How Do You Find the Laplace Inverse of \( \frac{40}{(s^2+4s+5)^2} \)?

[gethelpelectr]

Homework Statement

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

Homework 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.

The Attempt at a Solution

 

[dikmikkel]

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

 

[Ray Vickson]

Homework Statement

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

Homework 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.

The Attempt at a Solution

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

 

[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.

 

[Ray Vickson]

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

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