How do I use the convolution theorem to solve an initial value problem?

Question

I'm not quite sure how to use the convolution theorem for this problem. My attempt is that I took the laplace transform of both sides, however I'm confused as to what to do after. Thanks!

Answer 1

Note that using Laplace Transforms, we have

$$(s^2+4s+13)X(s)=F(s)$$

Hence, solving for $X(s)$ reveals

$$\begin{align} X(s)&=\frac{F(s)}{s^2+4s+13}\\\\ &=\frac{F(s)}{(s+2)^2+(3)^2} \\\\ &=\mathscr{L}\{f(t)\}\times \mathscr{L}\{e^{-2t}\sin(3t)u(t)\}\tag 1 \end{align}$$

Equation $(1)$ states that the Laplace Transform of $x(t)$ is equal to the product of the Laplace transform of $f(t)$ and the Laplace Transform of $h(t)=e^{-2t}\sin(3t)u(t)$.

The convolution theorem then guarantees that $x(t)=f(t)*h(t)$ and we can write

$$x(t)=\int_0^\infty f(t-u)e^{-2u}\sin(3u)\,du$$

as was to be shown!