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SECTION 15.1 Exact First-Order Equations 1093
SECTION 15.1 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 equa-
tions 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 first-
order 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 fsx, yd.
Definition of an Exact Differential Equation
The equation Msx, 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 deriv-
atives such that
f sx, yd 5 Msx, yd and f sx, yd 5 Nsx, yd.
x y
The general solution of the equation is fsx, yd 5 C.
From Section 12.3, you know that if f has continuous second partials, then
2 2
M5 f 5 f 5 N.
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 differen-
tial equation Msx, yd dx 1 Nsx, yd dy 5 0 is exact if and only if
M5N.
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
EXAMPLE1 Testing for Exactness
NOTE Every differential equation of 2 2
a. The differential equation sxy 1 xd dx 1 yx dy 5 0 is exact because
the form M N
2 2
Msxd dx 1 Nsyd dy 5 0 y 5 y fxy 1 xg 5 2xy and x 5 x fyx g 5 2xy.
is exact. In other words, a separable vari- But the equation sy2 1 1d dx 1 xy dy 5 0 is not exact, even though it is obtained
ables equation is actually a special type by dividing both sides of the first equation by x.
of an exact equation. b. The differential equation cos y dx 1 sy2 2 x sin yd dy 5 0 is exact because
M5 fcos yg 5 2sin y and N 5 fy2 2 x sin yg 5 2sin y.
y y x x
But the equation cos y dx 1 sy2 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 Msx, ydi 1 Nsx, ydj is the gradient of a poten-
tial function (Theorem 14.1). This means that a general solution fsx, 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.
EXAMPLE2 Solving an Exact Differential Equation
2 2
Solve the differential equation s2xy 2 3x d dx 1 sx 2 2yd dy 5 0.
Solution The given differential equation is exact because
M 2 N 2
y 5 y f2xy 2 3x g 5 2x 5 x 5 x fx 2 2yg.
The general solution, fsx, yd 5 C, is given by
f sx, yd 5 Msx, yd dx
E
2 2 3
5 s2xy 2 3x d dx 5 x y 2 x 1 gsyd.
E
In Section 14.1, you determined gsyd by integrating Nsx, yd with respect to y and
reconciling the two expressions for fsx, yd. An alternative method is to partially
differentiate this version of fsx, yd with respect to y and compare the result with
y C= 1000 Nsx, yd. In other words,
24 Nsx, yd
20 f sx, yd 5 fx2y 2 x3 1 gsydg 5 x2 1 g9syd 5 x2 2 2y.
16 C= 100 y y
12 g9syd 5 22y
8 C= 10
C= 1 Thus, g9syd 5 22y, and it follows that gsyd 5 2y2 1 C . Therefore,
1
x f sx, yd 5 x2y 2 x3 2 y2 1 C
−12 −8 −4 4 8 12 1
and the general solution is x2y 2 x3 2 y2 5 C. Figure 15.1 shows the solution curves
Figure 15.1 that correspond to C 5 1, 10, 100, and 1000.
SECTION 15.1 Exact First-Order Equations 1095
EXAMPLE3 Solving an Exact Differential Equation
TECHNOLOGY You can use a Find the particular solution of
graphing utility to graph a particular 2
solution that satisfies the initial condi- scos x 2 x sin x 1 y d dx 1 2xy dy 5 0
tion of a differential equation. In that satisfies the initial condition y 5 1 when x 5 p.
Example 3, the differential equation
and initial conditions are satisfied Solution The differential equation is exact because
when 2 1 x cos x 5 0, which
xy
implies that the particular solution M N
can be written as or
x 5 0 y x
! On a graphing
y 5 ± 2cos x.
calculator screen, the solution would 2
be represented by Figure 15.2 together y fcos x 2 x sin x 1 y g 5 2y 5 x f2xyg.
with the y-axis. Because Nsx, yd is simpler than Msx, yd, it is better to begin by integrating Nsx, yd.
4
f sx, yd 5 Nsx, yd dy 5 2xy dy 5 xy2 1 gsxd
E E
−12.57 12.57 Msx, yd
f sx, yd 5 fxy2 1 gsxdg 5 y2 1 g9sxd 5 cos x 2 x sin x 1 y2
x x
−4
Figure 15.2 g9sxd 5 cos x 2 x sin x
g
Thus, 9sxd 5 cos x 2 x sin x and
gsxd 5 scos x 2 x sin xd dx
E
5 x cos x 1 C
1
which implies that fsx, yd 5 xy2 1 x cos x 1 C , and the general solution is
1
xy2 1 x cos x 5 C. General solution
Applying the given initial condition produces
y ps1d2 1 p cos p 5 C
4
2 ( , 1)π which implies that C 5 0. Hence, the particular solution is
x xy2 1 x cos x 5 0. Particular solution
π 2π π π 2π π
3 3
− − − −2
−4 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 y2 1 cos x 5 0, and the y-axis is given
Figure 15.3 by x 5 0.
In Example 3, note that if z 5 fsx, yd 5 xy2 1 x cos x, the total differential of z
is given by
dz 5 f sx, yd dx 1 f sx, yd dy
x y
2
5 scos x 2 x sin x 1 y ddx 1 2xy dy
5 Msx, yd dx 1 Nsx, 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 fsx, 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 possi-
ble to make it exact by multiplying by an appropriate factor usx, yd, which is called an
integrating factor for the differential equation.
EXAMPLE4 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
2
2xy dx 1 x dy 5 0 Exact equation
is exact—the left side is the total differential of x2y.
b. If the equation
y dx 2 x dy 5 0 Not an exact equation
2
is multiplied by the integrating factor usx, yd 5 1yy , the resulting equation
1 dx 2 x dy 5 0 Exact equation
y y2
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 N sx, ydg 5 hsxd
Nsx, yd y x
s d
eh x dx
is a function of x alone, then e is an integrating factor.
2. If
1 fN sx, yd 2 M sx, ydg 5 ksyd
Msx, yd x y
s d
ek y dy
is a function of y alone, then e is an integrating factor.
STUDYTIP If either hsxd or ksyd 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.
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