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

by fysiikka111
Tags: inverse, laplace, transforms or s2
 P: 41 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
 HW Helper P: 1,584 when you differentiate the integral you take in differentiation under the integral sign: $$\frac{d}{ds}\int_{0}^{\infty}e^{-st}\sin tdt=\int_{0}^{\infty}\frac{d}{ds}\left( e^{-st}\sin t\right) dt$$ Differentiating the integrand: $$\frac{d}{ds}e^{-st}\sin t=-te^{-st}\sin t$$ As now we regard s as variable and t is fixed. So the above can be written as: $$(-t\sin t)e^{-st}$$ 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?
 P: 41 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.
 HW Helper P: 1,584 well done, got there in the end.

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