IVP’s With Step Functions — Question 4

Question 4 – Solution

Strategy. A decaying stable system can continue moving upward after its input stops. Differentiate the actual tail to locate the peak.

Step 1: Transform the pulse. The zero data give $$Y=\frac{1-e^{-2s}}{s(s+1)^2},\qquad \frac 1{s(s+1)^2}=\frac 1s-\frac 1{s+1}-\frac 1{(s+1)^2}.$$ Thus $r(t)=1-(1+t)e^{-t}$ is the unit-step response, with $r(0)=r'(0)=0$ and $r''+2r'+r=1$. The solution is $\boxed{y=r(t)-H_2(t)r(t-2)}$.

Step 2: Simplify the tail and match. For $t\ge 2$, $$y(t)=e^{-t}[(e^2-1)t-(e^2+1)].$$ At $2$ this gives $1-3e^{-2}=r(2)$; its derivative there is $2e^{-2}=r'(2)$. Hence $y,y'$ are continuous, with positive velocity even after the forcing is removed.

Step 3: Locate and certify the peak. For $0<t<2$, $y'=te^{-t}>0$. On the tail, $$y'=e^{-t}[2e^2-(e^2-1)t].$$ This derivative changes sign exactly once, at $$\boxed{t_*=\frac{2e^2}{e^2-1}>2,\qquad y(t_*)=(e^2-1)e^{-t_*}.}$$ Thus the peak is unique and global. The tail remains positive because its linear factor is already positive at $2$ and has positive slope, while $y(t)\to 0$. Damping does not imply an immediate drop in displacement at shutoff.

Step 4: Check the switch, area and domain. Since $y,y'$ are continuous, we have $\boxed{[y'']_2=-1}$. Both $y,y'$ start at zero and tend to zero. Integrating the equation over the half-line, including the matching boundary terms, yields $$\int_0^\infty y\,dt=\int_0^2 1\,dt=\boxed 2.$$ The nonzero $t e^{-t}$ tail gives exact domain $\boxed{s>-1}$. The $s=0$ singularity in the transform expression is removable, with $Y(0)=2$.