Page 263 - Schaum's Outline of Differential Equations

246 SOLUTIONS OF LINEAR DIFFERENTIAL EQUATIONS [CHAP. 24

Finally, using the method of partial fractions and taking the inverse transform, we obtain

24.11. Solve / - 5y = 0.
No initial conditions are specified. Taking the Laplace transform of both sides of the differential equation,
we obtain

Then, using Eq. (24.4) with C Q = y(0) kept arbitrary, we have

Taking the inverse Laplace transform, we find that

24.12. Solve /'- 3/ + 2y = e~ . x
c
No initial conditions are specified. Taking Laplace transforms, we have £(y"} - 3££{;y'} + 2££{;y} = t£(e~*), or
2
[s Y(s) - SC Q - ] - 3[sY( S) - C Q] + 2[Y(s)] = ll(s + 1)
Cl
Here c 0 and c 1 must remain arbitrary, since they represent y(0) and ;y'(0), respectively, which are unknown. Thus,

Using the method of partial fractions and noting that s 2 — 3s + 2 = (s — l)(s — 2), we obtain

where and

x
24.13. Solve /'- 3/ + 2y = e~ ; y(l) = 0, /(I) = 0.
The initial conditions are given at x = 1, not x = 0. Using the results of Problem 24.12, we have as the solution
to just the differential equation

Applying the initial conditions to this last equation, we find that d Q = — je and d l =^e~', hence,