How to solve this difference equations in the non-homogenous case Let $0<\alpha<1$ and $N\in\mathbb{N}$. I would like to solve the following difference equation
$$n=0: s_0 = 1+\alpha s_1 $$
$$n\in\{1,\dots,N-2\}: s_n = 1+\alpha s_{n+1}+(1-\alpha)s_n$$
$$n=N-1:s_{N-1}= 1+\alpha s_0 + (1-\alpha)s_{N-1}$$
and the boundary condition $s_N=0$. I was able to calculate the homogeneous solution, i.e. the solution to the equations:
$$n=0: s_0 = 1+\alpha s_1 $$
$$n\in\{1,\dots,N-2\}: s_n = \alpha s_{n+1}+(1-\alpha)s_n$$
$$n=N-1:s_{N-1}= 1+\alpha s_0 + (1-\alpha)s_{N-1}$$
to be $s_n = \frac{1}{\alpha},0<n<N$, $s_0=0$. Is this correct and how can I now solve this for a non-homogneous casse? To get the homogeneous solution I used that for $0<n<N$ we see that $s_n$ are all equal. Then I sued the equation for $n-1$ and $0$ to find the particular values for $s_0$ and all other values

$$n\in\{1,\dots,N-2\}: s_n = 1+\alpha s_{n+1}+(1-\alpha)s_n \\
\implies -1 = \alpha (s_{n+1}-s_n).
$$
Therefore $s_1, s_2, \ldots, s_{N-1}$ is an arithmetic sequence.
Can you end it now?
