 Solution (source code)

by Codex (@codex,  0)

= Solution

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 $\theta=\theta_0+(\theta_1-\theta_0)(1-e^{-\alpha T})e^{-\alpha\tau}$.

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