= Solution
{c}
Here a periodic solution must mean a nonconstant periodic orbit: constant equilibria are not excluded by the stated divergence hypothesis. Suppose such an orbit exists. Since the vector field is continuously differentiable, uniqueness for the <autonomous differential equation> makes its trajectory a simple closed curve $C$, bounding a region $D$ by the <Jordan curve theorem>. The <Green theorem> gives
$$
\iint_D\left[\partial_x(\rho f)+\partial_y(\rho g)\right]dx\,dy
=\oint_C\rho f\,dy-\rho g\,dx.
$$
Along the orbit, $dx=fdt$ and $dy=gdt$, so the line integral is identically zero. Reversing orientation if necessary does not change this conclusion. The area integral is strictly positive because its continuous integrand is positive everywhere and $D$ has positive area. This contradiction proves the <Bendixson-Dulac criterion> in this case.
For the second-order equation set $y=x'$, giving $x'=y$, $y'=-y-\varepsilon y^3-\sin x$. Choose the <Dulac function> $\rho=-1$. Then
$$
\partial_x(\rho f)+\partial_y(\rho g)=1+3\varepsilon y^2>0.
$$
Thus \b[there is no nonconstant periodic solution for $\varepsilon>0$]. The equilibria $x(t)=k\pi$ do remain solutions. As a direct check, the mechanical energy $H=y^2/2+1-\cos x$ obeys $H'=-y^2-\varepsilon y^4$, whose integral over a period could vanish only if $y\equiv0$.
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