I've encountered an IVP, and I need to solve it by applying the Laplace transform. The problem is:

$$y^{''}+4y=g(t), y(0)=y^{'}(0)=1$$

where $g(t)=\sin(t)$ for $0\leq{t}\leq\pi$ and $g(t)=0$ for $\pi\leq{t}$

Now, the Laplace transform is:

LHS: $L\{y^{''}\}+L\{4y\}=s^2Y(s)-sy(0)-y^{'}(0)+4Y(s)=(s^2+4)Y(s)-s-1$

RHS: $L\{g(t)\}=L\{\sin(t)-\sin(t)u(t-\pi)\}=\frac{1-e^{-{\pi}s}}{s^2+1}$

Then I get:

$$Y(s)=\frac{1-e^{-{\pi}s}}{(s^2+1)(s^2+4)}+\frac{s+1}{s^2+4}$$

Now I'm struggling with finding the inverse transform of $Y(s)$. Can anyone help me with this probelm. Thank you in advance!