Calculus 7eap1501 05

SECTION 15.
SECTION
15.1
Exact First-Order Equations
1093

Exact Differential Equations Integrating Factors
Exact Differential Equations

In Section 5.6, you studied applications of differential equations to growth and decay
problems. In Section 5.7, you learned more about the basic ideas of differential equations and studied the solution technique known as separation of variables. In this
chapter, you will learn more about solving differential equations and using them in
real-life applications. This section introduces you to a method for solving the firstorder differential equation
Msx, yd dx 1 Nsx, yd dy 5 0
for the special case in which this equation represents the exact differential of a
function z 5 f sx, yd.
Definition of an Exact Differential Equation
The equation M sx, yd dx 1 Nsx, yd dy 5 0 is an exact differential equation if
there exists a function f of two variables x and y having continuous partial derivatives such that
fx sx, yd 5 M sx, yd
and
fy sx, yd 5 Nsx, yd.
The general solution of the equation is f sx, yd 5 C.

From Section 12.3, you know that if f has continuous second partials, then
2f
2f
N
M
5
5
5
.
y
yx xy
x
This suggests the following test for exactness.
THEOREM 15.1
Test for Exactness
Let M and N have continuous partial derivatives on an open disc R. The differential equation M sx, yd dx 1 Nsx, yd dy 5 0 is exact if and only if
M N
5
.
y
x
Exactness is a fragile condition in the sense that seemingly minor alterations in

an exact equation can destroy its exactness. This is demonstrated in the following
example.
1094
CHAPTER 15
Differential Equations
EXAMPLE 1
Testing for Exactness
a. The differential equation sxy 2 1 xd dx 1 yx 2 dy 5 0 is exact because
NOTE Every differential equation of

the form
M
5
fxy 2 1 xg 5 2xy
y
y
M sxd dx 1 Ns yd dy 5 0
is exact. In other words, a separable variables equation is actually a special type
of an exact equation.
and
N
5
f yx 2g 5 2xy.
x
x
But the equation s y 2 1 1d dx 1 xy dy 5 0 is not exact, even though it is obtained

by dividing both sides of the first equation by x.
b. The differential equation cos y dx 1 s y 2 2 x sin yd dy 5 0 is exact because
M
5
fcos yg 5 2sin y
y
y
and
2
N
5
f y 2 x sin yg 5 2sin y.
x
x
But the equation cos y dx 1 s y 2 1 x sin yd dy 5 0 is not exact, even though it

differs from the first equation only by a single sign.
Note that the test for exactness of Msx, yd dx 1 Nsx, yd dy 5 0 is the same as the
test for determining whether Fsx, yd 5 M sx, yd i 1 N sx, yd j is the gradient of a potential function (Theorem 14.1). This means that a general solution f sx, yd 5 C to an
exact differential equation can be found by the method used to find a potential
function for a conservative vector field.
EXAMPLE 2
Solving an Exact Differential Equation
Solve the differential equation s2xy 2 3x 2d dx 1 s x 2 2 2yd dy 5 0.

Solution
The given differential equation is exact because
2
M
N
5
5
f2xy 2 3x 2g 5 2x 5
fx 2 2yg .
y
y
x
x
The general solution, f sx, yd 5 C, is given by
f sx, yd 5
5
C = 1000
E
E
M sx, yd dx
s2xy 2 3x 2d dx 5 x 2y 2 x 3 1 gs yd.
In Section 14.1, you determined gs yd by integrating Nsx, yd with respect to y and

reconciling the two expressions for f sx, yd. An alternative method is to partially
differentiate this version of f sx, yd with respect to y and compare the result with
Nsx, yd. In other words,
24
Nsx, yd
2
fysx, yd 5
f x y 2 x 3 1 gs ydg 5 x 2 1 g9 s yd 5 x 2 2 2y.
y
20
C = 100
16
g9 s yd 5 22y
12
8
C = 10
C=1
12
Figure 15.1
Thus, g9s yd 5 22y, and it follows that gs yd 5 2y 2 1 C1. Therefore,

x
12
f sx, yd 5 x 2 y 2 x 3 2 y 2 1 C1
and the general solution is x 2 y 2 x 3 2 y 2 5 C. Figure 15.1 shows the solution curves
that correspond to C 5 1, 10, 100, and 1000.
SECTION 15.1
EXAMPLE 3
TECHNOLOGY You can use a
graphing utility to graph a particular
solution that satisfies the initial condition of a differential equation. In
Example 3, the differential equation
and initial conditions are satisfied
when xy 2 1 x cos x 5 0, which
implies that the particular solution
can be written as x 5 0 or
y 5 !2cos x . On a graphing
calculator screen, the solution would
be represented by Figure 15.2 together
with the y-axis.
1095
Solving an Exact Differential Equation
Find the particular solution of
scos x 2 x sin x 1 y 2d dx 1 2xy dy 5 0

that satisfies the initial condition y 5 1 when x 5 p.
Solution
The differential equation is exact because

M
y
N
x
fcos x 2 x sin x 1 y 2g 5 2y 5 f2xyg.

y
x
Because N sx, yd is simpler than Msx, yd, it is better to begin by integrating Nsx, yd.
f sx, yd 5
12.57
Nsx, yd dy 5
2xy dy 5 xy 2 1 gsxd
Msx, yd
12.57
fxsx, yd 5
4
f xy 2 1 gsxdg 5 y 2 1 g9 sxd 5 cos x 2 x sin x 1 y 2

x
g9 sxd 5 cos x 2 x sin x
Figure 15.2
Thus, g9 sxd 5 cos x 2 x sin x and

gsxd 5
scos x 2 x sin xd dx
5 x cos x 1 C1
which implies that f sx, yd 5 xy 2 1 x cos x 1 C1 , and the general solution is
xy 2 1 x cos x 5 C.
General solution
Applying the given initial condition produces

y
p s1d 2 1 p cos p 5 C
4
2
which implies that C 5 0. Hence, the particular solution is
( , 1)
x
2
4
Figure 15.3
xy 2 1 x cos x 5 0.
Particular solution
The graph of the particular solution is shown in Figure 15.3. Notice that the graph
consists of two parts: the ovals are given by y 2 1 cos x 5 0, and the y-axis is given
by x 5 0.
In Example 3, note that if z 5 f sx, yd 5 xy 2 1 x cos x, the total differential of z
is given by
dz 5 fxsx, yd dx 1 fysx, yd dy
5 scos x 2 x sin x 1 y 2d dx 1 2xy dy
5 M sx, yd dx 1 N sx, yd dy.
In other words, M dx 1 N dy 5 0 is called an exact differential equation because
M dx 1 N dy is exactly the differential of f sx, yd.
1096
CHAPTER 15
Integrating Factors
If the differential equation Msx, yd dx 1 Nsx, yd dy 5 0 is not exact, it may be possible to make it exact by multiplying by an appropriate factor usx, yd, which is called an
integrating factor for the differential equation.
EXAMPLE 4
Multiplying by an Integrating Factor
a. If the differential equation

2y dx 1 x dy 5 0
Not an exact equation
is multiplied by the integrating factor usx, yd 5 x, the resulting equation

2xy dx 1 x 2 dy 5 0
Exact equation
is exactthe left side is the total differential of x 2 y.

b. If the equation
y dx 2 x dy 5 0
Not an exact equation
is multiplied by the integrating factor usx, yd 5 1yy 2, the resulting equation

1
x
dx 2 2 dy 5 0
y
y
Exact equation
is exactthe left side is the total differential of xyy.

Finding an integrating factor can be difficult. However, there are two classes of
differential equations whose integrating factors can be found routinelynamely,
those that possess integrating factors that are functions of either x alone or y alone.
The following theorem, which we present without proof, outlines a procedure for
finding these two special categories of integrating factors.
THEOREM 15.2
Integrating Factors
Consider the differential equation Msx, yd dx 1 Nsx, yd dy 5 0.

1. If
1
fM sx, yd 2 Nxsx, ydg 5 hsxd
Nsx, yd y
is a function of x alone, then eehsxd dx is an integrating factor.
2. If
1
fN sx, yd 2 Mysx, ydg 5 ks yd
Msx, yd x
is a function of y alone, then eeks yd dy is an integrating factor.
STUDY TIP If either hsxd or ks yd is constant, Theorem 15.2 still applies. As an aid to
remembering these formulas, note that the subtracted partial derivative identifies both the
denominator and the variable for the integrating factor.
SECTION 15.1
EXAMPLE 5
1097
Finding an Integrating Factor
Solve the differential equation s y 2 2 xd dx 1 2y dy 5 0.

The given equation is not exact because Mysx, yd 5 2y and Nxsx, yd 5 0.
However, because
Solution
Mysx, yd 2 Nxsx, yd 2y 2 0
5 1 5 hsxd
5
Nsx, yd
2y
it follows that eehsxd dx 5 ee dx 5 e x is an integrating factor. Multiplying the given
differential equation by e x produces the exact differential equation
s y 2e x 2 xe xd dx 1 2ye x dy 5 0
whose solution is obtained as follows.
f sx, yd 5
Nsx, yd dy 5
2ye x dy 5 y 2e x 1 gsxd
Msx, yd
fxsx, yd 5 y 2e x 1 g9 sxd 5 y 2e x 2 xe x
g9 sxd 5 2xe x
Therefore, g9 sxd 5 2xe x and gsxd 5 2xe x 1 e x 1 C1, which implies that
f sx, yd 5 y 2e x 2 xe x 1 e x 1 C1.
The general solution is y 2e x 2 xe x 1 e x 5 C, or y 2 2 x 1 1 5 Ce2x.
In the next example, we show how a differential equation can help in sketching a
force field given by Fsx, yd 5 Msx, yd i 1 Nsx, yd j.
EXAMPLE 6
Force field:
2y
F (x, y)
x2
y2
y2
x2
x
y2
Sketch the force field given by

j
Fsx, yd 5
Family of tangent curves to F:.
y2
Ce
Solution
x
3
2
3
Figure 15.4
At the point sx, yd in the plane, the vector Fsx, yd has a slope of
dy 2 s y 2 2 xdy!x 2 1 y 2 2 s y 2 2 xd
5
5
dx
2yy!x 2 1 y 2
2y
2y
y2 2 x
i2
j
2
!x 1 y
!x 2 1 y 2
2
by finding and sketching the family of curves tangent to F.
An Application to Force Fields
which, in differential form, is

2y dy 5 2 s y 2 2 xd dx
s y 2 2 xd dx 1 2y dy 5 0.
From Example 5, we know that the general solution of this differential equation is
y 2 2 x 1 1 5 Ce2x, or y 2 5 x 2 1 1 Ce2x. Figure 15.4 shows several representative curves from this family. Note that the force vector at sx, yd is tangent to the curve
passing through sx, yd.
1098
CHAPTER 15
LAB SERIES
Lab 20
E X E R C I S E S F O R S E C T I O N 15 .1
In Exercises 110, determine whether the differential equation
is exact. If it is, find the general solution.
1. s2x 2 3yd dx 1 s2y 2 3xd dy 5 0
17. y dx 2 sx 1 6y 2d dy 5 0
2. ye x dx 1 e x dy 5 0
18. s2x 3 1 yd dx 2 x dy 5 0
3. s3y 2 1 10xy 2d dx 1 s6xy 2 2 1 10x 2yd dy 5 0
19. s5x 2 2 yd dx 1 x dy 5 0
4. 2 coss2x 2 yd dx 2 coss2x 2 yd dy 5 0
20. s5x 2 2 y 2d dx 1 2y dy 5 0
5. s4x 3 2 6xy 2d dx 1 s4y 3 2 6xyd dy 5 0
21. sx 1 yd dx 1 tan x dy 5 0
6. 2y 2e xy dx 1 2xye xy dy 5 0
2
7.
9.
10.
22. s2x 2 y 2 1d dx 1 x 3 dy 5 0
1
sx dy 2 y dxd 5 0
x2 1 y2
8. e2sx
23. y 2 dx 1 sxy 2 1d dy 5 0
24. sx 2 1 2x 1 yd dx 1 2 dy 5 0
sx dx 1 y dyd 5 0
1y 2d
25. 2y dx 1 s x 2 sin!y d dy 5 0
1
s y 2 dx 1 x 2 dyd 5 0
sx 2 yd2
ey
26. s22y 3 1 1d dx 1 s3xy 2 1 x 3d dy 5 0
cos xy f ydx 1 sx 1 tan xyd dyg 5 0
In Exercises 11 and 12, (a) sketch an approximate solution of

the differential equation satisfying the initial condition by hand
on the direction field, (b) find the particular solution that satisfies the initial condition, and (c) use a graphing utility to graph
the particular solution. Compare the graph with the handdrawn graph of part (a).
Initial Condition
Differential Equation
11. s2x tan y 1 5d dx 1 sx 2 sec 2 yd dy 5 0
ys 2 d 5 py4
1
sx dx 1 y dyd 5 0
12.
!x 2 1 y 2
ys4d 5 3
2
x
Initial Condition
y
dx 1 flnsx 2 1d 1 2yg dy 5 0
x21
ys2d 5 4
ssin 3y dx 1 cos 3y dyd 5 0

