Lecture Notes in Differential Equations - Bruce E. Shapiro

356 LESSON 32. THE LAPLACE TRANSFORM
Since when t 0 = 0, e −t0s = e 0 = 1,
Example 32.20. Solve the initial value problem
The Laplace transform gives us
L[δ(t)] = 1 (32.231)
y ′′ + y = δ(y − π), y(0) = y ′ (0) = 0 (32.232)
s 2 Y (s) − sy(0) − y ′ (0) + Y (S) = e −πs (32.233)
But recall that (see (32.156))
(s 2 + 1)Y (s) = e −πs (32.234)
Y (s) = e−πs
1 + s 2 = (32.235)
L[f(t − a)U(t − a)] = e −as F (s) (32.236)
=⇒ f(t − a)U(t − a) = L −1[ e −as F (s) ] (32.237)
=⇒ f(t − π)U(t − π) = L −1[ e −aπ F (s) ] (32.238)
If we let F (s) = 1/(1 + s 2 ) then f(t) = sin t. Hence
[ ] e
U(t − π) sin t = L −1 −πs
1 + s 2
(32.239)
and therefore
y(t) = U(t − π) sin t (32.240)