Let me rephrase the system that you propose into a more concise setting. The density of random walkers (or the probability density of finding a walker in a given position $x$) follows a Fokker-Planck equation,
$$\partial_t \rho_t(x)=\mathcal{L}\rho_t(x).$$
I assume that the walkers are confined in a segment of length $2L$ centered in $x=0$, and that the diffusion is different in the left and right halves of the segment:
$$ \mathcal{L}\rho_t(x)=\begin{cases}D\partial_x^2 \rho_t(x),\quad x\in[0,L],\\
\tilde{D}\partial_x^2 \rho_t(x),\quad x\in[-L,0[. \end{cases}.$$
The solution to the Fokker-Planck equation has to fulfill the boundary conditions
$$ \int_{-L}^L\rho_t(x)\,dx =1,\quad\forall t \,(\text{normalization)},$$
and
$$ \lim_{x\to 0^+}\rho_t(x) = \lim_{x\to 0^-}\rho_t(x),\quad\forall t \,(\text{continuity)}.$$
Let us solve the stationary equation $\mathcal{L}\rho(x)=0$,
$$\rho(x) = \begin{cases} C_1 x+C_2,\quad x\in[0,L],\\
\tilde{C}_1 x+\tilde{C}_2,\quad x\in[-L,0[. \end{cases}.$$
Imposing the boundary conditions we can compute two of the four unknown constants: $\tilde{C}_2 = C_2$ by continuity, and $C_2= \frac{1}{2L}-\frac{C_1-\tilde{C}_1}{2}L^2$ by normalization. Interestingly, these boundary conditions are not enough to determine the stationary distribution. Indeed, we have also to give information about how the probability current behaves. Here is where equilibrium enters. Equilibrium distributions are a special case of stationary distributions in which the probability current equals zero. You can check that the only way that the stationary probability current equals zero for all $x$ is that $C_1=\tilde{C}_1=0$. Therefore, the equilibrium distribution for this process is the same one of the random walk with equal diffusion in the whole space,
$$ \rho(x) = \frac{1}{2L}, $$
so the walkers are expected to spend the same average time in $x<0$ and $x>0$ when the observation time is long enough.
