Solve $y'' + 3y'+2y = \cos(at)$ by first solving $y'' + 3y'+2y = \delta(t)$ $$y'' + 3y'+2y = \delta(t)$$
$r^2+3r+2=0 \implies r = -1,-2$
so homogeneous solution is $ y_h(t) = C_1e^{-t} + C_2e^{-2t} $
I'm really not sure how to find a particular solution here.
I feel we have to make use of $\int\limits_{-\infty}^{+\infty} \delta(t)\, dt = u(t)$, but I'm a bit stuck... I don't see how to change the given equation into a  one involving $\int\delta\,dt$. Help appreciated..
EDIT :  $\delta(t)$ is the dirac delta function
 A: I suppose you want to solve it with the green's method:
Let
$$L(D)=D^2+3D+2$$
so your equation is 
$$L(D)y=\cos(at)$$
And you want to find the green function:
$$L(D)G(t-\tau)=\delta(t-\tau)$$
Let
$$C(\omega)=\frac{1}{L(i \omega)}$$
then your green function is:
$$G(t-\tau)=\int_{-\infty}^{\infty} \mathrm{d} \omega \frac{C(\omega)}{2 \pi} e^{i \omega (t-\tau)}$$
$$G(t-\tau)=\int_{-\infty}^{\infty} \mathrm{d} \omega \frac{1}{2 \pi(-\omega^2+3i\omega+2)} e^{i \omega (t-\tau)}$$
I think you can calculate this integral with Residue's theorem. After that, the solution will be:
$$y(t)=\int_{-\infty}^{\infty}\mathrm{d} \tau \cos(a \tau) G(t-\tau)$$
Edit:
Calculating the green's function:
Let's take a look at the numerator. If we let $\omega=a+bi$, then we have that $e^{i \omega (t-\tau)}=e^{i(a+bi)(t-\tau)}=e^{ia(t-\tau)}e^{-b(t-\tau)}$.  We have 2 cases: $t-\tau>0$ and $t-\tau<0$. If $t-\tau>0$, then we will need the enclosing circle to be in the upper half, and if $t-\tau<0$, then in the lower half.
The roots of the denominator are at $\omega=i$ and $\omega=2i$, so they are both at the upper half. This means that $G(t-\tau)$ is zero if $t-\tau<0$. In the case of $t-\tau>0$, the integral is just $2 \pi i \sum \text{Res}$:
$$G(t-\tau)=\frac{1}{2\pi} 2 i \pi \left(\frac{e^{-(t-\tau)}}{-2i+3i}+\frac{e^{-2(t-\tau)}}{-4i+3i}\right)$$
$$G(t-\tau)=i\left(\frac{e^{-(t-\tau)}}{i}-\frac{e^{-2(t-\tau)}}{i}\right)$$
$$G(t-\tau)=e^{-(t-\tau)}-e^{-2(t-\tau)}$$
So the solution is:
$$y(t)=\int_{-\infty}^{\infty}\mathrm{d} \tau \cos(a \tau) G(t-\tau)$$
$$y(t)=\int_{-\infty}^{t}\mathrm{d} \tau \cos(a \tau) \left(e^{-(t-\tau)}-e^{-2(t-\tau)}\right)$$
And, if I didn't mess up the evaluation of the last integral, the result is
$$y(t)=\frac{a\sin(at)+\cos(at)}{a^2+1}-\frac{a\sin(at)+2\cos(at)}{a^2+4}$$
