Suppose that I have sequences $\{\alpha_n,\beta_n:n\in\mathbb{N}\}$ and consider the sequence $(x_n)$ defined recursively by:
$$\begin{cases}x_0=a\\ x_1=b\\ x_{n+1}=\alpha_n \cdot x_n + \beta_n\cdot x_{n-1}\end{cases}$$
Is there a way to write the sequence $(x_n)$ in closed form in terms of the elements $a,b,(\alpha_n),(\beta_n)$?
I supposed there should be a way to do this, but I have not been able to find it.
When $\alpha_n,\beta_n$ are constant, one can at least find a closed form in terms of powers of matrices by doing
$$\begin{bmatrix}x_{n+1}\\ x_n\end{bmatrix}=\begin{bmatrix}\alpha & \beta\\ 1 & 0\end{bmatrix}\begin{bmatrix}x_{n}\\ x_{n-1}\end{bmatrix}=\cdots =\begin{pmatrix}\alpha & \beta\\ 1 & 0\end{pmatrix}^ n\begin{bmatrix}x_{1}\\ x_0\end{bmatrix}=\begin{pmatrix}\alpha & \beta\\ 1 & 0\end{pmatrix}^ n\cdot \begin{pmatrix}b\\ a\end{pmatrix}$$
I guess one can do the same thing with the new problem obtaining: $$\begin{bmatrix}x_{n+1}\\ x_n\end{bmatrix}=\left(\begin{bmatrix}\alpha_n & \beta_n\\ 1 & 0\end{bmatrix}\cdots \begin{bmatrix}\alpha_2 & \beta_2\\ 1 & 0\end{bmatrix}\cdot \begin{bmatrix}\alpha_1 & \beta_1\\ 1 & 0\end{bmatrix}\right)\cdot \begin{pmatrix}b\\ a\end{pmatrix}$$ Is there a better way to do it? what is the closed form?