= Solution
For $\epsilon(n)=an+b$, <linearity of expectation> gives $\langle\epsilon(n)\rangle=am+b$. The mean therefore satisfies
$$
m'=\lambda+kb-(\beta-ak)m.
$$
If $\beta>ak$, the coefficient of $m$ is negative and the <stable equilibrium> is
$$
\boxed{m(t)\longrightarrow m_*=\frac{\lambda+kb}{\beta-ak}.}
$$
If $\beta<ak$, the positive feedback from batch addition dominates deaths and $m(t)$ grows exponentially at rate $ak-\beta$. At the threshold $\beta=ak$, it grows linearly as $m(t)=m(0)+(\lambda+kb)t$.
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