Recognitions:
Homework Help

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

Think of it like this, you know that:
$$\frac{1}{1-x}=1+x+x^{2}+x^{3}+x^{4}+\cdots$$
Now differentiate both sides to show that:
$$\frac{1}{(1-x)^{2}}=1+2x+3x^{2}+4x^{3}+\cdots$$
and now you know the power series for $$(1-x)^{2}$$ is just by differentiating the series.
 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
 Recognitions: Homework Help 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?
 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.
 Recognitions: Homework Help well done, got there in the end.