initial value problem: y'' + 4y = f(t) , y(0)= y'(0)=0. f(t) = { 0 if t 3} Solve the initial value problem: 

$$y'' + 4y = f(t) , y(0)= y'(0)=0. $$
where
$$ f(t) = \begin{cases} 0 &amp;t &lt; 3 \\ t &amp; t &gt; 3\end{cases} $$

I've solved for the homogeneous equation, $y'' + 4y = 0,$ to get the general solution, 
$y = A\cos(2t) + B \sin(2t)$ for some arbitrary constants $A$ and $B$. 

But, now where do I go next? 

Thanks!
A related problem. In fact, you need to solve these two differential equations

$$ \begin{cases} y'' + 4y = 0 &amp;t &lt; 3 \\ y'' + 4y = t &amp; t &gt; 3\end{cases} .$$
I would use a Laplace transform:

$$\scr{L}(y''+4y)=\scr{L}f(t)$$

$$s^2Y(s)+4Y(s)=F(s)$$

$$Y(s)=\frac{F(s)}{s^2+4}$$

$$y(t)=\scr{L}^{-1} \frac{f(s)}{s^2+4}$$

The Laplace transform of $f(t)$ is $\frac{1}{s^2}e^{-3s}$.

$$y(t)=\scr{L}^{-1}\frac{1}{s^2(s^2+4)}e^{-3s}$$

Using partial fraction decomposition:

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

$$\frac{1}{4}\scr{L}^{-1}\{\frac{1}{s^2}-\frac{1}{s^2+4}\}=\frac{1}{4}t-\frac{1}{8}\sin{2t}\space u(t)$$

$$\scr{L}^{-1}e^{-3s}=\delta{(t-3)}$$

The property of convolution states that:

$$f(t) * g(t)=\scr{L}\{f(s)g(s)\}$$

$$y(t)=\int_{-\infty}^\infty\delta{(\tau-3)}(\frac{1}{4}t-\frac{1}{4}\tau-\sin{(2(t-\tau))})u(t-\tau)d\tau$$
$$y(t)=(\frac{1}{4}t-\frac{3}{4}-\frac{1}{8}\sin(2t-6))u(t-3)$$