Nonhomogeneous Systems — Question 5

Question 5 – Solution

Strategy. Removing the common exponential reveals successive integrations along the nilpotent chain.

Step 1: Integrate along the chain. With $N=A+I=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, the transformed system is $Z'=NZ+(0,1)^T$, $Z(0)=0$. Thus $z_2'=1$, $z_1'=z_2$, giving $z_2=t$, $z_1=t^2/2$ and $$\boxed{x=\tfrac 12t^2e^{-t},\qquad y=te^{-t}.}$$ Substitution verifies the original equation and zero initial state.

Step 2: Reject an insufficient polynomial trial. The trial would give $Z=tv$, so its equation requires $v=tNv+(0,1)^T$ for every time. Matching coefficients forces $Nv=0$ and $v=(0,1)^T$, but $N(0,1)^T=(1,0)^T\ne 0$. These conditions contradict each other. A factor $t$ multiplying a constant vector is insufficient; the chain requires a quadratic term in the first component.

Step 3: Locate the separate response peaks. For $t\ge 0$, $x'=e^{-t}t(2-t)/2$ and $y'=e^{-t}(1-t)$. Their sign changes show $$\boxed{x_{\max}=2e^{-2}\text{ at }t=2,\qquad y_{\max}=e^{-1}\text{ at }t=1.}$$ The maxima occur at different states and different times. The figure marks these component peaks on the two time graphs.

Step 4: Interpret resonance without growth to infinity. Both components tend to zero, since every fixed polynomial times $e^{-t}$ vanishes. The homogeneous family is $e^{-t}(c_1+c_2t,c_2)^T$, whose largest degree is one; the forcing raises the first component’s degree to two. This is a resonant polynomial response, even though the negative exponential ultimately wins. Resonance describes an enhanced polynomial factor, not a universal guarantee of an unbounded solution. Adding any homogeneous solution still leaves forward decay.