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Put $A=g\sin \alpha >0$ and $v=\dot{x}$. Multiplying the equation by $xv$ givesso its first integral isThese 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 $x\downarrow 0$, 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 $\ddot{x}=A>0$.

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 $x$,Expanding the equation as $x\ddot{x}+v^2=Ax$ and substituting the first integral givesThus both the asymptotic phase plane trajectory and the limiting acceleration are independent of the initial data.