Introductory Differential Equations || Introduction to Differential Equations

C H A P T E R

1Introduction to Differential Equations

O U T L I N E

1.1 Introduction to DifferentialEquations: Vocabulary 3

Exercises 1.1 10

1.2 A Graphical Approach to Solutions:Slope Fields and Direction Fields 15

Exercises 1.2 19

Chapter 1 Summary: Essential Conceptsand Formulas 22

Chapter 1 Review Exercises 23

The purpose of Introductory DifferentialEquations is twofold. First, we introduce anddiscuss the topics covered in an undergraduatecourse in ordinary differential equations (ODEs).Second, we indicate how certain technologiessuch as computer algebra systems and graphingcalculators are used to enhance the study ofdifferential equations, not only by eliminatingsome of the computational difficulties that arisein the study of differential equations but alsoby overcoming some of the visual limitationsassociated with the solutions of differentialequations. The advantages of using technologysuch as graphing calculators and computeralgebra systems in the study of differentialequations are numerous, but perhaps the mostuseful is that of being able to produce thegraphics associated with solutions of differentialequations. This is particularly beneficial inthe discussion of applications because many

physical situations are modeled with differentialequations. For example, in Chapter 5, we seethat the motion of a pendulum can be modeledby a differential equation. When we solve theproblem of the motion of a pendulum, we usetechnology to watch the pendulum move. Thesame is true for the motion of a mass attachedto the end of a spring, as well as many otherproblems. In having this ability to use technol-ogy, the study of differential equations becomesmuch more meaningful as well as interesting.

Although this chapter is short in length, thevocabulary introduced here is used throughoutthe text. To a large extent, this chapter may beread quickly, but it is important to rememberthat subsequent chapters will take advantageof the terminology and techniques discussedhere. Any formal introduction to differentialequations should begin with German scientistGottfried Wilhelm Leibniz (1646-1716) and

1Introductory Differential Equations Copyright © 2014 Elsevier Inc. All rights reserved.http://dx.doi.org/10.1016/B978-0-12-417219-7.00001-6

2 1. INTRODUCTION TO DIFFERENTIAL EQUATIONS

British scientist Isaac Newton (1642-1727), theinventors of calculus. In integral calculus, welearn that the area under the graph of a smoothpositive function is given by a definite integral,but both Leibniz and Newton were moreconcerned with solving differential equationsthan finding areas. Many of the methods ofsolution we present in this text are from thegreat Swiss mathematician Leonhard Euler(1707-1783). Subsequently, many problems, suchas determining the motion of a plucked string,lead not only to ordinary and partial differentialequations but also to other areas of mathemat-ics. Indeed, differential equations are full of

Gottfried Wilhelm Leibniz (1646-1716)

Isaac Newton (1642-1727): The controversy as towho (Leibniz orNewton) discovered calculus firstbecame an obsession with Newton during thesecond half of his life.

rich and exciting applications; interesting ap-plications are included throughout the text tomotivate discussions, make the study of differ-ential equations more interesting and pertinentto the real world and indicate how some peopleuse differential equations beyond this course.However, mathematics is also interesting in itsown right; mathematical applications are alsoincluded throughout the text.

Isaac Newton: We hope this young Isaac Newtonwas not worried about who discovered calculusfirst.

Leonhard Euler (1707-1783)

1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS: VOCABULARY 3

1.1 INTRODUCTION TODIFFERENTIAL EQUATIONS:

VOCABULARY

We begin our study of differential equationsby explaining what a differential equation is.From our experience in calculus, we are familiarwith some differential equations. For example,suppose that the acceleration a(t) (measured inft./s2) of a falling object is a(t) = −32. Then,because a(t) = v′(t), where v(t) is the veloc-ity of the object (measured in ft./s), we havev′(t)= − 32 or

dvdt

= −32.

An equation like this involving a function of asingle variable is called an ordinary differentialequation (ODE). (If the equation involves partialderivatives, then it is called a partial differentialequation (PDE).) In this case, the function to bedetermined is v = v(t), which depends on thevariable t, representing time (measured in sec-onds). The goal in solving an ODE is to find afunction that satisfies the equation. We can solvethis ODE through integration:

v(t) =∫

a(t)dt =∫(−32)dt = −32t+ C,

where C is an arbitrary constant. This result in-dicates that v(t) = −32t + C is a solution ofthe ODE for any choice of the constant C. (Wecall this a general solution because it involves anarbitrary constant.) In fact, we have found everysolution of the ODE because each is expressed as−32t + C. Examples of solutions and the corre-sponding C values include v(t) = −32t (C = 0),v(t) = −32t + 32 (C = 32), and v(t) = −32t − 8(C = −8). This shows that there are an infinitenumber of solutions to the ODE.

We can verify that v(t) = −32t + C is a generalsolution of dv/dt = −32 through substitution:

dvdt

= ddt(−32t+ C) = −32.

When we substitute our solution into the leftside of the ODE, we obtain −32, the value onthe right side of the ODE. We have verified thatv(t) = −32t+ C satisfies the ODE for any choiceof C.

Many times we are given a particular condi-tion that the solution must satisfy. For example,suppose that the object considered earlier has aninitial velocity of −64 ft./s. In other words, thevelocity at time t = 0 seconds is representedwith the initial condition v(0) = −64. Therefore,we need to find a solution of the ODE that alsosatisfies the initial condition. We express thisinitial value problem (IVP) as

dvdt

= −32, v(0) = −64,

where we solve the IVP by first finding a generalsolution to the ODE and then by applying the initialcondition to determine the arbitrary constant. Whenwe substitute t = 0 into the general solutionv(t) = −32t + C, we obtain v(0) = −32 × 0 +C = C. We conclude that C = −64 so that v(0) =−64 is satisfied. This means that the solution tothe IVP is v(t) = −32t−64. Notice that unlike theODE, the IVP has only one solution.

Another application of differential equationsfound in calculus is finding a function whengiven (a) the slope of the line tangent to the graphof the function at any point (x, y) and (b) a pointon the graph.

For example, suppose that the slope of the tan-gent line at any point on the graph of a functiony = y(x) is given by

dydx

= 3x2 − 4x,

and further that the graph passes through thepoint (1, 4), which means that y(1) = 4. In thiscase, the ODE is given by dy/dx = 3x2 − 4x,where we need to find the function y = y(x) thatsatisfies the initial condition y(1) = 4. Therefore,we solve the IVP

dydx

= 3x2 − 4x, y(1) = 4.


As in the previous example, we find a generalsolution to dy/dx = 3x2−4x through integration.This yields

y = y(x) = x3 − 2x2 + C.

Now, when we apply the initial condition, wefind that

y(1) = 13 − 2 × 13 + C = −1 + C = 4

so that C = 5, which means that the solution tothe IVP is

y = y(x) = x3 − 2x2 + 5.

Figure 1.1 shows the graph of the solutiontogether with a portion of the tangent line atthe point (1, 4). Notice that the slope of the tan-gent line at the point (1, 4) is the value of dy/dxevaluated if x = 1: if x = 1, dy/dx = 3 ×12 − 4 × 1 = −1. This observation will beuseful in Section 1.2 in helping us better under-stand the behavior of solutions of differentialequations.

−1 −0.5 0.5 1 1.5 2x

1

2

3

4

5y

FIGURE 1.1 Solution to dy/dx = 3x2 −4x, y(1) = 4 alongwith a segment of the tangent line at (1, 4).

The previous examples are similar in that theyeach involve an ODE in which the highest orderderivative is the first derivative. We call equa-tions of this type first-order ODEs because theorder of a differential equation is the order ofthe highest order derivative appearing in theequation.

