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Integrating the first of Eqs. (19) gives
— (x ,y) = 2 x2 y + xy2 + h(y), (20)
where h is an arbitrary function of y only. To try to satisfy the second of Eqs. (19) we compute —y from Eq. (20) and set it equal to N, obtaining
x2 + 2xy + h'(y) = x2 + xy
h(y) = -\ x2 — xy. (21)
Since the right side of Eq. (21) depends on x as well as y, it is impossible to solve Eq. (21) for h(y). Thus there is no —(x, y) satisfying both of Eqs. (19).
Integrating Factors. It is sometimes possible to convert a differential equation that is not exact into an exact equation by multiplying the equation by a suitable integrating factor. Recall that this is the procedure that we used in solving linear equations in Section 2.1. To investigate the possibility of implementing this idea more generally, let us multiply the equation
M(x, y) dx + N(x, y) dy = 0 (22)
by a function f.i and then try to choose [i so that the resulting equation
li(x, y)M(x, y) dx + i(x, y)N(x, y) dy = 0 (23)
is exact. By Theorem 2.6.1 Eq. (23) is exact if and only if
(l M)y = (i N)x. (24)
Since M and N are given functions, Eq. (24) states that the integrating factor i must satisfy the first order partial differential equation
Mfy — Nfx + (My — Nx )l = °. (25) If a function i satisfying Eq. (25) can be found, then Eq. (23) will be exact. The solution of Eq. (23) can then be obtained by the method described in the first part of this section. The solution found in this way also satisfies Eq. (22), since the integrating factor i can be canceled out of Eq. (23).
A partial differential equation of the form (25) may have more than one solution; if this is the case, any such solution may be used as an integrating factor of Eq. (22). This possible nonuniqueness of the integrating factor is illustrated in Example 4.
Chapter 2. First Order Differential Equations
Unfortunately, Eq. (25), which determines the integrating factor x, is ordinarily at least as difficult to solve as the original equation (22). Therefore, while in principle integrating factors are powerful tools for solving differential equations, in practice they can be found only in special cases. The most important situations in which simple integrating factors can be found occur when x is a function of only one of the variables x or y, instead of both. Let us determine necessary conditions on M and N so that Eq. (22) has an integrating factor x that depends on x only. Assuming that x is a function of x only, we have
(XM)y = xMy, (xN)x = xNx + NdX .
Thus, if (x M) is to equal (x N)x, it is necessary that
dx = My - Nx
If (My — Nx)/ N is a function of x only, then there is an integrating factor x that also depends only on x; further, x(x) can be found by solving Eq. (26), which is both linear and separable.
A similar procedure can be used to determine a condition under which Eq. (22) has an integrating factor depending only on y; see Problem 23.
Find an integrating factor for the equation
(3xy + y2) + (x2 + xy) / = 0 (18)
and then solve the equation.
In Example 3 we showed that this equation is not exact. Let us determine whether it has an integrating factor that depends on x only. On computing the quantity (My — Nx)/ N, we find that
My(x, y) — Nx(x, y) 3x + 2y — (2x + y) 1
N(x, y) x2 + xy x.
Thus there is an integrating factor x that is a function of x only, and it satisfies the differential equation
d-x = x. (28)
x(x) = x. (29)
Multiplying Eq. (18) by this integrating factor, we obtain
(3x2 y + xy2) + (x2 + x2 y)/ = 0. (30)
The latter equation is exact and it is easy to show that its solutions are given implicitly by
x3 y + 2 x2 y2 = c. (31)
Solutions may also be readily found in explicit form since Eq. (31) is quadratic in y.
2.6 Exact Equations and Integrating Factors
You may also verify that a second integrating factor of Eq. (18) is
l(x, y) =
xy(2x + y)
and that the same solution is obtained, though with much greater difficulty, if this integrating factor is used (see Problem 32).
PROBLEMS Determine whether or not each of the equations in Problems 1 through 12 is exact. If it is exact, s find the solution.
1. (2x + 3) + (2y — 2)y = 0 2. (2x + 4y) + (2x — 2y)/ = 0
3. (3X2 — 2xy + 2) dx + (6y2 — x2 + 3) dy = 0
4. (2xy2 + 2 y) + (2x2 y + 2x)y = 0 5 dy= ax + by
dx bx + cy
dy ax — by
. dx ~~~
bx — cy
(ex sin y — 2y sin x) dx + (ex cos y + 2cosx) dy = 0 (ex sin y + 3y) dx — (3x — ex sin y) dy = 0 (yexy cos 2x — 2exy sin2x + 2x) dx + (xexy cos 2x — 3) dy = 0 (y/x + 6x) dx + (ln x — 2) dy = 0, x > 0
(x ln y + xy) dx + ( y ln x + xy) dy = 0; x > 0, y > 0
x dx ydy
(x2 + y2 )3/2 (x2 + y2)2'2
In each of Problems 13 and 14 solve the given initial value problem and determine at least approximately where the solution is valid.
13. (2x - y) dx + (2y - x) dy = 0, y(1) = 3
14. (9x2 + y - 1) dx - (4y - x) dy = 0, y(1) = 0
In each of Problems 15 and 16 find the value of b for which the given equation is exact and then solve it using that value of b.