A reactor initially operating at a power P_o is put on a period T such that the power can be approximated as P(t) = P_o exp (t / T) . Assuming that the coolant temperature is maintained at its initial value T_c(0) , solve Eq. (8.49) and show that the fuel temperature will be:
T_{f}(t)=T_{c}(0)+\frac{P_{o}R_{f}}{1+τ/T}\Bigl[\exp(t/T)+(\tau/T)\exp(-t/\tau)\Bigr]\,.
Rewrite Eq. (8.50) as
{\frac{d}{d t}}\theta(t)\!=\!{\frac{1}{M_{f}c_{f}}}P_{o}\exp(t/T)\!-\!{\frac{1}{\tau}}\theta(t)
where \theta(t)=T_{f}(t)-T_{c}(0) since T_c remains constant. Apply an integrating factor exp(t/τ):
\left[\frac{d}{d t}\theta(t)+\frac{1}{\tau}\theta(t)\right]\exp(t/\tau)=\frac{1}{M_{f}c_{f}}P_{o}\exp(t/T)\exp(t/\tau)
which is the same as
\frac{d}{d t}\Bigl[\theta(t)\exp(t/\tau)\Bigr]=\frac{1}{M_{f}c_{f}}P_{o}\exp\Bigl[(1/T+1/\tau)t\Bigr]
Integrate between 0 and t:
\theta(t)\exp(t/\tau)-\theta(0)=\frac{1}{M_{f}c_{f}}P_{o}\int_0^t exp[(1/T+1/\tau)t^{\prime}]d t^{\prime}
=\frac{1}{M_{f}c_{f}}P_{o}\left\{\frac{\exp\left[(1/T+1/\tau)t^{\prime}\right]-1}{(1/T+1/\tau)}\right\}
Solving for θ(t) we have
\theta(t)=\theta(0)\exp(-t\ /\tau)+{\frac{1}{M_{f}c_{f}}}P_{o}\left\{{\frac{\exp(t/T) – \exp(-t/\tau)}{(1/T+1/\tau)}}\right\}
Noting that the initial condition, Eq. (8.32) gives
\theta(0)=T_{f}(0)-T_{c}(0)=R_{f}P_{o}=\frac{\tau}{M_{f}c_{f}}P_{o} \ , \ \mathrm{since} \ \ \tau=M_{f}c_{f}R_{f},and that \theta(t)=T_{f}(t)-T_{c}(0) we obtain
T_{f}(t)=T_{c}(0)+\frac{\tau}{M_{f}c_{f}}P_{o}\exp(-t/\tau)+\frac{1}{M_{f}c_{f}}P_{o}\left\{\frac{\exp(t/T) – \exp(-t/\tau)}{(1/T+1/\tau)}\right\}
Then eliminating M_{f}c_{f}\,, we may reduce this equation to the form given in the problem