Can someone help solving this differential equation using Laplace transform? This is the equation and it's conditions:
$$ y''(t)+y(t)=e^{-2t}+\int_{0}^{t} \cos(t-s)y(s)ds, \quad t\geq0 $$
$$ y(0) = 0, \quad y'(0) = 1 $$
When I try solving it I get :
we set $Y$ to be the Laplace transform of $y(t)$
$$\mathcal L(y'')=s^2Y-sy(0)-y'(0)=s^2Y-1$$
$$\mathcal L(e^{-2t})=\frac{1}{s+2}$$
$$\mathcal L\left(\int_{0}^{t}\cos(t-s)y(s) \mathrm{d}s\right)=\frac{sY}{s^2+1}$$
Therefore when applying the Laplace transform we get:
$$Y(s)=\frac{(s+3)(s^2+1)}{\big((s^2+1)^2-s\big)(s+2)}$$
And here where I'm stuck I can't figure a way to apply the inverse transform.
Some additional information this exercise was into my math exam and no one could figure it out and since the results didn't came out yet I wanted to be sure if I was wrong or the exercise is unrealizable without computer.
 A: Starting from your expression for the partial fraction decomposition
\begin{align*}
\mathcal{L}[y](s) = \frac{(3 + s) (1 + s^2)}{(2 + s) (1 - s + 2 s^2 + s^4)}
\end{align*}
Now, Mathematica has a nice command to reduce the tedium of partial fractions, and of course I think we should use it:

Apart[((3 + s) (1 + s^2))/((2 + s) (1 - s + 2 s^2 + s^4)), s]

The result is
\begin{align*}
\frac{5}{27 (2 + s)} + \frac{38 - 3 s + 37 s^2 - 5 s^3}{
 27 (1 - s + 2 s^2 + s^4)}
\end{align*}
The first is easy, but the second is painful:
\begin{align*}
y(t) = \frac{5}{27}\exp(-2t)+ \frac{1}{27}\mathcal{L}^{-1}\left[ \frac{38 - 3 s + 37 s^2 - 5 s^3}{1 - s + 2 s^2 + s^4} \right](t)
\end{align*}
Once we find the roots of $1 - s + 2 s^2 + s^4$, we can do partial fraction decomposition, but the result is not enlightening. It is, as you say, not solvable without a computer, and with a computer is it not useful. Perhaps there is some other identity that allows this to be worked out nicely, but I am not personally aware of what it might be.
