsius per centimeter, where x is the distance in centimeters from the left end of thewire. This verbal statement is equivalent to the mathematical statement

dT

dxD 0:0002 0:01x:

Marginal cost

An economic decision may be based in part on the increment in cost that will beincurred if one more unit of some product is manufactured. Some economists call thiscost increment the marginal cost. It can usually be approximated by a derivative: theinstantaneous rate of change of the total cost function with respect to the number ofmanufactured units, say x, of the product. For this reason, many economists preferdefining the marginal cost (abbreviated MC) as this derivative. In this book, we usethe derivative definition of marginal cost. As an example, suppose it is stated that themarginal cost to manufacture x widgets9 is given by the function MC D 10xC 5000.We can express this statement more concisely with the differential equation

dC

dxD 10x C 5000;

where C.x/ denotes the total cost to produce x widgets.9A widget is a substitute for the name of some device or gadget, usually used when its real name is not

known or temporally forgotten, in other words, a thingamajig. Here we use it to mean a fictitious, manufac-tured product. In this way, we can easily fabricate marginal cost functions to illustrate the mathematics andeconomics, without also having to consider whether or not they represent reality.


Speed of a falling objectAn object is dropped from the top story of the Leaning Tower of Pisa, the famous

freestanding, eight-story bell tower of the cathedral of Pisa, Italy. Let s denote thedistance, in feet, the object has fallen t seconds after being dropped. The speed ofthe falling object is the rate with which its distance increases with time, or the timederivative Ps. According to the laws of elementary mechanics, the speed after t secondsis approximately 32t feet per second, provided the counteracting resistance of the airpushing upward on the object is negligible. Therefore, the speed of the falling object,up to the time it hits the ground, is modeled by the differential equation Ps D 32t , or

ds

dtD 32t:

Arc Length

A pair of equations, x D f .t/ and y D g.t/, will generate a curve in the xy-planewhen the independent variable t increases from a starting value to an end value .Under the assumption that the functions f and g have continuous derivatives, letsinvestigate how one goes about calculating the rate at which the length of the curveincreases with t .

Much of elementary calculus deals with graphs of smooth functions. A functionF.x/ defined for x , where and are real numbers, is said to be smoothif F 0.x/ exists and is continuous at every point of ; . The graph of y D F.x/when F is smooth is called a smooth curve. Each value of x corresponds to a point.x; y/ D .x;F.x// on the curve. As x increases from to , we can envision a pointparticle starting out at the initial point .;F.// of the curve and moving along it untilit ends up at the terminal point .;F.//.

Similarly, a pair of equations

x D f .t/; y D g.t/ (1.2.1)for t traces out a curve in the xy-plane as the independent variable t variesfrom t D to t D . Unlike before, the variable x is no longer an independentvariable; rather it, as does y, depends on the independent variable t . Each value of tcorresponds to a point .x; y/ D .f .t/;g.t// on the curve. The equations x D f .t/and y D g.t/ defining x and y are called parametric equations . The variable t is theparameter. A curve defined by (1.2.1) is said to be smooth if the derivatives of f andg exist, are continuous, and never simultaneously zero for all values of t in the interval; . Smoothness guarantees that

1. there is a unique tangent line at every point of the curve, and

2. .t/, the angle of inclination of the tangent line at the point .f .t/;g.t//, is de-fined for all values of t 2 ; and is a continuous function.

Consequently, a smooth curve has no corners or cusps. Moreover, no portion of thecurve is retraced as t increases. In other words, a point particle starting out at the


initial point .f ./;g.// will never stop and move in the reverse direction along thecurve before it reaches the terminal point .f ./;g.//. With that said, we are nowready to consider the length of a smooth curve, which is called its arc length, and todetermine the rate at which its arc length changes with respect to t .

Let s denote the arc length of a curve. For t1 , let s denote the length of theportion of the curve corresponding to values of t from t D t1 to t D t1 C t . Forvalues of t close to zero, the length s of this portion is approximately the length ofthe hypotenuse of the right triangle as illustrated in Figure 1.2

Fig. 1.2: Approximation of s

By the Pythagorean theorem,

s q.x/2 C .y/2 :

Since the difference quotient

x

tD f .t1 Ct/ f .t1/

t

is approximately equal to the derivative f 0.t1/ for values of t near zero, we canapproximate x with f 0.t1/t . Likewise, y g0.t1/t . Thus,

s q.f 0.t1/t/2 C .g0.t1/t/2 D t

q.f 0.t1//2 C .g0.t1//2;

ors

tq.f 0.t1//2 C .g.t1//2:

As t ! 0, this approximation becomes better and better. We conclude that the rateof change of the arc length s of the curve at t D t1 can be found by evaluating

ds

dtDq

f 0.t/2 C g0.t/2 Ds

dx

dt

2Cdy

dt

2(1.2.2)

at t D t1.In applications where t represents the time and .x; y/ D .f .t/;g.t// the position

of a moving body at time t , the derivative Ps given by (1.2.2) is the rate of change of thedistance of the body from its initial position, i.e., the speed of the particle.


