# Laplace Inverse of a solution

1. May 4, 2012

### gethelpelectr

1. The problem statement, all variables and given/known data

laplace inverse of (40/(s^2+4s+5)^2)?
2. Relevant equations

I completed the square in the denominator to get 40/((s+2)^2+1)^2
I know that I will get cosines and sines from the shape of it in the laplace inverse; however I'm stuck.

3. The attempt at a solution

2. May 4, 2012

### dikmikkel

I get the follwing result:
$20e^{-2t}(\sin(t)-t\cos(t))$
But couldn't find any formula, so i did it with Maple 15

3. May 4, 2012

### Ray Vickson

It will be a lot easier if you first factor the expression p = s^2 + 4s + 5, then convert your expression 1/p^2 to partial fractions.

RGV

4. May 4, 2012

### gethelpelectr

dikmikkel This answer is correct; however I don't know how to get there.
Ray Vickson, we are not used to getting complex numbers.

5. May 4, 2012

### Ray Vickson

You should get used to it; they are part of a standard toolkit and are used routinely in Physics, Engineering and Applied Math. However, if you don't want to use complex quantities you can use the convolution theorem instead: first get the inverse Laplace transform of 1/(s^2 + 4s + 5), then find that of 1/(s^2 + 4s + 5)^2 by convolution.

RGV

Last edited: May 4, 2012