introduction to differential equations · 2013-12-20 · abell-et al 04-ch01-p374935 2009/8/31...

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #1

CHAPTER 1

Introduction to Differential Equations

CONTENTS

1.1 Introduction to Differential Equations: Vocabulary.......................... 2

1.2 A Graphical Approach to Solutions: Slope Fields andDirection Fields .......................................................................... 15

Summary: Essential Concepts and Formulas.................................. 24

Review Exercises ....................................................................... 25

The purpose of Introductory Differential Equations with Boundary Value Problems is twofold.First, we introduce and discuss the topics on ordinary differential equations covered in anundergraduate course. Second, we indicate how certain technologies such as computer algebrasystems and graphing calculators are used to enhance the study of differential equations, notonly by eliminating some of the computational difficulties that arise in the study of differentialequations but also by overcoming some of the visual limitations associated with the solutionsof differential equations. The advantages of using technology such as graphing calculators andcomputer algebra systems in the study of differential equations are numerous, but perhapsthe most useful is that of being able to produce the graphics associated with solutions ofdifferential equations. This is particularly beneficial in the discussion of applications becausemany physical situations are modeled with differential equations. For example, in Chapter 5,we see that the motion of a pendulum can be modeled by a differential equation. When wesolve the problem of the motion of a pendulum, we use technology to watch the pendulummove. The same is true for the motion of a mass attached to the end of a spring, as wellas many other problems. In having this ability to use technology, the study of differentialequations becomes much more meaningful as well as interesting.

Gottfried Wilhelm Leibniz

(1646–1716)

Isaac Newton (1642–1727) The

controversy as to who (Leibniz or

Newton) discovered calculus first

became an obsession with Newton

during the second half of his life.

Although this chapter is short in length, the vocabulary introduced here is used throughoutthe text. To a large extent, this chapter may be read quickly; subsequent chapters will takeadvantage of the terminology and techniques discussed here. Any formal introduction todifferential equations should begin with German scientist Gottfried Wilhelm Leibniz (1646–1716) and British scientist Isaac Newton (1642–1727), the inventors of calculus. In integralcalculus, we learn that the area under the graph of a smooth positive function is given by adefinite integral, but both Leibniz and Newton were more concerned with solving differential

© 2010 Elsevier Inc. All rights reserved. 1

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2 CHAPTER 1: Introduction to Differential Equations

equations than finding areas. Many of the methods of solution that we present in this text arefrom the great Swiss mathematician Leonhard Euler (1707–1783).

Subsequently, many problems, such as determining the motion of a plucked string, not onlylead to ordinary and partial differential equations but also to other areas of mathematicsas well. Indeed, differential equations are full of rich and exciting applications; interestingapplications are included throughout the text to motivate discussions, make the study ofdifferential equations more interesting and pertinent to the real world, and indicate how somepeople use differential equations beyond this course. However, mathematics is also interestingin its own right; mathematical applications are also included throughout the text.

1.1 INTRODUCTION TO DIFFERENTIAL EQUATIONS:VOCABULARY

We begin our study of differential equations by explaining what a differential equation is. Fromour experience in calculus, we are familiar with some differential equations. For example,suppose that the acceleration a(t) (measured in ft/s2) of a falling object is a(t) = −32. Then,because a(t)= v′(t), where v(t) is the velocity of the object (measured in ft/s), we havev′(t) = −32 or

dv

dt= −32.

An equation like this involving a function of a single variable is called an ordinary differen-tial equation (ODE). (If the equation involves partial derivatives, then it is called a partialdifferential equation [PDE].) In this case, the function to be determined is v = v(t), whichdepends on the variable t, representing time (measured in seconds). The goal in solvingan ODE is to find a function that satisfies the equation. We can solve this ODE throughintegration:

v(t) =∫

a(t) dt =∫(−32) dt = −32t + C,

where C is an arbitrary constant. This result indicates that v(t) = −32t + C is a solutionof the ODE for any choice of the constant C. (We call this a general solution because itinvolves an arbitrary constant.) In fact, we have found every solution of the ODE becauseeach is expressed as −32t + C. Examples of solutions and the corresponding C valuesinclude v(t)= − 32t (C = 0), v(t)= − 32t + 32 (C = 32), and v(t) = −32t − 8 (C = −8).This shows that there are an infinite number of solutions to the ODE.

Isaac Newton We hope this young

Isaac Newton was not worried

about who discovered calculus

first.

Leonhard Euler (1707–1783)

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

dv

dt= d

dt(−32t + C) = −32

When we substitute our solution into the left side of the ODE, we obtain −32, the value onthe right side of the ODE. We have verified that v(t) = −32t + C satisfies the ODE for anychoice of C.

Many times we are given a particular condition that the solution must satisfy. For example,suppose that the object considered earlier has an initial velocity of −64 ft/s. In other words, the

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1.1 Introduction to Differential Equations: Vocabulary 3

velocity at time t = 0 seconds is represented with the initial condition v(0)= − 64. Therefore,we need to find a solution of the ODE that also satisfies the initial condition. We express thisinitial value problem (IVP) as

dv

dt= −32, v(0) = −64,

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

Another application of differential equations found in calculus is finding a function whengiven (1) the slope of the line tangent to the graph of the function at any point (x, y) and(2) a point on the graph.

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

dy

dx= 3x2 − 4x

and further that the graph passes through the point (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)that satisfies the initial condition y(1) = 4. Therefore, we solve the IVP

21 20.5 0.5 1 1.5 2x

1

2

3

4

5

y

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

dy

dx= 3x2 − 4x, y(1) = 4.

As in the previous example, we find a general solution to dy/dx = 3x2 − 4x throughintegration. This yields

y = y(x) = x3 − 2x2 + C.

