Differentiate the squared <Euclidean norm> and use the <skew-symmetric matrix> identity $A^T=-A$:
$$
\frac d{dt}\|\mathbf y(t)\|_2^2
=2\mathbf y^T\mathbf y'
=2\mathbf y^TA(\mathbf y)\mathbf y
=\mathbf y^T(A+A^T)\mathbf y=0.
$$
Thus $\|\mathbf y(t)\|_2^2$ is <constant>, and continuity of the nonnegative <square root> gives
$$
\boxed{\|\mathbf y(t)\|_2=\|\mathbf y_0\|_2\quad(t\geq0).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 341 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-341/1/a/solution