Tell me please if I'm doing this right way and if , so where my mistakes come from, because I ran out of ideas where to look at.
$a_0 = 10$
$a_1 = 10$
$a_n = 4a_{n-2} - 2^n \cdot n^2, \quad \text{for } n \geq 2$
Firstly I found homogeneous equation which came up to be
\begin{equation} \frac{15}{2} \cdot 2^n + \frac{5}{2} \cdot (-2)^n \end{equation}
After this I tried to find closed-from formula for this
\begin{equation} a_n^p=4a_{n-2}^p-2^n*n^2\\ a_n^p-4a_{n-2}^p=-2^n*n^2\\ \end{equation} and i let: \begin{equation} a_n^p=(An^2+Bn+C)*2^n\\ a_{n-1}^p=(A(n-1)2+B(n-1)+C)*2^{n-1}\\ a_{n-2}^p=(A(n-2)^2+B(n-2)+C)*2^{n-2}\\ \end{equation} after i got my n-2 and n element i wanted to find them
\begin{equation} (An^2+Bn+c)2^n-4(A(n-2)^2+B(n-2)+C)2^{n-2}=-2^nn^2\\ \end{equation} after i divide both sides by $2^n$ and multiply by $2^2$ i get that \begin{equation} 16An-16A+8B=-4n^2 \end{equation} which gives me after i put $n=2$ and $n=3$ \begin{equation} A=-5/4\\ B=-3/4 \end{equation} which in reality doesnt check if i plug any n's to my final result which looks like that \begin{equation} a_n=\frac{15}{2} \cdot 2^n + \frac{5}{2} \cdot -2^n(-\frac{5}{4}n^2-\frac{3}{4}n) \end{equation}
The general solution for the homogeneous equation is $a_n = a 2^n + b (-2)^{n}$. You do not want to plug in the initial values yet. You need to first find a particular solution.
Now you shouldn't get a particular solution of the form $(A n^2 + B n + C) 2^n$, because $2^n$ is a solution of the homogeneous equation. Instead you should try $a^p_n = (A n^3 + B n^2 + C n) 2^n$.
Then add the general solution of the homogeneous equation, and plug in the initial values to find $a$ and $b$.