= Finite-pulse response of a first-order linear equation
{title2=$y(t)=A\left[\frac{e^{\lambda(t-a)}-1}{\lambda}H(t-a)-\frac{e^{\lambda(t-b)}-1}{\lambda}H(t-b)\right]$}
Let $0<a<b$ and $y'-\lambda y=A[H(t-a)-H(t-b)]$, with zero initial data. For $\lambda\ne0$, 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 $Ae^{\lambda t}(e^{-\lambda a}-e^{-\lambda b})/\lambda$. 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.
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