I am used to finding periodic solution after finding the general solution. I am kinda stuck at its general solution. Don't understand how to go about. Any hint would do. Also is there a way to prove that this equation has a periodic solution without finding the general solution?
2026-03-25 17:45:08.1774460708
Solution to ODE $ x'= x^3+ \sin( t)$ and showing it has a periodic solution
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The cubic term provides positive feedback which causes exploding solutions in positive time direction with poles of the form $x(t)\approx\pm(2(t_*-t))^{-1/2}$.
Thus look at the reversed time direction. To have the more "intuitive" forward time direction consider $y(t)=x(-t)$. Then $$ y'(t)=-y(t)^3+\sin(t). $$ The same cubic term now provides negative feedback, moving large $y$ in the direction of the interval $[-1,1]$.
On the line $y=2$ we find that the slope $y'=-8+\sin(t)\le -7$ points down towards $0$, the same on the line $y=-2$ where $y'=8+\sin(t)\ge 7$ points upwards towards $0$. In conclusion, solutions starting in $[-2,2]$ stay inside that interval.
Let $f(a)$ denote the value $y(2\pi)$ of a solution for the initial value $y(0)=a$. If $\phi(t;t_0,y_0)$ is the flow, then $f(a)=ϕ(2\pi;0,a)$. $f$ is a continuous map from the interval $[-2,2]$ into itself. Thus there is at least one fixed point. The solution(s) to the fixed point(s) are $2\pi$-periodic.