E X AM P L E 1.1.1Determine the order of each of the following

differential equations: (a) dy/dx = x2/y2 cos y(b) uxx+uyy = 0 (e) (dy/dx)4 = y+x (d) d2x/dt2 +2dx/dt + 3x = sin t

Solution(a) This equation is first order because it in-

cludes only one first-order derivative, dy/dx. (b)This equation is classified as second order be-cause the highest order derivatives, uxx, repre-senting ∂2u/∂x2, and uyy representing ∂2u/∂y2,are of order two. The equation uxx + uyy = 0arises in many areas of study, which include fluidflows as well as electrostatic and gravitationalpotential, is often called Laplace’s equation afterPierre-Simon Laplace (also known as the Marquisde Laplace) (1749-1827) or the potential equation.Hence, Laplace’s equation is a second-order par-tial differential equation. (e) This is a first-orderequation because the highest order derivative isthe first derivative. Raising that derivative to thefourth power does not affect the order of the equa-tion. The expressions(

dydx

)4

andd4ydx4

do not represent the same quantities: (dy/dx)4

represents the derivative of y with respect to x,dy/dx, raised to the fourth power; d4y/dx4 repre-sents the fourth derivative of y with respect to x.(d) The highest order derivative is d2x/dt2, so theequation is second order.


Pierre-Simon Laplace (1749, Normandy, France-1827, Paris, France): Laplace’s name is probablymost remembered in reference to the Laplace trans-form that we will study in Chapter 8.

E X AM P L E 1.1.2(a) Show that y = c1 sin t + c2 cos t satisfies the

second-order ODE (ODE) y′′ +y = 0 where c1 andc2 are arbitrary constants. (b) Find the solution tothe IVP y′′ + y = 0, y(0) = 0, y′(0) = 1.

Solution(a) Differentiating, we obtain y′ = dy/dt =

c1 cos t − c2 sin t and y′′ = d2y/dt2 =−c2 cos t − c1 sin t. Therefore,

d2ydt2

+ y=−c1 sin t − c2 cos t + c1 sin t

+c2 cos t = 0,

so the function satisfies the ODE for anychoice of c1 and c2. We graph the solution forseveral values of these constants inFigure 1.2. Because the solution depends onat least one constant, we say that thefunctions shown in Figure 1.3 are membersof the family of solutions of the ODE.

(b) Evaluating the function at t = 0 yieldsy(0) = c1 sin 0+ c2 cos 0 = c2. Then, the initialcondition, y(0) = 0, indicates that c2 = 0.

1 2 3 4 5 6t

−1

−2

1

2

y

FIGURE 1.2 Graphs of y = c1 cos t + c2 sin t for variousvalues of c1 and c2.

Similarly, y′(0) = c1 cos 0 − c2 sin 0 = c1, soc1 = 1 so that the second initial condition,y′(0) = 1, is satisfied. Therefore, y(t) = sin tis the solution to the initial value problem(IVP). We graph this solution in Figure 1.3.Notice that the ODE has an infinite numberof solutions while the IVP has only oneunique solution. Also observe that thenumber of initial conditions for the initialvalue problem (IVP) matches the order ofthe ODE.

The next level of classification tells uswhetheran equation is linear or nonlinear in terms of thedependent variable. For an ODE, assuming thatthe independent variable is x and the dependentvariable is y, an ODE (of order n) is called linearif it can be written as

an(x)dnydxn

+ an−1(x)dn−1ydxn−1 + · · · + a2(x)

d2ydx2

+ a1(x)dydx

+ a0(x)y = f (x), (1.1)

where the functions ak(x), k = 0, 1, . . . , n, andf (x) are given and an(x) is not the zero func-tion. If f (x) is identically the zero function, thelinear equation (1.1) is said to be homogeneous.


2 4 6 8 10 12t

−1−0.5

0.51

y

FIGURE 1.3 Graph of y = y(t) = sin t.

You should verify that y(x) = 0, which we callthe trivial solution, is a solution to every linearhomogeneous equation. If f (x) is not identicallythe zero function, the linear equation (1.1) is saidto be nonhomogeneous.

In Chapter 4, we will learn that a generalsolution of the nth-order linear equation (1.1)is a solution that depends on n arbitraryconstants and includes all solutions of theequation.

If the equation under consideration cannotbe written in the form given by equation(1.1), the equation is said to be nonlinear.Therefore, some of the properties that lead toclassifying an equation as linear or nonlinearare powers of the dependent variable (or one ofits derivatives) and functions of the dependentvariable.

E X AM P L E 1.1.3Determine which of the following differential

equations are linear: (a)dydx

= x3, (b)d2udx2

+u = ex,

(c) (y− 1)dx + x cos ydy = 0, (d)d3ydx3

+ ydydx

= x,

(e)dydx

+ x2y = x, and (f)d2xdt2

+ sin x = 0.

Solution(a) This equation is linear because the nonlin-

ear term x3 is the function f (x) of the independentvariable in the general formula for a linear dif-ferential equation. (b) This equation is also linear.Using u as the name of the dependent variable

does not affect the linearity. (c) If y is the depen-dent variable, solving for dy/dx gives us

dydx

= 1 − yx cos y

.

Because the right side of this equation includes anonlinear function of y, the equation is nonlinear(in y). However, if x is the dependent variable,solving for dx/dy yields

dxdy

= cos y1 − y

x.

This equation is linear in the dependent variablex. (d) The coefficient of the term dy/dx is y insteadof an expression involving only the independentvariable x. Hence, this equation is nonlinear in thedependent variable y. (e) This equation is linear.The term x2 is the coefficient function. (f) For thisequation, note that x is the dependent variableand t is the independent variable. This equation,known as the pendulum equation because it modelsthe motion of a simple pendulum, is nonlinearbecause it involves a nonlinear function of thedependent variable x, sin x.

If an ODE has the form dy/dx = f (x), we canuse integration to determine y = y(x), althoughthe result may be in terms of integrals that can-not be evaluated using standard techniques ofintegration. The following examples of differ-ential equations of this type illustrate some ofthe typical methods of integration that may beencountered.


E X AM P L E 1.1.4Solve the following differential equations: (a)

dydx

= cos x, (b)dydx

= x√x2 + 1

, (c)dydx

= 1x2 + 16

,

(d)dydx

= xex, (e)ydx

= 14 − x2

, and (f)dydx

=sin x tan x.

SolutionIn each case, we integrate the indicated func-

tion and graph the solution for several valuesof the constant of integration. Each solution con-tains a constant of integration so there are in-finitely many solutions to each equation. (a) y =∫cos xdx = sin x + C (see Figure 1.4(a)). (b) To

evaluate y = ∫(x/√x2 + 1)dx, we let u = x2 +1 so

that du = 2xdx or (1/2)du = xdx. Then,

y = 12

∫2x√x2 + 1

dx

= 12

∫1√udu = 1

2

∫u−1/2 du = u1/2 + C

=√x2 + 1 + C (see Figure 1.4(b)).

(c) To integrate y = ∫(1/x2 + 16)dx, we use a

trigonometric substitution. Letting x = 4 tan θ ,−π/2 < θ < π/2, so that dx = 4 sec2 θ dθ gives us

y =∫

116 + (4 tan θ)2

4 sec2 θ dθ

=∫

116 + 16 tan2 θ

4 sec2 θ dθ

= 116

∫1

sec2 θ4 sec2 θ dθ

= 14

∫dθ = 1

4θ + C = 1

4tan−1

(x4

)+ C.

First, we use the identity 1 + tan2 θ = sec2 θand then resubstitute: x/4 = tan θ so θ =tan−1(x/4).

