(This question is about laplace transforms)
By making use of the convolution theorem show that the solution $y(t)$ to the ODE $$\ddot{y}(t)+4\dot{y}(t)+5y(t)=u(t), \quad y(0)=0,\quad \dot{y}(t)=0,$$ for any input $u(t)$ is $$y(t)=\int_0^te^{-2(t-r)}\sin{(t-r)}u(r)\, {\rm d}r.$$
So first we apply the Laplace transform to the ODE to yield $$\mathcal{Y}(s)=\frac{1}{s^2+4}\mathcal{U}(s).$$ Now we let $\mathcal{F}(s) = 1/(s^2+4)$. So we have that $\mathcal{Y}(s)=\mathcal{F}(s)\mathcal{U}(s)$. Thus by the convolution theorem $$y(t) = \int_0^tf(t-r)u(r)\, {\rm d}r.$$ Now, $$\mathcal{F}(s) = \frac{1}{s^2+4} \implies f(t) = \frac12\sin{(2t)}.$$ $$y(t) = \int_0^t \frac12\sin{(2(t-r))}u(r)\, {\rm d}r.$$ Which is clearly incorrect, what have I done wrong?
Error in finding the laplace transform. You should have $$\mathcal{Y}(s) = \frac{1}{s^2+4s+5}\mathcal{U}(s).$$