1.2.2 Simplest Type of Ordinary Differential EquationBefore we show examples of differential equations arising from proportions or fromNewtons second law of motion, let us first note the form of the previous equationsand then look into a method for finding their solutions. Each of them was the resultof knowing the rate at which a quantity changes. Their form is simply a derivativeequal to a constant or to an expression involving the independent variable but not thedependent one. In mathematical notation, this is expressed as

dy

dxD f .x/: (1.2.3)

The notation f .x/ conveys that f is a function of x but not of y while dy=dx tells usthat x is the independent variable and y is the dependent variable. It is important tokeep in mind that the right-hand side of (1.2.3) is a function of the independent variablealone. Whatever is said in this section about finding solutions of (1.2.3) does not carryover to equations of the form

dy

dxD f .y/;

where f is a function of the dependent variable alone, nor to equations of the form

dy

dxD f .x; y/;

where f is a function of both variables.Equations of the form shown in (1.2.3) are the simplest type of ordinary differential

equation in the sense that all that is required to find their solutions is direct integrationwith respect to x. A solution of (1.2.3) is any function whose derivative is the functionf . Calculus tells us that there are infinitely many such solutionsany two of whichdiffer by a mere constantand this infinite set is known as the indefinite integral ofthe function f , which is symbolized by

y DZ

f .x/ dx: (1.2.4)

That is to say, all of the solutions of (1.2.3) are the same as the set of all antiderivativesof the function f . For example, consider the differential equation

dy

dxD 2x:

In the general notation of (1.2.3), f .x/ D 2x. All solutions make up the set of allantiderivatives of 2x; namely, x2 C C as

d

dx.x2 CC / D 2x:

In other words, all solutions are computed from (1.2.4). Thus,

y DZ

f .x/ dx DZ2x dx D x2 C C:


Sometimes we may have to reach a little further back into the recesses of our memoriesof calculus to come up with an antiderivative, or we may have to consult a table ofintegrals. For example, the solutions of the differential equation

dy

dxD 1

1C x2

are the functions

y DZ

dx

1C x2 D tan1 x CC;

since

d

dx.tan1 x/ D 1

1C x2 :

Besides looking up formulas of integrals in tables, there are technological optionsas well: computer algebra systems,10 certain advanced hand-held calculators that arereally calculator-and-computer hybrids, etc. Even so, the formulas of the integralslisted in Table 1.4, known as basic integration formulas,11 ought to be memorized!They occur so often that their committal to memory will actually save time and effortjust think of the time it takes to locate a table and then look up the integral or to turn ona computer and enter the appropriate commands. Moreover, the basic formulas are sowell-known that it might prove somewhat embarrassing not to have them memorized.Would you not question an instructors competency if she or he had to consult a tableor computer or calculator for

Rx2 dx?

Table 1.4 contain a list of the basic integral formulas that will turn up regularlyin the text and problem sets in this book.12 For a function f .u/ listed in the table,Rf .u/ du is its indefinite integral. As you recall from calculus, the indefinite integral

is evaluated by finding an antiderivative of f .u/ and adding a constant of integration,say C , to it. The result is an expression that represents all of the antiderivatives off .u/. Equivalently, this represents all solutions of the differential equation

dy

duD f .u/:

10A computer algebra system or CAS is an interactive computer program that can carry out mathematicalcomputations with symbolic expressions, such as yielding the result x2 when it is given a command tocompute the integral of 2x. Some well-known computer algebra systems are Maple, Mathematica, Derive,and MATLAB. Even many calculator models have built-in software that perform symbolic manipulations.

11They are also called standard integral forms or elementary forms.12By memorizing the formulas in Table 1.4, a student will be able to work at least 90% of the problems in

this book without having to resort to tables, calculators, or computer algebra systems.


Table 1.4: Basic Integration Formulas

f.u/Rf.u/ du

k (k, a constant) kuC C

un (n 1) unC1

nC 1 C C

u1 D 1u

ln juj C C D(

lnuC C; if u > 0ln .u/C C; if u < 0

eu eu C C

cosu sinuC C

sinu cosuC C

sec2 u tanuC C

csc2 u cot uC C

secu ln j secuC tan uj C C

cscu ln j cscuC cotuj C C

secu tanu secuC C

cscu cotu cscuC C


f.u/Rf.u/du

1

1C u2 tan1 uC C

1p1 u2

sin1 uC C

1

upu2 1

sec1 juj C C

1.2.3 Simplest Types of Partial Differential EquationsIn the previous section, we saw that direct integration with respect to x solves ordinarydifferential equations of the type

dy

dxD f .x/:

Likewise, integration with respect to a single variable is all that is required to findsolutions of equations of the following two types:

@u

@xD f .x; y/ (1.2.5)

and@u

@yD f .x; y/: (1.2.6)

