Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 2 / 7B / ii

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. ThusEquivalently 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.