Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 14A / c / Solution

The second-order equation is equivalent to the planar system after setting $y=\dot{x}-ax+b{x}^3$. Any nonconstant periodic solution has a maximum $M$ and minimum $m$. At an extremum $\ddot{x}=x^3-x$, so $M>1$ and $m<-1$ are impossible. Equality at either endpoint would give one of the equilibrium points, and uniqueness would force a constant solution. Therefore a nonconstant periodic orbit lies in $-1<x<1$.

If $b/a<1$, throughout this striphas the sign of $a$ whenever $x\ne 0$. A nonconstant orbit spends a positive time away from $x=0$, so its integral cannot vanish, contradicting part (b). Thus there are no nonconstant periodic solutions. The constant equilibrium solutions of course remain; as usual the question's periodic-orbit assertion excludes equilibria.