= Solution
The second-order equation is equivalent to the planar system after setting $y=\dot x-ax+bx^3$. 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 strip
$$
(x-x^3)(ax-bx^3)=a x^2(1-x^2)\left(1-\frac ba x^2\right)
$$
has the sign of $a$ whenever $x\ne0$. A nonconstant orbit spends a positive time away from $x=0$, so its integral cannot vanish, contradicting part (b). Thus \b[there are no nonconstant periodic solutions]. The constant equilibrium solutions of course remain; as usual the question's periodic-orbit assertion excludes equilibria.
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