(d) To evaluate y = xex by hand, we use theIntegration by Parts formula with u = x and dv =ex dx. Then, du = dx and v = ex. This gives

y =∫

xex dx = xex −∫

ex dx = xex − ex + C.

(e) Use partial fractions to evaluate this integral.First, find the partial fraction decomposition of1/(4 − x2), which is determined by finding con-stants A and B that satisfy the following equation:

1(2 − x)(2 + x)

= A2 − x

+ B2 + x

.

These values are A = B = 1/4. (Why?) Thus,

y = 14

∫ (1

2 − x+ 1

2 + x

)dx

= 14[− ln |2 − x| + ln |2 + x|] + C

= 14ln∣∣∣∣2 + x2 − x

∣∣∣∣+ C (see Figure 1.5).

The Integration by Parts formula states that∫udv = u v− ∫

vdu.

Find C if y(1) = 2.

Notice that the solutions of dy/dx = 1/(4 − x2)are undefined if x = −2 or x = 2. This isbecause 1/(4−x2) is undefined at these two valuesof x. Later, we will discuss in greater detail therelationship between the differential equation andits solutions.

(f) In this case, y = ∫sin x tan xdx = (sin2 x/

cos x)dx because tan x = sin x/ cos x. Now use theidentity sin2 x + cos2 x = 1 or sin2 x = 1 − cos2 xand divide:

y =∫

sin2 xcos x

dx,

y =∫

1 − cos2 xcos x

dx,

y =∫(sec x− cos x) dx,

y = ln | sec x+ tan x| − sin x+ C.

For −π/2 < x < π/2,∫sec xdx = ln(sec x +

tan x)+ C.


2 4 6 8 10 12x

−4

−2

(a)

2

4

y

−2

−2

−4 2

(b)

4x

2

4

6

8y

FIGURE 1.4 (a) Graph of y = sin x+C for various valuesof C. (b) Graph of y =

√x2 + 1 + C for various values of C.

In Example 1.1.4, each solution is given as afunction y = y(x) of the independent variable.In these cases, the solution is said to be explicit.In solving some differential equations, however,

We will see that given an arbitrary differentialequation, constructing an explicit or implicitsolution is nearly always impossible. Conse-quently, although mathematicians were firstconcerned with finding analytic (explicit orimplicit) solutions to differential equations, theyhave since (frequently) turned their attention toaddressing properties of the solution and findingalgorithms to approximate solutions.

−4 −2 2 4x

−4

−2

2

4

y

FIGURE 1.5 Graph of y = 14 ln |(2 + x)/(2 − x)| + C for

various values of C.

we can find only an equation involving the in-dependent and dependent variables that the so-lution satisfies. In this case, we say that we havefound an implicit solution.

E X AM P L E 1.1.5Verify that the equation 2x2 +y2 −2xy+5x = 0

satisfies the following differential equation:

dydx

= 2y− 4x− 52y − 2x

.

SolutionWeuse implicit differentiation to compute y′ =

dy/dx if 2x2 + y2 − 2xy + 5x = 0:

4x + 2ydydx

− 2xdydx

− 2y + 5=0,

dydx(2y− 2x)=2y − 4x − 5,

dydx

= 2y − 4x − 52y − 2x

.

The equation 2x2 + y2 − 2xy + 5x = 0satisfies the differential equation dy/dx =(2y − 4x− 5)/(2y − 2x).


–6 –4 –2 2x

–6

–4

–2

2y

FIGURE 1.6 Graph of 2x2 + y2 − 2xy + 5x = 0.

Althoughwe cannot solve 2x2+y2−2xy+5x =0 for y as a function of x (see Figure 1.6), we candetermine the correspondingyvalue(s) for a givenvalue of x. For example, if x = −1, then

2+ y2 + 2y− 5 = y2 + 2y− 3 = (y+ 3)(y− 1) = 0.

Therefore, the points (−1,−3) and (−1, 1) lie onthe graph of 2x2+y2−2xy+5x = 0 (see Figure 1.7).

Find an equation of the line tangent to thegraph of 2x2 + y2 − 2xy + 5x = 0 at the

points (−1,−3) and (−1, 1).

In the same manner that we consider systemsof equations in algebra, we can also considersystems of differential equations. For example, ifx and y represent functions of t, we will learn inChapter 6 to solve the system of linear equations{

dx/dt = ax+ by,dy/dt = cx+ dy,

where a, b, c, and d represent constants anddifferentiation is with respect to t.Wewill seethat systems of differential equations arisenaturally inmanyphysical situations that aremodeled with more than one equation andinvolve more than one dependent variable.In addition, we will see that it is often use-ful to write a differential equation of ordergreater than one as a system of first-orderequations, especiallywhen the original equa-tion is nonlinear.

E X AM P L E 1.1.6(Duffing’s equation). Duffing’s equation is the

second-order nonlinear equation

d2xdt2

+ kdxdt

− x+ x3 = � cosωt, (1.2)

where k, �, and ω are positive constants.

Sources: See texts like Jordan and Smith’sNon-linear ODEs, [16].

Write Duffing’s equation as a system of first-order equations

SolutionLet y = x′. Then, y′ = x′′ and substituting into

Duffing’s equation gives us

x′′ + kx′ − x+ x3 = � cosωt,

y′ + ky− x+ x3 = � cosωt,

y′ = x− x3 − ky + � cosωt.

Thus, Duffing’s equation is equivalent to thenonlinear system

{x′ = y,

y′ = x− x3 − ky + � cosωt.

Note that a system of differential equationscan consist of more than two equations. For


–1.1 –1.05 –1 –0.95 –0.9–3.2

–3.1

–3

–2.9

–2.8

–1.15 –1.1 –1.05 –1 –0.95 –0.9 –0.85 –0.8

0.85

0.9

0.95

1

1.05

1.1

1.15

FIGURE 1.7 Notice that near the points (−1,−3) and (−1, 1) the implicit solution looks like a function. In fact, when wezoom in near the points (−1,−3) and (−1, 1), we see what appears to be the graph of a function.

example, the basic equations that describethe competition between two organisms, withpopulation densities x1 and x2, respectively, in achemostat are

Sources: See Smith and Waltman’s The Theory ofthe Chemostat [27], for a detailed discussion ofchemostat models.

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

S′ = 1 − S− m1Sa1 + S

x1 − m2Sa2 + S

x2

x′1 = x1

(m1Sa1 + S

− 1)

x′2 = x2

(m2Sa2 + S

− 1) , (1.3)

where ′ denotes differentiation with respect tot; S = S(t), x1 = x1(t), and x2 = x2(t). ForEquations (1.3), we remark that S denotes theconcentration of the nutrient available to thecompetitors with population densities x1 and x2.We investigate chemostat models in more detailin Chapter 7.

EXERCISES 1.1

Determine if each of the following equationsis an ODE or a partial differential equation. If theequation is an ODE, then determine (a) the orderof the ODE and (b) if the equation is linear ornonlinear.

1.d2ydx2

+ dydx

− 2y = x3

2. ydydx

+ y4 = sin x

3.∂2y∂t2

= c2∂2y∂x2

, c > 0 constant

4. y′′′ − 2y′′ + 5y′ + y = ex,(′ = d/dx; y = y(x))

5.(dydx

)2

+ y = 0

6. t2d2ydt2

+ tdydt

+ 2y = 0

7.1c2∂2z∂t2

= ∂2z∂x2

+ ∂2z∂y2

, (z = z(t, x, y))

8. uux + ut = 0, (u = u(t, x))

EXERCISES 1.1 11

9. x

(d2ydx2

)2

+ 2y = 2x

10.d2xdt2

+ 2 sin x = sin 2t, (x = x(t))11. ut + uux = σ uxx, σ constant12. (2x− 1)dx− dy = 013. (2t− y)dt− dy = 0

14.∂u∂x∂u∂y

= u, (u = u(x, y))

15. (2x− y)dx− ydy = 016. Write each of the following second-order

equations as a system of first-orderequations:

(a)d2xdt2

− dxdt

− 6x = 0

(b) 4d2xdt2

+ 4dxdt

+ 37x = 0

(c) Ld2xdt2

+ g sin x = 0, L, g positive

constants, x = x(t)

(d)d2xdt2

− μ(1 − x2)dxdt

+ x = 0, μ > 0constant

(e) td2xdt2

+ (b− t)dxdt

− ax = 0, a, b constants

In Exercises 17-28, verify that each of the givenfunctions is a solution to the correspondingdifferential equation. (A, B, and C representconstants.)

