= Solution
Take unilateral <Laplace transforms> of the cavity equation, keeping the initial operator:
$$
(s+\gamma/2)\widetilde a(s)=a(0)-\sqrt\gamma\,\widetilde b_{\mathrm{in}}(s).
$$
The input-output relation gives
$$
\widetilde b_{\mathrm{out}}(s)=\widetilde b_{\mathrm{in}}(s)+\sqrt\gamma\,\widetilde a(s)
=\frac{s-\gamma/2}{s+\gamma/2}\widetilde b_{\mathrm{in}}(s)+\frac{\sqrt\gamma\,a(0)}{s+\gamma/2}.
$$
A <transfer function> describes the forced zero-state part of the response. Thus the <passive cavity input-output transfer function> is
$$
\boxed{G(s)=\frac{s-\gamma/2}{s+\gamma/2}.}
$$
The omitted initial-state term is a decaying transient $\sqrt\gamma e^{-\gamma t/2}a(0)$, and is present for a general initial cavity state. This distinction avoids silently setting a physical annihilation operator equal to zero.
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