These are the simplest types of partial differential equations.A solution of (1.2.5) is a function whose partial derivative with respect to x is

equal to f .x; y/. Solving (1.2.5) means finding all of the functions that satisfy (1.2.5).Recall that partial differentiation with respect to x is simply ordinary differentiationwith respect to x with y held fixed. Solutions are found by reversing the operation ofdifferentiation, that is, by integration. Since differentiation with respect to x is carriedout by holding y fixed, the inverse operation of integration is carried out by integratingwith respect to x holding y fixed. We express this symbolically in this way:

u DZ

f .x; y/ dx:

As an example, let us find solutions of the partial differential equation

@u

@xD 2xy sinx: (1.2.7)


Integrating with respect to x, we obtain the solutions

u DZ.2xy sinx/ dx D x2y C cos x C C:

Or so we might think! But let us reconsider. Are these really all of the solutionsof (1.2.7)? When we think about it, the so-called constant of integration C is reallymore than just a constant hereany function that dependsonly on y should be includedtoo. The reason of course is that the partial derivative of a function of y (but not x)with respect to x is equal to 0; if g denotes such a function, we write

@

@xg.y/ D 0:

Therefore, the complete set of solutions of (1.2.7) include all functions of the form

u.x; y/ D x2y C cosx C g.y/:Finally, we check this statement by substituting x2yC cosxCg.y/ for the dependentvariable u to verify that equation (1.2.7) is satisfied

@u

@xD @

@x

x2y C cos xC g.y/ D 2xy sinx:

For an example of a partial differential equation of the form (1.2.6), let us replace@u=@x with @u=@y in (1.2.7) to obtain

@u

@yD 2xy sin x:

To reverse the partial differentiation this time, we have to integrate with respect to y.We then obtain the solutions

u.x; y/ DZ.2xy sinx/ dy D xy2 y sin x C h.x/;

where h denotes a function of x alone.As a final example, lets return to the copper plate shown in Figure 1.1. Instead of

being given a function that models the temperature variation of the plate and then beingasked to find the rates at which the temperature changes along the x and y directions,suppose that we are given one of these rates and asked to find the temperature function.

Example 1.9. The rate at which the temperature of the copper plate in Figure 1.1changes with respect to y is

@T

@yD 1 0:0004xy C=cm

Find out what can be said about the temperature function T itself.


Solution. Since the differentiation of the temperature function T is with respect to y,we must integrate the right-hand side of the equation with respect to y to find T . Thus,

T .x; y/ DZ.1 0:0004xy/ dy D y 0:0002xy2 C h.x/: (1.2.8)

Note however that this is all that we can say about the temperature function since weare given no other information. That is, we cannot completely determine it since thereis no other information that allows us to find the function h. To illustrate how wewould go about finding h had we more information, suppose that we also know thattemperatures along the y D 1 line are given by the function

'.x/ D 100 0:0002x 0:005x2:If (1.2.8) is to model the temperature at all points of the plate, then it follows thatT .x; 1/ D '.x/; accordingly,

1 0:0002x2C h.x/ D 100 0:0002x 0:005x2:Solving for h, we have

h.x/ D 99 0:005x2:Finally, substituting this into (1.2.8) gives the temperature function 13

T .x; y/ D 99C y 0:0002xy2 0:005x2:

1.2.4 Basic Integration TechniquesComplete familiarity with the standard integrals and a facility with integration tech-niques will be deciding factors in your success, or lack thereof, in solving differentialequations. Table 1.4 is a list of some standard integrals. This is not a complete list, butthese particular integrals are used throughout this book. Next to the functions in theleft-hand column are their antiderivatives in the right-hand column. These standard in-tegrals must be thoroughly memorized. Unfortunately, however, an integral involved insolving a differential equation probably will not look exactly like any of those listed inTable 1.4: in that case, substitution, integration by parts , and partial fractions becomeindispensable. They are techniques for transforming integrals to one of the standardintegrals. We review briefly these three integration techniques by presenting some ex-amples. If you find yourself rusty in the use of these techniques, it would be a verygood idea to refresh your memory by opening your favorite calculus book and gettingout past calculus class notes and worked-out problem sets.

Example 1.10. Find all solutions of the differential equation

dy

dxD 5

x2 C 9 :13Note this example is Example 1.8 worked backwards.


Solution. Referring to (1.2.3), f .x/ D 5=.x2C 9/. The solutions are given by (1.2.4).So

y DZ

f .x/ dx DZ

5

x2 C 9 dx D 5Z

1

9C x2 dx:

This integrand most nearly resembles the integrand 1=.1 C u2/ in Table 1.4. Hence,we need to rewrite the denominator of the integrand so that it leads off with the digit 1instead of with 9. Factoring out the 9 achieves this and we haveZ

1

9C x2 dx DZ

1

9

1C

x3

2 dx D 19Z

1

1C u2 3 du;

where the substitution u D x=3 was made. Since du=dx D 1=3, the differential dxmust be replaced with 3 du. Therefore, the solutions are

y D 53

Zdu

1C u2 D5

3tan1 uC C D 5

3tan1

x3

C C:

Example 1.11. Find the solutions ofdy

dxD 5x

x2 C 9 .

