Solutions to Systems — Question 1

Question 1 – Solution

Strategy. Check vector residuals first; then solve the triangular equations to establish completeness.

Step 1: Test all candidates. Direct differentiation gives $U'-AU=0$ and $V'-AV=0$. For the third candidate, $$W'=e^t(1,t+1)^T,\quad AW=e^t(1+t,t)^T, \quad\boxed{W'-AW=e^t(-t,1)^T.}$$ Its second residual component never vanishes, so $W$ is not a solution on any nonempty open interval.

Step 2: Prove that the valid family is complete. The matrix $\Phi=(U\ V)=e^t\begin{pmatrix}1&t\\0&1\end{pmatrix}$ has determinant $e^{2t}>0$. More explicitly, the second equation gives $y=c_2e^t$. Then $(e^{-t}x)'=c_2$, so $$\boxed{X=e^t(c_1+c_2t,c_2)^T=c_1U+c_2V.}$$ This derivation reaches every solution and introduces exactly two arbitrary constants. The family is valid on the whole real axis.

Step 3: Select and verify the initial-value solution. At $0$, the constants are $c_1=2$, $c_2=-1$, hence $\boxed{X=e^t(2-t,-1)^T}$. Its derivative is $e^t(1-t,-1)^T$, exactly $AX$, and its initial state is $(2,-1)^T$ as required.

Step 4: Interpret a component zero. Since $e^t\ne 0$, the first component first vanishes at $t=2$. There $X(2)=(0,-e^2)^T$ and $X'(2)=(-e^2,-e^2)^T$. The state is nonzero and continues moving; the first component crosses zero with negative slope. For a homogeneous linear system with continuous coefficients, only a full zero state at a time would force the identically zero solution by uniqueness. The plot displays both components.