Now, when we apply the initial condition, we find that

y(1) = 13 − 2 · 12 + C = −1 + C = 4

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

y = y(x) = x3 − 2x2 + 5.

Figure 1.1 shows the graph of the solution together with a portion of the tangent line atthe point (1, 4). Notice that the slope of the tangent line at the point (1, 4) is the value ofdy/dx evaluated 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 understand the behavior of solutions of differentialequations.

The previous two examples are similar as they each involve an ODE, in which the highestorder derivative is the first derivative. We call equations of this type first-order ordinarydifferential equations because the order of a differential equation is the order of the highestorder derivative appearing in the equation.

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■ Example 1.1.1Determine the order of each of the following differential equations: (a) dy/dx =x2/y2 cos y, (b) uxx + uyy = 0, (c) (dy/dx)4 = y + x, and (d) d2x/dt2 + 2dx/dt + 3x =sin t

Solution

(a) This equation is first-order because it includes only one first-order derivative, dy/dx.(b) This equation is classified as second order because the highest order derivatives, uxx

representing ∂2u/∂x2 and uyy representing ∂2u/∂y2, are of order two. The equation uxx +uyy = 0 arises in many areas of study, which include fluid flows as well as electrostatic andgravitational potential, and is often called Laplace’s equation after Pierre-Simon Laplace(also known as the Marquis de Laplace) (1749–1827) or the potential equation. Hence,Laplace’s equation is a second-order PDE. (c) This is a first-order equation because thehighest order derivative is the first derivative. Raising that derivative to the fourth powerdoes not affect the order of the equation. The expressions

(dy

dx

)4

andd4y

dx4

do not represent the same quantities: (dy/dx)4 represents the derivative of y with respectto x, dy/dx, raised to the fourth power; d4y/dx4 represents the fourth derivative of ywith respect to x. (d) The highest order derivative is d2x/dt2, so the equation is secondorder. ■

Pierre-Simon Laplace (1749,

Normandy, France–1827, Paris,

France). Laplace’s name is

probably most remembered in

reference to the Laplace transform

which we will study in Chapter 8.

1 2 3 4 5 6t

22

21

1

2

y

FIGURE 1.2Graphs of y = c1 cos t + c2 sin t for various values ofc1 and c2.

■ Example 1.1.2(a) Show that y = c1 sin t + c2 cos t satisfies the second-order ODE y′′ + y =0, where c1 and c2 are arbitrary constants. (b) Find the solution to the IVPy′′ + 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,

d2y

dt2+ y = −c1 sin t − c2 cos t + c1 sin t + c2 cos t = 0,

so the function satisfies the ODE for any choice of c1 and c2. We graphthe solution for several values of these constants in Fig. 1.2. Because the

solution depends on at least one constant, we say that the functions shown in Fig. 1.2 aremembers of the family of solutions of the ODE. (b) Evaluating the function at t = 0 yieldsv(0) = c1 sin 0 + c2 cos 0 = c2. Then, the initial condition, y(0) = 0, indicates that c2 = 0.Similarly, y′(0) = c1 cos 0 − c2 sin 0 = c1, so c1 = 1 so that the second initial condition,y′(0)= 1, is satisfied. Therefore, y(t)= sin t is the solution to the IVP. We graph thissolution in Fig. 1.3. Notice that the ODE has an infinite number of solutions, whereasthe IVP has only solution. Also observe that the number of initial conditions in the IVPmatches the order of the ODE. ■

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2 4 6 8 10 12t

2120.5

0.51

y

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

The next level of classification is based on the following definition. For an ODE, assumingthat the independent variable is x and the dependent variable is y, an ODE (of order n) iscalled linear if it can be written as

an(x)dny

dxn+ an−1(x)

dn−1y

dxn−1+ · · · + a2(x)

d2y

dx2+ a1(x)

dy

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

where the functions ak(x), k = 0, 1, . . . , n, and f (x) are given and an(x) is not the zerofunction. If f (x) is identically the zero function, the linear equation, Eq. (1.1), is said to behomogeneous. You should verify that y(x) = 0, which we call the trivial solution, is a solutionto every linear homogeneous equation. If f (x) is not identically the zero function, the linearequation, Eq. (1.1), is said to be nonhomogeneous.

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

If the equation under consideration cannot be written in the form given by Eq. (1.1), theequation is said to be nonlinear.

Therefore, some of the properties that lead to classifying an equation as linear or nonlinearare powers of the dependent variable (or one of its derivatives) and functions of the depen-dent variable. A similar classification is followed for PDEs. For PDE, the coefficients in alinear PDE are functions of the independent variables.

■ Example 1.1.3Determine which of the following differential equations are linear: (a) dy

dx = x3,

(b) d2udx2 + u = ex , (c) (y − 1) dx + x cos y dy = 0, (d) d3y

dx3 + y dydx = x, (e) dy

dx + x2y = x,

(f ) d2xdt2 + sin x = 0, (g) uxx + y uy = 0, and (h) uxx + u uy = 0

Solution

(a) This equation is linear because the nonlinear term x3 is the function f (x) of the inde-pendent variable in the general formula for a linear differential equation. (b) This equationis also linear. Using u as the name of the dependent variable does not affect the linearity.(c) If y is the dependent variable, solving for dy/dx gives us

dy

dx= 1 − y

x cos y.

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

dx

dy= cos y

1 − yx.

