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"...:tongue2:

 

[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:

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