Inverse laplace transform question

[cabellos]

How do i find the inverse laplace transform of s/(s-2)^2

 

[Karlisbad]

take the transform of f(t)=(1+t)e^{2t}

using the "shift" property so..

F(s+a)=L[e^{-at}f(t)] using

\frac{s+2}{s^{2}}=s^{-1}+2s^{-2}

the inverse of above is just 1+t then multiply it by exp(2t) and you get it without recalling "residue theorem"...

 

[EugP]

Use partial fractions:

\frac{s}{(s-2)^2} = \frac{A}{(s-2)^2} + \frac{B}{s-2}

To find A, multiply both sides by s-2 and evaluate at s=2:

s = A

A = 2

Now to find B, go back to the original expression again, and multiply (again) both sides by s-2,
then differentiate with respect to s and evaluate at s=2:

\frac{d}{ds}s = \frac{d}{ds}[B(s-2)]

B = 1

Now plug A and B into the original expression:

\frac{2}{(s-2)^2} + \frac{1}{s-2}

So the inverse Laplace would be:

L^{-1} = [2te^{2t}+e^{2t}] u(t)