Solve the differential equation $y''+4y=g(t)$ using Laplace transformation How can i solve the following differential equation using Laplace Transformation?
$$y'' + 4y = g(t)$$ 
$$Y(0)=0; Y'(0)=0$$
where: 
$$
 g(t) =
\begin{cases} 
0; &  0 \leq t <5  \\ 
(t-5)/5; &  5 \leq t <10 \\
1; &  t \geq 10 \\
\end{cases},
$$
 A: Hint: You could rewrite $g(t)$ using the Heaviside function $u(t-c)$. Observe we have
\begin{align}
g(t) = \frac{t-5}{5}[u(t-5)-u(t-10)]+u(t-10) = \frac{t-5}{5}u(t-5)+\left( 1-\frac{(t-10)+5}{5}\right)u(t-10)
\end{align}
which means
\begin{align}
(s^2+4)Y(s)=&\ \mathcal{L}\{y''+4y\} =\mathcal{L}\{g(t)\} = \frac{e^{-5s}}{5s^2}-\frac{e^{-10s}}{5s^2} \\
\implies&\ \ Y(s) = \frac{e^{-5s}}{5s^2(s^2+4)}- \frac{e^{-10s}}{5s^2(s^2+4)}
\end{align}
A: As already mentioned, solve these 3 equations \begin{align}y''+4y&=0 \\ y''+4y&=\frac{t-5}{5} \\ y''+4y&=1 \end{align}
with Laplace transformation and then glue the solutions together. $$\mathcal{L}_t\{y''+4y\}(s)=(s^2+4)\mathcal{L}_t\{y\}(s)$$ should be pretty standard, so it remains to calculate $\mathcal{L}_t(0), ~\mathcal{L}_t\{1\}, ~  \mathcal{L}_t\{\frac{t-5}{5}\}.$ Notice that
$$\mathcal{L}_t\{\frac{t-5}{5}\}(s)=\mathcal{L}_t\{\frac{t}{5}-1\}(s)=\frac{1}{5s^2}-\frac{1}{s}=\frac{1-5s}{5s^2}$$
So we get
\begin{align} \mathcal{L}_t\{y\}(s)&=0 \\ \mathcal{L}_t\{y\}(s)&=\frac{1}{s(s^2+4)} \\ \mathcal{L}_t\{y\}(s)&=\frac{1-5s}{5s^2(s^2+4)} \end{align}
Now use partial fraction decomposition and apply the inverse Laplace transform. Keeping in mind that
$$\mathcal{L}^{-1}_t\{\frac{1}{s^2} \}(t)=t, ~\mathcal{L}^{-1}_t\{\frac{1}{s} \}(t)=1, ~\mathcal{L}^{-1}_t\{\frac{1}{s^2+4} \}(t)=\frac{\sin(2t)}{2}, ~\mathcal{L}^{-1}_t\{\frac{s}{s^2+4} \}(t)=\cos(2t) $$
should do the job.
