Page 117 - Schaum's Outline of Differential Equations

100 THE METHOD OF UNDETERMINED COEFFICIENTS [CHAP. 11

and an assumed solution for (x + 1) cos x is given also by Eq. (11.5) as

(Note that we have used C and D in the last expression, since the constants A and B already have been used.)
We therefore take

= (AjX + A Q) sin x + (Bpc + B Q) cos x + (C±x + C Q) sin x + (D^ + D Q) cos x
y p
Combining like terms, we arrive at

as the assumed solution, where Ej = Aj + Cj and Fj = Bj + Dj (j = 0, 1).

11.12. Solve y -5y = (x- 1) sin x + (x + 1) cos x.
From Problem 10.3, y h = c^*. Using the results of Problem 11.11, we assume that

Thus,
Substituting these values into the differential equation and simplifying, we obtain

Equating coefficients of like terms, we have

Solving, we obtain E l = -2/13, E 0 = 71/338, F l = -3/13, and F 0 = -69/338. Then, from (1),

and the general solution is

11.13. Solve/ -5y = 3e*-2x+l.
From Problem 10.3, y h = c^. Here, we can write <j>(x) as the sum of two manageable functions:
x
ij>(x) = (3e") + (—2x+ 1). For the term 3e" we would assume a solution of the form Ae ; for the term —2x4- 1 we
would assume a solution of the form BIX + B 0. Thus, we try

Substituting (_/) into the differential equation and simplifying, we obtain

Equating coefficients of like terms, we find that A = —314, BI = 2/5, and B 0 = -3/25. Hence, (_/) becomes