## Related questions with answers

For solutions of $\overrightarrow{\textbf{x}}'=\textbf{A}\overrightarrow{\textbf{x}}$ with A skew-symmetric, the length of vector $\overrightarrow{\textbf{x}}$ is constant. What does this mean in terms of phase-plane trajectories? Use IDE's Matrix Element Input tool, or an open-ended graphic DE solver, to verify this for the following systems, and explain the role of the parameter k. (a)

$\overrightarrow{\textbf{x}}'=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\overrightarrow{\textbf{x}}$

(b)

$\overrightarrow{\textbf{x}}'=\begin{bmatrix}0&k\\-k&0\end{bmatrix}\overrightarrow{\textbf{x}},k>0$

.

Solution

Verified$(a)$ We have the system

$\overrightarrow{\textbf{x}}'=\begin{bmatrix}0&1\\-1&0\end{bmatrix}\overrightarrow{\textbf{x}}$

or

$\begin{align*}\begin{cases}x'=y\\y'=-x\end{cases}\end{align*}$

Trajectories of this system are circles centered around the origin and the length of the vector $\overrightarrow{\textbf{x}}$ is a constant. We can see trajectories on following photo:

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