Inverse Laplace transform of $\frac{s^2}{(s^2+a^2)^2}$ using frequency differentiation.

Question

I have already obtained that $\mathcal{L}^{-1}\left\{\frac{s}{(s^2+a^2)^2}\right\}=\frac{t \sin at}{2a}$.

I have tried using splitting $s \cdot \frac{s}{(s^2+a^2)^2}$ but I am stuck to solve.

Any hint for solving this question? Thanks.

(It is known that the answer is $\frac{at \cos at + \sin at}{2a}$

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Edit 1:

$\mathcal{L}[f']=s\mathcal{L}[f]-f(0)$
$f'(t) =\frac{at \cos at + \sin at}{2a}$
$\mathcal{L}\left[\frac{at \cos at + \sin at}{2a} \right] = s \mathcal{L} \left[ \frac{t \sin at}{2a} \right] = \frac{s^2}{(s^2+a^2)^2}$
$\frac{at \cos at + \sin at}{2a} = \mathcal{L}^{-1} \left[ \frac{s^2}{(s^2+a^2)^2} \right]$

Answer 1

For any map $f$, you have

$$s \cdot L(f) - f(0^+) = L(f^\prime).$$ Either you know this result, or you can find it back by integration by part.

Applying it to $f(t) = \frac{t \sin at}{2a}$ immediately provides the result you're looking for.