16. sx 2 1 y 2d dx 1 2xy dy 5 0
15.
usx, yd 5 x 2 y
29. s2y 5 1 x 2 yd dx 1 s2xy 4 2 2x 3d dy 5 0
usx, yd 5 x22 y23
30. 2y 3 dx 1 sxy 2 2 x 2d dy 5 0
(a)
1
x2
(b)
1
y2
(c)
1
xy
1
x2 1 y2
(d)
32. Show that the differential equation

is exact only if a 5 b. If a b, show that x m y n is an integrating factor, where
In Exercises 1316, find the particular solution that satisfies the

initial condition.
e 3x
28. s3y 2 1 5x 2 yd dx 1 s3xy 1 2x 3d dy 5 0
saxy 2 1 byd dx 1 sbx 2y 1 axd dy 5 0
Figure for 12
1
sx dx 1 y dyd 5 0
14. 2
x 1 y2
usx, yd 5 xy 2
y dx 2 x dy 5 0.
2
2
27. s4x 2 y 1 2y 2d dx 1 s3x 3 1 4xyd dy 5 0
31. Show that each of the following is an integrating factor for the
differential equation
In Exercises 2730, use the integrating factor to find the

general solution of the differential equation.
usx, yd 5 x22 y22
Figure for 11
13.
In Exercises 1726, find the integrating factor that is a function

of x or y alone and use it to find the general solution of the
differential equation.
ys0d 5 4
ys0d 5 p
ys3d 5 1
m52
2b 1 a
,
a1b
n52
2a 1 b
.
a1b
In Exercises 3336, use a graphing utility to graph the family of

tangent curves to the given force field.
33. Fsx, yd 5
34. Fsx, yd 5
y
!x 2 1 y 2
x
!x 2 1 y 2
i2
i2
x
!x 2 1 y 2
y
!x 2 1 y 2
35. Fsx, yd 5 4x 2 y i 2 2xy 2 1

36. Fsx, yd 5 s1 1
d i 2 2xy j
x2
x
j
y2
j
j
SECTION 15.1
In Exercises 37 and 38, find an equation for the curve with the
specified slope passing through the given point.
Slope
Point
37.
y2x
dy
5
dx 3y 2 x
s2, 1d
38.
22xy
dy
5
dx x 2 1 y 2
s0, 2d
C9 sxd
x dy
marginal cost
5
5
.
average cost
Csxdyx y dx
42. Programming Write a program for a graphing utility or

computer that will perform the calculations of Eulers Method
for a specified differential equation, interval, Dx, and initial
condition. The output should be a graph of the discrete points
approximating the solution.
Exercise 42 to approximate the solution of the differential equation over the indicated interval with the specified value of D x
and the initial condition, (b) solve the differential equation
analytically, and (c) use a graphing utility to graph the particular solution and compare the result with the graph of part (a).
Differential
Equation
Find the cost function if the elasticity function is

Esxd 5
1099
Eulers Method In Exercises 4346, (a) use the program of
39. Cost If y 5 Csxd represents the cost of producing x units in a

manufacturing process, the elasticity of cost is defined as
Esxd 5
20x 2 y
2y 2 10x
where Cs100d 5 500 and x 100.

40. Eulers Method Consider the differential equation
y9 5 Fsx, yd with the initial condition ysx0 d 5 y0. At any point
sxk , yk d in the domain of F, Fsxk , yk d yields the slope of the solution at that point. Eulers Method gives a discrete set of estimates of the y values of a solution of the differential equation
using the iterative formula
Interval
Dx
Initial
Condition
3
43. y9 5 x !
y
f1, 2g
0.01
ys1d 5 1
p
44. y9 5 s y 2 1 1d
4
f21, 1g
0.1
ys21d 5 21
45. y9 5
2xy
x2 1 y2
f2, 4g
0.05
ys2d 5 1
46. y9 5
6x 1 y 2
ys3y 2 2xd
f0, 5g
0.2
ys0d 5 1
47. Eulers Method Repeat Exercise 45 for Dx 5 1 and

discuss how the accuracy of the result changes.
48. Eulers Method Repeat Exercise 46 for Dx 5 0.5 and
discuss how the accuracy of the result changes.
yk11 5 yk 1 Fsxk , yk d Dx
where Dx 5 xk11 2 xk.
(a) Write a short paragraph describing the general idea of how
Eulers Method works.
(b) How will decreasing the magnitude of Dx affect the accuracy of Eulers Method?
41. Eulers Method Use Eulers Method (see Exercise 40) to
approximate ys1d for the values of Dx given in the table if
y9 5 x 1 !y and ys0d 5 2. (Note that the number of iterations
increases as Dx decreases.) Sketch a graph of the approximate
solution on the direction field in the figure.
Dx
0.50
0.25
0.10
Estimate of ys1d
The value of ys1d, accurate to three decimal places, is 4.213.
y
5
4
3
2
1
4 3 2 1
x
1
True or False? In Exercises 4952, determine whether the

statement is true or false. If it is false, explain why or give an
example that shows it is false.
49. The differential equation 2xy dx 1 s y 2 2 x 2d dy 5 0 is exact.
50. If M dx 1 N dy 5 0 is exact, then xM dx 1 xN dy 5 0 is also
exact.
51. If M dx 1 N dy 5 0 is exact, then f f sxd 1 M g dx 1 f gs yd 1
N g dy 5 0 is also exact.
52. The differential equation f sxd dx 1 gs yd dy 5 0 is exact.
1100
CHAPTER 15
SECTION
15.2
First-Order Linear Differential Equations

First-Order Linear Differential Equations Bernoulli Equations Applications

In this section, you will see how integrating factors help to solve a very important
class of first-order differential equationsfirst-order linear differential equations.
Definition of First-Order Linear Differential Equation

A first-order linear differential equation is an equation of the form
dy
1 Psxdy 5 Qsxd
dx
where P and Q are continuous functions of x. This first-order linear differential
equation is said to be in standard form.
To solve a first-order linear differential equation, you can use an integrating

factor usxd, which converts the left side into the derivative of the product usxdy. That
is, you need a factor usxd such that
dy
d fusxdyg
1 usxdPsxdy 5
dx
dx
usxdy9 1 usxdPsxdy 5 usxdy9 1 yu9sxd
usxdPsxdy 5 yu9sxd
usxd
Psxd 5
ln usxd 5
u9sxd
usxd
Psxd dx 1 C1
usxd 5 CeePsxd dx.

ANNA JOHNSON PELL WHEELER (18831966)
Anna Johnson Pell Wheeler was awarded a
masters degree from the University of Iowa
for her thesis The Extension of Galois Theory
to Linear Differential Equations in 1904.
Influenced by David Hilbert, she worked on
integral equations while studying infinite linear
spaces.
Because you dont need the most general integrating factor, let C 5 1. Multiplying the
original equation y9 1 Psxdy 5 Qsxd by usxd 5 eePsxddx produces
y9eePsxd dx 1 yPsxdeePsxd dx 5 QsxdeePsxd dx
d
yeePsxd dx 5 QsxdeePsxd dx.
dx
The general solution is given by
yeePsxd dx 5
QsxdeePsxd dx dx 1 C.
SECTION 15.2
THEOREM 15.3
1101
Solution of a First-Order Linear Differential Equation
An integrating factor for the first-order linear differential equation

y9 1 Psxdy 5 Qsxd
is usxd 5 eePsxd dx. The solution of the differential equation is
yeePsxd dx5
QsxdeePsxd dx dx 1 C.
STUDY TIP Rather than memorizing this formula, just remember that multiplication by the
integrating factor eePsxd dx converts the left side of the differential equation into the derivative
of the product yeePsxd dx.
EXAMPLE 1
Solving a First-Order Linear Differential Equation
Find the general solution of

xy9 2 2y 5 x2.
Solution
The standard form of the given equation is
y9 1 Psxdy 5 Qsxd
2
y9 2
y 5 x.
x
12
Standard form
Thus, Psxd 5 22yx, and you have
C=0
x
1
2
Figure 15.5
Integrating factor
Therefore, multiplying both sides of the standard form by 1yx2 yields

C=4
C=3
C=2
C=1
2
dx 5 2ln x 2
x
1
2
eePsxd dx 5 e2ln x 5 2.
x
Psxd dx 5 2
C = 1
C = 2
y9 2y 1
2 3 5
x2
x
x
d y
1
5
dx x2
x
y
1
5
dx
x2
x
3 4
y
5 ln x 1 C
x2
y 5 x2sln x 1 Cd.
||
||
General solution
Several solution curves sfor C 5 22, 21, 0, 1, 2, 3, and 4d are shown in Figure 15.5.
1102
CHAPTER 15
EXAMPLE 2
Solving a First-Order Linear Differential Equation

y9 2 y tan t 5 1,
p
p
< t < .
2
2
The equation is already in the standard form y9 1 Pstdy 5 Qstd. Thus,

Pstd 5 2tan t, and
Solution
Pstd dt 5 2
tan t dt 5 ln cos t
which implies that the integrating factor is

eePstd dt 5 e ln |cos t|
5 cos t .
C=2
C=1
C=0
C = 1
C = 2
2
Integrating factor
A quick check shows that cos t is also an integrating factor. Thus, multiplying
y9 2 y tan t 5 1 by cos t produces
d
f y cos tg 5 cos t
dt
y cos t 5
cos t dt
y cos t 5 sin t 1 C
y 5 tan t 1 C sec t.
Figure 15.6
General solution
Several solution curves are shown in Figure 15.6.
Bernoulli Equations
A well-known nonlinear equation that reduces to a linear one with an appropriate
substitution is the Bernoulli equation, named after James Bernoulli (16541705).
y9 1 Psxdy 5 Qsxd y n
Bernoulli equation
This equation is linear if n 5 0, and has separable variables if n 5 1. Thus, in the

following development, assume that n 0 and n 1. Begin by multiplying by y2n
and s1 2 nd to obtain
y2n y9 1 Psxd y12n 5 Qsxd
s1 2 nd y2n y9 1 s1 2 ndPsxd y12n 5 s1 2 ndQsxd
d 12n
f y g 1 s1 2 ndPsxd y12n 5 s1 2 ndQsxd
dx
which is a linear equation in the variable y12n. Letting z 5 y12n produces the linear
equation
dz
1 s1 2 ndPsxdz 5 s1 2 ndQsxd.
dx
Finally, by Theorem 15.3, the general solution of the Bernoulli equation is
y12nees12ndPsxd dx 5
s1 2 ndQsxdees12ndPsxd dx dx 1 C.
SECTION 15.2
EXAMPLE 3
1103
Solving a Bernoulli Equation
Find the general solution of y9 1 xy 5 xe2x y23.

2
For this Bernoulli equation, let n 5 23, and use the substitution
Solution
z 5 y4
z9 5 4y 3y9.
Let z 5 y12n 5 y12 s23d.

