I am having a hard time understanding these questions. I know I need to find the associated homogeneous recurrence relation first, then its characteristic equation. I cant figure out how to find the particular solution to the non homo recurrence relation though.
Ex: $$a_{n}= 4a_{n-1} + 4a_{n-2} + (n+1)2^n$$
My characteristic equation is $r^2-4r-4=0$ and $r=2(1+ \sqrt{2}), r=2(1- \sqrt{2})$. Next I need to guess some equation for my $f(n)=(n+1)2^n$ and plug it into the original to find some constants,, I am having the trouble here,, I dont understand how to come up with these guess equations.
I know the theorem that says the general solution (of the non homo recurrence relation) is the general solution of the associated recurrence relation + the particular solution: $a_{n}=a_{n}^{(h)} + a_{n}^{(p)}$
So far I have $a_{n}=A2(1+ \sqrt{2})^n + B2(1- \sqrt{2})^n + a_{n}^{(p)}$
Make sure you have enough functions to span the non-homogeneous term. In this case, the non-homogeneous term is $n2^n + 2^n$, so I would guess $a_n^{(p)} = An2^n + B2^n$. The reason this is enough is because the space of functions spanned by $\{n2^n, 2^n\}$ is invariant under the next operation, and it intersects the space of homogeneous solutions only at $0$.