Solve ordinary differential equation using Laplace transform

Question

I have trouble to solve the differential equation. I can write derivatives of Laplace transforms but I can't do anything $$ \ddot y(t)+3y(t)=\sin(t)\text{ with } y(0)=1,\,\dot y(0)=2 $$

Answer 1

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?