# Thread: [SOLVED] prove using laplace transforms

1. ## [SOLVED] prove using laplace transforms

cant figure out how to show that they are equal. can anyone help?

2. Originally Posted by mathprincess24

cant figure out how to show that they are equal. can anyone help?

$\mathcal{L}^{-1} \left(\frac{1}{s^2 + k^2} \right) = \frac{\sin (kt) - kt \cos (kt))}{2k^3}$
$\frac{2k^2}{(s^2 + k^2)^2}= \frac{s^2 + k^2 + k^2 - s^2}{(s^2 + k^2)^2}=\frac{1}{s^2 + k^2} + \frac{k^2 - s^2}{(s^2 + k^2)^2}.$ but we have that $\frac{k^2 - s^2}{(s^2 + k^2)^2}=\frac{d}{ds}\left(\frac{s}{s^2 + k^2} \right)=\frac{d}{ds} \mathcal{L}(\cos (kt))=\mathcal{L}(-t\cos(kt)). \ \ \color{red} (*)$

thus: $\frac{2k^2}{(s^2 + k^2)^2}= \frac{1}{k} \mathcal{L}( \sin(kt)) + \mathcal{L}(-t \cos(kt))=\mathcal{L} \left(\frac{\sin(kt) - kt \cos(kt)}{k} \right),$ and the claim follows.

$\color{red} (*)$: recall that in general: $\frac{d}{ds} \mathcal{L}(f(t))=\mathcal{L}(-tf(t)).$

3. thanks so much. that makes it a lot more clear now.