2nd order homogeneous repeated roots

Question

$$a_{n+2}-6a_{n+1}+9a_{n}= 3^n \quad n\geq 0 \quad a_{0}=2 \quad a_{1}=3$$ Got repeated roots of 3, so $a_{n}= A\cdot3^n+b\cdot (n\cdot3^n)$ how would i calculate A+B when there is $3^n$?

Edit: So i calculated $a_{n}=6\cdot3^n - n\cdot3^n$ can anyone check if thats right?

Thanks!

Answer 1

Your answer IS correct. To compute the constant, just try $n=0, 1$ (two equations for two unknowns).