17. dy/dx+ 2y = 0, y(x) = e−2x, y(x) = 5e−2x

18. dy/dx+ xy = 0, y(x) = e−x2/2

19. dy/dx+ y = sin x, y(x) =e−x − 1

2 cos x+ 12 sin x

20.d2ydt2

− dydt

− 12y = 0, y(t) = e4t, y(t) = e−3t

21. y′′ + 9y′ = 0, y = A + Be−9t (′ denotes d/dt)22. x′′ + 3x′ − 10x = 0, x(t) = Ae2t + Be−5t

23. x′′ + x = t cos t− cos t, x =A cos t+ B sin t+ 1

4 t2 sin t− 1

2 t sin t+ 14 t cos t

24. y′′ − 12y′ + 40y = 0, y = e6x cos 2x,y = e6x sin 2x (′ denotes d/dx)

25. y′′′ − 4y′ = 0, y = A + Be2x + Ce−2x

26. y′′′ − 2y′′ = 0, y = A + Bt+ Ce2t

27. x2y′′ − 12x y′ + 42y = 0, y = Ax6 + Bx7

28. t2y′′ + 3t y′ + 5y = 0, y = t−1(A cos(2 ln t)+ B sin(2 ln t)

)In Exercises 29-33, verify that the given equationsatisfies the differential equation. Use the equa-tion to determine y for the given value of x (or t).Confirm your result by graphing each equation usingappropriate technology.

29. dy/dx = −x/y, x2 + y2 = 16, x = 030. 3y(t2 + y)dt+ t(t2 + 6y)dy = 0,

t3y+ 3ty2 = 8, t = 231. dy/dx = −2y/x− 3, x3 + x2y = 100, x = 132. y cos tdt+ (2y+ sin t)dy = 0,

y2 + y sin t = 1, t = 033. (y/x+ cos y)dx+ (ln x− x sin y)dy = 0,

y ln x+ x cos y = 0, x = 1

In Exercises 34-43, use integration to find a solu-tion to the differential equation.

34. dy/dx = (x2 − 1)(x3 − 3x)3

35. dy/d = x sin x2

36. dy/dx = x/√x2 − 16

37. dy/dx = 1/(x ln x)38. dy/dx = x ln x39. dy/dx = xe−x

40.dydx

= −2(x+ 5)(x + 2)(x− 4)

41.dydx

= x− x2

(x + 1)(x2 + 1)

42. dy/dx =√x2 − 16/x

43. dy/dx = (4 − x2)3/2

44. dy/dx = 1/(x2 − 16)45. dy/dx = cos x cot x

46. dy/dx = sin3 x tan x


In Exercises 45-54, use the indicated conditionswith the indicated solution to determine the so-lution to the given problem.

47. dy/dx+ 2y = 0, y(0) = 2, y(x) = Ae−2x

48. dy/dt+ y = sin t, y(0) = −1,y(t) = Ae−t − 1

2 cos t + 12 sin t

49. y′′ − y′ − 12y = 0, y(0) = 1, y′(0) = −1,y = Ae4x + Be−3x

50. y′′ + 9y′ = 0, y(0) = 2, y′(0) = −1,y = A+ Be−9x

51. y′′′ − 2y′′ = 0, y(0) = 0, y′(0) = 1, y′′(0) = 3,y = A+ Bx+ Ce2x

52. y′′′ − 4y′ = 0, y(0) = 1, y′(0) = −1, y′′(0) = 0,y = A+ Be2x + Ce−2x

53. t2y′′ − 12t y′ + 42y = 0, y(1) = 0, y′(1) = −1,y(t) = At6 + Bt7

54. x2y′′ + 3xy′ + 5y = 0, y(1) = 0, y′(1) = 1,y(x) = x−1 (A cos(2 ln x)+ B sin(2 ln x)

)In Exercises 55-58, solve the IVP. Confirm yourresult by graphing each function on an appropriateinterval using appropriate technology.

55. dy/dx = 4x3 − x + 2, y(0) = 156. dy/dt = sin 2t− cos 2t, y(0) = 057. dy/dx = x−2 cos

(x−1), y(2/π) = 1

58. dy/dx = (ln x)/x, y(1) = 059. The velocity of a falling object with mass m

that is subjected to air resistanceproportional to the instantaneous velocity vof. the object is found by solvingt

¯he IVP{

mdv/dt = mg− cv,v(0) = v0,

where c > 0 is the

proportionality constant.(a) Given that a general solution to

mdv/dt = mg− cv isv(t) = mg/c+ Ke−ct/m, find the solutionof this IVP.

(b) Determine limt→∞ v(t).

60. The number of cells in a bacteria colonyafter t hours is determined by solving the

IVP

{dP/dt = kP,P(0) = P0.

(a) Given that a general solution ofdP/dt = kP is P(t) = Cekt, use the initialcondition to find C.

(b) Find the value of k so that thepopulation doubles in 8 h.

61. In 1840, the Belgian mathematician-biologistPierre F. Verhulst (1804-1849) developed thelogistic equation, dP/dt = rP − aP2, where rand a are positive constants, to predict thepopulation P(t) in certain countries.(a) Given that a general solution to this

equation is P(t) = r/(a + Ce−rt), find thesolution that satisfies P(0) = P0, whereP0 > 0 is constant.

(b) Determine limt→∞ P(t).

Pierre Francois Verhulst (1804-1849)

62. The differential equationdS/dt + 3S/(t + 100) = 0, where S(t) is thenumber of pounds of salt in a particular

EXERCISES 1.1 13

tank at time t, is used to approximate theamount of salt in the tank containing asaltwater mixture in which pure water isallowed to flow into the tank while themixture is allowed to flow out of the tank.If S(t) = 15, 000, 000/(t+ 100)3, show that Ssatisfies dS/dt + 3S/(t+ 100) = 0. What isthe initial amount of salt in the tank? Ast → ∞, what happens to the amount of saltin the tank?

63. The displacement (measured from x = 0) ofa mass attached to the end of a spring attime t is given by x(t) = 3 cos 4t+ 9

4 sin 4t.Show that x satisfies the ODE x′′ + 16x = 0.What is the initial position of the mass?What is the initial velocity of the mass?

64. Show that u(x, y) = ln√x2 + y2 satisfies

Laplace’s equation, uxx + uyy = 0.65. The temperature in a thin rod of length 2π

after tminutes at a position x between 0 and2π is given by u(x, t) = 3 − e−16kt cos 4x.Show that u satisfies ut = k uxx. What is theinitial temperature (t = 0) at x = π? Whathappens to the temperature at each point inthe wire as t → ∞?

66. The displacement u of a string of length 1 attime t position x, where x is measured fromx = 0, is given by u(x, t) = sinπx cos t.Show that u satisfies π2utt = uxx. What isthe value of u at the endpoints x = 0 andx = 1 for all values of t?

