= Solution
Since the <steady state> is strictly positive, the <positive part>[positive-part] <function> is differentiable in a neighbourhood of it. Write $p=p_*+q$. Retaining first-order terms gives
$$
q'(T)=-\alpha\mu p_*q(T)-\alpha(1+\mu p_*)q(T-1).
$$
Substitute the exponential mode $q=e^{\lambda T}$ to obtain
$$
\boxed{\lambda=-\beta-\gamma e^{-\lambda},\qquad\beta=\alpha\mu p_*,\quad\gamma=\alpha(1+\mu p_*)=\alpha/p_*.}
$$
Here the last equality uses the exact steady-state <equation>. For small $\mu$, $\beta=\alpha\mu+O(\alpha\mu^2)$ and $\gamma=\alpha(1+\mu)+O(\alpha\mu^2)$. Linearizing at the exact root avoids carrying a spurious constant residual from the printed approximate <equilibrium>.
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