DIFFERENTIAL EQUATIONS
441
46.69
Find a general solution of
y" + y = x sin x
(1)
As in Problem 46.65, the general solution of y" + y = 0 consists of all linear combinations of y, = cos x
and
y 2 = sinx,
and
W(y,,y 2 ) = l.
Then
2xsin2x-2x
2 ), by an integration by parts. Moreover,
by another integration by parts. Therefore, a particular solution is
y = yiU, + y 2 v 2 =
x sin 2x — 2x
2 ) + ksin x(sin 2x — 2x cos 2x) =
cos x(cos 2x + 2x sin 2x -2x ) +
tion is
(cos x + 2x sin x - 2x cos x) + C, cos x + C 2 sin x = (x/4) (sin x - x cos x) + D l cos x + D 2 sin x.
46.70
Solve the predator-prey system
(cos x sin 2x -
(cos x cos 2x + sin x sin 2x]
(cos x + 2x sin x - 2x cos x).
The general solusin x cos 2x)
x sin x cos x dx
46.72
46.71
For any function y of x, if x — e', then y also can be considered a function of /. Show that
and
Therefore,
Hence,
Solve
Let
x = e'.
By Problem 46.71,
and
Then (1) becomes
or, more simply,
The auxiliary equation is M
2 + 2a + l=0, with the repeated root « = -!. By Problem 46.48, the general
solution of (2) is y = (C, + C 2 Oe '. Since x = e', t = lnx. Hence,
46.73
Solve the linear third-order equation with constant coefficients y'" + 2y" - y' - 2y = 0.
46.74
Solve the linear third-order equation with constant coefficients y'" — 7y" + 16y' — 12y = 0.
The auxiliary equation is «
3 — 7«
2 + 16u - 12 = 0. 2 is a root. Division by u — 2 yields w
2 — 5u +
6 = (u- 2)(u - 3). Thus, 2 is a repeated root and 3 is a root. So, by an extension of Problem 46.48, the general
solution is C l e
3 ' + (C 2 + C 3 x)e
2 '.
The auxiliary equation is u
3 + 2u
2 — u -2 = 0. 1 is a root. Division by u — 1 yields u + 3u + 2 =
(u + l)(u + 2). Thus, the three roots are 1, -1, and -2. Hence, by an extension of Problem 46.48, the general
solution is C l e' + C 2 e~* + C 3 e~
2 *. (The method of variation of parameters also extends to inhomogeneous
equations of higher order.)
Differentiate (2) with respect to t:
Hence,
By
Problem 46.64, the general solution of this equation is
Then
By (2),