67. Find the value(s) of m so that y = xm is asolution of x2y′′ − 2xy′ + 2y = 0.

68. Find the value(s) of k so that y = ekt is asolution of y′′ − 3y′ − 18y = 0.

69. Use the fact that(d/dx)

(e2xy

) = e2x(dy/dx)+ 2e2xy and

integration to solve e2xdydx

+ 2e2xy = ex.

70. Use the fact that(d/dx)

(exy

) = ex(dy/dx)+ exy andintegration to solve ex(dy/dx)+ exy = xex.

71. The time-independent Schrödinger equation isgiven by

− h2

2md2ψ(x)dx2

+ U(x)ψ(x) = Eψ(x).

If U(x) = 0, find conditions on E so thatψ(x) = A sin(nπx/L) is a solution of thetime-independent Schrödingerequation.

The great physicist Erwin Schrödinger (1887-1961) received a Nobel prize for his work in 1933.

72. A singular solution of a differential equationis a solution that cannot be derived from thegeneral solution of the differential equation.Use implicit differentiation to show that−1/x+ 2/x2 + 1/y − 1/y2 = C is a general(implicit) solution of the differentialequation dy/dx = (x− 4)y3/[x3(y − 2)]. Isy = 0 a solution of this differentialequation? Is y = 0 a singular solution?

73. Show that x+ x2/y = C is a general(implicit) solution of the differentialequation dy/dx = (y2 + 2xy)/x2. Is y = 0 asolution of this differential equation? Isy = 0 a singular solution?

74. The current I(t) in an L-R circuit, whichcontains a resistor, an inductor, and avoltage source, satisfies the differentialequation RI + LdI/dt = E(t), where R and L


are constants representing the resistanceand the inductance and E(t) is the voltagesource. Is this equation linear or nonlinear?Determine the order of the equation.

In Exercises 75-77, (a) verify that the indicatedfunction is a solution of the given differentialequation and (b) graph the solution on the indi-cated interval(s).

75. xy′ + y = cos x, y = (sin x)/x;[−2π , 0) ∪ (0, 2π]

76. 16y′′ + 24y′ + 153y = 0, y = e−3t/4 cos 3t;[0, 3π/2]

77. x3y′′′ + x2y′′ + xy′ − 40y = 0,y = x−1 sin(3 ln x); (0,π]

78. (a) Verify that⎧⎪⎨⎪⎩x = e−t

(100

√3

3 sin√3t+ 20 cos

√3t)

y = e−t(− 40

√3

3 sin√3t+ 20 cos

√3t)

is a solution of the system of differential

equations

{dx/dt = 4y,

dy/dt = −x− 2y.(b) Graph x(t), y(t), and the parametric

equations

{x = x(t)y = y(t)

for 0 ≤ t ≤ 2π .

Throughout Introductory Differential Equations, weuse graphs of solutions of differential equations.In some cases, we are able to predict what thegraph of a solution should look like. If the graphof our proposed solution does not appear as pre-dicted, we know that either we made a mistakein constructing our proposed solution or our con-jecture about the general shape of the graph ofthe solution is wrong. In other cases, we will findthat it is easier to examine the graph of a solutionthan it is to examine the solution (if we are able toconstruct one in the first place).

79. (a) Show that (x2 + y2)2 = 5xy is an implicitsolution of

[4x(x2+y2)−5y]dx+[4y(x2+y2)−5x]dy=0.

(b) Graph (x2 + y2)2 = 5xy on the rectangle[−2, 2] × [−2, 2].

(c) Approximate all points on the graph of(x2 + y2)2 = 5xywith the x-coordinate 1.

(d) Approximate all points on the graph of(x2 + y2)2 = 5xywith the y-coordinate−0.319.

Often the calculus and algebra encountered insolving differential equations can be tedious, ifnot completely overwhelming or impossible. To-daymany sophisticated calculators and computeralgebra systems are capable of performing theintegration and algebraic simplification encoun-tered when solving many differential equations.Having access to these tools can be a great advan-tage: a large number of problems can be solvedquickly, we make conjectures as to the generalform of a solution to different forms of differentialequations, and these tools allow us to check andverify our work.

80. Solve dy/dx = sin4 x, y(0) = 0 and graphthe resulting solution on the interval [0, 4π].

81. Ageneral solution ofy(4) + 25

2 y′′ − 5y′ + 629

16 y = 0 is given byy = e−x/2(c1 cos 3x+ c2 sin 3x)+ex/2(c3 cos 2x+ c4 sin 2x), where c1, c2, c3,and c4 are constants. Solve the IVP⎧⎪⎨⎪⎩y(4) + 25

2 y′′ − 5y′ + 629

16 y = 0y(0) = 0, y′(0) = 1, y′′(0) = −1,y′′′(0) = 1

and

graph the resulting solution.82. Ageneral solution of the system{

dx/dt = 4ydy/dt = −4x

is

1.2 A GRAPHICAL APPROACH TO SOLUTIONS: SLOPE FIELDS AND DIRECTION FIELDS 15{x = −c1 cos 4t+ c2 sin 4ty = c2 cos 4t+ c1 sin 4t

, where c1 and

c2 are constants. Solve the IVP⎧⎪⎨⎪⎩dx/dt = 4ydy/dt = −4xx(0) = 4, y(0) = 0

and then graph x(t),

y(t), and the parametric equations{x = x(t),y = y(t).

83. Ageneral solution of the system{dx/dt = −5x+ 4ydy/dt = 2x+ 2y

is{x = −c1e3t − 4c2e−6t,y = 2c1e3t + c2e−6t,

where c1 and c2

are constants. Solve the IVP⎧⎪⎨⎪⎩x = −5x+ 4yy = 2x+ 2yx(0) = 4, y(0) = 0

and then graph x(t),

y(t), and the parametric equations{x = x(t),y = y(t).

1.2 A GRAPHICAL APPROACH TOSOLUTIONS: SLOPE FIELDS AND

DIRECTION FIELDS

• Systems of ODEs and Direction Fields• Relationship Between Systems of

First-Order and Higher Order Equations

Suppose that we are asked to solve the ODEdy/dx = e−x2 . In this case, we do not attemptto solve this equation through integration as wedid in the previous section because of the pres-ence of the function f (x) = e−x2 on the rightside of the ODE (see Exercise 21 at the end ofthis section). Instead, we can gain insight intothe behavior of solutions of this ODE through agraphical approach by considering the slope ofthe tangent line to solutions of the ODE. Recall

from Section 1.1 that the differential equationgives the slope of the tangent line to solutions ofthe ODE at the given point (x, y) in the xy-plane.Therefore, if we wish to determine the slopeof the tangent line to solution to the ODE thatpasses through (0, 1) at this point, we substitute(0, 1) into the right side of dy/dx = e−x2 . Be-cause the right side only depends on x, we haveslope

dydx

= e−(0)2 = 1.

In fact, the slope of the line tangent to solutionsat all points of the form (0, y) is 1. At the point(√ln 2, 4), we find that the slope is

dydx

= e−(√ln 2)2 = e− ln 2 = eln 2−1 = 12.

