= Solution
Introduce $H=x^2-y^2+y^4/2$. Along a solution,
$$
\dot H=2x(\alpha x-y+y^3)+(-2y+2y^3)(-x)=\boxed{2\alpha x^2}.
$$
At $\alpha=0$, $H$ is a <first integral>. The identity $H+1/2=x^2+(y^2-1)^2/2$ shows its <level sets> are bounded: neither $x$ nor $y$ can escape to infinity on a fixed level. Consequently \b[trajectories remain bounded as $t\to\infty$], including those starting far from the origin. This bound also prevents finite-time escape; no explicit solution is needed.
The <equilibrium points> are $(0,0)$ and $(0,\pm1)$. The <Jacobian matrix> at $(0,y_*)$ is
$$
J=\begin{pmatrix}\alpha&3y_*^2-1\\-1&0\end{pmatrix}.
$$
At the origin its <eigenvalues> are $(\alpha\pm\sqrt{\alpha^2+4})/2$, with opposite signs, so the origin is a <saddle point> in all three cases. An <eigenvector> for $\lambda$ lies on $y=-x/\lambda$; the stable direction has positive slope and the unstable direction negative slope. At $(0,\pm1)$ the <eigenvalues> are $(\alpha\pm\sqrt{\alpha^2-8})/2$.
When $\alpha=0$, these latter <eigenvalues> are $\pm i\sqrt2$. The nonlinear classification follows from the strict minima of $H$ at the two points: nearby <level sets> are closed curves without other equilibria, and the nonvanishing vector field makes them <periodic orbits>. Thus the two points are <center equilibria>. For $-1/2<H<0$, there are two separate closed ovals, one around each <center equilibrium>. The level $H=0$ consists of the <saddle point> and two <homoclinic orbits> forming a figure eight, explicitly
$$
x=\pm y\sqrt{1-y^2/2},\qquad |y|\le\sqrt2.
$$
For $H>0$, a single outer closed orbit encloses all three <equilibrium points>. Motion is clockwise, as the vector points downwards wherever $x>0$. The upper <homoclinic orbit> leaves the origin into $x<0,y>0$ and returns through $x>0,y>0$; the lower one has the reflected orientation.
When $\alpha=0.1$, the two nonzero <equilibrium points> have <eigenvalues> $0.05\pm i\sqrt{7.99}/2$ and are unstable <foci>. Nearby paths spiral outward clockwise. The <first integral> becomes a strictly increasing energy along every nonconstant trajectory: although $\dot H$ can momentarily vanish when $x=0$, it cannot vanish on an interval unless the solution is an equilibrium. Thus there are no <periodic orbits> or <homoclinic orbits>. The two stable separatrices of the <saddle point>, traced backwards, spiral towards the corresponding unstable <foci>. Its unstable separatrices move onto positive energy and escape outwards. All other nonequilibrium trajectories outside the <stable manifold> are unbounded forwards: a bounded forward limit set would have to lie in $x=0$, whose only invariant points are the three equilibria; the two unstable <foci> cannot attract a nonconstant trajectory, and convergence to the saddle occurs only on its <stable manifold>. Escape is not finite-time blow-up, since $\dot H\le2\alpha(H+1/2)$ bounds the energy on every finite interval.
When $\alpha=-0.1$, the <eigenvalues> at the nonzero points are $-0.05\pm i\sqrt{7.99}/2$, so they are <stable foci>. The energy decreases, keeping all forward trajectories in compact sublevel sets. A bounded limit set must lie in the invariant part of $x=0$, so trajectories converge to equilibria: generically to one of the two <stable foci>, exceptionally to the saddle along its <stable manifold>. The saddle's unstable separatrices spiral into the two <stable foci>; its <stable manifold> separates their attraction basins. Again, energy monotonicity excludes <periodic orbits> and <homoclinic orbits>. These three cases give the <quartic double-well phase portrait with linear damping> below.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-2-phase-portraits.png]
{title=Conservative double-well closed orbits and the outward or inward spirals produced by positive or negative alpha}
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