Solution. The difference between this integrand and the previous one is the x in thenumerator. What we should note is that aside from a numerical factor the numeratoris the derivative of the denominator. A standard integral results by merely substitutinganother variable for x2 C 9, say z. Then, dz D 2x dx and

y DZ

5x

x2 C 9 dx D5

2

Z2x dx

x2 C 9 D5

2

Zdz

zD 5

2ln jzj CC:

Thus, the solution of the differential equation is

y D 52

lnx2 C 9C C:


dxD x dx

x2 C 2x C 1 .

Solution. Since the independent variable is x and the right-hand side of the equationinvolves it but not y, we start off by indicating the solution of the equation is

y DZ

x

x2 C 2x C 1 dx:

Now we have to figure out how to carry out the integration. As a rule of thumb, wealways try substitution first. Clearly, the only possibility is u D x2C2xC5. This willnot work however since the differential is

du D .2x C 2/ dx D 2.x C 1/ dx;


but the numerator is not a constant multiple of x C 1. At this point, we may be at aloss of what to do nextuntil we notice that the denominator can be expressed as theperfect square .x C 1/2. Perhaps this slight change will allow us to integrate. Now theform of the right-hand side of

y DZ

x

.x C 1/2 dx

suggests that we try the substitution u D xC1. Then du D dx and x D u1. Hence,

y DZ

u 1u2

du DZ

u1 u2du D ln juj C 1

uC C:

Therefore,

y D ln jx C 1j C 1x C 1 C C:

Example 1.13. Find the solutions of dydx

D x dxx2 C 2xC 5 .

Solution. This is similar to Example 1.12; again it is clear that a direct substitution willnot work. Unlike the previous example though, the denominator is not a perfect square.However, we can change x2 C 2x into a perfect square by completing its square. Thisis accomplished by adding 1, which is obtained by squaring half the coefficient of x,namely 1. Thus,

x2 C 2xC 5 D .x2 C 2x C 1/ 1C 5 D .x C 1/2 C 4I

and soy D

Zx

.x C 1/2 C 4 dx:

Now let u D x C 1. Then

y DZ

u 1u2 C 4 du D

Zu

u2 C 4 du Z

1

u2 C 4 du

D 12

Z2u

u2 C 4 du1

4

Z1

1Cu2

2 duD 1

2lnu2 C 4

14

tan1u2

CK

D 12

ln .x2 C 2xC 5/ 14

tan1x C 12

CK:


dxD x cos 2x.


Solution. Once again, as the right-hand side of the equation depends only on x, solu-tions are given by

y DZx cos 2x dx:

Integration of the product of a power function and a sinusoidal function (sine or cosinefunction) calls for the method of integration by parts. Recall the integration by partsformula: Z

udv D uv Z

v du: (1.2.9)

To apply the formula toRx cos 2x dx, we let

u D x; dv D cos 2x dx

du D dx

sincedu

dxD 1

; v D 1

2sin 2x

as

Zcos 2x dx D 1

2sin 2x CK

:

Then

y D uv Z

v du D 12x sin 2x 1

2

Zsin 2x dx D x

2sin 2x 1

2

12

cos 2xC C;

and so all solutions are given by

y D x2

sin 2x C 14

cos 2x CC:

Learning when and how to use the integration by parts formula, to wit (1.2.7), isan important skill to master. If the original integral, denoted by

Rudv, is difficult or

impossible to integrate, the purpose of (1.2.7) is to present an alternative: the integralRv du. The art of integrating by parts is in the selection of u and dv so that

Rv du

is easier to integrate than isRudv. Achieving this is a matter of common sense, trial

and error, and experience. However, if you have some difficulty in selecting u and dvthe acronym LIATE14 is a useful device. LIATE is a mnemonic for remembering fivetypes of functions in the following order:

Logarithmic, Inverse trig, Algebraic, Trig, Exponential.

It is precisely this order that makes LIATE work. When the integrand is the productof any two of these types of functions and the substitution technique doesnt work, theintegration by parts formula should be tried. Let the type appearing first in the ordergiven by LIATE be u and the other one, along with the differential next to the integrand,be dv. Then apply the integration by parts formula. Hopefully,

Rv du will then be

easier to integrate thanRudv. To illustrate this, consider the integral

Rx cos 2x dx

from the previous example. Its integrand is the product of the two functions x andcos 2x, where x is an algebraic function and cos 2x is a trigonometric function. Thus,we let u D x because the type Algebraic comes before the type Trig in LIATE. So,dv D cos 2x dx. Lets illustrate LIATE again by solving the next equation.

14See Kasube [?, pp. 210211].



dxD sin1.2x/.

