Stuck on a particular 2nd order non-homogeneous ODE ($y'' + 4y' + 5y = t^2$) 
$y'' + 4y' + 5y = t^2$

So I solve for $r^2 + 4r + 5 = 0$ returning $r = 2 \pm 2i$. So $y_c = e^{-2t}(C_1\cos(2t) + C_2\sin(2t)$. For $y_p(t)$ I pick $At^2$. So $2A + 8At + 5At^2 = t^2$. I have been stuck on finding a particular solution for 30 minutes now.
My first question: Is there any particular method for finding $y_p(t)$?
Secondly, how to find $y_p(t)$ here?

You should use method of undetermined coefficients. 
Take
$$y_{p}(t)=at^2+bt+c$$
then
$$
y'_{p}(t)=2at+b
$$
$$
y''_{p}(t)=2a
$$
and substitue it in differential equation
$$2a+8at+4b+5at^2+5bt+5c=t^2$$
from equality of polynomials we have
$5a=1,8a+5b=0,2a+4b+5c=0$
from here we can find $a=\frac{1}{5},b=\frac{-8}{25},c=\frac{22}{125}$.
So 
$$
y_{p}(t)=\frac{1}{5}t^2+\frac{-8}{25}t+\frac{22}{125}
$$