Differentiate.
Multiplying the original equation by 4y3 produces

y9 1 xy 5 xe2x y23
2
4y3y9 1 4xy4 5 4xe2x
2
z9 1 4xz 5 4xe2x .
2
Original equation
Multiply both sides by 4y3.
Linear equation: z9 1 Psxdz 5 Qsxd
This equation is linear in z. Using Psxd 5 4x produces
Psxd dx 5
4x dx 5 2x 2
2
which implies that e2x is an integrating factor. Multiplying the linear equation by this
factor produces
z9 1 4xz 5 4xe2x
2
2
1 4xze2x 5 4xe x
d
2
2
fze2x g 5 4xe x
dx
z9e2x
ze2x 5
2
Linear equation
Exact equation
Write left side as total differential.
2
4xe x dx
ze2x 5 2e x 1 C
2
2
z 5 2e2x 1 Ce22x .
2
Integrate both sides.
Divide both sides by e2x .
Finally, substituting z 5 y4, the general solution is

y4 5 2e2x 1 Ce22x .
2
General solution
So far you have studied several types of first-order differential equations. Of

these, the separable variables case is usually the simplest, and solution by an integrating factor is usually a last resort.
Summary of First-Order Differential Equations
1.
2.
3.
4.
5.
6.
Method
Form of Equation
Separable variables:
Homogeneous:
Exact:
Integrating factor:
Linear:
Bernoulli equation:
Msxd dx 1 Ns yd dy 5 0
Msx, yd dx 1 Nsx, yd dy 5 0, where M and N are nth-degree homogeneous
Msx, yd dx 1 Nsx, yd dy 5 0, where Myy 5 Nyx
usx, ydMsx, yd dx 1 usx, ydNsx, yd dy 5 0 is exact
y9 1 Psxdy 5 Qsxd
y9 1 Psxd y 5 Qsxd yn
1104
CHAPTER 15
Applications
One type of problem that can be described in terms of a differential equation involves
chemical mixtures, as illustrated in the next example.
EXAMPLE 4
A Mixture Problem
4 gal/min
5 gal/min
A tank contains 50 gallons of a solution composed of 90% water and 10% alcohol.
A second solution containing 50% water and 50% alcohol is added to the tank at
the rate of 4 gallons per minute. As the second solution is being added, the tank
is being drained at the rate of 5 gallons per minute, as shown in Figure 15.7. Assuming
the solution in the tank is stirred constantly, how much alcohol is in the tank after
10 minutes?
Solution
Figure 15.7
Let y be the number of gallons of alcohol in the tank at any time t. You know
that y 5 5 when t 5 0. Because the number of gallons of solution in the tank at any
time is 50 2 t, and the tank loses 5 gallons of solution per minute, it must lose
150 52 t2y
gallons of alcohol per minute. Furthermore, because the tank is gaining 2 gallons of
alcohol per minute, the rate of change of alcohol in the tank is given by
dy
5
522
y
dt
50 2 t
dy
5
1
y 5 2.
dt
50 2 t
To solve this linear equation, let Pstd 5 5ys50 2 td and obtain
Pstd dt 1
5
dt 5 25 ln 50 2 t .
50 2 t
Because t < 50, you can drop the absolute value signs and conclude that
eePstd dt 5 e25 lns502td 5
1
s50 2 td5.
Thus, the general solution is

y
5
s50 2 td5
2
1
dt 5
1C
5
s50 2 td
2s50 2 td4
50 2 t
y5
1 Cs50 2 td5.
2
Because y 5 5 when t 5 0, you have

55
50
1 Cs50d5
2
20
5C
505
which means that the particular solution is

y5
50 2 t
50 2 t 5
2 20
.
2
50
Finally, when t 5 10, the amount of alcohol in the tank is

y5
50 2 10
50 2 10
2 20
2
50
5 13.45 gal
which represents a solution containing 33.6% alcohol.
SECTION 15.2
1105
In most falling-body problems discussed so far in the text, we have neglected air
resistance. The next example includes this factor. In the example, the air resistance on
the falling object is assumed to be proportional to its velocity v. If g is the gravitational constant, the downward force F on a falling object of mass m is given by the
difference mg 2 kv. But by Newtons Second Law of Motion, you know that
F 5 ma 5 msdvydtd, which yields the following differential equation.
m
dv
k
1 v5g
dt
m
dv
5 mg 2 kv
dt
EXAMPLE 5
A Falling Object with Air Resistance
An object of mass m is dropped from a hovering helicopter. Find its velocity as a

function of time t, assuming that the air resistance is proportional to the velocity of
the object.
Solution
The velocity v satisfies the equation
dv kv
1
5g
dt
m
where g is the gravitational constant and k is the constant of proportionality. Letting
b 5 kym, you can separate variables to obtain
dv 5 sg 2 bvd dt
dv
5
g 2 bv
dt
1
2 ln g 2 bv 5 t 1 C1
b
ln g 2 bv 5 2bt 2 bC1
|
|
|
|
g 2 bv 5 Ce2bt.
Because the object was dropped, v 5 0 when t 5 0; thus g 5 C, and it follows that
2bv 5 2g 1 ge2bt
v5
g 2 ge2bt mg
5
s1 2 e2ktymd.
b
k
NOTE Notice in Example 5 that the velocity approaches a limit of mgyk as a result of the
air resistance. For falling-body problems in which air resistance is neglected, the velocity
increases without bound.
E
S
A simple electrical circuit consists of electric current I (in amperes), a resistance

R (in ohms), an inductance L (in henrys), and a constant electromotive force E (in
volts), as shown in Figure 15.8. According to Kirchhoffs Second Law, if the switch
S is closed when t 5 0, the applied electromotive force (voltage) is equal to the sum
of the voltage drops in the rest of the circuit. This in turn means that the current I
satisfies the differential equation
Figure 15.8
dI
1 RI 5 E.
dt
1106
CHAPTER 15
EXAMPLE 6
An Electric Circuit Problem
Find the current I as a function of time t (in seconds), given that I satisfies the
differential equation LsdIydtd 1 RI 5 sin 2t, where R and L are nonzero constants.
Solution
In standard form, the given linear equation is
dI R
1
1 I 5 sin 2t.
dt
L
L
Let Pstd 5 RyL, so that eePstd dt 5 esRyLdt, and, by Theorem 15.3,
IesRyLdt 5
5
1
L
esRyLdt sin 2t dt
1
esRyLdtsR sin 2t 2 2L cos 2td 1 C.
4L2 1 R2
Thus, the general solution is

I 5 e2sRyLdt
I5
3 4L
1
esRyLdtsR sin 2t 2 2L cos 2td 1 C
1 R2
1
sR sin 2t 2 2L cos 2td 1 Ce2sRyLdt.
4L2 1 R2
E X E R C I S E S F O R S E C T I O N 15 . 2
True or False? In Exercises 1 and 2, determine whether the
1. y9 1 x!y 5 x2 is a first-order linear differential equation.
2. y9 1 xy 5 e x y is a first-order linear differential equation.
In Exercises 3 and 4, (a) sketch an approximate solution of the
differential equation satisfying the initial condition by hand on
the direction field, (b) find the particular solution that satisfies
the initial condition, and (c) use a graphing utility to graph the
particular solution. Compare the graph with the hand-drawn
graph of part (a).
In Exercises 512, solve the first-order linear differential

equation.
12
2
dy
6.
1
y 5 3x 1 1
dx 1 x 2
5.
dy
1
1
y 5 3x 1 4
dx
x
7. y9 2 y 5 cos x
8. y9 1 2xy 5 2x
9. s3y 1 sin 2xd dx 2 dy 5 0
10. s y 2 1dsin x dx 2 dy 5 0
11. sx 2 1dy9 1 y 5 x2 2 1
Initial Condition
12. y9 1 5y 5 e5x
dy
5 ex 2 y
dx
s0, 1d
4. y9 1 2y 5 sin x
s0, 4d
In Exercises 1318, find the particular solution of the differential equation that satisfies the boundary condition.
3.
13. y9 cos2 x 1 y 2 1 5 0
4
2
2
4
2
2
Figure for 3
2
2
4
Figure for 4
ys0d 5 5
14. x y9 1 2y 5 e
ys1d 5 e
15. y9 1 y tan x 5 sec x 1 cos x
ys0d 5 1
16. y9 1 y sec x 5 sec x
ys0d 5 4
1
y50
17. y9 1
x
12
ys2d 5 2
18. y9 1 s2x 2 1dy 5 0
ys1d 5 2
1yx 2
Boundary Condition
SECTION 15.2
In Exercises 1924, solve the Bernoulli differential equation.

19. y9 1 3x2 y 5 x2 y3
21. y9 1
20. y9 1 2xy 5 xy2
11x 2y 5 xy
22. y9 1
3
23. y9 2 y 5 x3!
y
11x 2y 5 x!y
24. yy9 2 2y2 5 ex
In Exercises 2528, (a) use a graphing utility to graph the

direction field for the differential equation, (b) find the
particular solutions of the differential equation passing through
the specified points, and (c) use a graphing utility to graph the
particular solutions on the direction field.
1107
33. Population Growth When predicting population growth,

demographers must consider birth and death rates as well as the
net change caused by the difference between the rates of immigration and emigration. Let P be the population at time t and let
N be the net increase per unit time resulting from the difference
between immigration and emigration. Thus, the rate of growth
of the population is given by
dP
5 kP 1 N,
dt
N is constant.
Solve this differential equation to find P as a function of time if

at time t 5 0 the size of the population is P0.
34. Investment Growth A large corporation starts at time t 5 0
to continuously invest part of its receipts at a rate of P dollars
per year in a fund for future corporate expansion. Assume that
the fund earns r percent interest per year compounded continuously. Thus, the rate of growth of the amount A in the fund is
given by
Points
25.
dy 1
2 y 5 x2
dx
x
s22, 4d, s2, 8d
26.
dy
1 2xy 5 x3
dx
s 0, 72 d, s 0, 2 12 d
27.
dy
1 scot xdy 5 x
dx
s1, 1d, s3, 21d
dA
5 rA 1 P
dt
28.
dy
1 2xy 5 xy2
dx
s0, 3d, s0, 1d
where A 5 0 when t 5 0. Solve this differential equation for A

as a function of t.
Electrical Circuits In Exercises 2932, use the differential

equation for electrical circuits given by
Investment Growth In Exercises 35 and 36, use the result of
dI
L 1 RI 5 E.
dt
35. Find A for the following.

(a) P 5 $100,000, r 5 6%, and t 5 5 years
In this equation, I is the current, R is the resistance, L is the

inductance, and E is the electromotive force (voltage).
29. Solve the differential equation given a constant voltage E0.
30. Use the result of Exercise 29 to find the equation for the current
if Is0d 5 0, E0 5 110 volts, R 5 550 ohms, and L 5 4 henrys.
When does the current reach 90% of its limiting value?
31. Solve the differential equation given a periodic electromotive
force E0 sin vt.
32. Verify that the solution of Exercise 31 can be written in the
form
I 5 ce2sRyLdt 1
Exercise 34.
E0
!R2 1 v2L2
sinsvt 1 fd
where f, the phase angle, is given by arctans2 vLyRd. (Note

that the exponential term approaches 0 as t `. This implies
that the current approaches a periodic function.)
(b) P 5 $250,000, r 5 5%, and t 5 10 years

36. Find t if the corporation needs $800,000 and it can invest
$75,000 per year in a fund earning 8% interest compounded
continuously.
37. Intravenous Feeding Glucose is added intravenously to
the bloodstream at the rate of q units per minute, and the body
removes glucose from the bloodstream at a rate proportional to
the amount present. Assume Qstd is the amount of glucose in
the bloodstream at time t.
(a) Determine the differential equation describing the rate of
change with respect to time of glucose in the bloodstream.
(b) Solve the differential equation from part (a), letting
Q 5 Q0 when t 5 0.
(c) Find the limit of Qstd as t
`.
38. Learning Curve The management at a certain factory has

found that the maximum number of units a worker can produce
in a day is 30. The rate of increase in the number of units N
produced with respect to time t in days by a new employee is
proportional to 30 2 N.
(a) Determine the differential equation describing the rate of
change of performance with respect to time.
(b) Solve the differential equation from part (a).
(c) Find the particular solution for a new employee who
produced ten units on the first day at the factory and 19
units on the twentieth day.
1108
CHAPTER 15
Mixture In Exercises 3944, consider a tank that at time

t 5 0 contains v0 gallons of a solution of which, by weight, q0
pounds is soluble concentrate. Another solution containing q1
pounds of the concentrate per gallon is running into the tank at
the rate of r1 gallons per minute. The solution in the tank is kept
well stirred and is withdrawn at the rate of r2 gallons per
minute.
39. If Q is the amount of concentrate in the solution at any time t,
show that
51. s1 1 y2d dx 1 s2x y 1 y 1 2d dy 5 0