Again, we obtain the same slope at all points(√ln 2, y) ≈ (0.832555, y) and (−√

ln 2, y) ≈(−0.832555, y). In Figure 1.8, we draw severalshort line segments using points of the form(0, y) for y = 0, ±1, ±2, ±3. Notice that we usea triangle with base length 1 and height length 1to assist in sketching the tangent lines with slope1 at these points. We use a triangle with baselength 2 and height length 1 to help us draw thelines of slope 1/2 that are tangent to solutionsat the points (−√

ln 2, y) for y = 0, ±1, ±2, ±3.By drawing a set of short line segments repre-senting the tangent lines to solutions of the ODEat numerous points in plane, we construct theslope field of the ODE. We show the slope fieldfor dy/dx = e−x2 on the square [−2, 2] × [−2, 2]in Figure 1.9(a). (Note that because constructinga slope field is time consuming, we usually leta computer algebra system do the work for us.)Observe that at each point along the y-axis, theslope is 1 aswe predicted earlier. Notice also thatthe slope appears to be zero for larger values of|x|. This because for large values of x (in absolutevalue), dy/dx ≈ 0 because limx→±∞ e−x2 = 0.Therefore, we expect solutions to “flatten out" asx increases.


x

y

(a)

1 2x

1

2

y

–2

(b)

–1

–1

–2

1 2x

1

2y

–2 –1

–1

–2(c)

FIGURE 1.8 (a) Several line segments in the slope field for dy/dx = e−x2 . (b) One view of the slope field for the equation.(c) A different view of the slope field for the equation.

1 2x

1

2

y

–2 –1

(a)

–1

–2

1

(b)

2x

1

2

y

–2 –1

–1

–2

1

(c)

2x

1

2y

–2 –1

–1

–2

FIGURE 1.9 (a) Slope field for dy/dx = e−x2 . (b) Using the slope field to sketch the solution to dy/dx = e−x2 , y(−2) = −1.(c) A different view of using the slope field to sketch the solution to dy/dx = e−x2 , y(−2) = −1.

We can use the slope field to investigate thesolution to an IVP such as

dydx

= e−x2 , y(−2) = −1

by starting at the point (−2,−1) and tracing thesolution by following the tangent slopes. Thissolution is sketched in Figure 1.9(b). Thus, al-though we were not able to determine explicitformulas for either the general solution of theODE or the solution to the IVP, we were able todetermine some properties of the solutions byusing the slope field of the differential equation.

Systems of ODEs and Direction Fields

We can also consider systems of differentialequations. In Chapter 6, we will learn how

to solve systems of first-order ODEs of theform {

dx/dt = ax+ by,

dy/dt = cx+ dy,

where a, b, c, and d are given constants. In thecase of this system, we solve for x = x(t) andy = y(t). For example, if we consider the system⎧⎨

⎩dx/dt = y,

dy/dt = −x,we can verify that the parametric equations{

x(t) = sin t

y(t) = cos t

1.2 A GRAPHICAL APPROACH TO SOLUTIONS: SLOPE FIELDS AND DIRECTION FIELDS 17

1

(a) (b)

2 3 4 5 6t

–1

–0.5

0.5

1

–1 –0.5 0.5 1 x

–1

–0.5

0.5

1y

FIGURE 1.10 (a) Graphs of x(t) = sin t and y(t) = cos t for 0 ≤ t ≤ 2π . (b) Graph of the parametric equations x(t) = sin tand y(t) = cos t for 0 ≤ t ≤ 2π .

satisfy the system because

dxdt

= ddt(sin t) = cos t = y and

dydt

= ddt(cos t) = − sin t = −x.

We can graph each function separately as wedo in Figure 1.10(a). (The graph of x(t) is the solidcurve; that of y(t) is dashed.) Another option is tograph them as a pair of parametric equations asin Figure 1.10(b). As we recall, the graph of thispair of parametric equations is a circle of radius1 centered at the origin because x2 = sin2 t andy2 = cos2 t, so that x2 + y2 = sin2 t + cos2 t =1. However, we must indicate the orientation ofthe curve (the direction of increasing parametervalue t). For this pair of parametric equations, wefind that at t = 0, x(0) = sin 0 = 0 and y(0) =cos 0 = 1. Therefore, the point (0, 1) correspondsto t = 0. Similarly, the point (1, 0) correspondsto t = π/2 because x(π/2) = sinπ/2 = 1 andy(π/2) = cosπ/2 = 0. This means that thesolution moves from (0, 1) to (1, 0) as t increases.To determine if the orientation is clockwise orcounter-clockwise, we test a t-value between 0and π/2. Choosing t = π/4, we find the x(π/4) =sinπ/4 = 1/

√2 and y(π/4) = cosπ/4 = 1/

√2

so the orientation is clockwise. The parametricequations {x(t) = sin t, y(t) = cos t} satisfythe IVP {

dx/dt = y, x(0) = 0

dy/dt = −x, y(0) = 1

because they satisfy the system of differentialequations as well as the two initial conditions.Notice that in the case of an IVP involving a sys-tem of differential equations, an initial conditionis given for each of the variables x and y thatdepend on t. In the parametric plot, the solutionto this IVPpasses through the point (x(0), y(0)) =(0, 1).

Another way to view a system of two ODEsis through the use of a direction field, which issimilar to a slope field. For example, for the first-order system {

dx/dt = ax+ by,dy/dt = cx+ dy,

we first write it as a first-order equation with

dydx

= dy/dtdx/dt

= cx+ dyax+ by

.

Observe that the procedure described here can be

used for any systemof the form

{dx/dt = f (x, y),

dy/dt = g(x, y).

Then, we can consider the slope field associatedwith this differential equation. For example, ifwe refer back to the system{

dx/dt = y,dy/dt = −x,

we obtain the first-order equation dy/dx =−x/y, so we can determine the slope of tangent


lines to solutions at points in the xy-plane. Forexample, at the point (1/

√2, 1/

√2), the solution

to this system that passes through this pointhas slope (−1/

√2)/(1/

√2) = −1. In a similar

manner, we can find the slope at other pointsin the plane. However, as we mentioned in ourearlier discussion, we must indicate the orien-tation when we graph parametric equations,so we consider the vector 〈dx/dt, dy/dt〉 =dx/dt i + dy/dt j with components from thesystem of differential equations. In the caseof this system, we consider 〈dx/dt, dy/dt〉 =〈y,−x〉. At the point (1/

√2, 1/

√2), we obtain

the vector 〈1/√2,−1/√2〉. This means that the

solution through (1/√2, 1/

√2) has tangent vec-

tor 〈1/√2,−1/√2〉. The direction field is made

up of tangent vectors such as 〈1/√2,−1/√2〉

to solutions at points in the plane, so it issimilar to the slope field for dy/dy = −x/yshown in Figure 1.11(a), except that vectors areused to indicate the orientation of solutions. InFigure 1.11(b), we show the direction field forthis system. The vectors in the direction fieldindicate that solutions to this system are circlesin the xy-plane that are directed clockwise. Wegraph several solutions along with the direction

Generally, we will use standard mathematical no-tation throughout Introductory Differential Equa-tions so i = 〈1, 0〉 and j = 〈0, 1〉.

field in Figure 1.11(c). A collection of solutions inthe xy-plane is called the phase portrait of the sys-tem. Notice that at points in the first quadrant,where x > 0 and y > 0, dx/dt = y > 0 anddy/dt = −x < 0. This means that x(t) increasesand y(t) decreases along solutions in the firstquadrant. In the second quadrant, where x < 0and y > 0, dx/dt = y > 0 and dy/dt = −x > 0.Therefore, x(t) and y(t) increase along solutionsin the second quadrant.We can performa similaranalysis for points in the other two quadrants.

