Solutions to Systems — Question 10

Question 10 – Solution

Strategy. An invertible matrix of solution columns determines its coefficient matrix through its derivative, while its determinant alone does not determine lengths.

Step 1: Recover and verify the matrix. The determinant of $\Phi$ is $1$, so $$\Phi^{-1}=\begin{pmatrix}1+t^2&-t\\-t&1\end{pmatrix},\qquad \boxed{A=\Phi'\Phi^{-1}=\begin{pmatrix}-t&1\\1-t^2&t\end{pmatrix}.}$$ Multiplication gives $A\Phi=\begin{pmatrix}0&1\\1&2t\end{pmatrix}=\Phi'$. Any coefficient matrix satisfying the column equations must equal $\Phi'\Phi^{-1}$, proving uniqueness at every real time.

Step 2: Select the constants for the IVP. The full solution is $\boxed{X=\Phi(t)c}$ with constant $c\in\mathbb R^2$. Indeed $(\Phi^{-1}X)'=0$ for any solution, proving completeness. At time $1$, $$c=\begin{pmatrix}2&-1\\-1&1\end{pmatrix}(2,3)^T=(1,1)^T.$$ Thus $\boxed{X=(1+t,1+t+t^2)^T}$, which equals $(2,3)^T$ at $1$. Its derivative follows either directly or from the verified column identity.

Step 3: Rule out an uncoupled alternative. The first component of the first column is the constant $1$. Its equation would force $a(t)=0$ throughout the interval. The first component of the second column is $t$, whose derivative is $1$; with $a=0$ its proposed equation would instead give derivative $0$. This is impossible on any nonempty interval. The coupling is essential.

Step 4: Separate independence from length preservation. Here $\det\Phi=1$ and $\operatorname{tr}A=-t+t=0$, consistent with the constant determinant. But the second solution column has squared length $$\boxed{t^2+(1+t^2)^2=1+3t^2+t^4,}$$ which is not constant. A nonzero determinant ensures independence of the columns; even a constant determinant does not force individual solution lengths to stay fixed. The explicit column supplies a counterexample.