= Solution
For $i\geq1$, the stationarity equation is
$$
\pi_i=a\pi_{i-1}+(1-a-b)\pi_i+b\pi_{i+1},
$$
which rearranges to the stated recurrence. At zero,
$$
\pi_0=(1-a)\pi_0+b\pi_1,
\qquad\boxed{b\pi_1=a\pi_0.}
$$
Define the adjacent probability current $J_i=a\pi_i-b\pi_{i+1}$. The recurrence says $J_i=J_{i-1}$, and the boundary equation says $J_0=0$. Hence $\pi_{i+1}=(a/b)\pi_i$ for all $i$. Since $\rho=a/b<1$, <normalization> gives
$$
\boxed{\pi_i=(1-\rho)\rho^i,\qquad\rho=\frac ab,\quad i\geq0.}
$$
The adjacent balance equations also verify stationarity directly. If $a=0$, this formula means the point mass at zero; otherwise the <birth-death chain> is irreducible and this is its unique stationary <probability distribution>.
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