Relationship Between Systems ofFirst-Order and Higher Order Equations

Again, consider the system

{dx/dt = y,dy/dt = −x. If

we differentiate the equation dx/dt = y with re-spect to t, we obtain d2x/dt2 = dy/dt. Therefore,if we equate dy/dt = −x and dy/dt = d2x/dt2,we have d2x/dt2 = −x or d2x/dt2 + x = 0.We say that the system of two first-order ODEsis equivalent to the second-order ODE d2x/dt2 +x = 0. Often, however, we begin with a second-order ODE of the form d2x/dt2 + bdx/dt+ c x =f (t) and would like to write the equation as asystem of first-order ODEs. We do this by lettingdx/dt = y and by differentiating this equationwith respect to t to obtain d2x/dt2 = dy/dt.Solving the second-order ODE for d2x/dt2, wefind d2x/dt2 = −bdx/dt − c x + f (t) so thatdy/dt = −bdx/dt − c x + f (t). Replacing dx/dt

(a)

–10 –5 5 10x

–10

–5

5

10

y

(b)

–10 –5 5 10x

–10

–5

5

10

y

(c)

–10 5 5 10x

–10

–5

5

10

y

FIGURE 1.11 (a) Slope field for dy/dx = −x/y. (b) Direction field for dx/dt = y, dy/dt = −x. (c) Direction field fordx/dt = y, dy/dt = −x and several solution curves.

EXERCISES 1.2 19

in this equation with y, we obtain the followingequivalent system of first-order ODEs:{

dx/dt = y,dy/dt = −b y− c x+ f (t).

By writing the second-order ODE as a systemof first-order ODEs, we investigate the behaviorof the solution of the second-order ODE by ob-serving the behavior in the direction field andphase portrait of the corresponding system. Asimilar procedure is used to transform an ODEof degree greater than two into a system of first-order ODEs.

EXERCISES 1.2

In Exercises 1-4, use the slope field to deter-mine if the indicated path is that of a solution tothe differential equation.

1.dydx

= −y/x –2 –1 1 2x

–2

–1

1

2

y

2.dydx

= x2 + y2 –1.5 –1 –0.5 0.5 1 1.5x

–1.5

–1

–0.5

0.5

1

1.5y

3.dydx

= x/y2 1.5 1 0.5 0.5 1 1.5 2

x

–2

–1.5

–1

–0.5

0.5

1

1.5

2y

4.dydx

= x2 − y –2 –1.5 –1 –0.5 0.5 1 1.5 2x

–2

–1.5

–1

–0.5

0.5

1

1.5

2y

In Exercises 5-8, use the slope field to sketch thesolutions of the differential equation that passthrough the given points.

5.dydx

= x/y –2 –1.5 –1 –0.5 0.5 1 1.5 2x

–2

–1.5

–1

–0.5

0.5

1

1.5

2y

6.dydx

= x2 − y –2 –1.5 –1 –0.5 0.5 1 1.5 2x

–2

–1.5

–0.5

0.5

1

1.5

2y


7.dydx

= x − y –2 –1.5 –1 –0.5 0.5 1 1.5 2x

–2

–1.5

–1

–0.5

0.5

1

1.5

2y

8.dydx

= y2 − x2 –2 –1.5 –1 –0.5 0.5 1 1.5 2x

–2

–1.5

–1

–0.5

0.5

1

1.5

2y

9. Graph the slope field for dy/dt = sin y.Determine limt→∞ y(t) if y(0) = y0 where(a) y0 = −3π/2; (b) y0 = −π/2; (c) y0 = π/2;(d) y(0) = 3π/2.

10. Graph the slope field for dy/dx = sin x.Does limx→∞ y(x) exist for any initialcondition y(0) = y0? Solve the ODE andfind limx→∞ y(x). Does this match thegraphical result?

11. Graph the slope field for dy/dx = e−x. Doeslimx→∞ y(x) exist for any initial conditiony(0) = y0? Solve the ODE and findlimx→∞ y(x). Does this match the graphicalresult?

12. Graph the slope field for dy/dx =1/(x2 + 1). Does limx→∞ y(x) exist for anyinitial condition y(0) = y0? Solve the ODEand find limx→∞ y(x). Does this match thegraphical result?

In Exercises 13-14, use the direction field of thegiven system to sketch the graph of the solutionthat satisfies the indicated initial conditions. De-termine limt→∞ x(t) and limt→∞ y(t) in each case(if they exist).

13. System:

{dx/dt = ydy/dt = x

; (a) x(0) = 0,

y(0) = 2; (b) x(0) = −2, y(0) = 0; (c)x(0) = −2, y(0) = 2;

–4 –2 2 4x

–4

–2

2

4

y

14. System:

{dx/dt = 3x− 2ydy/dt = 4x− y

; x(0) = 0,

y(0) = 5;

–15 –10 –5 5 10 15x

–15

–10

–5

5

10

15

y

In Exercises 15-20, write the second-order equa-tion as a system of first-order equations.

15. d2x/dt2 + 4x = 016. d2x/dt2 − 5 dx/dt = 017. d2x/dt2 + 4 dx/dt+ 13x = 018. x′′ − 6x′ + 7x = 0 (′ = d/dt; x = x(t))19. x′′ + 16x = sin t20. x′′ + 4x′ + 13x = e−t21. (a) Use a computer algebra system to solve

dy/dx = e−x2 ; (b) Graph the solution to theIVP dy/dx = e−x2 , y(0) = a fora = −2, −1, 0, 1, 2. (c) Graph the slope fieldof the differential equation together withthe solutions in (b). (d) Do the solutions

EXERCISES 1.2 21

appear to match the results described at thebeginning of the section?

22. (a) Use a computer algebra system to graphthe slope field for dy/dx = sin(2x− y). (b)Use the slope fie to graph the solution y(x)that satisfies the initial condition y(0) = 5.Does limx→∞ y(x)appear to exist?

23. Consider the systems{dx/dt = −ydy/dt = x

and

{dx/dt = −y,dy/dt = −x.

(a) For any given initial conditions, in abrief paragraph explain why you thinkthat the solutions of the systems aresimilar or different.

(b) Figure 1.12 shows the direction fieldassociated with each system. Use thedirection field to help you graph thesolutions that satisfy these initialconditions(i) x(0) = 0.5, y(0) = 0;(ii) x(0) = −0.25, y(0) = 0;(iii) x(0) = 0, y(0) = 0.75; and(iv) x(0) = 0, y(0) = −0.5.

(c) How do your graphs affect yourconjecture in (a)?

24. Consider the systems{dx/dt = x/2dy/dt = y

and

{dx/dt = −x/2,dy/dt = −y.

(a) For any given initial conditions, in a briefparagraph explain why you think that thesolutions of the systems are similar ordifferent.

(b) Figure 1.13 shows the direction fieldassociated with each system. Use thedirection field to help you graph thesolutions that satisfy these initialcondition(i) x(0) = 0.5, y(0) = 0.25;(ii) x(0) = −0.25, y(0) = −0.5;(iii) x(0) = −0.5, y(0) = 0.75; and(iv) x(0) = 0.75, y(0) = −0.5.

(c) How do your graphs affect yourconjecture in (a)?

25. (Competing Species) The system of

equations,

{dx/dt = x(a − b1x− b2y)

dy/dt = y(c − d1x− d2y)where a, b1, b2, c, d1, and d2 representpositive constants, can be used to modelthe size of the population of two species,represented by x(t) and y(t), competing fora common food supply.(a) Figure 1.14(a) shows the direction field

for the system if a = 1, b1 = 2, b2 = 1,c = 1, d1 = 0.75, and d2 = 2. (i) Use thedirection field to graph various solutionsif both x(0) and y(0) are positive. (ii) Use

–1 –0.5 0.5 1x

–1

–0.5

0.5

1

y

–1 –0.5 0.5 1x

–1

–0.5

0.5

1

y

FIGURE 1.12 Figure for Exercise 23.


–1 –0.5 0.5 1x

–1

–0.5

0.5

1

y

–1 –0.5 0.5 1x

–1

–0.5

0.5

1

y

FIGURE 1.13 Figure for Exercise 24.

the direction field and your graphs toapproximate limt→∞ x(t) andlimt→∞ y(t).

