= Solution
Differentiate the proposed <first integral> along the vector field:
$$
\dot K=2x\left(\frac\alpha2x+y-2y^3\right)+(2y-4y^3)(-x)
=\boxed{\alpha x^2}.
$$
At $\alpha=0$, $K$ is conserved and the system is an <inverted quartic oscillator phase portrait>. Its <equilibria> are $O=(0,0)$ and $S_\pm=(0,\pm1/\sqrt2)$. The <Jacobian matrix> is
$$
J=\begin{pmatrix}0&1-6y^2\\-1&0\end{pmatrix}.
$$
At the origin its <eigenvalues> are $\pm i$. Moreover $K$ has a strict local minimum there, and its nearby positive contours are closed, so the origin is a genuine <center equilibrium>, not merely an inconclusive linear centre. The arrows circulate clockwise: on the positive $x$-axis, $\dot y<0$. At either $S_\pm$, the <eigenvalues> are $\pm\sqrt2$, giving <saddle equilibria>. The stable eigendirection is $\delta x=\sqrt2\delta y$, and the unstable one is $\delta x=-\sqrt2\delta y$.
The special energy level factors as
$$
K=\frac14\quad\Longleftrightarrow\quad
x^2=(y^2-\tfrac12)^2,\qquad
\boxed{x=\pm(y^2-\tfrac12).}
$$
Between the two saddles, its left arc has $x<0$ and travels upward from $S_-$ to $S_+$; its right arc has $x>0$ and travels downward from $S_+$ to $S_-$. These two <heteroclinic orbits> bound the central eye. Trajectories inside the eye are closed <periodic orbits> and do not tend to the centre as $t\to\infty$.
More explicitly, a level $K=C$ has
$$
x^2=y^4-y^2+C.
$$
For $0<C<1/4$, its bounded central component is periodic, while its exterior components turn at $|y|=\sqrt{(1+\sqrt{1-4C})/2}$ and then escape. For $C>1/4$, $x$ never vanishes and the two branches run from one vertical infinity to the other. Levels $C\le0$ have only exterior nonconstant components, apart from the origin at $C=0$.
On the upper outer <separatrix>, the $x>0$ branch arrives from positive infinity and approaches $S_+$ in forward infinite time; the $x<0$ branch leaves $S_+$ towards positive infinity. On the lower outer <separatrix>, the $x<0$ branch arrives from negative infinity and approaches $S_-$, while the $x>0$ branch leaves towards negative infinity. Approaching a saddle along its <stable manifold> is exponential at rate $\sqrt2$. A trajectory close to but off a <separatrix> can spend a long time near the saddle before completing a <periodic orbit> or escaping.
The escape is \b[<finite-time blowup>], not an asymptote at $t=\infty$. On any escaping level, $|x|=\sqrt{y^4-y^2+C}\sim y^2$, so the remaining travel time is bounded by a convergent integral proportional to $\int^\infty dy/y^2$. Thus an outward upper branch has $y\sim(T-t)^{-1}$, $x\sim-(T-t)^{-2}$; an outward lower branch has $y\sim-(T-t)^{-1}$, $x\sim(T-t)^{-2}$. The statement about approaching from infinity likewise refers to a finite past endpoint. Only <equilibria>, periodic trajectories and trajectories tending to saddles have a future defined for all time in this conservative portrait.
For small positive $\alpha$, the <equilibria> remain in place, but the origin's <eigenvalues> become $\alpha/4\pm i\sqrt{1-\alpha^2/16}$, so it becomes a repelling <focus>. The saddles retain one <eigenvalue> of each sign, $\alpha/4\pm\sqrt{2+\alpha^2/16}$. Since $\dot K\ge0$, and its time integral is strictly positive on every nonconstant trajectory, trajectories cross $K$-contours towards larger values. Near the centre this means outward spiralling. No nonconstant <periodic orbit> can remain, and the former saddle-to-saddle connections at equal $K=1/4$ are broken. Stable saddle trajectories approach from $K<1/4$; unstable ones depart into $K>1/4$. Generic central trajectories eventually escape, while the exceptional stable saddle manifolds still approach their saddles.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ia/paper-2-quartic-phase-portrait.png]
{title=Conservative quartic phase portrait with heteroclinic separatrices, and outward contour crossing under small positive antidamping}
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