This equation is linear in the dependent variable x. (d) The coefficient of the term dy/dx isy instead of an expression involving only the independent variable x. Hence, this equationis nonlinear in the dependent variable y. (e) This equation is linear. The term x2 is thecoefficient function. (f) For this equation, note that x is the dependent variable; t is theindependent variable. This equation, known as the pendulum equation because it modelsthe motion of a simple pendulum, is nonlinear because it involves a nonlinear function of thedependent variable x, sin x. (g) This PDE is linear because the coefficient of uy (u = u(x, y)is the dependent variable) is multiplied by a function of one of the independent variables(x and y), which in this case is f (x, y) = y. (h) Here, we have a product of u (the dependentvariable) and one of its derivatives, so the equation is nonlinear. ■

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

■ Example 1.1.4Solve the following differential equations:

(a) dydx = cos x, (b) dy

dx = x√x2+1

, (c) dydx = 1

x2+16, (d) dy

dx = xex , and (e) dydx = 1

4−x2

Solution

In each case, we integrate the indicated function and graph the solution for several valuesof the constant of integration. Each solution contains a constant of integration so there areinfinitely many solutions to each equation. (a) y = ∫

cos xdx = sin x + C [see Fig. 1.4(a)].

2 4 6 8 10 12x

24

22

2

4

y

24 22

22

2 4x

2

4

6

8y

(b)(a)

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

√x2 + 1 + C for various values of C.

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(b)To evaluate y = ∫ x√x2+1

dx, we let u = x2 + 1 so that du = 2x dx or 12 du = x dx.Then,

y = 1

2

∫2x√

x2 + 1dx = 1

2

∫1√u

du = 1

2

∫u−1/2 du = u1/2 + C

=√

x2 + 1 + C. [See Fig. 1.4(b).]

(c) To integrate y = ∫ 1x2+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 =∫

1

16 + (4 tan θ)24 sec2 θ dθ

=∫

1

16 + 16 tan2 θ4 sec2 θ dθ = 1

16

∫1

sec2 θ4 sec2 θ dθ

= 1

4

∫dθ = 1

4θ+ C = 1

4tan−1

(x

4

)+ C.

First, we use the identity

1 + tan2 θ = sec2 θ and then

resubstitute: x/4 = tan θ so

θ = tan−1(x/4).

Graph y = 14 tan−1 ( x

4

)+ C if

C = −1, 0, and 1.(d) To evaluate y = xex by hand, we use the integration by parts formula with u = x anddv = 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 decom-position of 1/(4 − x2), which is determined by finding constants A and B that satisfy theequation

1

(2 − x)(2 + x)= A

2 − x+ B

2 + x.

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

y = 1

4

∫ (1

2 − x+ 1

2 + x

)dx = 1

4[− ln |2 − x| + ln |2 + x|] + C

= 1

4ln

∣∣∣∣2 + x

2 − x

∣∣∣∣+ C. [See Fig. 1.5.]

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

The integration by parts formula

states that∫

udv = uv − ∫ vdu.

Find C if y(1) = 2.

In Example 1.1.4, each solution is given as a function y = y(x) of the independent variable.In these cases, the solution is said to be explicit. In solving some differential equations,however, we can find only an equation involving the independent and dependent vari-ables that the solution satisfies. In this case, we say that we have found an implicitsolution.

We will see that given an arbitrary

differential equation, constructing

an explicit or implicit solution is

nearly always impossible.

Consequently, although

mathematicians were first

concerned with finding analytic

(explicit or implicit) solutions to

differential equations, they have

since (frequently) turned their

attention to addressing properties

of the solution and finding

algorithms to approximate

solutions.

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4x

22

24

24

2

y

222

4

FIGURE 1.5Graph of y = 1

4 ln∣∣∣ 2+x

2−x

∣∣∣+ C for various values of C.

■ Example 1.1.5Verify that the equation 2x2 + y2 − 2xy + 5x = 0 satisfies the differentialequation

dy

dx= 2y − 4x − 5

2y − 2x.

Solution

We use implicit differentiation to compute y′ = dy/dx if 2x2 + y2 − 2xy +5x = 0:

4x + 2ydy

dx− 2x

dy

dx− 2y + 5 = 0

dy

dx(2y − 2x) = 2y − 4x − 5

dy

dx= 2y − 4x − 5

2y − 2x

The equation 2x2 + y2 − 2xy + 5x = 0 satisfies the differential equation

dy

dx= 2y − 4x − 5

2y − 2x.

Although we cannot solve 2x2 + y2 − 2xy + 5x = 0 for y as a function of x (seeFig. 1.6), we can determine the corresponding y value(s) for a given value of x. Forexample, 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 on the graph of 2x2 + y2 − 2xy +5x = 0 (see Fig. 1.7). ■

x

26

24

22

22 22426

2

y

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

In the same manner that we consider systems of equations in algebra, we canalso consider systems of differential equations. For example, if x and y representfunctions of t, we will learn in Chapter 6 to solve the system of linear equations

{dx/dt = ax + by

dy/dt = cx + dy,

where a, b, c, and d represent constants and differentiation is with respect to t.We will see that systems of differential equations arise naturally in many physicalsituations that are modeled with more than one equation and involve more thanone dependent variable. In addition, we will see that it is often useful to write adifferential equation of order greater than one as a system of first-order equations,especially when the original equation is nonlinear.

Find an equation of the

line tangent to the graph of

2x2 + y2 − 2xy + 5x = 0 at the points

(−1, −3) and (−1, 1).

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21.1 21.05 21 20.95 20.923.2

23.1

23

22.9

22.8

21.1521.121.05 21 20.9520.920.8520.8

0.85

0.9

0.95

1

1.05

1.1

1.15

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

■ Example 1.1.6 (Duffing’s Equation)Duffing’s equation is the second-order nonlinear equation

d2x

dt2+ k

dx

dt− x + x3 = � cosωt, (1.2)

where k, �, and ω are positive constants. Sources: See texts like Jordan and

Smith’s Nonlinear Ordinary

Differential Equations [16].Write Duffing’s equation as a system of first-order equations.