Solution. Solutions are all of the antiderivatives of sin1.2x/ indicated by the indefiniteintegral

y.x/ DZ

sin1.2x/ dx:

Obviously, the substitution u D 2x wont lead anywhere unless the formula forZsin1 udu

is already available.15 At first, it looks as though the integration by parts technique isalso useless here by virtue of the absence of a product of two functions. Nevertheless,there is oneif we use the artifice of writing the integrand as the product: 1sin1.2x/.Then u D sin1.2x/, since the constant function 1 belongs to the Algebraic type andis preceded by the Inverse Trig type in LIATE. This forces dv to be the rest of theintegrand or 1 and the differential dx; that is, dv D 1 dx D dx. Hence,

du D u0.x/ dx Dd

dxsin1.2x/

dx D 2p

1 .2x/2dxI v D x:

By (1.2.9), the solutions are

y.x/ DZ

sin1.2x/ u

dxdvD1dx

D sin1.2x/ u

xv

Z

xv

2p1 .2x/2 dx

du

D x sin1.2x/

2

8Z 8x dxp

1 4x2D x sin1.2x/C 1

2

p1 4x2 C C:

Example 1.16. Find the solutions of dydx

D 6x2 9 .

Solution. Although the right-hand side of this equation does not appear in Table 1.4,the factorability of x2 9 is the tip-off that we can integrate using the method ofpartial fractions. Generally speaking, a quadratic polynomial, such as x2 9, is said tobe reducible over the reals or factorable when it can be expressed as a product of twolinear polynomials with real coefficients. Otherwise, it is said to be irreducibleover thereals. For the sake of brevity, we will omit the phrase over the reals. Thus, x2 9 is

15Of course, this integral could be found in some handbook or with a computer algebra system or asophisticated calculator. But the point of this section is to review integration and to hone the skills alreadyacquired in a calculus course to solve relatively simple integrals without having to resort to integral tables orto technology.


reducible since it reduces to the product of the linear polynomials x3 and xC3. Thatis, it is equal to the product .x 3/.xC 3/. On the other hand, x2C 9 is irreducible,16since it is just not possible to write it as a product of two linear polynomials with realcoefficients. An important result of algebra states that it is always possible to expressa nonconstant polynomial with real coefficients as a product of linear and irreduciblequadratic polynomials with real coefficients. So, as a rule of thumb, when faced withintegrating a rational function ,17 first factor its denominator into a product of linear andirreducible quadratic polynomialsunless the numerator is a constant multiple of thederivative of the denominator; in which case the method of substitution will work, asdemonstrated in Example 1.11.

Now lets apply the method of partial fractions to evaluate the integralZ6

x2 9 dx:

Its integrand is a rational function with a reducible denominator. After factoring thedenominator, we expand the integrand writing it as a sum of two simpler rational func-tions with linear polynomials as their denominators:

6

.x 3/.x C 3/ DA

x 3 CB

x C 3 : (1.2.10)

The fractionsA

x 3 andB

x C 3are called partial fractions because their denominators contain part of the originaldenominator x29 but not all of it. The expansion of the integrand in (1.2.10) is knownas its partial fraction expansion . The coefficients A and B are called undeterminedcoefficients until values are found making (1.2.10) an identity. To determine the valuesof A and B, clear out the denominators in (1.2.10) by multiplying both sides by thedenominator of the integrand with the result

A.x C 3/CB.x 3/ D 6:

Now we can quickly find the value of A by setting x D 3 since this eliminates the terminvolving B:

x D 3 ) 6AC 0B D 6 ) A D 1:Likewise, setting x D 3 eliminates the term involving A and yields B D 1. Thus,

6

x2 9 D1

x 3 1

xC 3 :

16That is, it is irreducible over the reals (meaning the set of real numbers). However, it is reducible overthe set of complex numbers since x2 C 9 D .x 3i/.x C 3i/, where i is defined by i2 D 1.

17Recall that a rational function is the quotient of two polynomials. Constants, such as 6 and , are alsoconsidered polynomials.


Now integration is a piece of cake:

y.x/ DZ

6

x2 9 dx DZ

dx

x 3 Z

dx

x C 3D ln jx 3j ln jx C 3j C C D ln

x 3xC 3

C C:

Lets review next the method of partial fractions when the denominator of the ra-tional function is the product of a linear factor and an irreducible quadratic factor.


dxD 18

x3 C 9x .

Solution. The method of substitution is clearly of no help in integrating the right-handside directly due to the absence of the factor

d

dx.x3 C 9x/ D 3x2 C 9

in the numerator. Lets see how the method of partial fractions fares. First, we mustfactor the denominator completely. Factoring out an x, we have

x3 C 9x D x.x2 C 9/:No more factoring is possible, since the quadratic factor is irreducible. Since the de-nominator consists of two factors, the partial fraction expansion of the integrand con-sists of two fractions. The form of partial fractions can be summed up by the dictum:

Put constants over linear factors and linear factors over irreduciblequadratic factors.

