Show that the general solution to the recurrence equation $a_n+ba_{n-1} + ca_{n-2} = 0$ can be written as $a_n = Ar_1^n + Br_2^n$ if the roots $r_1$ and $r_2$ of the characteristic equation are unequal. Also account for the case where the roots are equal.
So my attempt is as follows:
Making the ansatz above, the homogenous equation becomes \begin{align*} aAr_1^n + aBr_2^n + bAr_1^{n-1} + bBr_2^{n-1} + cAr_1^{n-2} + cBr_2^{n-2} &= \\ Ar_1^{n-2}(ar_1^2 + br_1^{1} + cr_1) + r_2^{n-2}B(ar_2^2 + br_2^{1} + cr_2) &= 0 \end{align*} since we know that both of the paranthesis are zero, by the characteristic polynomial. I'm not quite sure how to do the latter part, and I'm not entirely sure about my proof either, since I never seem to use the fact that the roots are distinct.
If $r_1=r_2$, then you have obtained a solution with just $1$ arbitrary constant, since you can replace $A+B$ by a constant $C$.
In the equal roots case you must use $(An+B)(ar^{n+2}+br^{n+1}+cr^{n})$.