= Solution
The forcing is a unit pulse on $0<t<1$. For $t<0$, the homogeneous equation and continuity with $y(0)=0$ give $y=0$. On $0<t<1$, integration with $y(0)=0$ gives $y=1-e^{-t}$. At the end of the pulse its value is $1-e^{-1}$, and the subsequent homogeneous decay must start at that value. Thus
$$
\boxed{y(t)=\begin{cases}
0,&t\leq0,\\
1-e^{-t},&0\leq t\leq1,\\
(1-e^{-1})e^{-(t-1)},&t\geq1.
\end{cases}}
$$
Equivalently it is the difference of two translated step responses, $(1-e^{-t})H(t)-(1-e^{-(t-1)})H(t-1)$. The solution is continuous, rises concavely to its maximum $1-e^{-1}$ at $t=1$, and then decays exponentially to zero. Its derivative has jumps at the pulse edges; those are permitted because the right-hand side has step discontinuities.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ia/paper-2-pulse-response.png]
{title=Continuous response to a unit-duration forcing pulse, rising to one minus exp(-1) and then decaying}
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