Page 246 - Schaum's Outline of Differential Equations

CHAP. 22] INVERSE LAPLACE TRANSFORMS 229

Therefore,

22.13. Use partial fractions to decompose

To the linear factors s — 2 and s + 1, we associate respectively the fractions Al(s - 2) and Bl(s + 1). We set

and, upon clearing fractions, obtain

To find A and B, we use the alternative procedure suggested in Problem 22.11. Substituting s = —l and then s = 2
into (_/), we immediately obtain A = 5/3 and B = -2/3. Thus,

22.14. Use partial fractions to decompose

3
2
3
Note that s - s - 2 factors into (s - 2)(s + 1). To the factor j = (s - O) , which is a linear polynomial raised to
3
the third power, we associate the sum A^s + A 2/i 2 + A 3/s . To the linear factors (s - 2) and (s + 1), we associate the
fractions Bl(s - 2) and Cl(s + 1). Then

or, clearing fractions,
3
2
8 = AiS (s - 2)0 + 1) + A 2s(s - 2)0 + 1) + A 3(s - 2)0 + 1) + B^(s + 1) + Cs (s - 2)
Letting s = —1,2, and 0, consecutively, we obtain, respectively, C = 8/3, B = 1/3, and A 3 = —4. Then choosing 5=1
and s = —2, and simplifying, we obtain the equations A 1+A 2 = —l and 2A 1 — A 2 = —S, which have the solutions
A 1 = —3 and A 2 = 2. Note that any other two values for s (not —1, 2, or 0) will also do; the resulting equations may
be different, but the solution will be identical. Finally,

22.15. Find

No function of this form appears in Appendix A. Using the results of Problem 22.13 and Property 22.1, we
obtain