= Solution
{c}
Where $y\ne\beta$, taking the quotient of the two <ordinary differential equations> gives
$$
\frac{dy}{dx}=\frac{x^3-x}{y-\beta},\qquad
\frac{(y-\beta)^2}{2}=\frac{x^4}{4}-\frac{x^2}{2}+\text{constant}.
$$
Choose the <first integral>
$$
\boxed{h(x,y)=\frac{(y-\beta)^2}{2}+\frac{x^2}{2}-\frac{x^4}{4}.}
$$
The <chain rule> checks conservation even at points excluded in taking that quotient:
$$
\dot h=(x-x^3)(y-\beta)+(y-\beta)(x^3-x)=0.
$$
Thus each trajectory is contained in a <connected component> of a <level set> of $h$. At $(0,\beta)$ the <gradient> vanishes and the <Hessian matrix> is the identity, so $h$ has a strict <local minimum> there. To make the closed-trajectory argument explicit, put $z=y-\beta$ and $0<E<1/4$. The bounded component of $h=E$ is
$$
z=\pm\sqrt{2E-x^2+x^4/2},\qquad
|x|\leq r_E:=\sqrt{1-\sqrt{1-4E}}<1.
$$
The two arcs join at $(\pm r_E,0)$ to form a <simple closed curve> that is a <regular level set>. The <gradient> cannot vanish on it: the only <critical point of a dynamical system> in $|x|<1$ is the centre, whose level is $0$.
The <vector field> is tangent and nowhere zero on this <compact set>, so its speed has a positive minimum. It therefore traverses the curve in finite time and returns to its initial point, giving a <periodic orbit>. These ovals shrink to the centre as $E\downarrow0$. This is the <planar first integral criterion for a centre> and proves the required nonlinear assertion; it also gives <Lyapunov stability> without <asymptotic stability>.
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