52. s1 1 2e2x1yd dx 1 e2x1y dy 5 0
dy
50
53. y cos x 2 cos x 1
dx
54. sx 1 1d
dy
5 ex 2 y
dx
55. sx2 1 cos yd
dy
5 22xy
dx
56. y9 5 2x!1 2 y2
57. s3y2 1 4xyd dx 1 s2xy 1 x2d dy 5 0
r2Q
dQ
1
5 q1r1.
dt
v0 1 sr1 2 r2dt
58. sx 1 yd dx 2 x dy 5 0
40. If Q is the amount of concentrate in the solution at any time t,

write the differential equation for the rate of change of Q with
respect to t if r1 5 r2 5 r.
41. A 200-gallon tank is full of a solution containing 25 pounds of
concentrate. Starting at time t 5 0, distilled water is admitted
to the tank at a rate of 10 gallons per minute, and the
well-stirred solution is withdrawn at the same rate.
59. s2y 2 e xd dx 1 x dy 5 0
60. s y2 1 xyd dx 2 x2 dy 5 0
61. sx2 y4 2 1d dx 1 x3 y3 dy 5 0
62. ydx 1 s3x 1 4yd dy 5 0
63. 3ydx 2 sx2 1 3x 1 y2d dy 5 0
64. x dx 1 s y 1 eydsx2 1 1d dy 5 0
(a) Find the amount of concentrate Q in the solution as a

function of t.
(b) Find the time at which the amount of concentrate in the
tank reaches 15 pounds.
(c) Find the quantity of the concentrate in the solution as
t `.
42. Repeat Exercise 41, assuming that the solution entering the
tank contains 0.05 pound of concentrate per gallon.
43. A 200-gallon tank is half full of distilled water. At time t 5 0,
a solution containing 0.5 pound of concentrate per gallon enters
the tank at the rate of 5 gallons per minute, and the well-stirred
mixture is withdrawn at the rate of 3 gallons per minute.
(a) At what time will the tank be full?
(b) At the time the tank is full, how many pounds of concentrate will it contain?
44. Repeat Exercise 43, assuming that the solution entering the
tank contains 1 pound of concentrate per gallon.
In Exercises 4548, match the differential equation with its
solution.
Solution
45. y9 2 2x 5 0
(a) y 5 Ce x
46. y9 2 2y 5 0
(b) y 5
2 12
47. y9 2 2xy 5 0
(c) y 5 x2 1 C
48. y9 2 2xy 5 x
(d) y 5 Ce2x
1 Ce x
In Exercises 4964, solve the first-order differential equation by

any appropriate method.
49.
dy e2x1y
5 x2y
dx
e
50.
dy
x11
5
dx ys y 1 2d
SECTION PROJECT
Weight Loss A persons weight depends on both the amount

of calories consumed and the energy used. Moreover, the amount
of energy used depends on a persons weightthe average
amount of energy used by a person is 17.5 calories per pound
per day. Thus, the more weight a person loses, the less energy
the person uses (assuming that the person maintains a constant
level of activity). An equation that can be used to model weight
loss is
C
17.5
2
w
1dwdt2 5 3500
3500
where w is the persons weight (in pounds), t is the time in days,
and C is the constant daily calorie consumption.
(a) Find the general solution of the differential equation.
(b) Consider a person who weighs 180 pounds and begins a diet
of 2500 calories per day. How long will it take the person to
lose 10 pounds? How long will it take the person to lose 35
pounds?
(c) Use a graphing utility to graph the solution. What is the
limiting weight of the person?
(d) Repeat parts (b) and (c) for a person who weighs 200
pounds when the diet is started.
SECTION 15.3
SECTION
15.3
Second-Order Homogeneous Linear Equations
1109

Second-Order Linear Differential Equations
Higher-Order Linear Differential Equations Applications
Second-Order Linear Differential Equations

In this section and the following section, we discuss methods for solving higher-order
linear differential equations.
Definition of Linear Differential Equation of Order n

Let g1, g2, . . . , gn and f be functions of x with a common (interval) domain. An
equation of the form
ysnd 1 g sxdysn21d 1 g sxdysn22d 1 . . . 1 g sxdy9 1 g sxdy 5 f sxd
1
n21
is called a linear differential equation of order n. If f sxd 5 0, the equation is

homogeneous; otherwise, it is nonhomogeneous.
NOTE
Notice that this use of the term homogeneous differs from that in Section 5.7.
We discuss homogeneous equations in this section, and leave the nonhomogeneous case for the next section.
The functions y1, y2, . . . , yn are linearly independent if the only solution of the
equation
C1 y1 1 C2 y2 1 . . . 1 Cn yn 5 0
is the trivial one, C1 5 C2 5 . . . 5 Cn 5 0. Otherwise, this set of functions is
linearly dependent.
EXAMPLE 1
Linearly Independent and Dependent Functions
a. The functions y1sxd 5 sin x and y2 5 x are linearly independent because the only
values of C1 and C2 for which
C1 sin x 1 C2x 5 0
for all x are C1 5 0 and C2 5 0.
b. It can be shown that two functions form a linearly dependent set if and only if one
is a constant multiple of the other. For example, y1sxd 5 x and y2sxd 5 3x are
linearly dependent because
C1x 1 C2s3xd 5 0
has the nonzero solutions C1 5 23 and C2 5 1.
1110
CHAPTER 15
The following theorem points out the importance of linear independence in

constructing the general solution of a second-order linear homogeneous differential
equation with constant coefficients.
THEOREM 15.4
Linear Combinations of Solutions
If y1 and y2 are linearly independent solutions of the differential equation

y0 1 ay9 1 by 5 0, then the general solution is
y 5 C1 y1 1 C2 y2
where C1 and C2 are constants.
Proof We prove this theorem in only one direction. If y1 and y2 are solutions, you can
obtain the following system of equations.
y10 sxd 1 ay19 sxd 1 by1sxd 5 0

y20 sxd 1 ay29 sxd 1 by2sxd 5 0
Multiplying the first equation by C1, multiplying the second by C2, and adding the
resulting equations together produces
fC1 y10 sxd 1 C2 y20 sxdg 1 afC1 y19sxd 1 C2 y29sxdg 1 bfC1 y1sxd 1 C2 y2sxdg 5 0
which means that
y 5 C1 y1 1 C2 y2
is a solution, as desired. The proof that all solutions are of this form is best left to a
full course on differential equations.
Theorem 15.4 states that if you can find two linearly independent solutions, you
can obtain the general solution by forming a linear combination of the two solutions.
To find two linearly independent solutions, note that the nature of the equation
y0 1 ay9 1 by 5 0 suggests that it may have solutions of the form y 5 emx. If so, then
y9 5 memx and y0 5 m2emx. Thus, by substitution, y 5 emx is a solution if and only if
y0 1 ay9 1 by 5 0
1 amemx 1 bemx 5 0
emxsm2 1 am 1 bd 5 0.
m2emx
Because emx is never 0, y 5 emx is a solution if and only if

m2 1 am 1 b 5 0.
Characteristic equation
This is the characteristic equation of the differential equation

y0 1 ay9 1 by 5 0.
Note that the characteristic equation can be determined from its differential equation
simply by replacing y0 with m2, y9 with m, and y with 1.
SECTION 15.3
E X P L O R AT I O N
For each differential equation below
find the characteristic equation. Solve
the characteristic equation for m, and
use the values of m to find a general
solution to the differential equation.
Using your results, develop a general
solution to differential equations with
characteristic equations that have
distinct real roots.
(a) y0 2 9y 5 0
(b) y0 2 6y9 1 8y 5 0
EXAMPLE 2
1111
Characteristic Equation with Distinct Real Roots
Solve the differential equation

y0 2 4y 5 0.
Solution
In this case, the characteristic equation is
m2 2 4 5 0
Characteristic equation
so m 5 2. Thus, y1 5 em1x 5 e2x and y2 5 em 2x 5 e22x are particular solutions of

the given differential equation. Furthermore, because these two solutions are linearly
independent, you can apply Theorem 15.4 to conclude that the general solution is
y 5 C1e2x 1 C2e22x.
General solution
The characteristic equation in Example 2 has two distinct real roots. From
algebra, you know that this is only one of three possibilities for quadratic equations.
In general, the quadratic equation m2 1 am 1 b 5 0 has roots
m1 5
2a 1 !a2 2 4b
2
and
m2 5
2a 2 !a2 2 4b
2
which fall into one of three cases.

1. Two distinct real roots, m1 m2
2. Two equal real roots, m1 5 m2
3. Two complex conjugate roots, m1 5 a 1 bi and m2 5 a 2 bi
In terms of the differential equation y0 1 ay9 1 by 5 0, these three cases correspond
to three different types of general solutions.
THEOREM 15.5
Solutions of y 0 1 ay9 1 by 5 0
The solutions of
y0 1 ay9 1 by 5 0
fall into one of the following three cases, depending on the solutions of the
characteristic equation, m2 1 am 1 b 5 0.
1. Distinct Real Roots If m1 m2 are distinct real roots of the characteristic
equation, then the general solution is
y 5 C1em1x 1 C2em2x.
2. Equal Real Roots If m1 5 m2 are equal real roots of the characteristic
equation, then the general solution is
y 5 C1em1x 1 C2xem1x 5 sC1 1 C2xdem1x.
3. Complex Roots If m1 5 a 1 bi and m2 5 a 2 bi are complex roots of the
characteristic equation, then the general solution is
y 5 C1eax cos bx 1 C2eax sin bx.
1112
CHAPTER 15
EXAMPLE 3
Characteristic Equation with Complex Roots
Find the general solution of the differential equation

y0 1 6y9 1 12y 5 0.
Solution
The characteristic equation
m2 1 6m 1 12 5 0
has two complex roots, as follows.
26 !36 2 48
2
26 !212
5
2
5 23 !23
5 23 !3i
m5
Thus, a 5 23 and b 5 !3, and the general solution is

y 5 C1e23x cos!3x 1 C2e23x sin!3x.
NOTE In Example 3, note that although the characteristic equation has two complex roots, the
solution of the differential equation is real.
EXAMPLE 4
Characteristic Equation with Repeated Roots

y0 1 4y9 1 4y 5 0
subject to the initial conditions ys0d 5 2 and y9s0d 5 1.
Solution
m2
The characteristic equation

1 4m 1 4 5 sm 1 2d2 5 0
has two equal roots given by m 5 22. Thus, the general solution is
y 5 C1e22x 1 C2xe22x.
General solution
Now, because y 5 2 when x 5 0, we have

2 5 C1s1d 1 C2s0ds1d 5 C1.
Furthermore, because y9 5 1 when x 5 0, we have
y9 5 22C1e22x 1 C2s22xe22x 1 e22xd
1 5 22s2ds1d 1 C2f22s0ds1d 1 1g
5 5 C2.
Therefore, the solution is
y 5 2e22x 1 5xe22x.
Particular solution
Try checking this solution in the original differential equation.
SECTION 15.3
1113
Higher-Order Linear Differential Equations

For higher-order homogeneous linear differential equations, you can find the general
solution in much the same way as you do for second-order equations. That is, you begin
by determining the n roots of the characteristic equation. Then, based on these n roots,
you form a linearly independent collection of n solutions. The major difference is that
with equations of third or higher order, roots of the characteristic equation may occur
more than twice. When this happens, the linearly independent solutions are formed by
multiplying by increasing powers of x, as demonstrated in Examples 6 and 7.
EXAMPLE 5
Solving a Third-Order Equation
Find the general solution of y999 2 y9 5 0.

Solution
The characteristic equation is
m3 2 m 5 0
msm 2 1dsm 1 1d 5 0
m 5 0, 1, 21.
Because the characteristic equation has three distinct roots, the general solution is
y 5 C1 1 C2e2x 1 C3e x.
EXAMPLE 6
General solution
Find the general solution of y999 1 3y0 1 3y9 1 y 5 0.

