= Solution
Integrate the first equation over $(1-\varepsilon,1+\varepsilon)$. The integral of the ordinary cooling term tends to zero as $\varepsilon\downarrow0$, while the <Dirac delta function> integrates to one. Thus the <jump condition for an impulse> is
$$
\boxed{T_1(1^+)-T_1(1^-)=-1.}
$$
After the jump there is no further forcing, so <Newton's law of cooling> gives
$$
T_1(t)-T_\infty=[(T_0-T_\infty)e^{-a}-1]e^{-a(t-1)}.
$$
Equivalently,
$$
\boxed{T_1(t)=T_\infty+(T_0-T_\infty)e^{-at}-e^{-a(t-1)},\qquad t>1.}
$$
The last term has jump minus one at time one and then decays at the same exponential rate as the unforced temperature.
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