Solutions to Systems — Question 5

Question 5 – Solution

Strategy. Subtract one particular solution to expose the homogeneous family; keep the forcing when testing superposition.

Step 1: Verify and complete the affine family. Here $P'=(0,-2e^{-t})^T=AP+f$. For any other solution $X$, the difference $H=X-P$ satisfies $H'=AH$, so $H=(c_1e^{-t},c_2e^{-2t})^T$. Conversely each such sum solves the forced equation. Thus the proposed family is complete.

Step 2: Test a weighted combination. For $Z=\alpha U+\beta V$, $$Z'-AZ=(\alpha+\beta)f(t).$$ Because the first component of $f$ is always $1$, equality to $f$ holds exactly when $\boxed{\alpha+\beta=1}$. In particular an average is a solution, while a sum generally is not. The difference of two forced solutions solves the homogeneous equation.

Step 3: Impose the zero initial state. The constants are $c_1=-1$, $c_2=-2$, yielding $$\boxed{X(t)=(1-e^{-t},\,2e^{-t}-2e^{-2t})^T.}$$ At $0$ this is zero, but $X'(0)=(1,2)^T=f(0)$. A zero initial state therefore does not imply a zero solution in this nonhomogeneous problem.

Step 4: Locate the peak and interpret the limit. For $y=2e^{-t}(1-e^{-t})$, $y'=2e^{-t}(2e^{-t}-1)$. It is positive before $\ln 2$ and negative after, so the unique maximum on $t\ge 0$ is $\boxed{y(\ln 2)=1/2}$. The state tends to $(1,0)^T$. However the constant vector $(1,0)^T$ has zero derivative while $A(1,0)^T+f(t)=(0,2e^{-t})^T\ne 0$ at finite times. A limiting state need not itself be a constant solution of a time-dependent forced equation. The plot marks the actual second-component peak.