Solution

Let 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 the nonlinear system

{x′ = y

y′ = x − x3 − ky +� cosωt

■

Note that a system of differential equations can consist of more than two equations. Forexample, the basic equations that describe the competition between two organisms, with

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population densities x1 and x2, respectively, in a chemostat are

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

S′ = 1 − S − m1S

a1 + Sx1 − m2S

a2 + Sx2

x′1 = x1

(m1S

a1 + S− 1

)

x′2 = x2

(m2S

a2 + S− 1

),

(1.3)

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

Sources: See Smith and Waltman’s

The Theory of the Chemostat [27]

for a detailed discussion of

chemostat models.

Exercises 1.1

Determine if each of the following equations is an ODE or a PDE. If the equation is an ODE,then determine (a) the order of the ODE and (b) if the equation is linear or nonlinear.

1.d2y

dx2+ dy

dx− 2y = x3

2. ydy

dx+ y4 = sin x

3.∂2y

∂t2= c2 ∂

2y

∂x2, c > 0 constant

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

5.(

dy

dx

)2+ y = 0

6. t2d2y

dt2+ t

dy

dt+ 2y = 0

7.1

c2

∂2z

∂t2= ∂2z

∂x2+ ∂2z

∂y2(z = z(t, x, y))

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

9. x

(d2y

dx2

)2

+ 2y = 2x

10.d2x

dt2+ 2 sin x = sin 2t (x = x(t))

11. ut + u ux = σ uxx , σ constant

12. (2x − 1) dx − dy = 0

13. (2t − y) dt − dy = 0

14.∂u

∂x

∂u

∂y= u, (u = u(x, y))

15. (2x − y) dx − y dy = 0

16. Write each of the following second-order equations as a system of first-order equations.

a.d2x

dt2− dx

dt− 6x = 0

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b. 4d2x

dt2+ 4

dx

dt+ 37x = 0

c. Ld2x

dt2+ g sin x = 0, L, g positive constants, x = x(t)

d.d2x

dt2−μ(1 − x2)

dx

dt+ x = 0, μ > 0 constant

e. td2x

dt2+ (b − t)

dx

dt− ax = 0, a, b constants

In Exercises 17–28, verify that each of the given functions is a solution to the correspondingdifferential equation. (A, B, and C represent constants.)

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 − 12 cos x + 1

2 sin x

20.d2y

dt2− dy

dt− 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 + 14 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 equation satisfies the differential equation. Use theequation to determine y for the given value of x (or t). Confirm your result by graphing eachequation using appropriate technology.

29. dy/dx = −x/y, x2 + y2 = 16, x = 0

30. 3y(t2 + y)dt + t(t2 + 6y)dy = 0, t3y + 3ty2 = 8, t = 2

31. dy/dx = −2y/x − 3, x3 + x2y = 100, x = 1

32. y cos t dt + (2y + sin t) dy = 0, y2 + y sin t = 1, t = 0

33. (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 solution to the differential equation.

34. dy/dx = (x2 − 1)(x3 − 3x)3

35. dy/dx = x sin x2

36. dy/dx = x/√

x2 − 16

37. dy/dx = 1/(x ln x)

38. dy/dx = x ln x

39. dy/dx = xe−x

40.dy

dx= −2(x + 5)

(x + 2)(x − 4)

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #12


41.dy

dx= 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)

In Exercises 45–54, use the indicated conditions with the indicated solution to determine thesolution to the given problem.

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

46. dy/dt + y = sin t, y(0) = −1, y(t) = Ae−t − 12 cos t + 1

2 sin t

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

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

49. y′′ + 9y = 0, y(0) = 0, y′(0) = 1, y(x) = A cos 3x + B sin 3x

50. y′′ + 9y = 0, y(0) = 0, y′(0) = 0, y(t) = A cos 3t + B sin 3t

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 your result by graphing each function on anappropriate interval using appropriate technology.

55. dy/dx = 4x3 − x + 2, y(0) = 1

56. dy/dt = sin 2t − cos 2t, y(0) = 0

57. dy/dx = x−2 cos(x−1), y(2/π) = 1

58. dy/dx = (ln x)/x, y(1) = 0

59. The velocity of a falling object with mass m that is subjected to air resistance proportionalto the instantaneous velocity v of the object is found by solving the IVP⎧⎨⎩

m dv/dt = mg − cv

v(0) = v0, where c > 0 is the proportionality constant.

a. Given that a general solution to m dv/dt = mg − cv is v(t) = mg/c + Ke−ct/m, find thesolution of this IVP.

b. Determine limt→∞ v(t).

60. The number of cells in a bacteria colony after t hours is determined by solving the IVP⎧⎨⎩

dP/dt = kP

P(0) = P0.

a. Given that a general solution of dP/dt = kP is P(t) = Cekt , use the initial condition tofind C.

b. Find the value of k so that the population doubles in 8 hours.

61. In 1840, the Belgian mathematician-biologist Pierre F. Verhulst (1804–1849) developed thelogistic equation, dP/dt = rP − aP2, where r and a are positive constants, to predict thepopulation P(t) in certain countries.

Pierre Francois Verhulst

(1804–1849)

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #13


a. Given that a general solution to this equation is P(t) = ra+Ce−rt , find the solution that

satisfies P(0) = P0, where P0 > 0 is constant.b. Determine limt→∞ P(t).

62. The differential equation dS/dt + 3S/(t + 100) = 0, where S(t) is the number of pounds ofsalt in a particular tank at time t, is used to approximate the amount of salt in the tankcontaining a salt–water mixture, in which pure water is allowed to flow into the tank whilethe mixture is allowed to flow out of the tank. If S(t) = 15, 000, 000/(t + 100)3, show thatS satisfies dS/dt + 3S/(t + 100) = 0. What is the initial amount of salt in the tank? Ast → ∞, what happens to the amount of salt in the tank?

