= Solution
Put $A=g\sin\alpha>0$ and $v=\dot x$. Multiplying the equation by $xv$ gives
$$
\frac d{dt}\frac{(xv)^2}2=Ax^2\dot x=\frac d{dt}\frac{Ax^3}3,
$$
so its <first integral> is
$$
\boxed{x^2v^2=\frac{2A}3x^3+c},\qquad v=\sqrt{\frac{2A}3x+\frac c{x^2}}.
$$
These level curves give the positive <phase plane> for the <avalanche front with linear mass entrainment>. For $c<0$, the curve starts at $v=0$, $x=(-3c/(2A))^{1/3}$ and rises. For $c>0$, it diverges as $x\downarrow0$, has a minimum at $x=(3c/A)^{1/3}$, and then rises. The $c=0$ curve is $v=\sqrt{2Ax/3}$. Forward-time arrows have increasing $x$; at a zero-velocity endpoint the equation gives $\ddot x=A>0$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-4-avalanche-phase-plane.png]
{title=Avalanche phase trajectories approaching the linear-entrainment asymptote}
Every forward-moving trajectory has $x\to\infty$: after it has entered $v>0$, a bounded limiting $x$ would force a zero of the displayed positive-branch speed ahead of it, but its only possible zero is the lower endpoint. For large $x$,
$$
\frac{v}{\sqrt{2Ax/3}}=\sqrt{1+\frac{3c}{2Ax^3}}\longrightarrow1,\qquad v-\sqrt{2Ax/3}\longrightarrow0.
$$
Expanding the equation as $x\ddot x+v^2=Ax$ and substituting the <first integral> gives
$$
\boxed{\ddot x=\frac A3-\frac c{x^3}\longrightarrow\frac13g\sin\alpha}.
$$
Thus both the asymptotic <phase plane> trajectory and the limiting <acceleration> are independent of the initial data.
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