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. Can I please have help solving the following? I am not quite sure how to approach this problem and was wondering if someone could please help. Thanks!
Consider the system $$x'=-y-x^2$$ $$y'=x.$$ Find a reversible $S$, identify Fix$(S)$ and show that the origin is a nonlinear center. 
 A: The system $S$ given by
$\dot x = -y - x^2, \tag 1$
$\dot y = x, \tag 2$
has precisely one fixed point, $(0, 0)$, since
$0 = -y - x^2, \tag 3$
$0 = x, \tag 4$
has exactly the one solution
$x = y = 0. \tag 5$
The Jacobean matrix $J_S(x, y)$ of $S$ is 
$J_S(x, y) = \begin{bmatrix} \dfrac{\partial \dot x}{\partial x} & \dfrac{\partial \dot x}{\partial y}\\
\dfrac{\partial \dot y}{\partial x} & \dfrac{\partial \dot y}{\partial y} \end{bmatrix} = \begin{bmatrix} -2x & -1 \\ 1 & 0 \end{bmatrix}; \tag 6$
thus
$J_S(0, 0) = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag 7$
the characteristic polynomial of this matrix is
$\lambda^2 + 1 = \det(J_S(0, 0) - \lambda I) = \det \left (\begin{bmatrix} -\lambda & -1 \\ 1 & -\lambda \end{bmatrix} \right ), \tag 8$
and so the eigenvalues of $J_S(0, 0)$, which are the roots of this polynomial, are
$\lambda = \pm i; \tag 9$
since the eigenvalues are purely imaginary, the point $(0, 0)$ is classified as a center of the system $S$.
To reverse the system $S$, that is, to write the equations for a system $S_R$ the trajectories of which run opposite to those of $S$, we merely need to negate both $\dot x$ and $\dot y$; thus the equations of $S_R$ are
$\dot x = y + x^2, \tag{10}$
$\dot y = -x; \tag{11}$
the analysis of $S_R$ parallels that of $S$; the principal fact of import is that we have
$J_{S_R}(0, 0) = -J_S(0, 0) =  \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \tag{12}$
with the same characteristic polynomial $\lambda^2 + 1$, and hence the same eigenvalues $\pm i$ as given in (9).  Thus $S_R$ is also possessed of a center at the origin.