63. The displacement (measured from x = 0) of a mass attached to the end of a spring at time t

is given by x(t) = 3 cos 4t + 94 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 t minutes at a position x between 0 and 2πis given by u(x, t) = 3 − e−16kt cos 4x. Show that u satisfies ut = k uxx . What is the initialtemperature (t = 0) at x = π? What happens to the temperature at each point in the wireas t → ∞?

66. The displacement u of a string of length � at time t position x, where x is measured fromx = 0 is given by u(x, t) = sin πx cos t. Show that u satisfies π2utt = uxx . What is the value ofu at the endpoints x = 0 and x = 1 for all values of t?

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

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

69. Use the fact that ddx

(e2xy

)= e2x dy

dx + 2e2xy and integration to solve e2x dydx + 2e2xy = ex .

70. Use the fact that ddx

(exy) = ex dy

dx + exy and integration to solve ex dydx + exy = xex .

The great physicist Erwin

Schrödinger (1887–1961) received a

Nobel prize for his work in 1933.

71. The time-independent Schrödinger equation is given by

− h2

2m

d2ψ(x)

dx2+ U(x)ψ(x) = Eψ(x).

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

72. A singular solution of a differential equation is 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 differential equation dy/dx = (x − 4)y3/[x3(y − 2)]. Is y = 0 asolution of this differential equation? Is y = 0 a singular solution?

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

74. The current I(t) in an L-R circuit, which contains a resistor, an inductor, and a voltagesource, satisfies the differential equation RI + LdI/dt = E(t), where R and L are constantsrepresenting the resistance and the inductance and E(t) is the voltage source. Is thisequation linear or nonlinear? Determine the order of the equation.

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #14


In Exercises 75–77, (a) verify that the indicated function is a solution of the given differentialequation and (b) graph the solution on the indicated interval(s).

Throughout Introductory

Differential Equations with

Boundary Value Problems, we use

graphs of solutions of differential

equations. In some cases, we are

able to predict what the graph of a

solution should look like. If the

graph of our proposed solution

does not appear as predicted, we

know that either we made a

mistake in constructing our

proposed solution or our

conjecture about the general

shape of the graph of the solution

is wrong. In other cases, we will

find that it is easier to examine the

graph of a solution than it is to

examine the solution (if we are

able to construct one in the first

place).

Often the calculus and algebra

encountered in solving differential

equations can be tedious, if not

completely overwhelming or

impossible. Today many

sophisticated calculators and

computer algebra systems are

capable of performing the

integration and algebraic

simplification encountered when

solving many differential

equations. Having access to these

tools can be a great advantage: a

large number of problems can be

solved quickly, we make

conjectures as to the general form

of a solution to different forms of

differential equations, and these

tools allow us to check and verify

our work.

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√

33 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.

79. a. Show that (x2 + y2)2 = 5xy is an implicit solution 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 = 5xy with x-coordinate 1.

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

80. Solve dy/dx = sin4 x, y(0) = 0, and graph the resulting solution on the interval [0, 4π].81. A general solution of y(4)+ 25

2 y′′ − 5y′ + 62916 y = 0 is given by y = 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)+ 252 y′′ − 5y′ + 629

16 y = 0

y(0) = 0, y′(0) = 1, y′′(0) = −1, y′′′(0) = 1and graph the resulting solution.

82. A general solution of the system

⎧⎨⎩

dx/dt = 4y

dy/dt = −4xis

⎧⎨⎩

x = −c1 cos 4t + c2 sin 4t

y = c2 cos 4t + c1 sin 4t,

where c1 and c2 are constants. Solve the IVP

⎧⎪⎪⎨⎪⎪⎩

dx/dt = 4y

dy/dt = −4x

x(0) = 4, y(0) = 0

and then graph x(t),

y(t), and the parametric equations

⎧⎨⎩

x = x(t)

y = y(t).

83. A general solution of the system

⎧⎨⎩

dx/dt = −5x + 4y

dy/dt = 2x + 2yis

⎧⎨⎩

x = −c1e3t − 4c2e−6t

y = 2c1e3t + c2e−6t ,

where c1 and c2 are constants. Solve the IVP

⎧⎪⎪⎨⎪⎪⎩

x = −5x + 4y

y = 2x + 2y

x(0) = 4, y(0) = 0

and then graph x(t),

y(t), and the parametric equations

⎧⎨⎩

x = x(t)

y = y(t).

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #15

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

1.2 A GRAPHICAL APPROACH TO SOLUTIONS: SLOPEFIELDS AND DIRECTION FIELDS

■ Systems of ODE and Direction Fields

■ Relationship Between Systems of First-Order and Higher Order Equations

Suppose that we are asked to solve the ODE dy/dx = e−x2. In this case, we do not attempt

to solve this equation through integration as we did in the previous section because of thepresence of the function f (x) = e−x2

on the right side of the ODE. (See Exercise 21 at theend of this section.) Instead, we can gain insight into the behavior of solutions of this ODEthrough a graphical approach by considering the slope of the tangent line to solutions of theODE. Recall from Section 1.1 that the differential equation gives the slope of the tangentline to solutions of the ODE at the given point (x, y) in the xy-plane. Therefore, if we wish todetermine the slope of the tangent line to the solution to the ODE that passes through (0, 1)at this point, we substitute (0, 1) into the right side of dy/dx = e−x2

. Because the right sideonly depends on x, we have slope

dy

dx= e−(0)2 = 1.

In fact, the slope of the line tangent to solutions at all points of the form (0, y) is 1. At thepoint (

√ln 2, 4), we find that the slope is

dy

dx= e−(√ln 2)2 = e− ln 2 = eln 2−1 = 1

2.

