A solution of the second-order differential equation $$ x''+x-x^3=0 $$ satisfies the initial condition $x(0)=0$ and $x'(0)=v_0$. For what value of $v_0$ is the solution periodic?
I have tried using the phase-plane method, starting by writing $$ x'=y \tag{1}$$ $$ y'=x^3-x \tag{2}$$ So I have $$ \frac{dy}{dx}=\frac{y'}{x'}=\frac{x^3-x}{y} $$ $$ y^2=\frac{x^4}{2}-x+c= $$ For the initial equation $x(0)=0$ and $x'(0)=v_0$, it follows that $c=v_0^2$ and I have $$ y^2= \frac{x^4}{2}-x^2+v_0^2 = \frac{(x^2-1)^2+2v_0^2-1}{2} \tag{3}$$ But I'm stuck on this. How can I determine the value $v_0$ that makes this solution a periodic solution? What condition ensures that this is a periodic solution? Or generally, what is the general idea to find $v_0$ that makes this solution a periodic solution?
Edit: Fix the typo $(x-1)^2 \rightarrow (x^2-1)^2$
Equation $(3)$ seems to say it all, provided you fix the typo to $(x^2-1)^2$ because if $v_0>\frac1{\sqrt2}$ then $y=x^{\prime}>0$, so $x$ just keeps on going. If $v_0=\frac1{\sqrt2}$, then when $y=x^{\prime}=0$, $x=1$, so $y^{\prime}=x^{\prime\prime}=0$ also, so (unstable) equilibrium is reached, so the system never turns around. Or, put another way, $$\frac1{\sqrt2}dt=\frac{dx}{1-x^2}$$ So it takes infinite time to reach $x=1$. If $v_0<\frac1{\sqrt2}$, then the turnaround point is reached in finite time and there is acceleration back to the starting point at that point and the system is conservative, so it's periodic. The same analysis works if $v_0>-\frac1{\sqrt2}$, so the range of periodicity seems to be $v_0\in(-\frac1{\sqrt2},\frac1{\sqrt2})$, but I seem to have been batting $0.000$ in dynamical systems lately so you might want to check over my statements carefully.
EDIT: Might as well add an improved phase portrait.