Solving $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$ I'm trying to solve the recurrence $a_n=3a_{n-1}-2a_{n-2}+3$ for $a_0=a_1=1$.  First I solved for the homogeneous equation $a_n=3a_{n-1}-2a_{n-2}$ and got $\alpha 1^n+\beta 2^n=a_n^h$.  Solving this gives $a_n^h =1$.  The particular solution, as I understand, will be $a_n^*=B$ since $f(n)=3\times 1^n$.  But then I get $B=a^*=3a^*_{n-1}-2a^*_{n-2}+3=B+3$.  This has to be a mistake, but I don't see what I did wrong.

Hint: Look for a particular solution of the shape $a_n^\ast=Bn$. You should get $B=-3$.
