Page 115 - Schaum's Outline of Differential Equations

98 THE METHOD OF UNDETERMINED COEFFICIENTS [CHAP. 11

2
3
11.6. Solve y - 6y + 25y = 50t -36t - 63t + 18.
Again by Problem 9.10,

Here <j>(t) is a third-degree polynomial in t. Using (11.1) with t replacing x, we assume that

Consequently,

and

Substituting these results into the differential equation, we obtain

or, equivalently,

Equating coefficients of like powers of t, we have

Solving these four algebraic equations simultaneously, we obtain A 3 = 2, A 2 = 0, A 1 = —3, and A Q = 0, so that (_/)
becomes

The general solution is

x
11.7. Solve /" - 6y" + 11/ - 6y = 2xe~ .
1
3
From Problem 10.1, y h = c^ + c^e * + c 3 e *. Here $(x) = e^p^x), where a = —1 and p n(x) = 2x, a first-degree
polynomial. Using Eq. (11.4), we assume that y p = e^(A lx + A 0), or

Thus,

Substituting these results into the differential equation and simplifying, we obtain

Equating coefficients of like terms, we have

from which Aj = -1/12 and A 0 = -13/144.
Equation (_/) becomes

and the general solution is