 Solution

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 $\varphi$ as the pre-shutdown value in the initial conditions. Neglect post-shutdown nuclear fission and neutron capture. ThenMultiply this first-order linear differential equation by the integrating factor $e^{{\lambda }_{X}t}$ and integrate:With the preceding steady initial conditions, ${\lambda }_I{I}_0=G={\gamma }_I{\Sigma }_f\varphi$ and $X_0=G/({\lambda }_X+\varphi {\sigma }_{aX})$, this isIf ${\lambda }_X+\varphi {\sigma }_{aX}\gg |{\lambda }_X-{\lambda }_I|$, the coefficient of the first exponential term is small compared with the coefficients of the feeding term, yieldingThis 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 $\dot{X}(0)=G-{\lambda }_X{X}_0$. For equal nuclear decay constants, the continuous limit of the exact solution is $X(t)=(X_0+{\lambda }_I{I}_{0}t)e^{-{\lambda }_{I}t}$.