Solution
The characteristic equation is
m3 1 3m2 1 3m 1 1 5 0
sm 1 1d3 5 0
m 5 21.
Because the root m 5 21 occurs three times, the general solution is
y 5 C1e2x 1 C2 xe2x 1 C3x2e2x.
EXAMPLE 7
General solution
Solving a Fourth-Order Equation
Find the general solution of ys4d 1 2y0 1 y 5 0.

Solution
The characteristic equation is as follows.
m4 1 2m2 1 1 5 0
sm2 1 1d2 5 0
m 5 i
Because each of the roots m1 5 a 1 bi 5 0 1 i and m2 5 a 2 bi 5 0 2 i occurs
twice, the general solution is
y 5 C1 cos x 1 C2 sin x 1 C3x cos x 1 C4x sin x.
General solution
1114
CHAPTER 15
Applications
l = natural
length
y = displacement
m
A rigid object of mass m attached to the end

of the spring causes a displacement of y.
Figure 15.9
One of the many applications of linear differential equations is describing the motion
of an oscillating spring. According to Hookes Law, a spring that is stretched (or
compressed) y units from its natural length l tends to restore itself to its natural length
by a force F that is proportional to y. That is, Fs yd 5 2ky, where k is the spring
constant and indicates the stiffness of the given spring.
Suppose a rigid object of mass m is attached to the end of a spring and causes a
displacement, as shown in Figure 15.9. Assume that the mass of the spring is
negligible compared with m. If the object is pulled down and released, the resulting
oscillations are a product of two opposing forcesthe spring force Fsyd 5 2ky and
the weight mg of the object. Under such conditions, you can use a differential
equation to find the position y of the object as a function of time t. According to
Newtons Second Law of Motion, the force acting on the weight is F 5 ma, where
a 5 d 2 yydt2 is the acceleration. Assuming that the motion is undampedthat is,
there are no other external forces acting on the objectit follows that msd 2 yydt2d 5
2ky, and you have
1 2
d2y
k
1
y 5 0.
dt2
m
EXAMPLE 8
Undamped motion of a spring
Undamped Motion of a Spring
Suppose a 4-pound weight stretches a spring 8 inches from its natural length. The
weight is pulled down an additional 6 inches and released with an initial upward
velocity of 8 feet per second. Find a formula for the position of the weight as a
function of time t.
By Hookes Law, 4 5 ks 23 d, so k 5 6. Moreover, because the weight w is
4
given by mg, it follows that m 5 wyg 5 32
5 18 . Hence, the resulting differential
equation for this undamped motion is
Solution
d2y
1 48y 5 0.
dt2
Because the characteristic equation m2 1 48 5 0 has complex roots m 5 0 4!3i,
the general solution is
y 5 C1e0 cos 4!3 t 1 C2e0 sin 4!3 t 5 C1 cos 4!3 t 1 C2 sin 4!3 t.
Using the initial conditions, you have
1
1
5 C1s1d 1 C2s0d
C1 5
2
2
y9std 5 24!3 C1 sin 4!3 t 1 4!3 C2 cos 4!3 t
1
2!3
8 5 24!3
s0d 1 4!3 C2s1d
C2 5
.
2
3
12
Consequently, the position at time t is given by

y5
1
2!3
cos 4!3 t 1
sin 4!3 t.
2
3
y s0d 5 12
y9s0d 5 8
SECTION 15.3
1115
Suppose the object in Figure 15.10 undergoes an additional damping or frictional

force that is proportional to its velocity. A case in point would be the damping force
resulting from friction and movement through a fluid. Considering this damping
force, 2psdyydtd, the differential equation for the oscillation is
m
d2y
dy
5 2ky 2 p
dt2
dt
or, in standard linear form,
1 2
d2y
p dy
k
1
1 y 5 0.
dt2
m dt
m
Damped motion of a spring
A damped vibration could be caused by

friction and movement through a liquid.
Figure 15.10
In Exercises 14, verify the solution of the differential equation.
Solution
31. Consider the differential equation y0 1 100y 5 0 and the

solution y 5 C1 cos 10x 1 C2 sin 10x. Find the particular
solution satisfying each of the following initial conditions.
1. y 5 sC1 1 C2xde23x
y0 1 6y9 1 9y 5 0
(a) ys0d 5 2, y9s0d 5 0
2. y 5 C1e2x 1 C2e22x
y0 2 4y 5 0
(b) ys0d 5 0, y9s0d 5 2
3. y 5 C1 cos 2x 1 C2 sin 2x
y0 1 4y 5 0
(c) ys0d 5 21, y9s0d 5 3
4. y 5 e2x sin 3x
y0 1 2y9 1 10y 5 0
In Exercises 530, find the general solution of the linear

5. y0 2 y9 5 0
6. y0 1 2y9 5 0
7. y0 2 y9 2 6y 5 0
8. y0 1 6y9 1 5y 5 0
9. 2y0 1 3y9 2 2y 5 0
10. 16y0 2 16y9 1 3y 5 0
32. Determine C and v such that y 5 C sin!3 t is a particular

solution of the differential equation y0 1 vy 5 0, where
y9s0d 5 25.
In Exercises 3336, find the particular solution of the linear
33. y0 2 y9 2 30y 5 0
34. y0 1 2y9 1 3y 5 0
ys0d 5 1, y9s0d 5 24
ys0d 5 2, y9s0d 5 1
11. y0 1 6y9 1 9y 5 0
12. y0 2 10y9 1 25y 5 0
13. 16y0 2 8y9 1 y 5 0
14. 9y0 2 12y9 1 4y 5 0
15. y0 1 y 5 0
16. y0 1 4y 5 0
17. y0 2 9y 5 0
18. y0 2 2y 5 0
19. y0 2 2y9 1 4y 5 0
20. y0 2 4y9 1 21y 5 0
21. y0 2 3y9 1 y 5 0
22. 3y0 1 4y9 2 y 5 0
23. 9y0 2 12y9 1 11y 5 0
24. 2y0 2 6y9 1 7y 5 0
Think About It In Exercises 37 and 38, give a geometric

argument to explain why the graph cannot be a solution of the
differential equation. It is not necessary to solve the differential
equation.
26. ys4d 2 y0 5 0
37. y0 5 y9
25.
ys4d
2y50
36. y0 1 2y9 1 3y 5 0
35. y0 1 16y 5 0
ys0d 5 0, y9s0d 5 2
ys0d 5 2, y9s0d 5 1
38. y0 5 2 12 y9
27. y999 2 6y0 1 11y9 2 6y 5 0
28. y999 2 y0 2 y9 1 y 5 0
29. y999 2 3y0 1 7y9 2 5y 5 0
30. y999 2 3y0 1 3y9 2 y 5 0
3 2 1
3 2
x
1
1116
CHAPTER 15
Vibrating Spring In Exercises 3944, describe the motion of

a 32-pound weight suspended on a spring. Assume that the
weight stretches the spring 23 foot from its natural position.
39. The weight is pulled 12 foot below the equilibrium position and
released.
40. The weight is raised 32 foot above the equilibrium position and
released.
41. The weight is raised 23 foot above the equilibrium position
1
and started off with a downward velocity of 2 foot per second.
42. The weight is pulled 21 foot below the equilibrium position and started off with an upward velocity of
foot per second.
1
2
released. The motion takes place in a medium that furnishes a
1
damping force of magnitude 8 speed at all times.
1
44. The weight is pulled 2 foot

below the equilibrium position and released. The motion takes
place in a medium that furnishes a damping force of magnitude
1
4 v at all times.
||
Vibrating Spring In Exercises 4548, match the differential

equation with the graph of a particular solution. [The graphs
are labeled (a), (b), (c), and (d).] The correct match can be made
by comparing the frequency of the oscillations or the rate at
which the oscillations are being damped with the appropriate
coefficient in the differential equation.
(a)
(b)
x
1
y0 1 ay9 1 by 5 0
has complex roots given by m1 5 a 1 bi and m2 5 a 2 bi,
show that
y 5 C1ea x cos bx 1 C2eax sin b x
is a solution.
True or False? In Exercises 5154, determine whether the

51. y 5 C1e3x 1 C2e23x is the general solution of y0 2 6y9 1
9 5 0.
52. y 5 sC1 1 C2 xdsin x 1 sC3 1 C4xdcos x is the general solution of ys4d 1 2y0 1 y 5 0.
53. y 5 x is a solution of an y snd 1 an21y sn21d 1 . . . 1 a1 y9 1
a0 y 5 0 if and only if a1 5 a0 5 0.
54. It is possible to choose a and b such that y 5 x 2e x is a solution
of y0 1 ay9 1 by 5 0.
The Wronskian of two differentiable functions f and g, denoted
by W( f, g), is defined as the function given by the determinant
Wx f, gc 5
y
3
50. If the characteristic equation of the differential equation
| |
f
f9
g
.
g9
The functions f and g are linearly independent if there exists at

least one value of x for which Wx f, gc 0. In Exercises 5558,
use the Wronskian to verify the linear independence of the two
functions.
55. y1 5 eax
56. y1 5 eax
y2 5 ebx, a b
(c)
(d)
y
3
y
3
46. y0 1 25y 5 0
48. y0 1 y9 1
37
4
y50
49. If the characteristic equation of the differential equation

y0 1 ay9 1 by 5 0
has two equal real roots given by m 5 r, show that
y 5 C1erx 1 C2 xerx
is a solution.
cos bx, b 0
y2 5 x2
x 2 y0 1 axy9 1 by 5 0,
x > 0
where a and b are constants.
47. y0 1 2y9 1 10y 5 0
y2 5
58. y1 5 x
45. y0 1 9y 5 0
sin bx
eax
y2 5 xeax
59. Eulers differential equation is of the form
x
2
57. y1 5
eax
(a) Show that this equation can be transformed into a secondorder linear equation with constant coefficients by using the
substitution x 5 et.
(b) Solve x 2 y0 1 6xy9 1 6y 5 0.
60. Solve
y0 1 Ay 5 0
where A is constant, subject to the conditions ys0d 5 0 and
yspd 5 0.
SECTION 15.4
SECTION
15.4
Second-Order Nonhomogeneous Linear Equations
1117

Nonhomogeneous Equations Method of Undetermined Coefficients
Variation of Parameters
Nonhomogeneous Equations
In the preceding section, we represented damped oscillations of a spring by the homogeneous second-order linear equation
1 2
d 2y
p dy
k
1
1 y 5 0.
dt 2
m dt
m
Free motion
The Granger Collection
This type of oscillation is called free because it is determined solely by the spring and
gravity and is free of the action of other external forces. If such a system is also
subject to an external periodic force such as a sin bt, caused by vibrations at the opposite end of the spring, the motion is called forced, and it is characterized by the
nonhomogeneous equation
SOPHIE GERMAIN (17761831)
Many of the early contributors to calculus

were interested in forming mathematical
models for vibrating strings and membranes,
oscillating springs, and elasticity. One of these
was the French mathematician Sophie
Germain, who in 1816 was awarded a prize by
the French Academy for a paper entitled
Memoir on the Vibrations of Elastic Plates.
1 2
d 2y
p dy
k
1
1 y 5 a sin bt.
dt 2
m dt
m
Forced motion
In this section, you will study two methods for finding the general solution of a
nonhomogeneous linear differential equation. In both methods, the first step is to find
the general solution of the corresponding homogeneous equation.
y 5 yh
General solution of homogeneous equation
Having done this, you try to find a particular solution of the nonhomogeneous
equation.
y 5 yp
Particular solution of nonhomogeneous equation
By combining these two results, you can conclude that the general solution of the
nonhomogeneous equation is y 5 yh 1 yp, as stated in the following theorem.
THEOREM 15.6
Solution of Nonhomogeneous Linear Equation
Let
y0 1 ay9 1 by 5 Fsxd
be a second-order nonhomogeneous linear differential equation. If yp is a particular solution of this equation and yh is the general solution of the corresponding
homogeneous equation, then
y 5 yh 1 yp
is the general solution of the nonhomogeneous equation.
1118
CHAPTER 15
Method of Undetermined Coefficients

