Finite-pulse response of a first-order linear equation ($y(t)=A\left[\frac{e^{\lambda (t-a)}-1}{\lambda }H(t-a)-\frac{e^{\lambda (t-b)}-1}{\lambda }H(t-b)\right]$)

Let $0<a<b$ and $y^{\prime }-\lambda y=A[H(t-a)-H(t-b)]$, with zero initial data. For $\lambda \ne 0$, the Laplace transform time-shift rule gives the displayed response, since the transform of the input is $A(e^{-as}-e^{-bs})/s$ and the equation supplies the factor $(s-\lambda {)}^{-1}$. Equivalently, integrate the impulse response $e^{\lambda t}H(t)$ over the active pulse. Before $a$ the solution is zero; during the pulse it is $A(e^{\lambda (t-a)}-1)/\lambda$; after $b$ it is $A{e}^{\lambda t}(e^{-\lambda a}-e^{-\lambda b})/\lambda$. The solution is continuous at both switching times. At $\lambda =0$ take the limit, giving linear growth during the pulse and the constant $A(b-a)$ afterward. Positive $\lambda$ gives growing memory, while negative $\lambda$ gives decay after forcing stops.