Page 114 - Schaum's Outline of Differential Equations

CHAP. 11] THE METHOD OF UNDETERMINED COEFFICIENTS 97

11.4. Solve
From Problem 9.10,

Here <j>(f) has the form displayed in Case 3 with the independent variable t replacing x, ki = 2, k 2 = —1, and /? = j.
Using (11.3), with t replacing x, we assume that

Consequently,

and

Substituting these results into the differential equation, we obtain

or, equivalently

Equating coefficients of like terms, we have

It follows that A = 56/663 and B = -20/663, so that (1) becomes

The general solution is

11.5. Solve y - 6y + 25y = 64e~'.
From Problem 9.10,

Here <j>(t) has the form displayed in Case 2 with the independent variable t replacing x, k= 64 and a = —1. Using
(11.2), with t replacing x, we assume that

Consequently, y p = —Ae~' and y p = Ae~'. Substituting these results into the differential equation, we obtain

or, equivalently, 32Ae~' = 64e~'. It follows that 32A = 64 or A = 2, so that (_/) becomes y p = 2e~'. The general
solution is