Solutions to Systems — Question 3

Question 3 – Solution

Strategy. Derive the determinant law by the product rule, and construct the columns by solving the triangular equations.

Step 1: Differentiate the determinant. Substitution gives $$W'=(2tu_1+u_2)v_2+u_1(-tv_2) -(-tu_2)v_1-u_2(2tv_1+v_2)=tW.$$ The off-diagonal terms cancel. This is the trace law here, since $\operatorname{tr}A=t$, obtained directly rather than assumed.

Step 2: Apply the determinant’s initial value. At $0$, $W(0)=1$, and solving $W'=tW$ gives $\boxed{W(t)=e^{t^2/2}}$. The proposed $e^{t^2}$ has derivative $2t e^{t^2}$ and fails the required equation except at the isolated time $0$; matching the initial value is insufficient.

Step 3: Construct and verify the matrix. Put $J(t)=\int_0^t e^{-3s^2/2}\,ds$. The second equation gives $y=c_2e^{-t^2/2}$, and the first gives $(e^{-t^2}x)'=c_2e^{-3t^2/2}$. Therefore $$\boxed{\Phi(t)=\begin{pmatrix}e^{t^2}&e^{t^2}J(t)\\0&e^{-t^2/2}\end{pmatrix}.}$$ Since $J(0)=0$, $\Phi(0)=I$. The derivative of its upper-right entry is $2t e^{t^2}J+e^{-t^2/2}$, precisely the required first-row expression. The remaining entries satisfy their rows directly. Its determinant agrees with Step 2 and never vanishes, so this is a principal fundamental matrix.

Step 4: Interpret the selected solution. The initial vector $(0,1)^T$ selects the second column: $X=(e^{t^2}J(t),e^{-t^2/2})^T$. Its second component strictly decreases for $t>0$, even while $W$ grows. A determinant concerns a pair of independent columns, not the separate monotonicity of each component. The nonzero determinant guarantees a complete independent family throughout the real axis.