Nonhomogeneous Systems — Question 7

Question 7 – Solution

Strategy. Boundedness can require an exact cancellation of an unstable mode rather than follow from bounded forcing alone.

Step 1: Solve and isolate the growing coefficient. Integrating factors give $$\boxed{x=(p+\tfrac 12)e^t-\tfrac 12e^{-t},\qquad y=1+(q-1)e^{-t}.}$$ The second component is bounded for every $q$. The first is bounded exactly when $\boxed{p=-1/2}$, so the full bounded family has that fixed first initial value and arbitrary $q$.

Step 2: Check the limit against the original equation. For bounded data, $X(t)\to(0,1)^T$. However the constant function $X=(0,1)^T$ has zero derivative, whereas the original right side is $(e^{-t},0)^T\ne 0$ at finite times. It is the limiting equilibrium of the limiting equation, not an exact constant solution of the stated time-dependent equation.

Step 3: Recover the cancellation from a future integral. Solving the first equation with no growing term gives $$\boxed{x(t)=-\int_t^\infty e^{t-s}e^{-s}\,ds=-\tfrac 12e^{-t}.}$$ At zero it selects $p=-1/2$. This is a boundedness condition imposed over the whole future, used to choose an initial value. It does not replace the ordinary forward initial-value evolution once that value is fixed.

Step 4: Measure sensitivity to the unstable mode. Changing $p$ by $\delta$ adds $\delta e^t$ to the first component. Its absolute difference starts below one and grows strictly, first reaching one at $\boxed{t=\ln(1/|\delta|)}$. The input remains bounded throughout, but every such perturbed response is unbounded. The unstable homogeneous coefficient, not the input size alone, determines the failure. Exact cancellation selects a special family; it does not make the underlying system stable.