Nonhomogeneous Systems — Question 1

Question 1 – Solution

Strategy. A constant particular solution shifts the equilibrium; the translated state evolves through the familiar homogeneous modes.

Step 1: Translate to the forced equilibrium. Solving $-2x+y+3=0$, $x-2y=0$ gives $\boxed{X_*=(2,1)^T}$. With $Z=X-X_*$, the equation becomes $Z'=AZ$. The origin is not an equilibrium of the forced system.

Step 2: Resolve the transient coefficients. The eigenpairs of $A$ are $-1,(1,1)^T$ and $-3,(1,-1)^T$. Thus every solution is $X_*+c_1e^{-t}(1,1)^T+c_2e^{-3t}(1,-1)^T$. The initial state requires $c_1=-3/2$, $c_2=-1/2$, giving $$\boxed{x=2-\tfrac 32e^{-t}-\tfrac 12e^{-3t},\qquad y=1-\tfrac 32e^{-t}+\tfrac 12e^{-3t}.}$$

Step 3: Check monotonicity and the limit. Here $x'=\tfrac 32(e^{-t}+e^{-3t})>0$ for $t\ge 0$, while $y'=\tfrac 32(e^{-t}-e^{-3t})>0$ for $t>0$ and $y'(0)=0$. Both start at zero and tend to $2$ and $1$, respectively. Their monotone approach excludes overshoot; the initial velocity is $(3,0)^T$, as the original equation requires. The two horizontal dotted lines in the figure are component limits, not additional time-dependent solutions.

Step 4: Identify the affine superposition rule. If $U,V$ solve the same forced equation, then $(\alpha U+\beta V)'-A(\alpha U+\beta V)=(\alpha+\beta)(3,0)^T$. It is a solution exactly when $\boxed{\alpha+\beta=1}$. In particular the zero function is not a solution, and the sum of two solutions generally fails. Differences of solutions are homogeneous; adding any homogeneous solution to one particular solution gives the complete affine family derived above.