Lecture Notes in Differential Equations - Bruce E. Shapiro

351
The inverse transform is
[ ]
f(t) = L −1 1
s(s + 1)
[ ] [ ]
1 1
= L −1 L −1
s s + 1
(32.178)
(32.179)
= 1 · e −1·t (32.180)
= e −t (32.181)
Example 32.17. Find a function f(t) whose Laplace Transform is
F (s) =
The inverse transform is
[
]
f(t) = L −1 s
(s 2 + 4s − 5) 2
Using partial fractions,
[
= L −1 s
s 2 + 4s − 5 ·
[
= L −1 s
(s + 5)(s − 1) ·
[
= L −1 s
(s + 5)(s − 1)
s
(s 2 + 4s − 5) 2 (32.182)
]
1
s 2 + 4s − 5
1
(s + 5)(s − 1)
]
· L −1 [
1
(s + 5)(s − 1) = −1 6 ·
s
(s + 5)(s − 1) = 5 6 ·
]
]
1
(s + 5)(s − 1)
1
s + 5 + 1 6 ·
1
s + 5 + 1 6 ·
1
s − 1
1
s − 1
The inverse transforms of (32.187) and (32.188) are thus
[
]
L −1 1
= − 1 [ ] 1
(s + 5)(s − 1) 6 L−1 + 1 [ ] 1
s + 5 6 L−1 s − 1
(32.183)
(32.184)
(32.185)
(32.186)
(32.187)
(32.188)
(32.189)
= − 1 6 e−5t + 1 6 et (32.190)
[
]
L −1 s
(s + 5)(s − 1)
= 5 [ ] 1
6 L−1 + 1 [ ] 1
s + 5 6 L−1 s − 1
(32.191)
= 5 6 e5t + 1 6 et (32.192)