Nonhomogeneous Systems — Question 2

Question 2 – Solution

Strategy. Match polynomial coefficients before imposing initial data; stability controls differences of responses, not the size of an unbounded input.

Step 1: Match the polynomial coefficients. Substituting $X_p=ut+v$ gives $u=A(ut+v)+(t,1)^T$. The coefficient of $t$ requires $Au=-(1,0)^T$, hence $u=(1,0)^T$. The constant equation $u=Av+(0,1)^T$ gives $v=(0,1)^T$. Thus $\boxed{X_p=(t,1)^T}$. A constant trial cannot cancel the forcing’s nonzero coefficient of $t$.

Step 2: Complete the family and solve the IVP. The homogeneous solution is $e^{-t}(c_1+c_2t,c_2)^T$. Initial zero gives $c_1=0$, $c_2=-1$, so $$\boxed{x=t(1-e^{-t}),\qquad y=1-e^{-t}.}$$

Step 3: Check the original equation and asymptotics. We have $x'=1-e^{-t}+te^{-t}$ and $y'=e^{-t}$. The right sides $-x+y+t$ and $-y+1$ give those same expressions. The initial values vanish. Moreover $x-t=-te^{-t}\to 0$ and $y\to 1$. The first component tracks a growing ramp; its tracking error decays, but its value does not tend to a finite constant.

Step 4: Separate stability from bounded response. On $t\ge 0$, $y$ is bounded between zero and one, while $x\to\infty$. For any two solutions, their difference has the form $e^{-t}(d_1+d_2t,d_2)^T\to 0$. Thus initial-data effects disappear even though the common forced response grows. Negative eigenvalues of $A$ guarantee decay of the homogeneous transient; they do not turn an unbounded polynomial forcing into a bounded steady state. A complete answer therefore describes both the particular response and the transient.