Again, we obtain the same slope at all points (√

ln 2, 4) ≈ (0.832555, 4) and (−√ln 2, 4) ≈

(−0.832555, 4). In Fig. 1.8, we draw several short line segments using points of the form(0, y) for y = 0, ±1, ±2, ±3. Notice that we use a triangle with base length 1 and heightlength 1 to assist in sketching the tangent lines with slope 1 at these points. We use a triangle

22 21 1 2x

22

(b)

21

1

2

y

(a)

x

y

FIGURE 1.8(a) Several line segments in the slope field for dy/dx = e−x2

. (b) Slope field for the equation.

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #16


(b)

22 21.5 21 20.5 0.5 1 1.5 2x

22

21.5

21

20.5

0.5

1

1.5

2y

(a)

22 21 1 2x

22

21

1

2

y

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.

with base length 2 and height length 1 to help us draw the lines of slope 1/2 that are tangentto solutions at the points (−√

ln 2, 4) for y = 0, ±1, ±2, ±3. By drawing a set of short linesegments representing the tangent lines to solutions of the ODE at numerous points in theplane, we construct the slope field of the ODE. We show the slope field for dy/dx = e−x2

on the square [−2, 2] × [−2, 2] in Fig. 1.9(a). (Note that because constructing a slope fieldis time consuming, we usually let a computer algebra system do the work for us.) Observethat at each point along the y-axis, the slope is 1 as we predicted earlier. Notice also thatthe slope appears to be zero for larger values of |x|. This is because for large values of x (inabsolute value), dy/dx ≈ 0 because limx→±∞ e−x2 = 0. Therefore, we expect solutions to“flatten out” as x increases.

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

dy

dx= e−x2

, y(−2) = −1

by starting at the point (−2, −1) and tracing the solution by following the tangent slopes.This solution is sketched in Fig. 1.9(b). Thus, although we did not determine a formula forsolutions to the ODE or the IVP, we were able to determine some properties of the solutionsby using the slope field of the differential equation.

Systems of ODE and Direction FieldsWe can also consider systems of differential equations. In Chapter 6, we will learn how tosolve systems of first-order ODEs of the form{

dx/dt = ax + by

dy/dt = cx + dy,

where a, b, c, and d are given constants. In the case 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,

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #17


1 2 3 4 5 6t

21

20.5

0.5

1

(a) (b)

2120.5 0.5 1x

21

20.5

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.

we can verify that the parametric equations

{x(t) = sin t

y(t) = cos t

satisfy the system because

dx

dt= d

dt(sin t) = cos t = y and

dy

dt= d

dt(cos t) = −sin t = −x.

We can graph each function separately as we do in Fig. 1.10(a). (The graph of x(t) is the solidcurve; that of y(t) is dashed.)Another option is to graph them as a pair of parametric equationsas in Fig. 1.10(b). As we recall, the graph of this pair of parametric equations is a circle ofradius 1 centered at the origin because x2 = sin2 t and y2 = cos2 t, so that x2 + y2 = sin2 t +cos2 t = 1. However, we must indicate the orientation of the curve (the direction of increasingparameter value t). For this pair of parametric equations, we find that at t = 0, x(0)= sin 0 = 0and y(0)= cos 0 = 1. Therefore, the point (0, 1) corresponds to t = 0. Similarly, the point(1, 0) corresponds to t =π/2 because x(π/2)= sin π/2 = 1 and y(π/2)= cosπ/2 = 0. Thismeans that the solution moves from (0, 1) to (1, 0) as t increases. To determine if the orientationis clockwise or counter-clockwise, we test a t-value between 0 and π/2. Choosing t =π/4,we find the x(π/4)= sin π/4 = 1/

√2 and y(π/4)= cosπ/4 = 1/

√2, so the orientation is

clockwise. The parametric equations {x(t)= sin t, y(t)= cos t} satisfy the IVP

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

dy/dt = −x, y(0) = 1

because they satisfy the system of differential equations as well as the two initial conditions.Notice that in the case of an IVP involving a system of differential equations, an initialcondition is given for each of the variables x and y that depend on t. In the parametric plot,the solution to this IVP passes through the point (x(0), y(0)) = (0, 1).

Another way to view a system of two ODEs is 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,

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #18


we first write it as a first-order equation with

dy

dx= dy/dt

dx/dt= cx + dy

ax + by.

Then, we can consider the slope field associated with this differential equation. For example,if we refer back to the system {

dx/dt = y

dy/dt = −x,

Observe that the procedure

described here can be used for any

system of the form

⎧⎨⎩

dx/dt = f (x, y)

dy/dt = g(x, y).

Generally, we will use standard

mathematical notation throughout

Modern Differential Equations so

i =< 1, 0 > and j =< 0, 1 >.

we obtain the first-order equation dy/dx = −x/y, so we can determine the slope of tangentlines to solutions at points in the xy-plane. For example, at the point (1/

√2, 1/

√2), the

solution to this system that passes through this point has slope −1/√

21/

√2

= −1. In a similar

manner, we can find the slope at other points in the plane. However, as we mentioned in ourearlier discussion, we must indicate the orientation when we graph parametric equations, sowe consider the vector< dx/dt, dy/dt >= dx/dt i + dy/dt j with components from the sys-tem of differential equations. In the case of 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 vector<1/√

2, −1/√

2>. The directionfield is made up of tangent vectors such as <1/

√2, −1/

√2> to solutions at points in the

plane, so it is similar to the slope field for dy/dy = −x/y shown in Fig. 1.11(a), except thatvectors are used to indicate the orientation of solutions. In Fig. 1.11(b), we show the directionfield for this system. The vectors in the direction field indicate that solutions to this systemare circles in the xy-plane that are directed clockwise. We graph several solutions along withthe direction field in Fig. 1.11(c). A collection of solutions in the xy-plane is called the phaseportrait of the system. Notice that at points in the first quadrant, where x > 0 and y > 0,

210 25 5 10x

210

25

5

10

y

210 25 5 10x

y

210

25

5

10

210 25 5 10x

210

25

5

10

y

(b) (c)(a)

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

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #19


dx/dt = y > 0 and dy/dt = −x < 0. This means that x(t) increases and y(t) decreases alongsolutions in the first quadrant. In the second quadrant, where x < 0 and y > 0, dx/dt = y > 0,and dy/dt = −x > 0. Therefore, x(t) and y(t) increase along solutions in the second quadrant.We can perform a similar analysis for points in the other two quadrants.