As a result, the form of the expansion is

18

x3 C 9x DA

xC Bx C C

x2 C 9 :

We clear out the denominators of the expansion by multiplying both of its sides byx.x2 C 9/, obtaining

A.x2 C 9/C .Bx C C /x D 18:Grouping like terms together gives

.ACB/x2 C Cx C 9A D 18:This equation is satisfied for all values of x if A, B, and C have values satisfying theequations:

ACB D 0I C D 0I 9A D 18:Consequently, A D 2; B D 2; C D 0 and so

y DZ

18

x3 C 9x dx DZ

2

xdx

Z2x

x2 C 9 dx D 2 ln jxj ln jx2 C 9j C C:


Therefore, the solutions are y D ln

x2

x2 C 9C C .

We conclude this section with an example of solving a partial differential equation.

Example 1.18. Find the solutions of @z@x

D y sec .3x/.Solution. The equation itself indicates that the variable z is dependent on both x and y.The nature of the partial derivative indicates that we have to integrate with respect tox: The equation itself indicates that the variable z is dependent on both x and y. Thenature of the partial derivative indicates that we have to integrate with respect to x:

z DZy sec.3x/ dx D y

Zsec.3x/ 1

3 3 dx D y

3

Zsec v dv;

where v D 3x. From Table 1.4, we have

z D y3

ln j secv C tan vj C g.y/:

Therefore, the solutions of the partial differential equation are

z D y3

ln j sec.3x/C tan.3x/j C g.y/:

1.2.5 Proportions Involving Rates of ChangeThe statement u is (directly) proportional to v means that

u D kvfor some constant k. The parentheses enclosing the word directly indicate that itsuse is optional; it is frequently omitted. The constant k is called the constant of pro-portionality. That u is proportional to v is also indicated by writing

u / v:Example 1.19. The assertion that a citys average daily garbage collection G is pro-portional to its population p translates to the mathematical statement:

G / p;which means

G D kpfor some constant k. If there is any truth to this statement, then the value of k wouldhave to be determined experimentally by comparing the amount of garbage and thepopulation of a city on a given day.


Example 1.20. Consider the wry comment of some good-natured hostess, as shewatched a guests toast land on her newly laid carpet, that the likelihood of toast land-ing jelly-side down is directly proportional to the cost of the carpet. With L denotingthe likelihood, or probability, of this happening and C the cost, this comment translatesto the mathematical statement:

L D C;where is the constant of proportionality.18

Example 1.21. Finally, in a more serious vein, we mention a postulate by the famousphysicist Max Planck that aided in the development of a field of physics called quantummechanics. In 1900, in an attempt to reconcile the theory of black body radiation withexperimental results, Planck postulated that energy is not radiated continuously butrather in discrete amounts called quanta and that a quantum of energy E is proportionalto the frequency of the radiation.19 That is,

E D hwhere h denotes the proportionality constant. By adjusting the value of h, he was ableto reconcile his theory with experimental results. In fact, h is now called Plancksconstant. Its current accepted value is 6:63 1034 joulesecond.

If two variables u and v are not directly proportional but are related by the equation

u D kv

for some constant k, then we say u is inversely proportional to v or u varies in-versely with v.

Example 1.22. In the kinetic theory of gases, there is the empirical result known asBoyles Law. It states:

The volume V of a certain amount of gas confined to a container and heldat a constant temperature is inversely proportional to the pressure P onthe gas.

Mathematically, this is expressed as

V D kP

or PV D k;

where k is the constant of proportionality.

18The letter is the lower case Greek letter gamma.19The letter is the lower case Greek letter nu.


Example 1.23. A Wall Street rule of thumb is that airline stock prices increase when-ever petroleum stock prices go down and vice versa. Hence,

AP D C;where A and P are the prices of the airline and petroleum stocks, respectively, and Cis the constant of proportionality.

Sometimes it must seem to the owner of a carespecially a new carthat thelikelihood that a car gets a dent is inversely proportional to its age.

Finally we observe that even someone disinclined to mathematics (and therebyless fortunate) is tempted at times to invoke mathematics to make a point. Take forexample this statement by David M. Knight: The fundamental rule is that our abilityto recognize the voice of God is in inverse proportion to our attachment to the thingsof this world. 20

Since this book is about ordinary differential equations, the proportional relation-ships that we consider from now on will involve ordinary derivatives. The followingexamples are taken from physics and chemistry.

