= Solution
During the pulse, the <photon number density>[photon density] reaches its maximum when $\dot n=0$ with $n>0$, so $N^*=N_{\rm th}^*$. Evaluating <1/c/i> at this inversion gives
$$
n_{\max}=\frac{f_cN_i^*}{\beta}\left[1-\frac{1+\ln r}{r}\right].
$$
The ideal instantaneous output <power> is $P(t)=V_c\hbar\omega_Ln(t)/\tau_c$. Therefore
$$
\boxed{P_{\max}=\frac{N_i^*V_g\hbar\omega_L}{\beta\tau_c}
\frac{r-1-\ln r}{r}.}
$$
For $r\gg1$, almost all the extractable <energy> is stored briefly as intracavity <photons>, and $P_{\max}\simeq E_{\max}/\tau_c$. The leading rise while the inversion remains near its initial value is $n\propto e^{(r-1)t/\tau_c}$, with short growth time $\tau_c/(r-1)$. Depletion then takes the inversion below threshold. At late times $N^*\simeq N_f^*$ and
$$
n(t)\propto\exp\left[-\left(1-\frac{N_f^*}{N_{\rm th}^*}\right)\frac{t}{\tau_c}\right],
$$
which approaches a tail with decay time $\tau_c$ at large $r$. The pulse is consequently asymmetric, with a rapid rise and a slower approximately exponential trailing edge. The negligible-seed build-up delay depends on the seed strength. The pumping and spontaneous population decay have been neglected over the short pulse; parasitic <optical cavity>[cavity] loss multiplies useful <power> by the output fraction described in <1/b/ii>.
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