Find the solution of the IVP using Laplace transforms

Question

The equation is as such:

$y''+y=t\sin t$; $y(0)=1, y'(0)=2$

I took the Laplace transform of both sides to yield

$F(S)(s^{2}+1)-(s+2)=\frac{-2s}{(s^{2}+1)^{2}}$, and then

$F(s)=\frac{s}{s^{2}+1}+\frac{2}{s^{2}+1}+\frac{-2s}{(s^{2}+1)^{3}}$, and finally

$F(s)=\mathcal{L}\{\cos t\}+2\mathcal{L}\{\sin t\}+\frac{-2s}{(s^{2}+1)^{3}}$.

I can't figure out what would be the function whose Laplace transform is given by $\frac{-2s}{(s^{2}+1)^{3}}$, any help would be appreciated.

Answer 1

By using the convolution theorem we have $$\frac{-2s}{(s^{2}+1)^{3}}=\frac{-2s}{(s^{2}+1)^{2}}*\frac{1}{(s^{2}+1)}=h*g$$ where $$g(t)=\sin t$$ and $$H(s)=\frac{-2s}{(s^{2}+1)^{2}}=\frac{d}{ds}\frac{1}{(s^{2}+1)}\\ h(t)=-t\sin t$$

Answer 2

The ILT of that third term is the sum of the residues of the function times $e^{s t}$ at the poles $s=i$ and $s=-i$. because these are triple poles, each of the residues is a second derivative. For example:

$$\operatorname*{Res}_{s=i} \frac{-2 s \, e^{s t}}{(s^2+1)^3} = \frac1{2!} \left [\frac{d^2}{ds^2} \frac{-2 s\, e^{s t}}{(s+i)^3}\right ]_{s=i} = \frac18 t (t+i) e^{i t}$$ $$\operatorname*{Res}_{s=-i} \frac{-2 s\, e^{s t}}{(s^2+1)^3} = \frac1{2!} \left [\frac{d^2}{ds^2} \frac{-2 s\, e^{s t}}{(s-i)^3}\right ]_{s=-i} = \frac18 t (t-i) e^{-i t}$$

The ILT is the sum of these terms, or

$$\frac14 t^2 \cos{t} - \frac14 t \sin{t}$$