Inverse Laplace Transform(s/((s^2)+1)^2

hunt_mat

Think of it like this, you know that:
[tex]
\frac{1}{1-x}=1+x+x^{2}+x^{3}+x^{4}+\cdots
[/tex]
Now differentiate both sides to show that:
[tex]
\frac{1}{(1-x)^{2}}=1+2x+3x^{2}+4x^{3}+\cdots
[/tex]
and now you know the power series for [tex](1-x)^{2}[/tex] is just by differentiating the series.

fysiikka111

So I know that differentiating both sides of the sin(at) LT will give me the square that I need in the denominator, but then I'm stuck with the d/ds on the sin side. I'm not sure what to do after this.
Thanks

hunt_mat

when you differentiate the integral you take in differentiation under the integral sign:
[tex]
\frac{d}{ds}\int_{0}^{\infty}e^{-st}\sin tdt=\int_{0}^{\infty}\frac{d}{ds}\left( e^{-st}\sin t\right) dt
[/tex]
Differentiating the integrand:
[tex]
\frac{d}{ds}e^{-st}\sin t=-te^{-st}\sin t
[/tex]
As now we regard s as variable and t is fixed. So the above can be written as:
[tex]
(-t\sin t)e^{-st}
[/tex]
and so you're taking the Laplace transform of something different now. That something is the answer to your question. Is this making sense now?

fysiikka111

I think I got it now.
After differentiating both sides and dividing by -2:
s/(s^2+a^2)^2=L{tsin at/2}
Hence,
L-1{s/(s^2+1)^2}=tsint/2
However, how do we know that "a" isn't -1, since (-1)^2=1.

hunt_mat

well done, got there in the end.