SECTION 15.1
SECTION
15.1
Exact First-Order Equations
1093
Exact First-Order Equations
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
y
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
−8
Figure 15.1
−4
Thus, g9s yd 5 22y, and it follows that gs yd 5 2y 2 1 C1. Therefore,
x
4
8
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.
4
f sx, yd 5
−12.57
Exact First-Order Equations
E
Nsx, yd dy 5
E
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
E
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
(π , 1)
x
−3π
−2π
−π −2
−4
Figure 15.3
π
2π
3π
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
Differential Equations
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 exact—the 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 exact—the 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 routinely—namely,
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
Exact First-Order Equations
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
E
Nsx, yd dy 5
E
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
i
y2
x2
x
y2
Sketch the force field given by
j
Fsx, yd 5
Family of tangent curves to F:.
y2
x
Ce
1
x
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
2
1
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.
y
3
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
Differential Equations
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 1–10, 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.
2
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
2
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
1
y
−4
4
4
2
2
x
−2
2
−4
4
−2
−4
−4
4
Differential Equation
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 Þ b, show that x m y n is an integrating factor, where
In Exercises 13–16, 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
x
−2
In Exercises 27–30, use the integrating factor to find the
general solution of the differential equation.
usx, yd 5 x22 y22
y
Figure for 11
13.
In Exercises 17–26, 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 33–36, 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
1
35. Fsx, yd 5 4x 2 y i 2 2xy 2 1
36. Fsx, yd 5 s1 1
d i 2 2xy j
x2
2
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 Euler’s 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
Euler’s Method In Exercises 43–46, (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
Exact First-Order Equations
20x 2 y
2y 2 10x
where Cs100d 5 500 and x ≥ 100.
40. Euler’s 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. Euler’s 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. Euler’s Method Repeat Exercise 45 for Dx 5 1 and
discuss how the accuracy of the result changes.
48. Euler’s 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
Euler’s Method works.
(b) How will decreasing the magnitude of Dx affect the accuracy of Euler’s Method?
41. Euler’s Method Use Euler’s 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
2
True or False? In Exercises 49–52, 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.