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 $$y = (\frac{B}{c}) t^{2} + (\frac{cb}{B}-2A)t + \frac{a - \frac{B}{c} - \frac{ACb}{B}+2A^{2}}{B} + y_{c}?$$