How to solve $y''+Ay'+By=a+bt+ct^{2}$? I know that the solution of this second-order non-homogeneous will be of the form $y = y_{p} + y_{c}$.
First I find the solution to the homogeneous DE: $y''+Ay'+By=0$, with the characteristic equation $r^{2}+Ar+B=0$. However, it's hard to actually get the roots since I don't know what $A$ and $B$ are. 
Any ideas on how to overcome this problem?
EDIT: By using the quadratic formula: $r = \frac{-A\pm \sqrt{A^{2}-4B}}{2}$. 
So we don't really know about the form of the complementary solution, $y_{c}$.
Now, we find the particular solution of $y''+Ay'+By=a+bt+ct^{2}$.
Let $y_{p} = A_{1}t^{2}+B_{1}t+C_{1}$.
So $y'_{p} = 2A_{1}t+B_{1}$ and $y''_{p} = 2A_{1}$.
We substitute $y_{p}$ and it's first and second derivatives back into the non-homogeneous DE to obtain:
$t^{2}(A_{1}B)+t(2AA_{1}+BB_{1})+(2A_{1}+AB_{1}+BC_{1})=ct^{2}+bt+a$
So $$A_{1}B=c;$$
$$2AA_{1}+BB_{1}=b;$$
$$2A_{1}+AB_{1}+BC_{1}=a.$$
From the above equations I obtain $A_{1} = \frac{B}{c}$, $B_{1} = \frac{cb}{B}-2A$, and $C_{1} = \frac{a - \frac{B}{c} - \frac{ACb}{B}+2A^{2}}{B}$.
So the particular solution, $$y_{p} = (\frac{B}{c}) t^{2} + (\frac{cb}{B}-2A)t + \frac{a - \frac{B}{c} - \frac{ACb}{B}+2A^{2}}{B}$$.  
Then how do I write down expression for the general solution, $y$.
Is it just $$y = (\frac{B}{c}) t^{2} + (\frac{cb}{B}-2A)t + \frac{a - \frac{B}{c} - \frac{ACb}{B}+2A^{2}}{B} + y_{c}?$$

to find the particular soultion,we assume
$$y_p=A_1t^2+B_1t+C_1$$
$$y'_p=2A_1t+B_1$$
$$y''_p=2A_1$$
substitute in D.E to get
$$2A_1+A(2A_1t+B_1)+B(A_1t^2+B_1t+C_1)=a+bt+ct^2$$
$$BA_1=c$$
$$A_1=c/B$$
so the $c$,$B$ are known 
$$2AA_1+BB_1=b$$
$$B_1=(b-2A(c/B))/B$$
$$2A_1+AB_1+BC_1=a$$
$$C_1=(a-2A_1-AB_1)/B$$
If you want to find the complementary solution
$$r^2+Ar+B=0$$
$$r=0.5A\mp \sqrt{0.25A^2-B}$$ 
$$r=\alpha \pm \beta $$
$$y_c=K_1e^{\alpha + \beta}+K_2e^{\alpha - \beta}$$
$$y_c=e^\alpha(K_1e^\beta+K_2e^{-\beta})$$
If the $\beta$ is complex part, you can change the exponential to $sin$ and $cos$
