Nonhomogeneous Systems — Question 8

Question 8 – Solution

Strategy. Match the state at the switch, then track the stored second-component response as it continues to feed the first.

Step 1: Solve and match the two intervals. For $0\le t\le 1$, integration gives $x=1-(1+t)e^{-t}$, $y=1-e^{-t}$. Set $a=1-2/e$, $b=1-1/e$, the switch values. For $\tau=t-1\ge 0$, the homogeneous continuation is $$\boxed{x=e^{-\tau}(a+b\tau),\qquad y=b e^{-\tau}.}$$ At $\tau=0$ these match the first formulas, so both components are continuous.

Step 2: Compare one-sided derivatives. Before the switch, $(x',y')=(te^{-t},e^{-t})$. Thus $X'(1^-)=(1/e,1/e)^T$. Immediately after it, $X'(1^+)=(-a+b,-b)^T=(1/e,-1+1/e)^T$. The derivative $x'$ is continuous; $y'$ has jump $-1$. A finite step input changes a derivative, not the state itself.

Step 3: Find the delayed first-component peak. Before switching, $x'>0$ for $t>0$. Afterward, $x'=e^{-\tau}(b-a-b\tau)$, with $b-a=1/e$. It changes sign once, at $\tau=1/(e-1)$, hence $$\boxed{t_{\max}=1+\frac 1{e-1},\qquad x_{\max}=(1-1/e)e^{-1/(e-1)}.}$$ The still-positive $y$ initially exceeds $x$ after the input ends, so $x'=-x+y$ remains positive until the two components become equal.

Step 4: Integrate the total response. Both components tend to zero exponentially times a linear factor, so their integrals converge. Integrating $y'=-y+u$ over $[0,\infty)$ and using its zero endpoints gives $\int y=\int u=1$. Integrating $x'=-x+y$ similarly gives $\boxed{\int_0^\infty x(t)\,dt=\int_0^\infty y(t)\,dt=1}$. Equal total areas do not mean equal graphs or simultaneous peaks. The figure marks the switch and the delayed maximum of $x$.