You already know how to find the solution yh of a linear homogeneous differential
equation. The remainder of this section looks at ways to find the particular solution
yp. If Fsxd in
consists of sums or products of xn, emx, cos bx, or sin bx, you can find a particular
solution yp by the method of undetermined coefficients. The gist of this method is to
guess that the solution yp is a generalized form of Fsxd. Here are some examples.
1. If Fsxd 5 3x 2, choose yp 5 Ax 2 1 Bx 1 C.
2. If Fsxd 5 4xex, choose yp 5 Axex 1 Bex.
3. If Fsxd 5 x 1 sin 2x, choose yp 5 sAx 1 Bd 1 C sin 2x 1 D cos 2x.
Then, by substitution, determine the coefficients for the generalized solution.
EXAMPLE 1
Find the general solution of the equation

y0 2 2y9 2 3y 5 2 sin x.
Solution
To find yh, solve the characteristic equation.
2 2m 2 3 5 0
sm 1 1dsm 2 3d 5 0
m 5 21 or
m2
Thus, yh 5 C1
e2x
1 C2
e3x.
m53
Next, let yp be a generalized form of 2 sin x.
yp 5 A cos x 1 B sin x
yp9 5 2 A sin x 1 B cos x
yp0 5 2 A cos x 2 B sin x
Substitution into the original differential equation yields
y0 2 2y9 2 3y 5 2 sin x
2A cos x 2 B sin x 1 2A sin x 2 2B cos x 2 3A cos x 2 3B sin x 5 2 sin x
s24A 2 2Bdcos x 1 s2A 2 4Bdsin x 5 2 sin x.
By equating coefficients of like terms, you obtain
24A 2 2B 5 0 and
with solutions A 5
yp 5
1
5
2A 2 4B 5 2
and B 5 2 25 . Therefore,
1
2
cos x 2 sin x
5
5
and the general solution is

y 5 yh 1 yp
5 C1e2x 1 C2e3x 1
1
2
cos x 2 sin x.
5
5
SECTION 15.4
1119
In Example 1, the form of the homogeneous solution

yh 5 C1e2x 1 C2e3x
has no overlap with the function Fsxd in the equation
However, suppose the given differential equation in Example 1 were of the form
y0 2 2y9 2 3y 5 e2x.
Now, it would make no sense to guess that the particular solution were y 5 Ae2x,
because you know that this solution would yield 0. In such cases, you should alter
your guess by multiplying by the lowest power of x that removes the duplication. For
this particular problem, you would guess
yp 5 Axe2x.
EXAMPLE 2

y0 2 2y9 5 x 1 2ex.
Solution
The characteristic equation m2 2 2m 5 0 has solutions m 5 0 and m 5 2.
Thus,
yh 5 C1 1 C2e 2x.
Because Fsxd 5 x 1 2ex, your first choice for yp would be sA 1 Bxd 1 Cex. However,
because yh already contains a constant term C1, you should multiply the polynomial
part by x and use
yp 5 Ax 1 Bx 2 1 Cex
yp9 5 A 1 2Bx 1 Ce x
yp0 5 2B 1 Cex.
Substitution into the differential equation produces
y0 2 2y9 5 x 1 2ex
s2B 1 Cexd 2 2sA 1 2Bx 1 Cexd 5 x 1 2ex
s2B 2 2Ad 2 4Bx 2 Cex 5 x 1 2ex.
Equating coefficients of like terms yields the system
2B 2 2A 5 0,
24B 5 1,
with solutions A 5 B 5
2 14
2C 5 2
and C 5 22. Therefore,
1
1
yp 5 2 x 2 x 2 2 2ex
4
4
y 5 yh 1 yp
5 C1 1 C2e 2x 2
1
1
x 2 x 2 2 2e x.
4
4
1120
CHAPTER 15
In Example 2, the polynomial part of the initial guess
sA 1 Bxd 1 Cex
for yp overlapped by a constant term with yh 5 C1 1 C2e2x, and it was necessary to
multiply the polynomial part by a power of x that removed the overlap. The next
example further illustrates some choices for yp that eliminate overlap with yh.
Remember that in all cases the first guess for yp should match the types of functions
occurring in Fsxd.
EXAMPLE 3
Choosing the Form of the Particular Solution
Determine a suitable choice for yp for each of the following.

a. y0 5 x 2
b. y0 1 2y9 1 10y 5 4 sin 3x
c. y0 2 4y9 1 4 5 e 2x
yh
C1 1 C2x
C1e2x cos 3x 1 C2e2x sin 3x
C1e 2x 1 C2 xe 2x
Solution
a. Because Fsxd 5 x 2, the normal choice for yp would be A 1 Bx 1 Cx 2. However,

because yh 5 C1 1 C2x already contains a linear term, you should multiply by x 2
to obtain
yp 5 Ax 2 1 Bx3 1 Cx 4.
b. Because Fsxd 5 4 sin 3x and each term in yh contains a factor of e2x, you can
simply let
yp 5 A cos 3x 1 B sin 3x.
c. Because Fsxd 5 e 2x, the normal choice for yp would be Ae 2x. However, because
yh 5 C1e 2x 1 C2xe 2x already contains an xe 2x term, you should multiply by x 2 to
get
yp 5 Ax 2e 2x.
EXAMPLE 4

y999 1 3y0 1 3y9 1 y 5 x.
Solution
From Example 6 in the preceding section, you know that the homogeneous
solution is
yh 5 C1e2x 1 C2 xe2x 1 C3 x 2e2x.
Because Fsxd 5 x, let yp 5 A 1 Bx and obtain yp9 5 B and yp0 5 0. Thus, by substitution, you have
s0d 1 3s0d 1 3sBd 1 sA 1 Bxd 5 s3B 1 Ad 1 Bx 5 x.

Thus, B 5 1 and A 5 23, which implies that yp 5 23 1 x. Therefore, the general
solution is
y 5 yh 1 yp
5 C1e2x 1 C2xe2x 1 C3x 2e2x 2 3 1 x.
SECTION 15.4
1121
The method of undetermined coefficients works well if Fsxd is made up of polynomials or functions whose successive derivatives have a cyclic pattern. For functions such
as 1yx and tan x, which do not have such characteristics, it is better to use a more
general method called variation of parameters. In this method, you assume that yp
has the same form as yh, except that the constants in yh are replaced by variables.
To find the general solution to the equation y0 1 ay9 1 by 5 Fsxd, use the
following steps.
1. Find yh 5 C1y1 1 C2y2.
2. Replace the constants by variables to form yp 5 u1y1 1 u2 y2.
3. Solve the following system for u19 and u29.
u19 y1 1 u 29 y2 5 0
u19 y19 1 u 29 y29 5 Fsxd
4. Integrate to find u1 and u2. The general solution is y 5 yh 1 yp.
EXAMPLE 5

y0 2 2y9 1 y 5
ex
,
2x
x > 0.
Solution The characteristic equation m2 2 2m 1 1 5 sm 2 1d2 5 0 has one

solution, m 5 1. Thus, the homogeneous solution is
yh 5 C1y1 1 C2 y2 5 C1ex 1 C2xex.

Replacing C1 and C2 by u1 and u2 produces
yp 5 u1y1 1 u2 y2 5 u1e x 1 u 2xe x.
The resulting system of equations is
u19 e x 1 u29 xe x 5 0
ex
u19 e x 1 u29 sxe x 1 e xd 5 .
2x
Subtracting the second equation from the first produces u29 5 1ys2xd. Then, by
substitution in the first equation, you have u19 5 2 12 . Finally, integration yields
u1 5 2
1
x
dx 5 2
2
2
and
u2 5
1
2
1
1
dx 5 ln x 5 ln !x.
x
2
From this result it follows that a particular solution is

1
yp 5 2 xe x 1 s ln !x dxe x
2
y 5 C1e x 1 C2xe x 2
1 x
xe 1 xe x ln !x.
2
1122
CHAPTER 15
EXAMPLE 6

y0 1 y 5 tan x.
Because the characteristic equation m2 1 1 5 0 has solutions m 5 i, the
homogeneous solution is
Solution
yh 5 C1 cos x 1 C2 sin x.
Replacing C1 and C2 by u1 and u2 produces
yp 5 u1 cos x 1 u2 sin x.
The resulting system of equations is
u19 cos x 1 u29 sin x 5 0
2u19 sin x 1 u29 cos x 5 tan x.
Multiplying the first equation by sin x and the second by cos x produces
u19 sin x cos x 1 u29 sin2x 5 0
2u19 sin x cos x 1 u29 cos2x 5 sin x.
Adding these two equations produces u29 5 sin x, which implies that
sin2 x
cos x
cos2 x 2 1
5
cos x
5 cos x 2 sec x.
u19 5 2
Integration yields
u1 5
scos x 2 sec xd dx
5 sin x 2 ln sec x 1 tan x

and
u2 5
sin x dx
5 2cos x
so that
yp 5 sin x cos x 2 cos x ln sec x 1 tan x 2 sin x cos x

5 2cos x ln sec x 1 tan x

y 5 yh 1 yp
5 C1 cos x 1 C2 sin x 2 cos x ln sec x 1 tan x .
SECTION 15.4
1123
In Exercises 14, verify the solution of the differential equation.
Solution
1. y 5 2s
e2x
2. y 5 s 2 1
2 cos xd
y0 1 y 5 10e2x
1
2x
y0 1 y 5 cos x
dsin x
3. y 5 3 sin x 2 cos x ln sec x 1 tan x

4. y 5 s 5 2 ln sin x dcos x 2 x sin x
y0 1 y 5 tan x
6. y0 2 2y9 2 3y 5 x 2 2 1
7. y0 1 y 5 x3
8. y0 1 4y 5 4
ys0d 5 1, y9s0d 5 6
10. y0 2 9y 5 5e3x
11. y0 2 10y9 1 25y 5 5 1 6ex

12. 16y0 2 8y9 1 y 5 4sx 1 exd
13. y0 1 y9 5 2sin x
ys0d 5 0, y9s0d 5 23
24. y0 1 y 5 sec x tan x
25. y0 1 4y 5 csc 2x
26. y0 2 4y9 1 4y 5 x 2e2x
27. y0 2 2y9 1 y 5 ex ln x
28. y0 2 4y9 1 4y 5
e2x
x
Electrical Circuits In Exercises 29 and 30, use the electrical
5. y0 2 3y9 1 2y 5 2x
ys0d 5 1, y9s0d 5 0
23. y0 1 y 5 sec x
y0 1 y 5 csc x cot x
In Exercises 520, solve the differential equation by the method

of undetermined coefficients.
9. y0 1 2y9 5 2ex
In Exercises 2328, solve the differential equation by the

method of variation of parameters.
14. y0 1 y9 2 2y 5 3 cos 2x
ys0d 5 21, y9s0d 5 2
15. y0 1 9y 5 sin 3x
16. y0 1 4y9 1 5y 5 sin x 1 cos x
circuit differential equation
_+
_ +
_+
d 2q
R dq
1
1
1
1
q5
Extc
dt2
L dt
LC
L
where R is the resistance (in ohms), C is the capacitance (in
farads), L is the inductance (in henrys), Estd is the electromotive
force (in volts), and q is the charge on the capacitor (in
coulombs). Find the charge q as a function of time for the
electrical circuit described. Assume that qx0c 5 0 and q9x0c 5 0.
29. R 5 20, C 5 0.02, L 5 2
Estd 5 12 sin 5t
30. R 5 20, C 5 0.02, L 5 1
Estd 5 10 sin 5t
17. y999 2 3y9 1 2y 5 2e22x
Vibrating Spring In Exercises 3134, find the particular
18. y999 2 y0 5 4x 2
solution of the differential equation
ys0d 5 1, y9s0d 5 1, y0s0d 5 1

19. y9 2 4y 5 xex 2 xe4x
ys0d 5
1
3
20. y9 1 2y 5 sin x
y
1p2 2 5 52
21. Think About It

(a) Explain how, by observation, you know that a particular
solution of the differential equation y0 1 3y 5 12 is
yp 5 4.
(b) Use the explanation of part (a) to give a particular solution
of the differential equation y0 1 5y 5 10.
(c) Use the explanation of part (a) to give a particular solution
of the differential equation y0 1 2y9 1 2y 5 8.
w
w
y0 xtc 1 by9xtc 1 kyxtc 5 Fxtc
g
g
for the oscillating motion of an object on the end of a spring. Use
a graphing utility to graph the solution. In the equation, y is the
displacement from equilibrium (positive direction is downward) measured in feet, and t is time in seconds (see figure). The
constant w is the weight of the object, g is the acceleration due
to gravity, b is the magnitude of the resistance to the motion, k
is the spring constant from Hookes Law, and Fxtc is the acceleration imposed on the system.
31.
(b) Use the explanation of part (a) to find a particular solution

of the differential equation y0 1 5y 5 10 cos x.
(c) Compare the algebra required to find particular solutions in
parts (a) and (b) with that required if the form of the particular solution were yp 5 A cos x 1 B sin x.
1 48y 5 24
32 s48 sin 4td
ys0d 5 14 , y9s0d 5 0
32.
2
32 y0
2
1 4y 5 32
s4 sin 8td
ys0d 5 14 , y9s0d 5 0
33.
2
32 y0
2
1 y9 1 4y 5 32
s4 sin 8td
ys0d 5 14 , y9s0d 5 23
22. Think About It

(a) Explain how, by observation, you know that a form
of a particular solution of the differential equation
y0 1 3y 5 12 sin x is yp 5 A sin x.
24
32 y0
34.
4
32 y0
1 12 y9 1 25
2 y 5 0
ys0d 5 12 , y9s0d 5 24
l = natural
length
y = displacement
m
Spring displacement
1124
CHAPTER 15
35. Vibrating Spring Rewrite yh in the solution for Exercise 31

by using the identity
a cos vt 1 b sin vt 5 !a2 1 b2 sinsvt 1 fd
39. Solve the differential equation
where f 5 arctan ayb.
x 2y0 2 xy9 1 y 5 4x ln x
36. Vibrating Spring The figure shows the particular solution

of the differential equation
4
25
y0 1 by9 1
y50
32
2
given that y1 5 x and y2 5 x ln x are solutions of the corresponding homogeneous equation.