1.2.6 Newtons Law of CoolingIt is patently clear to anyone that an ice-cold can of soda left on a patio will eventuallywarm up to the outside temperature and a cup of hot chocolate set on a kitchen tablewill cool down to room temperature. Experimental evidence indicates that for moderatetemperature differences between a body and its surroundings, the rate of change of thetemperature T of the body is proportional to the difference in the temperatures of thebody and its surroundings. Expressed in the language of calculus, this statement takeson the form

dT

dt/ T Ta

ordT

dtD c.T Ta/ (1.2.11)

where Ta is the ambient temperature, or the temperature of the surroundings, andc is the constant of proportionality. This model of temperature change is known asNewtons law of cooling . Clearly, the temperature of a body is decreasing when T >Ta but increasing when T < Ta. Or, from our knowledge of calculus, the derivativedT=dt is negative when T > Ta but positive when T < Ta. In either case, the constantc in (1.2.11) must be negative. As it is customary in the sciences to keep physicalconstants and parameters positive, we replace c with k, where k > 0. Accordingly,(1.2.11) becomes

dT

dtD k.T Ta/ .k > 0/: (1.2.12)

Note that Newtons law of cooling does not result from attempting to explain the phys-ical processes taking place, such as heat transfer between a body and its surroundings

20See Knight [?, p. 80]. This unconventional example stems from Fr. Knights tongue-in-cheek questionabout the importance of mathematics in Gods ultimate plans for humankind.


by conduction, convection, and radiation. So, in a sense all of the unexplained physicalprocesses are represented by the constant of proportionality k. This is asking a lot of aconstant and would explain the experimental evidence that Newtons law of cooling isnot always valid, such as in the case of extreme temperature differences.

1.2.7 Rates of Chemical ReactionsThe gas ethane is a hydrocarbon: a compound consisting of only carbon and hydrogen.An ethane molecule is composed of two carbon atoms and six hydrogen atoms, soits molecular formula is C2H6. Ethane decomposes when it is strongly heated in theabsence of air. Under certain experimental conditions, it has been observed that therate at which this decomposition takes place is proportional to the concentration ofethane. 21 In chemistry, it is customary to denote the concentration of a compound byenclosing its formula with brackets: thus, [C2H6] denotes the concentration of ethane.In this notation, the observation that ethane decomposes at a rate proportional to itsconcentration can be expressed as:

Rate of decomposition of ethane D k C2H6;where the constant of proportionality k is called a reaction rate constant . It is a pa-rameter, positive in value, which must be determined experimentally. Since ethanedecomposes, the rate of change in its concentration is a negative quantitythus, thederivative d[C2H6]=dt is equal to the negative of the rate of decomposition:

d C2H6dt

D k C2H6 : (1.2.13)Since the mathematical manipulation of bracketed formulas is cumbersome when solv-ing differential equations involving them, it is convenient to replace them with lowercase letters. For example, by letting x.t/ D C2H6 denote the concentration of ethaneat time t , equation (1.2.13) simplifies to

dx

dtD kx:

As for a second example, we consider one of several chemical reactions that havebeen conjectured by some scientists to take place in the earths stratosphere to explainthe alleged depletion of the ozone layer in the polar regions. 22 In this reaction, an oxy-gen atom (O) collides with an ozone molecule (O 3) to produce two oxygen molecules(O2):

OC O3 ! 2O2:It is believed that the rate at which the concentration of oxygen molecules increases isproportional to the product of the concentrations of oxygen atoms and ozone molecules.Translating this statement in the language of mathematics, we obtain the differentialequation:

d O2dt

D KOO3;21See Atkins [?, p. 131].22See Atkins [?, p. 140].


where K is the constant of proportionality. Since the derivative is positive, K must bepositive of course, its value can only be determined experimentally.

1.2.8 Newtons Second Law of MotionImportant applications of differential equations are found in the branch of physicscalled classical mechanics, which is the study of the motion of bodies and particles.A body possesses both mass and extent, whereas a particle is the idealized notion of ageometric point possessing mass. Classical mechanics is based on three famous lawsof motion formulated by Sir Isaac Newton (1642-1747); so it is also called Newtonianmechanics. A proper study of the laws of motion rightfully belongs to a physics or en-gineering course on classical mechanics. Nevertheless, the secondof these laws, knownas Newtons second law of motion , is a virtual treasure-trove of differential equationsproblems that are instructive and motivational, yet not too difficult. For this reason, weuse it in this book as our primary source for problems and examples and to demonstratethe indispensability of differential equations for modeling physical phenomenon. Wewill discuss Newtons second law of motion in the remainder of this section, but onlybriefly. It is the fundamental law of classical mechanics. It may be only a slight ex-aggeration to say that any classical mechanics course is essentially a study of how toapply the second law to a variety of situations.

A body accelerates when external forces act on it provided they do not cancel eachother out. We can add these forces, but we have to take into account that they may actin different directions. A directed quantity, of which a force is an example, that is char-acterized by both direction and magnitude is called a vector.23 Some other examples ofvectors are displacements and velocities. The (vector) addition of two or more vectorsresults in a single vector, called the vector sum, that is equivalent to all the other vectorsacting concurrently. The vector sum of all the external forces acting on a body is calledthe resultant force or resultant. It is the single force that can replace the original set ofexternal forces and still cause the body to move in precisely the same way.

