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# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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Solve the differential equation
(y cos x + 2xey) + (sin x + x2ey — 1) y = 0. (15)
It is easy to see that
My(x, y) = cos x + 2xey = Nx(x, y),
so the given equation is exact. Thus there is a — (x, y) such that
—x(x, y) = y cos x + 2xey,
—y(x, y) = sin x + x2 ey — 1.
Integrating the first of these equations, we obtain
— (x, y) = ysinx + x2ey + h(y). (16)
Setting —y = N gives
—y(x, y) = sin x + x2 ey + h'(y) = sin x + x2 ey — 1.
Thus h'(y) = — 1 and h(y) = —y. The constant of integration can be omitted since any solution of the preceding differential equation is satisfactory; we do not require the most general one. Substituting for h(y) in Eq. (16) gives
— (x, y) = y sin x + x2 ey — y.
Hence solutions ofEq. (15) are given implicitly by
y sin x + x2ey — y = c. (17)
2.6 Exact Equations and Integrating Factors
93
Solve the differential equation
(3xy + y2) + (x2 + xy) y = 0. (18)
Here,
My(x, y) = 3x + 2y, Nx(x, y) = 2x + y;
since My = Nx, the given equation is not exact. To see that it cannot be solved by the procedure described previously, let us seek a function — such that
—x(x, y) = 3xy + y2, —y(x, y) = x2 + xy. (19)
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 — ó from Eq. (20) and set it equal to N, obtaining
3
x2 + 2xy + h'(y) = x2 + xy
2 or
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.³ and then try to choose /ã 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
(³ M)y = (f N)x. (24)
Since M and N are given functions, Eq. (24) states that the integrating factor ³ must satisfy the first order partial differential equation
Mfy — Nfx + (My — Nx )f = °. (25) If a function ³ 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 ³ 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 ofEq. (22). This possible nonuniqueness of the integrating factor is illustrated in Example 4.
94
Chapter 2. First Order Differential Equations
EXAMPLE
4
Unfortunately, Eq. (25), which determines the integrating factor f, 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 ³ 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 ³ that depends on x only. Assuming that ³ is a function of x only, we have
(³ M)y = ³ My, (f N)x = ³ Nx + Ndf .
Thus, if (³ M) is to equal (³ N)x, it is necessary that
Ô = My - Nx
dx N
³. (26)
If (My — Nx)/ N is a function of x only, then there is an integrating factor ³ that also depends only on x; further, /j,(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.
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