Inverse laplace transform of h(s) = 1/(s+4)^2

Inverse Laplace of will be

Step-by-step explanation:

We have to find the inverse Laplace transform of

We know that of is

As in H(s) there is square of

So i inverse Laplace there will be multiplication of t

So the inverse Laplace of  will be

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23