Simple inverse using Laplace transform

Question

I have the following excercise. I looked at the Laplace transform table for said transform, but I can't find any that looks similar to this. Help, please?

$$ \mathcal{L} ^ {-1} \left[ \frac{s} {((s + 3) ^ 2 + 2)} \right] $$

Thanks.

Answer 1

Hint:

If you let me call: $F(s) = \frac{s}{(s+3)^2+2}$, then remember that:

• You can rearrange $F(s)$ as follows:

$$F(s) = F_1(s) + F_2(s) = \frac{s+3}{(s+3)^2+2} - \frac{3}{\sqrt{2}}\frac{\sqrt{2}}{(s+3)^2+(\sqrt{2})^2},$$

• Remember that: $\mathcal{L}^{-1}_t\left[ \frac{a}{s^2+a^2}\right] = \sin{at},$ and $\mathcal{L}^{-1}_t\left[ \frac{s}{s^2+a^2}\right] = \cos{at}$.

• Also recall that $\mathcal{L}^{-1}[G(s-b)] = e^{bt} g(t)$ if $G(s) = \mathcal{L}[g(t)]$.

Use this to conclude that:

$$f(t) = \mathcal{L}^{-1}[F(s)] = e^{-3t} \cos{\sqrt{2} t} - \frac{3}{\sqrt{2}} e^{-3t} \sin{\sqrt{2} t}. $$

Cheers!