= Solution
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 $\begin{pmatrix}0&1\\-4x^3&0\end{pmatrix}$. At $(-1,0)$ its <eigenvalues> are $\pm2$, so this is a <saddle point>. At $(1,0)$ they are $\pm2i$: the <linearization> has a <centre equilibrium>.
The requested first-order approximation follows by putting $x=1+u$, $y=v$: $\dot u=v+O(\|(u,v)\|^2)$ and $\dot v=-4u+O(\|(u,v)\|^2)$. The linearized system conserves $4u^2+v^2$, giving the approximate ellipses
$$
\boxed{4(x-1)^2+y^2=\text{constant}.}
$$
However, \b[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 function
$$
H=\frac{X^2+Y^2}{2}+\frac7{12}X^3+\frac12XY^2+\frac13Y^3
+\frac14X^4-\frac3{16}X^3Y+\frac3{16}XY^3.
$$
Expanding the vector field through cubic order and differentiating gives
$$
\boxed{\frac{dH}{d\tau}=\frac3{16}r^4+O(r^5)>0\quad\text{for sufficiently small }r>0.}
$$
The 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.
\Image[/past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2005-linear-centre-nonlinear-focus.png]
{title=The linearized ellipses hide the nonlinear outward spiral}
{description=Original fourth-order Runge-Kutta integration of the printed vector field, compared with its linearization from the same initial point. The local polynomial calculation, rather than the plot, proves the weak instability.}
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