I'm struggling with an inverse laplace transform:

1/((s^2+1)^2)

What I'd normally do is to use the convolution theorem, and get y(t)=sin^2(t), but I'm being a bit of a retard, and can't seem to get my answer to fit with the suggested solution.

Question

I'm struggling with an inverse laplace transform:

1/((s^2+1)^2)

What I'd normally do is to use the convolution theorem, and get y(t)=sin^2(t), but I'm being a bit of a retard, and can't seem to get my answer to fit with the suggested solution.

anonymous · Answer

inverse laplace of $$1\div(s ^{2}+1)$$ is sint
using convolution we need to find
$$\int\limits_{0}^{t} siny \sin(t-y) dy$$

anonymous · Answer

sinAsinB = 1/2[cos(A - B) - cos(A + B)]

anonymous · Answer

As I suspected, I'm a retard. I forgot to calculate the convolution after I had found the inverse transform. Thanks!

anonymous · Answer

welcome