Can someone please tell me why our solution implies the origin is a nonlinear center? Thank you for your time and help!
Consider the system $$x'=-y-x^2$$ $$y'=x.$$ Find a reversor $S$, identify Fix$(S)$ and show that the origin is a nonlinear center.
$\textbf{Solution:}$ $S(x,y) = \begin{pmatrix} -x \\ y \end{pmatrix}$ is a reversor for our system. Sending $x\to -x, y\to y,$ and $t\to -t$ we get $-(-x') = -y-(-x)^2 \implies x' = -y - x^2$ and $-y' = -x \implies y'=x.$ If $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -x \\ y \end{pmatrix}$ then $\begin{pmatrix} x \\ y \end{pmatrix} \in$ Fix$(S)$. So, Fix$(S)$ $$= \left \{\begin{pmatrix} 0 \\ y \end{pmatrix}\colon y \in \mathbb{R} \right\}.$$
A movement integral can be obtained as follows. From
$$ \frac{dx}{dy}=-\frac{x^2+y}{x}\Rightarrow \frac 12 (x^2)'+x^2+y=0 $$
or
$$ x^2 = C_0e^{-2y}-y+\frac 12 $$
so with $C_0=-\frac 12$ we can determine a set of closed level curves around the origin, characterizing a center.