IVP’s With Step Functions — Question 1

Question 1 – Solution

Strategy. Keep the initial response separate from the two delayed step responses, then compare the three monotonic pieces.

Step 1: Transform and invert. The initial term gives $(s+2)Y-1=3(e^{-s}-e^{-3s})/s$, hence $$Y=\frac 1{s+2}+\frac{3(e^{-s}-e^{-3s})}{s(s+2)},\qquad s>0.$$ Since $\mathcal L^{-1}\{1/[s(s+2)]\}=(1-e^{-2t})/2$, $$\boxed{y=e^{-2t}+\tfrac 32H_1(t)(1-e^{-2(t-1)}) -\tfrac 32H_3(t)(1-e^{-2(t-3)}).}$$

Step 2: Match the pieces. Set $M=\tfrac 32+(e^{-2}-\tfrac 32)e^{-4}$. The same solution is $$y(t)=\begin{cases} e^{-2t},&0\le t<1,\\ \tfrac 32+(e^{-2}-\tfrac 32)e^{-2(t-1)},&1\le t<3,\\ M e^{-2(t-3)},&t\ge 3. \end{cases}$$ The matching values are $y(1)=e^{-2}$ and $y(3)=M$. Each piece satisfies $y'+2y=0,3,0$, respectively, and $y(0)=1$. These matching conditions also give uniqueness interval by interval.

Step 3: Identify the corners. Writing $[v]_a=v(a+)-v(a-)$, the equation and continuity imply $$[y']_1=3,\qquad [y']_3=-3,\qquad [y]_1=[y]_3=0.$$ The step factors multiply responses that vanish at their activation times, so no jump in $y$ is introduced.

Step 4: Compare all possible maxima. The solution decreases on $[0,1]$, increases on $[1,3]$ toward $3/2$, then decreases to zero. Moreover $M>\tfrac 32-\tfrac 32e^{-4}>1$, since $e^4>3$. Thus the unique global maximum is $\boxed{y(3)=M}$. Its nonzero tail is exactly a positive multiple of $e^{-2t}$, so the exact real transform domain is $\boxed{s>-2}$; the displayed formula extends removably across $s=0$.