= Solution
On the active interval the excess temperature satisfies $\theta_2'+a\theta_2=-1$. Multiplication by the <integrating factor> $e^{at}$ and integration from time one gives
$$
e^{at}\theta_2(t)-e^a\theta_2(1)=-\int_1^t e^{as}\,ds
=-\frac{e^{at}-e^a}{a}.
$$
There is no impulse in this equation, so <continuity> supplies $\theta_2(1)=(T_0-T_\infty)e^{-a}$. Therefore
$$
\boxed{T_2(t)=T_\infty-\frac1a+
e^{-at}\left(T_0-T_\infty+\frac{e^a}{a}\right),\qquad1<t<2.}
$$
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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