(b) Figure 1.14(b) shows the direction fieldfor the system if a = 1, b1 = 1, b2 = 1,c = 0.67, d1 = 0.75, and d2 = 1. (i) Usethe direction field to graph varioussolutions if both x(0) and y(0) arepositive. (ii) Use the direction field andyour graphs to determine the fate of thespecies with population y(t). Whathappens to the species with popula-tion x(t)?

CHAPTER 1 SUMMARY: ESSENTIALCONCEPTS AND FORMULAS

Differential Equation (DE) An equation thatcontains the derivative or differentials of oneor more dependent variables with respect toone or more dependent variables.Ordinary Differential Equation (ODE) If adifferential equation contains only ordinaryderivatives (of one or more dependentvariables) with respect to a singleindependent variable, the equation is calledan ODE.

Partial Differential Equation (PDE) Adifferential equation that contains the partialderivatives or differentials of one or moreindependent variables with respect to morethan one independent variable is called apartial differential equation.Linear ODE A linear ODE is an equation thatcan be written in the forman(x)y(n) + an−1(x)y(n−1) + · · · + a2(x)y′′ +a1(x)y+ a0y = f (x), where y = y(x).Order of an Equation The order of thehighest order derivative in a differentialequation is called the order of the equation.Solution A solution of a differential equationon a given interval is a function that iscontinuous on the interval and has all thenecessary derivatives that are present in thedifferential equation such that whensubstituted into the equation yields anidentity for all values on theinterval.Explicit Solution A solution given as afunction of the independent variable.Implicit Solution A solution given as arelation such as f (x, y) = 0, f (t, x) = 0, orf (t, y) = 0.Trivial Solution y = 0 is always a solution ofthe nth-order linear homogeneousequation.

CHAPTER 1 REVIEW EXERCISES 23

0.2(a) 0.4 0.6 0.8 1x

0.2

0.4

0.6

0.8

1y

0.2(b)

0.4 0.6 0.8 1x

0.2

0.4

0.6

0.8

1y

FIGURE 1.14 (a) Figure for Exercise 25 (a). (b) Figure forExercise 25 (b).

Slope Field Acollection of line segments thatindicate the slope of the tangent line to thesolution(s) of a differentialequation.Phase Portrait A collection of graphs of

solutions to the system

{dx/dt = f (x, y)dy/dt = g(x, y)

in

the xy-plane.Direction Field A collection of vectors thatindicate the slope and direction of thetangent line to the solutions of a system ofdifferential equations.

CHAPTER 1 REVIEW EXERCISES

In Exercises 1-5, determine (a) if the equationis an ordinary differential or partial differentialequation; (b) the order of the differential equa-tion; and (c) if the equation is linear or nonlinear.

1. dy/dt = y2. a ux + ut = 0, u = u(x, t), a constant3. d2y/dx2 + 2 dy/dx+ y = 04. mx′′ + kx = sin t, m and k positive constants;

x = x(t)

5.∂φ

∂x∂2φ

∂x2= ∂2φ

∂y2

In Exercises 6-13, verify that the given functionis a solution of the corresponding differentialequation. (A and B denote constants.)

6. dy/dx+ y cos x = 0, y = e− sin x

7. dy/dx− y = sin x, y = (ex − cos x− sin x)/28. y′′ + 4y′ − 5y = 0, y = e−5x, y = ex

9. y′′ − 6y′ + 45y = 0, y = e3x(cos 6x− sin 6x)10. xy′′ − xy′ − 16y = 0, y = Ax5 + Bx−3

11. x2y′′ + 3xy′ + 2y = 0,y = x−1(cos(ln x)− sin(ln x))

12. d2y/dx2 + 2 dy/dx+ 2y = x, y = (x − 1)/213. y′′ − 7y′ + 12y = 2, y = Ae3x + Be4x + 1/6

In Exercises 14 and 15, verify that the given im-plicit function satisfies the differential equation.

14. (2x− 3y)dx+ (2y− 3x)dy = 0,x2 − 3xy+ y2 = 1

15. (y cos(xy)+ sin x)dx+ x cos(xy)dy = 0,sin(xy)− cos x = 0

In Exercises 16-19, find a solution of the differen-tial equation.

16. dy/dx = xe−x2

17. dy/dx = x2 sin x

18.dydx

= 2x2 − x+ 1(x − 1)(x2 + 1)

19. dy/dx = x2/√x2 − 1

Exercises 20-22, use the indicated initial orboundary conditions with the given general


solution to determine the solution(s) to the givenproblem.

20. dy/dx+ 2y = x2, y(0) = 1,y = 1

4 − 12x+ 1

2x2 + Ae−2x

21. y′′ + 4y = t, y(0) = 1, y(π/4) = π/16,y = t/4 +A cos 2t+ B sin 2t

22. x2y′′ + 5xy′ + 4y = 0, y(1) = 1, y′(1) = 0,y = Ax−2 + Bx−2 ln x

Exercises 23 and 24, solve the initial-value prob-lem. Graph the solution on an appropriate inter-val.

23. dy/dx = cos2 x sin x, y(0) = 024. dy/dx = 1

3 (4x− 9)(x− 3)−2/3, y(0) = 025. The temperature on the surface of a steel

ball at time t is given by u(t) = 70e−kt + 30(in oF) where k is a positive constant. Showthat u satisfies the first-order equationdu/dt = −k(u− 30). What is the initialtemperature (t = 0) on the surface of theball? What happens to the temperature ast → ∞?

26. The displacement (measured from x = 0) ofa mass attached to the end of a spring at

time t is given by x(t) = 14e−t(cos

√35t+

9√35

sin√35t). Show that x satisfies the

ODE x′′ + 2x′ + 36x = 0. What is the initialdisplacement of the mass? What is theinitial velocity of the mass?

27. For a particular wire of length 1 ft., thetemperature at time t hours at a position ofx feet from the end (x = 0) of the wire isestimated by u(x, t) = e−π2kt sinπx− e−4π2kt

sin 2πx. Show that u satisfies the equationut = k uxx. What is the initial temperature(t = 0) at x = 1? What happens to thetemperature at each point in the wire ast → ∞?

28. The height u of a long string at time t andposition xwhere x is measured from the

–3 –2 –1 1 2 3

(a)

x

–1

–0.5

0.5

1

1.5

2y

–1 –0.5 0.5

(b)

1x

–1

–0.5

0.5

1

y

FIGURE 1.15 (a) Figure for Exercise 30. (b) Figure forExercise 31.

middle of the string (x = 0) is given byu(x, t) = sin x cos 2t. Show that u satisfiesthe wave equation utt = 4uxx. What is theinitial height (t = 0) at x = 0?

29. Show that u(x, y) = tan−1(y/x) satisfiesLaplace’s equation uxx + uyy = 0.

30. The slope field of y′ = 2(y− y2) is shown inFigure 1.15(a). Sketch the graph of thesolution that satisfies the indicated initialcondition. Also, determine if limx→∞ y(x)exists. (a) y(0) = 0.5, (b) y(0) = 1.5, (c)y(0) = −0.5.

CHAPTER 1 REVIEW EXERCISES 25

31. The direction field for{dx/dt = x, dy/dt = 2y} is shown inFigure 1.15(b). Is it possible to select initialconditions so that limt→∞ x(t) = 0 andlimt→∞ y(t) = 0? (Briefly explain.)

32. The functionJ1(x) = ∑∞

k=0((−1)k/k!(k + 1)!22k+1)x2k+1 iscalled the Bessel function of order 1. Verifythat J1(x) is a solution of Bessel’s equation oforder 1, x2y′′ + xy+ (

x2 − 1)y = 0.

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