$a_n = 4a_{n-1} - 4a_{n-2} + (n^2 + 1)2^n$
a) Find the general solution of the associated homogeneous equation.
b) Find the solution of the non-homogeneous relation, $a_0 = 0, a_1 = 1$
My work:
part (a):
$a_n - 4a_{n-1} + 4a_{n-2} = (n^2 + 1)2^n$
$a_n - 4a_{n-1} + 4a_{n-2} = 0$
$n^2 - 4n + 4 = 0$
$(n - 2)^2 = 0$
$n = 2$
So, $a_n = A(2)^n + B(2)^nn$
Is my part (a) correct?
part (b):
The reasonable result of the particular solution will be $(An^2 + Bn + C)2^n$
But and then, I don't know how to continue even I have substituted the result into the equation.
Inserting your solution into the homogeneous equation one gets: \begin{align} 0 &= A 2^n + B n 2^n - 4(A 2^{n-1} + B (n-1) 2^{n-1}) + 4 (A 2^{n-2} + B (n-2) 2^{n-2} \\ &=A 2^n + B n 2^n - 2A 2^n- 2 B(n-1) 2^n + A 2^n + B(n-2) 2^n \\ &= (A + n B - 2A - 2 B(n-1) + A + B(n-2)) 2^n \\ &= 0 \cdot 2^n \\ &= 0 \end{align} So your a) part looks fine.
Regarding b):
The general solution will be $$ a_n = (A + B n) 2^n + b_n $$ where $b_n$ is some particular solution. The conditions require $$ 0 = a_0 = A + b_0 \\ 1 = a_1 = 2(A+B) + b_1 $$ so the asked for solution is $$ a_n = (-b_0 + ( (1-b_1)/2+b_0) n) 2^n + b_n $$ This leaves the problem of finding one particular solution $b_n$. Any will do. Assuming $$ b_n = p(n) 2^n $$ for some order $N$ polynomial $p$ seems to be a possible way. I end up with a fourth order polynomial after some lengthy calculation.