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Find all $2\pi$ periodic solutions (either constant or non-constant) of the nonautonomous equation

$\dot{x}=x(1+\cos(t))-x^3$.

I know that the only equilibria is $x=0$ which is a source.

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    $\begingroup$ But $x=\pm1$ are not equilibria, are they? $\endgroup$ – Did Nov 20 '16 at 19:24
  • $\begingroup$ You're right, then is $x=0$ the only periodic orbit? $\endgroup$ – BronchoX Nov 20 '16 at 19:28
  • $\begingroup$ Why should they? $\endgroup$ – Did Nov 20 '16 at 19:36
  • $\begingroup$ I don't understand your question $\endgroup$ – BronchoX Nov 20 '16 at 19:40
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    $\begingroup$ @Did Btw, me neither understand your approach. Since the question has already been answered, could you shed more light on your idea? $\endgroup$ – Evgeny Nov 25 '16 at 13:17
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You can solve this as Bernoulli equation, set $u=r^{-2}$ then $$ u'=-2r^{-3}r'=-2(1+\cos t)u+2 $$ which now can be nicely solved as a first order linear ODE.

$A(t)=e^{2(t+\sin t)}$, then $(A(t)u(t))'=2A(t)$, $$ A(2\pi)u(2\pi)=u(0)+\int_0^{2\pi}A(s)ds $$ So for $u(2\pi)=u(0)$ you need $$ u(0)=\frac{\int_0^{2\pi}e^{2(s+\sin s)}ds}{e^{4\pi}-1}=\frac12e^{2\sin(\theta·2\pi)},\quad\theta\in(0,1) $$ which gives exactly one positive radius.

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  • $\begingroup$ Great answer, its so clear! $\endgroup$ – BronchoX Feb 7 '18 at 3:14

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