Newtons second law of motion is an equation relating the mass of a body, theresultant force acting on the body, and its acceleration. It is derived from the followingthree experimental observations:

1. A body accelerates in the direction of the resultant force.

2. The magnitude of the acceleration of a body of constant mass is proportional tothe magnitude of the resultant force.

3. For a constant resultant force, the magnitude of the acceleration is inversely pro-portional to the mass of the body.

Letting m denote the mass of the body, a its acceleration, and F the resultant of all theexternal forces acting on the body, we can encapsulate the above observations with theproportionality statement:

a / Fm:

23Directed quantities also have to obey certain rules of combination before they can legitimately be con-sidered vectors.


Thus, we end up with the vector equation

ma D F; (1.2.14)where is a constant of proportionality. The letters for the acceleration and forceare in boldface to indicate that they are vector quantities. However, the letter m isnot in boldface since mass is a scalar, a quantity that has magnitude but no directionassociated with it.

If units of measurement (such as time, mass, and length) are chosen so that thevalue of is 1, then (1.2.14) simplifies to

F D ma: (1.2.15)In words:

The resultant force acting on a body is equal to the product of the massof the body and its acceleration .

This is Newtons second law of motion for the accelerated motion of a body subjectedto external forces when its mass remains constant.

Since acceleration is the instantaneous rate at which velocity changes with time,Newtons second law can be written as

F D mdvdt

; (1.2.16)

where v is the velocity, that is, the instantaneous rate at which the position of a bodychanges with time t . This statement of Newtons second law is only valid if the massof a body is constant. Note that this is a first-order differential equation.

The momentum of a body is defined as the vector quantity mv. If the mass of abody changes, then we have to use the general form of Newtons second law of motion :

The resultant force acting on a body is equal to the rate of change of themomentum of the body.

This is expressed mathematically by the vector equation

F D ddt

.mv/: (1.2.17)

Observe that (1.2.17) reduces to (1.2.16) when m is constant.For simple situations where the only forces causing motion of a body act either

in one direction or in the opposite direction, such as forces acting vertically, directedupward or downward, or forces acting horizontally, directed to the right or to the left, apositive or negative sign is affixed to the magnitude of the force to indicate its direction.For forces acting horizontally, we use the convention that positive forces are directedto the right whereas negative forces are directed to the left. Similarly, for forces actingvertically, we will regard upward as positive and downward as negative. Then wemerely have to add forces algebraically to obtain the resultant force. As a result, wecan write (1.2.15) without using boldface letters:

F D ma: (1.2.18)


Let us say a few words about units of measurement for the quantities m, a, andF . Unfortunately there is no single set of units in use today. There is the metricsystem and then there are systems involving English units, such as the foot and thepound. Generally speaking, all of the systems of units have one thing in common:units are chosen so that the constant of proportionality in (1.2.14) is equal to 1. Sucha set of units is known as absolute units or consistent units. One set of consistentunits uses the second for time and the metric units kilogram for mass and meter forlength. Consequently, velocity and acceleration inherit the units meters per second andmeters per second per second, respectively. Setting D 1, m D 1 kg (kilogram), anda D 1 m=s2 (meter per second per second) in (1.2.18), we obtain the force

F D .1 kg/ 1

m

s2

D 1 kg m

s2:

This unit of force is called the newton. In other words, a force of 1 newton acting ona body with a mass of 1 kilogram will accelerate it at a rate of 1 meter per second persecond. For more details, consult a textbook on classical mechanics, such as Osgood[?, pp. 5152] or Becker [?, p. 25].

We will use Newtons second law shortly (see Example 1.24) to model the verticalmotion of a body of constant mass near the earths surface. All bodies in the universeare subject to the gravitational attractive force of the earth. In fact, according to New-tons law of universal gravitation, every body in the universe exerts a gravitational forceon every other body in the universe. This includes everybody too! The force that theearth exerts on a body is called the force due to gravity. Experimentally it is found thatwhen various bodiesthose whose motion is not appreciably affected by the resistiveforce of air24 and other factors25are released from the same point above the earthssurface, each of them falls with the same acceleration. This constant acceleration iscalled the acceleration due to gravity .

Actually the acceleration due to gravity varies from place to place. However, closeto the earths surface it is nearly constant. Its precise value, as well as approximatevalues, is denoted by the letter g. At sea level and mid-latitudes, measurements showthat g is approximately 32:2 ft/s2 in the system of units known as the U. S. CustomarySystem.26 In another set of units called the SI units, short for the French Syste`meInternational dUnites and commonly referred to as the metric system, g is approxi-mately 9:81 m/s2. In the SI system, g is defined precisely as 9.80665 m/s 2 and in theU.S. Customary System as 32:174049 ft/s2. These precise values are referred to as thestandard acceleration of gravity (or standard gravity).

The U.S. Customary System can be confusing at times. For instance, consider theword pound. It can designate a unit of ma

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