40. True or False? yp 5 2e2x cos e2x is a particular solution
of the differential equation
1
ys0d 5 , y9s0d 5 24
2
y0 2 3y9 1 2y 5 cos e2x.
for values of the resistance component b in the interval f0, 1g.

(Note that when b 5 12 , the problem is identical to that of
Exercise 34.)
(a) If there is no resistance to the motion sb 5 0d, describe the
motion.
(b) If b > 0, what is the ultimate effect of the retarding force?
(c) Is there a real number M such that there will be no oscillations of the spring if b > M ? Explain your answer.
y
b=0
b=
1
2
b=1
Generated by Maple
37. Parachute Jump The fall of a parachutist is described by

the second-order linear differential equation
dy
w d 2y
5w
2k
g dt2
dt
where w is the weight of the parachutist, y is the height at time
t, g is the acceleration due to gravity, and k is the drag factor of
the parachute. If the parachute is opened at 2000 feet,
ys0d 5 2000, and at that time the velocity is y9s0d 5 2100 feet
per second, then for a 160-pound parachutist, using k 5 8, the
differential equation is
25y0 2 8y9 5 160.
Using the given initial conditions, verify that the solution of the
differential equation is
y 5 1950 1 50e21.6t 2 20t.
38. Parachute Jump Repeat Exercise 37 for a parachutist who

weighs 192 pounds and has a parachute with a drag factor of
k 5 9.
SECTION 15.5
SECTION
15.5
Series Solutions of Differential Equations
1125

Power Series Solution of a Differential Equation Approximation by Taylor Series
Power Series Solution of a Differential Equation

We conclude this chapter by showing how power series can be used to solve certain
types of differential equations. We begin with the general power series solution
method.
Recall from Chapter 8 that a power series represents a function f on an interval of
convergence, and that you can successively differentiate the power series to obtain a
series for f 9, f 0, and so on. These properties are used in the power series solution
method demonstrated in the first two examples.
EXAMPLE 1
Power Series Solution
Use a power series to solve the differential equation y9 2 2y 5 0.

Assume that y 5 oan xn is a solution. Then, y9 5 onan x n21. Substituting
for y9 and 22y, you obtain the following series form of the differential equation.
(Note that, from the third step to the fourth, the index of summation is changed to
ensure that xn occurs in both sums.)
Solution
y9 2 2y 5 0
`
nan xn21 2 2
n51
oax
n
50
n50
o na x
n
n21
n51
o sn 1 1da
o 2a x
n
n50
n
n11x
n50
o 2a x
n
n50
Now, by equating coefficients of like terms, you obtain the recursion formula
sn 1 1dan11 5 2an, which implies that
an11 5
2an
,
n11
n 0.
This formula generates the following results.

a0
a1
a0
2a0
a2
22a0
2
a3
23a0
3!
a4
24a0
4!
a5
25a0
5!
. . .
. . .
Using these values as the coefficients for the solution series, you have
y5
` s2xdn
2na0 n
x 5 a0
5 a0e 2x.
n50 n!
n50 n!
`
1126
CHAPTER 15
In Example 1, the differential equation could be solved easily without using a

series. The differential equation in Example 2 cannot be solved by any of the methods
discussed in previous sections.
EXAMPLE 2
Power Series Solution
Use a power series to solve the differential equation y0 1 xy9 1 y 5 0.

Solution
Assume that
oa x
n
is a solution. Then you have
n50
y9 5
o na x
n
n21
o na x ,
xy9 5
n51
y0 5
n51
o nsn 2 1da x
n
n22
n52
Substituting for y0, xy9, and y in the given differential equation, you obtain the following series.
`
o nsn 2 1da x
n
n22
n52
o na x
n
n50
oax
n
50
n50
o nsn 2 1da x
n
n22
52
n52
o sn 1 1da x
n
n50
To obtain equal powers of x, adjust the summation indices by replacing n by n 1 2 in

the left-hand sum, to obtain
`
sn 1 2dsn 1 1dan12 xn 5 2
n50
o sn 1 1da x .
n
n50
By equating coefficients, you have sn 1 2dsn 1 1dan12 5 2 sn 1 1dan, from which

you obtain the recursion formula
an12 5 2
a
sn 1 1d
an 5 2 n ,
sn 1 2dsn 1 1d
n12
n 0,
and the coefficients of the solution series are as follows.

a0
2
a
a
a4 5 2 2 5 0
4
2?4
a
a0
a6 5 2 4 5 2
6
2?4?6
a1
3
a
a
a5 5 2 3 5 1
5
3?5
a
a1
a7 5 2 5 5 2
7
3?5?7
a2 5 2
a2k 5
a3 5 2
s21d 0
s21d 0
5 k
2 ? 4 ? 6 . . . s2kd
2 sk!d
ka
ka
a2k11 5
s21dk a1
3 ? 5 ? 7 . . . s2k 1 1d
Thus, you can represent the general solution as the sum of two seriesone for the
even-powered terms with coefficients in terms of a0 and one for the odd-powered
terms with coefficients in terms of a1.
y 5 a0 1 2
5 a0
x2
x4
x3
x5
1
2 . . . 1 a1 x 2 1
2. . .
2
2?4
3
3?5
`
s21dkx 2k
s21dkx 2k11
1 a1
k
. . . s2k 1 1d
k50 2 sk!d
k50 3 ? 5 ? 7
`
The solution has two arbitrary constants, a0 and a1, as you would expect in the
general solution of a second-order differential equation.
SECTION 15.5
1127
Approximation by Taylor Series

A second type of series solution method involves a differential equation with initial
conditions and makes use of Taylor series, as given in Section 8.10.
EXAMPLE 3
Approximation by Taylor Series
Use a Taylor series to find the series solution of

y9 5 y 2 2 x
given the initial condition y 5 1 when x 5 0. Then, use the first six terms of this
series solution to approximate values of y for 0 x 1.
Solution
Recall from Section 8.10 that, for c 5 0,
y 5 ys0d 1 y9s0dx 1
x
0.0
1.0000
0.1
1.1057
0.2
1.2264
0.3
1.3691
0.4
1.5432
0.5
1.7620
0.6
2.0424
0.7
2.4062
0.8
2.8805
0.9
3.4985
1.0
4.3000
y0 s0d 2 y999s0d 3 . . .
x 1
x 1
.
2!
3!
Because ys0d 5 1 and y9 5 y 2 2 x, you obtain the following.

y9 5
y9 5
y999 5
ys4d 5
ys5d 5
y2 2 x
2yy9 2 1
2yy0 1 2s y9d2
2yy999 1 6y9y0
2yys4d 1 8y9y999 1 6s y0 d2
ys0d 5
y9s0d 5
y0 s0d 5
y999s0d 5
ys4ds0d 5
ys5ds0d 5
1
1
22151
21254
8 1 6 5 14
28 1 32 1 6 5 66
Therefore, you can approximate the values of the solution from the series
y0 s0d 2 y999s0d 3 ys4ds0d 4 ys5ds0d 5 . . .
x 1
x 1
x 1
x 1
2!
3!
4!
5!
1
4
14 4 66 5 . . .
5 1 1 x 1 x 2 1 x3 1
x 1
x 1
.
2
3!
4!
5!
y 5 ys0d 1 y9s0dx 1
Using the first six terms of this series, you can compute values for y in the interval
0 x 1, as shown in the table at the left.
In Exercises 16, verify that the power series solution of the
differential equation is equivalent to the solution found using
the techniques in Sections 5.7 and 15.115.4.
In Exercises 11 and 12, find the first three terms of each of the
power series representing independent solutions of the differential equation.
1. y9 2 y 5 0
2. y9 2 ky 5 0
11. sx 2 1 4dy0 1 y 5 0
3. y0 2 9y 5 0
4. y0 2 k 2y 5 0
5. y0 1 4y 5 0
6. y0 1 k 2y 5 0
In Exercises 710, use power series to solve the differential

equation and find the interval of convergence of the series.
7. y9 1 3xy 5 0
9. y0 2 xy9 5 0
8. y9 2 2xy 5 0
10. y0 2 xy9 2 y 5 0
12. y0 1 x 2y 5 0
In Exercises 13 and 14, use Taylors Theorem to find the series

solution of the differential equation under the specified initial
conditions. Use n terms of the series to approximate y for the
given value of x and compare the result with the approximation
given by Eulers Method for Dx 5 0.1.
13. y9 1 s2x 2 1dy 5 0, ys0d 5 2, n 5 5, x 5 12 ,
14. y9 2 2xy 5 0, ys0d 5 1, n 5 4, x 5 1
1128
CHAPTER 15
15. Investigation Consider the differential equation y0 1

9y 5 0 with initial conditions ys0d 5 2 and y9s0d 5 6.
(a) Find the solution of the differential equation using the
techniques of Section 15.3.
(b) Find the series solution of the differential equation.
(c) The figure shows the graph of the solution of the
differential equation and the third-degree and fifth-degree
polynomial approximations of the solution. Identify each.
y
In Exercises 17 and 18, use Taylors Theorem to find the series

solution of the differential equation under the specified initial
conditions. Use n terms of the series to approximate y for the
given value of x.
1
17. y0 2 2xy 5 0, ys0d 5 1, y9s0d 5 23, n 5 6, x 5 4
1
18. y0 2 2xy9 1 y 5 0, ys0d 5 1, y9s0d 5 2, n 5 8, x 5 2
In Exercises 1922, verify that the series converges to the given

function on the indicated interval. (Hint: Use the given differential equation.)
3
2
19.
xn
5 ex, s2 `, `d
n50 n!
Differential equation: y9 2 y 5 0
x
1 2
20.
16. Consider the differential equation y0 2 xy9 5 0 with the initial
conditions
and
ys0d 5 0
y9s0d 5 2. (See Exercise 9.)
Differential equation: y0 1 y 5 0
21.
s21dnx 2n11
5 arctan x, s21, 1d
2n 1 1
n50
`
Differential equation: sx 2 1 1dy0 1 2xy9 5 0
(a) Find the series solution satisfying the initial conditions.

(b) Use a graphing utility to graph the third-degree and fifthdegree series approximations of the solution. Identify the
approximations.
s21dnx2n
5 cos x, s2 `, `d
s2nd!
n50
`
22.
s2nd!x2n11
o s2 n!d s2n 1 1d 5 arcsin x, s21, 1d
n50
Differential equation: s1 2 x 2dy0 2 xy9 5 0
(c) Identify the symmetry of the solution.

