A 2d recurrence equation representing a step-free continuous-time Markov process on N^2 with frequency-dependent rates $\textbf{Background:}$ Consider a particle in $\mathbb{N}^2$ starting at a point $(x,y)$ that can only move one step at a time along the lattice. The particle moves each up or down in continuous time with rate $y$, and each left or right with rate $x$. It will therefore be moving faster, the further away from the origin that it is. Furthermore, once it hits an axis, it cannot leave that axis, so the axes act as a kind of absorbing boundary. I am wondering about the expected time (or, in an ideal world, the distribution) for the particle to hit either axis for the first time. 
$\textbf{Question:}$ Write $v(x,y)$ for the expected time for the particle to hit either axis given it began at $(x,y)$. What is $v(x,y)$? We can write the recurrence equation
$$ v(x,y) = \frac{1}{2(x+y)} + \frac{x}{2(x+y)} (v(x-1,y) + v(x+1,y)) + \frac{y}{2(x+y)} (v(x,y-1) + v(x,y+1))$$
for $x,y>0$, and with boundary conditions $v(x,0) = v(0,y) = 0 \; \; \; \forall \;  x,y>0$. 
In particular, if we let $x=y$, the equation can be written as 
$$ \frac{1}{2x} + (v(x+1,x) - v(x,x)) = (v(x,x) - v(x-1,x))$$
which feels, to me, like it might be a good place to start.
$\textbf{What I've got so far}:$ 


*

*A poor upper bound on $v(x,y)$ that shows it's finite. 

*The particle must hit one of the axes eventually, and can't hit both at the same time. The probability that the particle hits the x-axis before the y-axis, starting at the point $(x,y)$, is $x/(x+y)$ .

*If one defines the same process in 1d, the expected time to hit the origin given a start point of $x$ is clearly infinite. But the expected hitting time of either $0$ or $A$, starting at $0<x<A$, is $x(H_A - H_x)$, where $H_m$ is the $m$-th harmonic number.

 A: Some experimenting reveals that the hitting time $t_1(x)$ of the 1D process started at $x\in\mathbb{N}$ (in the notation of OP's comment) has a simple cdf,
$$\mathbb{P}(t_1(x) < t ) = \left(\frac{t}{t+1}\right)^x.$$
I don't know an easy combinatorial proof, but one may check explicitly that it solves
$$\mathbb{P}(t_1(x) < t ) = \int_0^t \mathrm{d}s\, 2x\,e^{-2x s} \tfrac12\left[\mathbb{P}(t_1(x+1) < t-s )+\mathbb{P}(t_1(x-1) < t-s )\right].$$
Edit: The fact that $\mathbb{P}(t_1(x) < t ) = \mathbb{P}(t_1(1) < t )^x$ can easily be understood in the light of James Martin's answer. For the whole population to die out before time $t$, the descendants of each of the $x$ initial individuals have to die out before time $t$. 
It follows immediately that
$$\mathbb{P}(\min(t_1(x),t_2(y))>t) = \left[1-\left(\frac{t}{t+1}\right)^{x}\right]\left[1-\left(\frac{t}{t+1}\right)^{y}\right].$$
Hence
\begin{align*}v(x,y) &= \mathbb{E}(\min(t_1(x),t_2(y))) \\
&= \int_0^\infty \mathrm{d}t\left[1-\left(\frac{t}{t+1}\right)^{x}\right]\left[1-\left(\frac{t}{t+1}\right)^{y}\right]\\
&=1 - x\,H_x-y\, H_y+(x+y)H_{x+y-1} < \infty,
\end{align*}
where $H_n$ is the $n$th harmonic number.