Relationship Between Systems of First-Order and HigherOrder Equations

Again, consider the system

{dx/dt = y

dy/dt = −x. If we differentiate the equation dx/dt = y with

respect 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 ODEs is equivalent to the second-order ODE d2x/dt2 + x = 0. Often, however, webegin with a second-order ODE of the form d2x/dt2 + b dx/dt + c x = f (t) and would liketo write the equation as a system of first-order ODEs. We do this by letting dx/dt = y andby differentiating this equation with respect to t to obtain d2x/dt2 = dy/dt. Solving thesecond-order ODE for d2x/dt2, we find d2x/dt2 = −b dx/dt − c x + f (t) so that dy/dt =−b dx/dt − c x + f (t). Replacing dx/dt with y in this equation, we obtain the followingequivalent system of first-order ODEs:

{dx/dt = y

dy/dt = −by − cx + f (t)

By writing the second-order ODE as a system of first-order ODEs, we investigate the behaviorof the solution of the second-order ODE by observing the behavior in the direction field andphase portrait of the corresponding system. A similar 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 determine if the indicated path is that of a solution tothe differential equation.

1.dy

dx= −y/x

22 21 1 2x

2

1

1

2

y

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #20


2.dy

dx= x2 + y2

21.5 21 20.5 0.5 1 1.5x

21.5

21

20.5

0.5

1

1.5y

3.dy

dx= x/y

2221.52120.5 0.5 1 1.5 2x

22

21.5

21

20.5

0.5

1

1.5

2y

4.dy

dx= x2 − y

2221.5 2120.5 0.5 1 1.5 2x

22

21.5

21

20.5

0.5

1

1.5

2y

In Exercises 5–8, use the slope field to sketch the solutions of the differential equations thatpass through the given points.

5.dy

dx= x/y

22 21.5 21 20.5 0.5 1 1.5 2x

22

21.5

21

20.5

0.5

1

1.5

2y

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #21


6.dy

dx= x2 − y

22 21.5 21 20.5 0.5 1 1.5 2x

22

21.5

21

20.5

0.5

1

1.5

2y

7.dy

dx= x − y

22 21.5 21 20.5 0.5 1 1.5 2x

22

21.5

21

20.5

0.5

1

1.5

2y

8.dy

dx= y2 − x2

2221.52120.520.5

21

21.5

22

0.5

0.5

1

1.5

2y

x1 1.5 2

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; and (d) y(0) = 3π/2.

10. Graph the slope field for dy/dx = sin x. Does limx→∞ y(x) exist for any initial conditiony(0) = y0? Solve the ODE and find limx→∞ y(x). Does this match the graphical result?

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

12. Graph the slope field for dy/dx = 1/(x2 + 1). Does limx→∞ y(x) exist for any initial conditiony(0) = y0? Solve the ODE and find limx→∞ y(x). Does this match the graphical result?

In Exercises 13–14, use the direction field of the given system to sketch the graph of thesolution that satisfies the indicated initial conditions. Determine limt→∞ x(t) and limt→∞ y(t)in each case (if they exist).

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #22


13. System:

⎧⎨⎩

dx/dt = y

dy/dt = x; (a) x(0) = 0, y(0) = 2; (b) x(0) = −2, y(0) = 0; (c) x(0) = −2, y(0) = 2.

22

22

2

2 4

4

24

24x

y

14. System:

⎧⎨⎩

dx/dt = 3x − 2y

dy/dt = 4x − y; x(0) = 0, y(0) = 5.

215 210 25 5 10 15x

215

210

25

5

10

15

y

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

15. d2x/dt2 + 4x = 0

16. d2x/dt2 − 5 dx/dt = 0

17. d2x/dt2 + 4 dx/dt + 13x = 0

18. x′′ − 6x′ + 7x = 0 (′ = d/dt; x = x(t))

19. x′′ + 16x = sin t

20. x′′ + 4x′ + 13x = e−t

21. (a) Use a computer algebra system to solve dy/dx = e−x2; (b) Graph the solution to the IVP

dy/dx = e−x2, y(0) = a for a = −2, −1, 0, 1, 2. (c) Graph the slope field of the differential

equation together with the solutions in (b). (d) Do the solutions appear to match the resultsdescribed at the beginning of the section?

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

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #23


21 20.5 0.5 1x

21

20.5

0.5

1

y

21 20.5 0.5 1x

21

20.5

0.5

1

y

FIGURE 1.12Figure for Exercise 23.

23. Consider the systems

⎧⎨⎩

dx/dt = −y

dy/dt = xand

⎧⎨⎩

dx/dt = −y

dy/dt = −x.

a. For given initial conditions, in a brief paragraph explain why you think that thesolutions of the systems are similar or different.

b. Figure 1.12 shows the direction field associated with each system. Use the directionfield to help you graph the solutions that satisfy these initial conditions(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 your conjecture in (a)?

24. Consider the systems

⎧⎨⎩

dx/dt = x/2

dy/dt = yand

⎧⎨⎩

dx/dt = −x/2

dy/dt = −y.

a. For given initial conditions, explain why you think that the solutions of the systems aresimilar or different in a brief paragraph.

b. Figure 1.13 shows the direction field associated with each system. Use the directionfield to help you graph the solutions that satisfy these initial conditions(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 your conjecture 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 represent positive constants can be used to model the size of the populationof two species, represented by x(t) and y(t), competing for a common food supply.

