 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ia/paper-2/5a/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

On the active interval the excess temperature satisfies

θ_{2}^{'} + a θ_{2} = - 1

e^{a t}

and integration from time one gives

e^{a t} θ_{2} (t) - e^{a} θ_{2} (1) = - \int_{1}^{t} e^{a s} d s = - \frac{e ^{a t} - e ^{a}}{a} .

There is no impulse in this equation, so continuity supplies

θ_{2} (1) = (T_{0} - T_{\infty}) e^{- a}

. Therefore

T_{2} (t) = T_{\infty} - \frac{1}{a} + e^{- a t} (T_{0} - T_{\infty} + \frac{e ^{a}}{a}), 1 < t < 2.

The equivalent correction to the common unforced temperature is

- (1 - e^{- a (t - 1)}) / a

, which vanishes continuously at the beginning of the interval.

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