Lecture Notes in Differential Equations - Bruce E. Shapiro

348 LESSON 32. THE LAPLACE TRANSFORM
Inverting the transform,
[ ]
1
f(t) = L −1 [F (s)] = L −1 s − 2 + 4
(s − 2)
[ ] [ ]
2
1
= L −1 + 4L −1 1
s − 2 (s − 2)
[ ] [ ]
2
1 1
= e 2t L −1 + 4e 2t L −1
s
t 2
(32.140)
(32.141)
(32.142)
= e 2t · 1 + 4e 2t · t (32.143)
= (1 + t)e 2t ] (32.144)
Example 32.14. Solve y ′ +4y = e −4t , y(0) = 2, using Laplace Transforms.
Applying the transform,
L[y ′ ] + 4L[y] = L [ e −4t] (32.145)
sY (s) − y(0) + 4Y (s) = 1
s + 4
(32.146)
(s + 4)Y (s) = 2 + 1
s + 4
(32.147)
Solving for Y (s),
Y (s) = 2
s + 4 + 1
(s + 4) 2 (32.148)
y(t) = L −1 [Y (s)] (32.149)
[ ]
2
= L −1 s + 4 + 1
(s + 4) 2 (32.150)
[ ] [ ]
2
= L −1 + L −1 1
s + 4 (s + 4) 2 (32.151)
[ ] [ ]
1 1
= 2e −4t L −1 + e −4t L −1
s
s 2 (32.152)
= 2e −4t · 1 + e −4t · t (32.153)
= (2 + t)e −4t (32.154)