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?

How to find the particular and general solution to $y''+4y'+5y=t^2$.
Given $y''+4y'+5y=t^2$, we must assume this is a non-homogeneous second order ODE.
We from this there are two general forms to solve this problem, undetermined coefficients and variation of parameters. Here we will use undetermined coefficients.
We must now also assume that the Method of Undetermined Coefficients is a systematic way to determine the general form/type of the particular solution $y_p$ based on the non-homogeneous term (forcing function) $f(x)$ in the given ODE.
First, we should find the homogeneous solution so we can apply this to the general solution later on. We do this by using $λ$ in place of the $y$ and using exponents in place of the order of the derivative and setting the equation equal to zero. After doing this we will solve for $λ$.
$y''+4y'+5y=t^2$
$λ^2+4λ+5=0$
We have to use the quadratic formula in order to solve for $λ$.
We know the quadratic formula is,
$\frac{-b±\sqrt{(b)^2-4ac}}{2a}$
Thus,
$\frac{-(4)±\sqrt{(4)^2-4(1)(5)}}{2(1)}=\frac{-4±\sqrt{(16-20)}}{2}=\frac{-4±2i}{2}=-2±i$
From this we can write our homogeneous solution:
$y_h(t)=c_1e^{-2t}\cos(t)+c_2e^{-2t}\sin(t)$
Now we must find and use a trial solution, $y_t$. We know the forcing function is $t^2$. Thus the basic trial function for a function with the exponential $2$, we know the trial must be a polynomial function.
$y_t=at^2+bt+c$
From here, the derivative and second derivative must be taken.
$y'_t=2at+b$
$y''_t=2a$
Now, plugging in these results into our original ODE we should get:
$2a+4(2at+b)+5(at^2+bt+c)=t^2$
$5at^2+t[8a+5b]+[2a+4b+c]=t^2$
Here, we can equate the coefficients in order to solve for the unknowns.
$x^2: 5a=1$
$a=\frac{1}{5}$
$x: 8a+5b=0$
$8(\frac{1}{5})+5b=0$
$b=\frac{-8}{25}$
constants:$2a+4b+c=0$
$2(\frac{1}{5})+4(\frac{-8}{25})+c=0$
$c=\frac{22}{25}$
Thus,
$a=\frac{1}{5}, b=\frac{-8}{25}, \text{ and } c=\frac{22}{25}$
We can then plug these into our trial solution to form our particular solution:
$y_p=\frac{1}{5}t^2-\frac{8}{25}t+\frac{22}{25}$
To find the general solution, we know that:
$y_g(t)=y_h(t)+y_p(t)$
Thus, our general solution is:
$y_g(t)=c_1e^{-2t}\cos(t)+c_2e^{-2t}\sin(t)+\frac{1}{5}t^2-\frac{8}{25}t+\frac{22}{25}$
