= Solution
Use <neutron fluence> $\Phi$ as the independent variable and abbreviate $a=\sigma_{a\rm Th}$, $b=\sigma_{a\rm U}$ and $c=\sigma_{c\rm Th}$. Substituting the parent solution into the second <linear differential equation> gives
$$
\frac{dN_{\rm U}}{d\Phi}+bN_{\rm U}=cN_{\rm Th0}e^{-a\Phi}.
$$
The <integrating factor> is $e^{b\Phi}$. Integrating from the initial state yields
$$
\boxed{N_{\rm U}(\Phi)=N_{\rm U0}e^{-b\Phi}+\frac{cN_{\rm Th0}}{b-a}\left(e^{-a\Phi}-e^{-b\Phi}\right),\qquad b\ne a.}
$$
The first term is survival of the initial population; the second accumulates production at earlier exposures, weighted by subsequent survival. For $b=a$, the continuous <limit> is $(N_{\rm U0}+cN_{\rm Th0}\Phi)e^{-a\Phi}$. The same formula applies to a time-varying prescribed flux through $\Phi(t)$; intervals of zero flux leave both populations unchanged.
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