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Since the exponential factor is positive, equilibria must have $y=0$ and $x^4=1$. Thus $\boxed{(x,y)=(-1,0),(1,0)}$. At either point the Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -4x^3& 0\end{array}\right)$. At $(-1,0)$ its eigenvalues are $\pm 2$, so this is a saddle point. At $(1,0)$ they are $\pm 2i$: the linearization has a centre equilibrium.

The requested first-order approximation follows by putting $x=1+u$, $y=v$: $\dot{u}=v+O(\Vert (u,v){\Vert }^2)$ and $\dot{v}=-4u+O(\Vert (u,v){\Vert }^2)$. The linearized system conserves $4u^2+v^2$, giving the approximate ellipsesHowever, the printed nonlinear system has a weak repelling focus at $(1,0)$, rather than a centre. A centre in the linearization need not be a nonlinear centre. Here that distinction can be proved explicitly, without relying on the numerical drift in the figure.

Put $X=2(x-1)$, $Y=y$, $\tau =2t$ and $r^2=X^2+Y^2$. The weak focus of an exponential polynomial planar system has the local positive functionExpanding the vector field through cubic order and differentiating givesThe cubic correction cancels the cubic terms in the derivative of $H$, and the quartic correction leaves the positive radial term $3r^4/16$. Since $H=r^2/2+O(r^3)$ is positive and the angular velocity stays near $-1$, small nonzero trajectories spiral outward. Equivalently the reversed system has a strict local Lyapunov function and an asymptotically stable equilibrium. In particular, a periodic orbit sufficiently near the equilibrium is impossible, because $H$ would have to increase strictly and return to its starting value. This proves the failure of the printed centre assertion while retaining its correct linearized ellipse calculation.

For the linearized system a centre occurs when $\boxed{{\lambda }_{1,2}=\pm i\beta ,\beta \ne 0}$. Its real solutions are periodic ellipses after an invertible linear coordinate change.

As explained by a centre in the linearization need not be a nonlinear centre, for a nonlinear system these eigenvalues alone do not guarantee a centre. For example $\dot{X}=-Y-X(X^2+Y^2)$, $\dot{Y}=X-Y(X^2+Y^2)$ has eigenvalues $\pm i$ at zero but polar equations $\dot{r}=-r^3$, $\dot{\theta }=1$, so trajectories spiral inward. Removing the cubic terms gives a genuine centre with the same Jacobian. Thus there is no eigenvalue-only sufficient condition for a nonlinear centre in the stated generality.

For the linearized system a centre occurs when $\boxed{{\lambda }_{1,2}=\pm i\beta ,\beta \ne 0}$. Its real solutions are periodic ellipses after an invertible linear coordinate change.

As explained by a centre in the linearization need not be a nonlinear centre, for a nonlinear system these eigenvalues alone do not guarantee a centre. For example $\dot{X}=-Y-X(X^2+Y^2)$, $\dot{Y}=X-Y(X^2+Y^2)$ has eigenvalues $\pm i$ at zero but polar equations $\dot{r}=-r^3$, $\dot{\theta }=1$, so trajectories spiral inward. Removing the cubic terms gives a genuine centre with the same Jacobian. Thus there is no eigenvalue-only sufficient condition for a nonlinear centre in the stated generality.

A centre equilibrium of the linearized real planar system hasAfter a real invertible linear change of coordinates, the system is a constant angular rotation. Its nonzero orbits are ellipses of period $2\pi /|\omega |$, so it is stable but not asymptotically stable. This is a classification of the linearization: a centre in the linearization need not be a nonlinear centre. In particular, imaginary eigenvalues alone do not prove a nonlinear centre equilibrium, which requires a punctured neighbourhood filled with periodic orbits. Part (c) illustrates exactly this limitation.

This real planar dynamical system has a saddle point at $(-1,0)$ and a repelling weak focus at $(1,0)$. The latter has purely imaginary linearized eigenvalues $\pm 2i$, so its linearization has closed elliptical orbits. Nonlinear terms nevertheless produce outward drift.

Set $X=2(x-1)$, $Y=y$, $\tau =2t$ and $r^2=X^2+Y^2$. The cubic expansion isThe linear rotation acts on scalar functions by $L=Y{\partial }_X-X{\partial }_Y$. For $H_2=r^2/2$, its cubic derivative is $X{Y}^2-3X^{2}Y/4-Y^3/2$. The homogeneous correction $H_3=7X^3/12+X{Y}^2/2+Y^3/3$ satisfies $L{H}_3$ equal to the negative of that derivative. The remaining quartic derivative of $H_2+H_3$ is $3X^2{Y}^2/2-X^{3}Y$. Withdirect differentiation givesfor all sufficiently small nonzero $r$. Meanwhile $H_2+H_3+H_4=r^2/2+O(r^3)>0$. In reversed time it is a strict Lyapunov function, proving asymptotic stability of the reversed system and hence repulsion forward in time. The angular speed stays close to $-1$, so the nearby nonzero trajectories spiral rather than approaching along a fixed direction. This explicit calculation illustrates why a centre in the linearization need not be a nonlinear centre.