Determine the general solution of the inhomogenous system (Difference&Differential Equations) Determine the general solution of the system
$y(n+1)=\begin{bmatrix} 5 & 1\\ -1 & 3 \\ \end{bmatrix} y(n) + 4^n \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix},\quad  n \in \mathbf{N}$
The homogenous part of finding general solution is easy, however I am kinda lost in finding the general solution of this system with the +B[n] part.

$\begin{bmatrix} a_{1}\\ b_{1} \\ \end{bmatrix}
=\begin{bmatrix} 5 & 1\\ -1 & 3 \\ \end{bmatrix} \begin{bmatrix} a_0\\ b_0 \\ \end{bmatrix} +  \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}=
\begin{bmatrix} 5a_0 + b_0+1\\ -a_0+3b_0 -1 \end{bmatrix} 
$
$\begin{bmatrix} a_{2}\\ b_{2} \\ \end{bmatrix}
=\begin{bmatrix} 5 & 1\\ -1 & 3 \\ \end{bmatrix} \begin{bmatrix} 5a_0 + b_0+1\\ 3b_0-a_0 -1 \end{bmatrix}   + 4 \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}=
4\begin{bmatrix}   6 a_0 + 2 b_0+2\\ - ( 2a_0 - 2b_0+2) \end{bmatrix} 
$
$\begin{bmatrix} a_{3}\\ b_{3} \\ \end{bmatrix}=
4^2\begin{bmatrix} 7 a_0+3 b_0+3\\ - (3 a_0-b_0+3) \\ \end{bmatrix}
$
$\begin{bmatrix} a_{4}\\ b_{4} \\ \end{bmatrix}=
4^3\begin{bmatrix}  8 a_0 + 4b_0+4\\ - ( 4 a_0+4) \\ \end{bmatrix}
$
$\begin{bmatrix} a_{5}\\ b_{5} \\ \end{bmatrix}=
4^4\begin{bmatrix}   9 a_0 + 5 b_0+5\\ - (  5 a_0 + b_0+5) \\ \end{bmatrix}
$
$\vdots$
$\begin{bmatrix} a_{n}\\ b_{n} \\ \end{bmatrix}
=4^{n-1}\begin{bmatrix} 4+n& n \\ -n&4-n \\ \end{bmatrix}\begin{bmatrix} a_0\\ b_0 \\ \end{bmatrix} +
4^{n-1}n \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}
$
