Page 245 - Schaum's Outline of Differential Equations

228 INVERSE LAPLACE TRANSFORMS [CHAP. 22

Then,

22.11. Use partial function to decompose

2
To the linear factor s + 1, we associate the fraction Al(s + 1); whereas to the quadratic factor s + 1, we asso-
ciate the fraction (Bs + C)i(s 2 + 1). We then set

Clearing fractions, we obtain

2
2
or s (0) + 5(0) + 1 = s (A + B) + s(B + C) + (A + C)
Equating coefficients of like powers of s, we conclude that A + B = 0, B + C = 0, and A + C = 1. The solution of
this set of equations is A = -|, B = —-|, and C = -j. Substituting these values into (_/), we obtain the partial-fractions
decomposition

The following is an alternative procedure for finding the constants A, B, and C in (_/). Since (2) must hold for
all s, it must in particular hold s = —1. Substituting this value into (2), we immediately find A = ^. Equation (2) must
also hold for s = 0. Substituting this value along with A = -| into (2), we obtain C = -|. Finally, substituting any other
value of s into (2), we find that B = - -|.

22.12. Use partial fractions to decompose

2
2
To the quadratic factors s + 1 and s + 4s + 8, we associate the fractions (As + B)l(s 2 + 1) and
(Cs + D)/(s 2 + 4s + 8). We set

and clear fractions to obtain
l = (As + B)(s 2 + 4s + 8) + (Cs + D)(s 2 + 1)
3
2
3
2
or 5 (0) + s (G) + 5(0) + 1 = s (A + C) + s (4A + B + D) + s(SA + 4B + C) + (SB + D)
Equating coefficients of like powers of s, we obtain A + C = 0, 4A + B + D = 0, SA + 4B + C = 0, and SB + D = 1.
The solution of this set of equation is