= Solution
\b[The general solution contains both decay of the initial xenon and continued feeding from the initial iodine.] Use $t=0$ for removal of the flux, and retain $\phi$ as the pre-shutdown value in the initial conditions. Neglect post-shutdown <nuclear fission> and <neutron capture>. Then
$$
I(t)=I_0e^{-\lambda_It},
\qquad
\dot X+\lambda_XX=\lambda_II_0e^{-\lambda_It}.
$$
Multiply this <first-order linear differential equation> by the <integrating factor> $e^{\lambda_Xt}$ and integrate:
$$
\boxed{
X(t)=X_0e^{-\lambda_Xt}
+\frac{\lambda_II_0}{\lambda_X-\lambda_I}
\left(e^{-\lambda_It}-e^{-\lambda_Xt}\right).
}
$$
With the preceding steady initial conditions, $\lambda_II_0=G=\gamma_I\Sigma_f\phi$ and $X_0=G/(\lambda_X+\phi\sigma_{aX})$, this is
$$
X(t)=G\left[
\frac{e^{-\lambda_Xt}}{\lambda_X+\phi\sigma_{aX}}
+\frac{e^{-\lambda_It}-e^{-\lambda_Xt}}{\lambda_X-\lambda_I}
\right].
$$
If $\lambda_X+\phi\sigma_{aX}\gg|\lambda_X-\lambda_I|$, the coefficient of the first exponential term is small compared with the coefficients of the feeding term, yielding
$$
\boxed{
X(t)\simeq
\frac{\gamma_I\Sigma_f\phi}{\lambda_X-\lambda_I}
\left(e^{-\lambda_It}-e^{-\lambda_Xt}\right).
}
$$
This approximation neglects the finite initial xenon population and is consequently not a small relative error at $t=0$, where its value is zero while the exact $X_0$ is positive. The exact expression supplies the initial value and satisfies $\dot X(0)=G-\lambda_XX_0$. For equal <nuclear decay constants>, the continuous limit of the exact solution is $X(t)=(X_0+\lambda_II_0t)e^{-\lambda_It}$.
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