Consider a double sequence $a_{m,n}$ defined by the difference equation
$$ a_{m,n} = \alpha(m,n) a_{m,n-1} + \beta(m,n) a_{m-1,n} $$
with known boundary conditions $a_{m,0}$ and $a_{0,n}$, and where $\alpha,\beta : {\Bbb Z}^2 \to {\Bbb R}$. How to obtain a closed-form solution for $a_{m,n}$ in terms of $\alpha,\beta$?
Example: $\alpha=\mathrm{const.}$ and $\beta(m,n)=m$.
Hint.
Assuming $\alpha(m,n)=\alpha, \beta(m,n) = m$ we have
$$ a_{m,n} = \alpha a_{m,n-1}+m a_{m-1,n} $$
or by the characteristic functions method with $u(x,y) = \sum_{m,n}a_{m,n}x^my^n$
$$ u(x,y) - \alpha x u(x,y) - \partial_x u(x,y) = \phi_0(x,y) $$
with $\phi_0(x,y)$ to account for the initial conditions. After solving for $u(x,y)$ we can recover the $a_{m,n}$ after a series expansion.
I hope now it completes it for you now. I am not very good in recurrence relations but I tried. If it does not work, let me know I will delete my answer.
Interpret the recurrence as a weighted lattice–path problem.
Consider the integer lattice $\mathbb{Z}_{\ge 0}^2. $
To compute $a_{m,n}$, the recurrence allows two possibilities:
$ (m,n-1) \rightarrow (m,n) $ with weight $\alpha(m,n), $
$ (m-1,n) \rightarrow (m,n)$ with weight $\beta(m,n). $
Thus the value $ a_{m,n}$ is obtained by summing weighted contributions coming from these two predecessor points.
This is the key idea.
Here is the next hint, building directly on the lattice--path picture you already have.
Hint 2. Unroll the recurrence completely. Any value $a_{m,n}$ can be written as a sum over all monotone lattice paths from the boundary to $(m,n)$.
More precisely, every path $\gamma$ from either $(k,0)\to(m,n)$ or $(0,\ell)\to(m,n)$, using only steps $(1,0)$ and $(0,1)$, contributes a weight $ \prod_{(i,j)\in\gamma} \begin{cases} \alpha(i,j) & \text{if the step is }(0,1),\\ \beta(i,j) & \text{if the step is }(1,0), \end{cases} $ multiplied by the boundary value at the starting point.
So $a_{m,n}$ is a path sum: $ a_{m,n} =\sum_{\gamma:(k,0)\to(m,n)} a_{k,0}\,w(\gamma) +\sum_{\gamma:(0,\ell)\to(m,n)} a_{0,\ell}\,w(\gamma). $
The recurrence disappears; what remains is combinatorics plus products of $\alpha,\beta$.
The next move after this hint is to reorganize these sums, either by fixing the number of horizontal steps first or by grouping paths with the same step pattern.
Fix $k\ge 0$ and consider monotone lattice paths $\gamma:(k,0)\to(m,n).$ Such a path consists of exactly $ m-k$ horizontal steps and $n$ vertical steps.
Hence these paths are in bijection with subsets $S\subseteq{1,2,\dots,m-k+n} \text{ satisfying } |S|=n, \text{ where } S \text{ records the positions of the vertical steps.} $
$ \text{For a given } S, \text{ the corresponding path } \gamma_S$ is uniquely determined, and its weight is
$w(\gamma_S)$
$\prod_{r=1}^{m-k+n} \begin{cases} \alpha(i_r,j_r) & \text{if } r\in S,\ \beta(i_r,j_r) & \text{if } r\notin S, \end{cases} $
$ \text{where } (i_r,j_r) \text{ is the lattice point entered at the } r\text{-th step.} $
Therefore the path sum can be written explicitly as
$\sum_{\gamma:(k,0)\to(m,n)} w(\gamma)$
$\sum_{\substack{S\subseteq{1,\dots,m-k+n}\ |S|=n}} \prod_{r=1}^{m-k+n} \begin{cases} \alpha(i_r,j_r) & r\in S,\ \beta(i_r,j_r) & r\notin S. \end{cases} $
An analogous expression holds for paths starting at $(0,\ell)$.
Then use induction on $m+n$
One way to write down this summation of weights along paths should be \begin{align*} a_{m,n} &= \sum_{1 \leq m_0 \leq \dots \leq m_n = m} \left(a_{m_0,0} \prod_{1 \leq s \leq n} \left(a(m_{s-1}, s) \prod_{m_{s-1} < k \leq m_s} b(k,s)\right)\right) \\ &+ \sum_{1 \leq n_0 \leq \dots \leq n_m = n} \left(a_{0,n_0} \prod_{1 \leq s \leq m} \left(b(s, n_{s-1}) \prod_{n_{s-1} < k \leq n_s} a(s, k)\right)\right), \end{align*} not sure if that helps you.
