Page 113 - Schaum's Outline of Differential Equations

96 THE METHOD OF UNDETERMINED COEFFICIENTS [CHAP. 11

or, equivalently,

Equating the coefficients of like powers of x, we obtain

Solving this system, we find that A 2 = -2, A 1 = 2, and A 0 = -3. Hence (_/) becomes

and the general solution is

3x
11.2. Solve/'-/ -2y = e .
x
2x
From Problem 9.1, y h = c^ + C 2e . Here <j>(x) has the form displayed in Case 2 with k = 1 and a = 3. Using
(112), we assume that

X
3
Thus, y' p = 3A6 * and y" p = 9A^ . Substituting these results into the differential equation, we have

>x
It follows that 4A = 1, or A = ^, so that (_/) becomes y p =±e . The general solution then is

11.3. Solve /' -y' -2y = sin 2x.
x
2x
Again by Problem 9.1, y h = c-\e~ + C 2e . Here <j>(x) has the form displayed in Case 3 with k 1 = 1, k 2 = 0, and
/?= 2. Using (11.3), we assume that

Thus, y' p = 2A cos 2x — 2B sin 2x and y'p = —4A sin 2x — 4B cos 2x. Substituting these results into the differential
equation, we have

or, equivalently,
(-6A + 2B) sin2^ + (-6B - 2A) cos2x = (l) sin 2x + (0) cos 2x
Equating coefficients of like terms, we obtain

Solving this system, we find that A = -3/20 and B = 1/20. Then from (_/),

and the general solution is