Having
$a_n = 3a_{n-1} + 2a_{n-2} + 3·2^{2n-1}$
$a_1 = 12$
$a_0 = 0$
I solved the homogeneous part and got:
$a^{{h}}_n = 1/12·2^n - 1/12·1^n$
This is the particular solution that I need to solve:
$a^{p}_n = C·3·2^{2n-1}$
but the exponent 2n-1 confuses me. I'm don't know how to solve it.
I appreciate any help
We try for a particular solution of the shape $c\cdot 2^{2n-1}$. Substituting in our recurrence, we have $$c\cdot 2^{2n-1}=3c\cdot 2^{2n-3}+2c\cdot 2^{2n-5}+3\cdot 2^{2n-1}.$$ Dividing by $2^{2n-5}$ we get $$16c=12c+2c+48.$$
Remark: The algebra may feel simpler if we look for a solution of the shape $d\cdot 2^{2n}$, that is, $d\cdot 4^n$. We get $d\cdot 4^{n}=3d\cdot 4^{n-1}+2d\cdot 4^{n-2}+\frac{3}{2}\cdot 4^n$. Divide through by $4^{n-2}$.