I want to solve the following exercise:
Determine the stability properties of the following solution: $$ (\cos t, \sin t) \text{ of }\,\begin{align}\dot{x} &= -y(x^2 + y^2)^{-1/2}\\\dot{y} &= x(x^2 + y^2)^{-1/2}\end{align} $$
What I've tried: I tried to expand this system around its solution $(\cos t, \sin t)$ and examine the eigenvalues of the coefficient matrix. We have \begin{align} f(x) = \begin{pmatrix}-y(x^2 + y^2)^{-1/2}\\x(x^2 + y^2)^{-1/2}\end{pmatrix} \end{align} and thus \begin{align} \dfrac{\partial f(x,y)}{\partial(x,y)} = \begin{bmatrix}\dfrac{xy}{(x^2 + y^2)^{3/2}} & -\dfrac{x^2}{(x^2 + y^2)^{3/2}}\\\dfrac{y^2}{(x^2 + y^2)^{3/2}} & -\dfrac{xy}{(x^2+ y^2)^{3/2}}\end{bmatrix} \end{align} The linear expansion is given by \begin{align} \begin{bmatrix}\dot{x}\\\dot{y}\end{bmatrix} = \dfrac{\partial f(\cos t, \sin t)}{\partial (x,y)}\begin{bmatrix}x\\ y\end{bmatrix} + \text{ higher order terms} \end{align} where $$ \dfrac{\partial f(\cos t, \sin t)}{\partial (x,y)} = \begin{bmatrix} \cos t\sin t & -\cos^2t\\\sin^2t & -\cos t \sin t\end{bmatrix} $$ Unfortunately the coefficient matrix $\dfrac{\partial f(\cos t, \sin t)}{\partial (x,y)}$ is singular (its determinant is zero) and the critical point is degenerate. I therefore cannot say anything about the stability of the solution based on the eigenvalues. I've looked at the answer in my textbook and apparently this solution is stable. I just don't know how to show it.
Question: How do I show that this solution is stable?