Solve ordinary differential equation using Laplace transform I have trouble to solve the differential equation. I can write derivatives of Laplace transforms but I can't do anything 
$$
\ddot y(t)+3y(t)=\sin(t)\text{ with } y(0)=1,\,\dot y(0)=2
$$
 A: I think I have a solution. Define the Laplace transform as 
\begin{equation*}
\mathcal{L}_t[f(t)](s)=\int^{\infty}_{0}f(t)e^{-st}dt.
\end{equation*}
Applying this to both sides of the equation gives 
\begin{equation*}
\mathcal{L}_t[y''(t)+3y(t)](s)=\mathcal{L}_t[\sin(t)](s)\\
\Rightarrow \mathcal{L}_t[y''(t)](s)+3(\mathcal{L}_t[y(t)](s)).
\end{equation*}
Using the Laplace transform identity for double derivatives gives 
\begin{equation*}
3(\mathcal{L}_t[y(t)](s))-(sy(0))+s^2\mathcal{L}_t[y(t)](s)-y'(0)=\mathcal{L}_t[\sin(t)](s)\\
\Rightarrow 3(\mathcal{L}_t[y(t)](s))+s^2(\mathcal{L}_t[y(t)](s))-sy(0)-y'(0)=\frac{1}{s^2+1}\\
\Rightarrow (s^2+3)(\mathcal{L}_t[y(t)](s))-sy(0)-y'(0)=\frac{1}{s^2+1}\\
\Rightarrow \mathcal{L}_t[y(t)](s)=\frac{y(0)s^3+y(0)s+y'(0)+y'(0)s^2+1}{s^4+4s^2+3}\\
\Rightarrow \mathcal{L}_t[y(t)](s)=\frac{1}{2(s^2+1)}-\frac{1}{2(s^2+3)}+\frac{sy(0)}{s^2+3}+\frac{y'(0)}{s^2+3}.
\end{equation*}
Now we compute inverse Laplace transforms term-by-term to get:
\begin{equation*}
y(t)=\frac{\sin(t)}{2}-\frac{\sin(\sqrt{3}t)}{2\sqrt{3}}+y(0)\cos(\sqrt{3}t)+\frac{y'(0)\sin(\sqrt{3}t)}{\sqrt{3}}
\end{equation*}
Apply the initial conditions. Does that help?