23. Find the first six terms in the series solution of Airys equation
y0 2 xy 5 0.
R E V I E W E X E R C I S E S F O R C H A P T E R 15
In Exercises 14, classify the differential equation according to
type and order.
1.
2u
2u
5 c2 2
2
t
x
(a) Sketch several solution curves for the differential equation

on the direction field.
(b) Find the general solution of the differential equation.
Compare the result with the sketches from part (a).
2. yy0 5 x 1 1
4. s y0 d2 1 4y9 5 0
3. y0 1 3y9 2 10 5 0
6.
dy
5 !1 2 y 2
dx
In Exercises 5 and 6, use the given differential equation and its

direction field.
y
2
5.
dy y
5
dx x
y
3
2
1
4 3 2 1
2
3
4
x
1
(a) Sketch several solution curves for the differential equation

on the direction field.
(b) When is the rate of change of the solution greatest? When is
it least?
(c) Find the general solution of the differential equation.
Compare the result with the sketches from part (a).
REVIEW EXERCISES
In Exercises 710, match the differential equation with its

solution.
Solution
7. y9 2 4 5 0
(a) y 5 C1e2x 1 C2e22x
8. y9 2 4y 5 0
(b) y 5 4x 1 C
9. y0 2 4y 5 0
(c) y 5 C1 cos 2x 1 C2 sin 2x
10. y0 1 4y 5 0
(d) y 5 Ce4x
In Exercises 1132, find the general solution of the first-order

dy y
2 5 2 1 !x
11.
dx x
dy
1 xy 5 2y
12.
dx
2y y9
5
13. y9 2
x
x
dy
3
2 3x2y 5 ex
14.
dx
15.
dy y x
2 5
dx x y
16.
dy 3y
1
2 25 2
dx
x
x
17. s10x 1 8y 1 2d dx 1 s8x 1 5y 1 2d dy 5 0

18. s y 1 x3 1 xy 2d dx 2 xdy 5 0
19. s2x 2 2y3 1 yd dx 1 sx 2 6xy 2d dy 5 0
2 2
3
3 4
20. 3x y dx 1 s2x y 1 x y d dy 5 0
21. dy 5 s y tan x 1 2ex d dx
22. ydx 2 s x 1 !xy d dy 5 0
23. sx 2 y 2 5d dx 2 sx 1 3y 2 2d dy 5 0
24. y9 5 2x!1 2 y 2
25. x 1 yy9 5 !x 2 1 y 2
26. xy9 1 y 5 sin x
1129
37. s1 1 yd ln s1 1 yd dx 1 dy 5 0
ys0d 5 2
38. s2x 1 y 2 3d dx 1 sx 2 3y 1 1d dy 5 0
ys2d 5 0
39. y9 5 x 2y 2 2 9x 2
ys0d 5
3s1 1 ed
12e
40. 2xy9 2 y 5 x3 2 x
ys4d 5 2
In Exercises 41 and 42, find the orthogonal trajectories of the
given family and sketch several members of each family.
41. sx 2 Cd2 1 y 2 5 C 2
42. y 2 2x 5 C
43. Snow Removal Assume that the rate of change in the
number of miles s of road cleared per hour by a snowplow is
inversely proportional to the height h of snow.
(a) Write and solve the differential equation to find s as a function of h.
(b) Find the particular solution if s 5 25 miles when
h 5 2 inches and s 5 12 miles when h 5 10 inches
s2 h 15d.
44. Growth Rate Let x and y be the sizes of two internal organs
of a particular mammal at time t. Empirical data indicate that
the relative growth rates of these two organs are equal, and
hence we have
27. yy9 1 y 2 5 1 1 x 2
1 dx 1 dy
5
.
x dt
y dt
28. 2xdx 1 2ydy 5 sx 2 1 y 2d dx
Solve this differential equation, writing y as a function of x.
29. s1 1 x 2d dy 5 s1 1 y 2d dx
30. x3yy9 5 x4 1 3x 2y 2 1 y4
31. xy9 2 ay 5 bx4
32. y9 5 y 1 2xs y 2 ex d
In Exercises 3340, find the particular solution of the differential equation that satisfies the boundary condition.
33. y9 2 2y 5 ex
ys0d 5 4
2y
5 2x9y5
34. y9 1
x
ys1d 5 2
35. xdy 5 sx 1 y 1 2d dx
ys1d 5 10
36. yexydx 1 xexy dy 5 0
ys22d 5 25
45. Population Growth The rate of growth in the number N of

deer in a state park varies jointly over time t as N and L 2 N,
where L 5 500 is the estimated limiting size of the herd. Write
N as a function of t if N 5 100 when t 5 0 and N 5 200 when
t 5 4.
46. Population Growth The rate of growth in the number N of
elk in a game preserve varies jointly over time t (in years) as N
and 300 2 N where 300 is the estimated limiting size of the
herd.
(a) Write and solve the differential equation for the population
model if N 5 50 when t 5 0 and N 5 75 when t 5 1.
(b) Use a graphing utility to graph the direction field of the
differential equation and the particular solution of part (a).
(c) At what time is the population increasing most rapidly?
(d) If 400 elk had been placed in the preserve initially, use the
direction field to describe the change in the population over
time.
1130
CHAPTER 15
47. Slope The slope of a graph is given by y9 5 sin x 2 0.5y.

Find the equation of the graph if the graph passes through the
point s0, 1d. Use a graphing utility to graph the solution.
48. Investment Let Astd be the amount in a fund earning interest
at an annual rate r compounded continuously. If a continuous
cash flow of P dollars per year is withdrawn from the fund, the
rate of change of A is given by the differential equation
In Exercises 6164, find the particular solution of the differential equation that satisfies the initial conditions.
Initial Conditions
61. y0 2 y9 2 6y 5 54
ys0d 5 2, y9s0d 5 0
62. y0 1 25y 5 ex
ys0d 5 0, y9s0d 5 0
63. y0 1 4y 5 cos x
ys0d 5 6, y9s0d 5 26
dA
5 rA 2 P
dt
64. y0 1 3y9 5 6x
ys0d 5 2, y9s0d 5 10
3
where A 5 A0 when t 5 0. Solve this differential equation for

A as a function of t.
Vibrating Spring In Exercises 65 and 66, describe the motion

of a 64-pound weight suspended on a spring. Assume that the
weight stretches the spring 43 feet from its natural position.
49. Investment A retired couple plans to withdraw P dollars per

year from a retirement account of $500,000 earning 10%
compounded continuously. Use the result of Exercise 48 and a
graphing utility to graph the function A for each of the following continuous annual cash flows. Use the graphs to describe
what happens to the balance in the fund for each of the cases.
released.
(a) P 5 $40,000
67. Investigation
(b) P 5 $50,000
(c) P 5 $60,000
50. Investment Use the result of Exercise 48 to find the time
necessary to deplete a fund earning 14% interest compounded
continuously if A0 5 $1,000,000 and P 5 $200,000.
In Exercises 5154, find the particular solution of the differential equation that satisfies the initial conditions. Use a graphing
utility to graph the solution.
released. The motion takes place in a medium that furnishes a
damping force of magnitude 18 speed at all times.
Initial Conditions
51. y0 2 y9 2 2y 5 0
ys0d 5 0, y9s0d 5 3
52. y0 1 4y9 1 5y 5 0
ys0d 5 2, y9s0d 5 27
53. y0 1 2y9 2 3y 5 0
ys0d 5 2, y9s0d 5 0
54. y0 1 2y9 1 5y 5 0
ys1d 5 4, ys2d 5 0
In Exercises 5560, find the general solution of the second-order

The differential equation
8
8
y0 1 by9 1 ky 5
Fstd,
32
32
1
ys0d 5 ,
2
y9s0d 5 0
models the motion of a weight suspended on a spring.

(a) Solve the differential equation and use a graphing utility to
graph the solution for each of the assigned quantities for b,
k, and Fstd.
(i) b 5 0, k 5 1, Fstd 5 24 sin p t
(ii) b 5 0, k 5 2, Fstd 5 24 sins 2!2 td

(iii) b 5 0.1, k 5 2, Fstd 5 0
(iv) b 5 1, k 5 2, Fstd 5 0
(b) Describe the effect of increasing the resistance to motion b.
(c) Explain how the motion of the object would change if a
stiffer spring (increased k) were used.
(d) Matching the input and natural frequencies of a system is
known as resonance. In which case of part (a) does this
occur, and what is the result?
68. Think About It Explain how you can find a particular solution of the differential equation
55. y0 1 y 5 x3 1 x
56. y0 1 2y 5 e 2x 1 x
57. y0 1 y 5 2 cos x
58. y0 1 5y9 1 4y 5 x 2 1 sin 2x
59. y0 2 2y9 1 y 5
2xex
60. y0 1 2y9 1 y 5
1
x 2ex
y0 1 4y9 1 6y 5 30
by observation.
In Exercises 69 and 70, find the series solution of the differential equation.
69. sx 2 4dy9 1 y 5 0
70. y0 1 3xy9 2 3y 5 0

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SECTION 15.

Exact First-Order Equations

Exact First-Order Equations

Exact Differential Equations

fy sx, yd 5 Nsx, yd.

The general solution of the equation is f sx, yd 5 C.

Test for Exactness

Exactness is a fragile condition in the sense that seemingly minor alterations in

Testing for Exactness

a. The differential equation sxy 2 1 xd dx 1 yx 2 dy 5 0 is exact because

NOTE Every differential equation of

But the equation s y 2 1 1d dx 1 xy dy 5 0 is not exact, even though it is obtained

But the equation cos y dx 1 s y 2 1 x sin yd dy 5 0 is not exact, even though it

Solving an Exact Differential Equation

Solve the differential equation s2xy 2 3x 2d dx 1 s x 2 2 2yd dy 5 0.

The given differential equation is exact because

In Section 14.1, you determined gs yd by integrating Nsx, yd with respect to y and

Thus, g9s yd 5 22y, and it follows that gs yd 5 2y 2 1 C1. Therefore,

Solving an Exact Differential Equation

Find the particular solution of

scos x 2 x sin x 1 y 2d dx 1 2xy dy 5 0

The differential equation is exact because

fcos x 2 x sin x 1 y 2g 5 2y 5 f2xyg.

Exact First-Order Equations

f xy 2 1 gsxdg 5 y 2 1 g9 sxd 5 cos x 2 x sin x 1 y 2

Thus, g9 sxd 5 cos x 2 x sin x and

Applying the given initial condition produces

which implies that C 5 0. Hence, the particular solution is

Multiplying by an Integrating Factor

a. If the differential equation

Not an exact equation

is multiplied by the integrating factor usx, yd 5 x, the resulting equation

is exactthe left side is the total differential of x 2 y.

Not an exact equation

is multiplied by the integrating factor usx, yd 5 1yy 2, the resulting equation

is exactthe left side is the total differential of xyy.

Consider the differential equation Msx, yd dx 1 Nsx, yd dy 5 0.

Exact First-Order Equations

Finding an Integrating Factor

Solve the differential equation s y 2 2 xd dx 1 2y dy 5 0.

Sketch the force field given by

Family of tangent curves to F:.

by finding and sketching the family of curves tangent to F.

An Application to Force Fields

which, in differential form, is

3. s3y 2 1 10xy 2d dx 1 s6xy 2 2 1 10x 2yd dy 5 0

5. s4x 3 2 6xy 2d dx 1 s4y 3 2 6xyd dy 5 0

26. s22y 3 1 1d dx 1 s3xy 2 1 x 3d dy 5 0

cos xy f ydx 1 sx 1 tan xyd dyg 5 0

In Exercises 11 and 12, (a) sketch an approximate solution of

ssin 3y dx 1 cos 3y dyd 5 0

32. Show that the differential equation

In Exercises 1316, find the particular solution that satisfies the

28. s3y 2 1 5x 2 yd dx 1 s3xy 1 2x 3d dy 5 0

saxy 2 1 byd dx 1 sbx 2y 1 axd dy 5 0

27. s4x 2 y 1 2y 2d dx 1 s3x 3 1 4xyd dy 5 0

In Exercises 2730, use the integrating factor to find the

usx, yd 5 x22 y22

In Exercises 1726, find the integrating factor that is a function