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #24


21 20.5 0.5 1x

21

20.5

0.5

1

y

21 20.5 0.5 1x

21

20.5

0.5

1y

FIGURE 1.13Figure for Exercise 24.

0.2 0.4 0.6

(a)

0.8 1x

0.2

0.4

0.6

0.8

1y

(b)

0.2 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 for Exercise 25(b).

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 the direction field to graph various solutions if both x(0)and y(0) are positive. (ii) Use the direction field and your graphs to approximatelimt→∞ x(t) and limt→∞ y(t).

b. Figure 1.14(b) shows the direction field for the system if a = 1, b1 = 1, b2 = 1, c = 0.67,d1 = 0.75, and d2 = 1. (i) Use the direction field to graph various solutions if both x(0)and y(0) are positive. (ii) Use the direction field and your graphs to determine the fate ofthe species with population y(t). What happens to the species with population x(t)?

SUMMARY: ESSENTIAL CONCEPTS AND FORMULAS

Differential Equation (DE) An equation that contains the derivative or differentials ofone or more dependent variables with respect to one or more dependent variables.

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #25

Review Exercises 25

Ordinary Differential Equation (ODE) If a differential equation contains only ordi-nary derivatives (of one or more dependent variables) with respect to a singleindependent variable, the equation is called an ordinary differential equation.

Partial Differential Equation (PDE) A differential equation that contains the partialderivatives or differentials of one or more independent variables with respect to morethan one independent variable is called a partial differential equation.

Linear Ordinary Differential Equation A linear ODE is an equation that can be writ-ten in the form an(x)y(n)+ an−1(x)y(n−1)+ · · · + a2(x)y′′ + a1(x)y′ + a0(x)y = f (x),where y = y(x).

Order of an Equation The order of the highest order derivative in a differential equationis called the order of the equation.

Solution A solution of a differential equation on a given interval is a function that iscontinuous on the interval and has all the necessary derivatives that are present in thedifferential equation such that when substituted into the equation yields an identity forall values on the interval.

Explicit Solution A solution given as a function of the independent variable.Implicit Solution A solution given as a relation such as f (x, y)= 0, f (t, x) = 0, or

f (t, y)= 0.Trivial Solution y = 0 is always a solution of the nth-order linear homogeneous equation.Slope Field A collection of line segments that indicate the slope of the tangent line to the

solution(s) of a differential equation.

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 Acollection of vectors that indicate the slope and direction of the tangentline to the solutions of a system of differential equations.

REVIEW EXERCISES

In Exercises 1–5, determine (a) if the equation is an ODE or PDE; (b) the order of thedifferential equation; and (c) if the equation is linear or nonlinear.

1. dy/dt = y

2. aux + ut = 0, u = u(x, t), a constant

3. d2y/dx2 + 2 dy/dx + y = 0

4. m x′′ + 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 function is 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)/2

8. 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

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #26


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)/2

13. y′′ − 7y′ + 12y = 2, y = Ae3x + Be4x + 1/6

In Exercises 14 and 15, verify that the given implicit function satisfies the differentialequation.

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 differential equation.

16. dy/dx = xe−x2

17. dy/dx = x2 sin x

18.dy

dx= 2x2 − x + 1

(x − 1)(x2 + 1)

19. dy/dx = x2/√

x2 − 1

In Exercises 20–22, use the indicated initial or boundary conditions with the given generalsolution to determine the solution(s) to the given problem.

20. dy/dx + 2y = x2, y(0) = 1, y = 14 − 1

2 x + 12 x2 + 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

In Exercises 23 and 24, solve the IVP. Graph the solution on an appropriate interval.

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

24. dy/dx = 13 (4x − 9)(x − 3)−2/3, y(0) = 0

25. 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. Show that u satisfies the first-order equationdu/dt = −k(u − 30). What is the initial temperature (t = 0) on the surface of the ball? Whathappens to the temperature as t → ∞?

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

is given by x(t) = 14 e−t

(cos

√35t + 9√

35sin

√35t)

. Show that x satisfies the ODE

x′′ + 2x′ + 36x = 0. What is the initial displacement of the mass? What is the initial velocityof the mass?

27. For a particular wire of length 1 foot, the temperature at time t hours at a position of x feetfrom the end (x = 0) of the wire is estimated by u(x, t) = e−π2kt sin πx − e−4π2kt sin 2πx.Show that u satisfies the equation ut = k uxx . What is the initial temperature (t = 0) atx = 1? What happens to the temperature at each point in the wire as t → ∞?

28. The height u of a long string at time t and position x where x is measured from the middleof the string (x = 0) is given by u(x, t) = sin x cos 2t. Show that u satisfies the waveequation utt = 4uxx . What is the initial height (t = 0) at x = 0?

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

ABELL-ET AL 04-Ch01-P374935 2009/8/31 11:53 #27

Review Exercises 27

(a)

23 22 21 1 2 3x

21

20.5

0.5

1

1.5

2y

(b)

21 20.5 0.5 1x

21

20.5

0.5

1

y

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

30. The slope field of y′ = 2(y − y2) is shown in Fig. 1.15(a). Sketch the graph of the solutionthat satisfies the indicated initial condition. Also, determine if limx→∞ y(x) exists.(a) y(0) = 0.5; (b) y(0) = 1.5; (c) y(0) = −0.5.

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

32. The function J1(x) = ∑∞k=0

(−1)k

k!(k + 1)!22k+1x2k+1 is called the Bessel function of order 1.

Verify that J1(x) is a solution of Bessel’s equation of order 1, x2y′′ + xy + (x2 − 1)y = 0.

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