Problem 14. 9-14

We have:

We solve this and find

We can then use the inverse Laplace transform, and then we get:

h(t) = (1 – 2te ^-t – e ^-2t)u(t)

Initial and final value:

The limit of sH(s) when s approaches infinity is 0, and when we look at h(t) we get h(0) = 0.

The limit of sH(s) when s approaches 0 is 1, and when we look at h(t) we get the limit of h(t) when t approaches infinity is 1 also.

Thus the initial and final value theorems are verified here.