Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 2 / 8C (Differential Equations)

During warming, $\dot{\theta }=\alpha ({\theta }_1-\theta )$ and $\theta ={\theta }_1-({\theta }_1-{\theta }_0)e^{-\alpha t}$. During cooling, with $\tau =t-T$, $\dot{\theta }=-\alpha (\theta -{\theta }_0)$ and $\theta ={\theta }_0+({\theta }_1-{\theta }_0)(1-e^{-\alpha T})e^{-\alpha \tau }$.

$N(2T)/N(0)=\exp [-{\int }_0^{2T}\beta (\theta (t))dt]$. Evaluation gives the implicit equation

For $x=\alpha T\ll 1$, $e^{-x}(1-e^{-x})=x-\frac{3}{2}{x}^2+O(x^3)$. Hence $\frac{3}{2}{\beta }_{\max }\alpha T^2\sim \log 100$ and