Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 2 / 2A / Solution

Set $x=e^z>0$. The chain rule gives $\ddot{x}=x(\ddot{z}+{\dot{z}}^2)$, so the transformed equation is the forced harmonic oscillator $\ddot{x}+x=-1$, with $x(0)=1$ and $\dot{x}(0)=V$. ThereforeThis real solution is defined only where its logarithm has positive argument. Put $R=\sqrt{V^2+4}$ and $\delta =\arctan (V/2)$. Then $x+1=R\cos (t-\delta )$, and the connected interval containing zero is $|t-\delta |<\arccos (1/R)$.

Since $\dot{z}=\dot{x}/x$, a stationary point requires $\dot{x}=0$. The positive branch cannot contain the oscillator minimum $x=-R-1$. It contains the maximum $x=R-1$, attained first at $t=\delta$. HenceAt the endpoints where $x\downarrow 0$, $z$ tends to minus infinity; the oscillator solution cannot be continued through those zeros as a finite real $z$.