Question

find Laplace Inverse of : 1/(s^2+6s+13)^2

Answer 1

Use the fact that L^-1(b/(s+a)^2 + b^2) = e^(-at)sin(bt)
Now it's been squared, use convolution integral
L(f)L(g) = L(f*g)
where f = e^(-at)sin(bt)
& g= e^(-at)sin(bt)

I know there's a better method, so i'll come back...

Answer 2

i know a lot better method ... it's called wolfram :P
http://www.wolframalpha.com/input/?i=laplace+inverse+1%2F%28s^2%2B6s%2B13%29^2+
( though currently not working )

Answer 3

Okay this shuld help... Find laplace transform of sinbt, cosbt, tsinbt and tcosbt... and then u have to look for combinations which will give you the desired result... If u need the full solution, pls tell me.


@experimentX : Lolz

Answer 4

inverse transform are sure a hell of trouble ... :D