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Given a governing ODE such as $$m\ddot{x}=x^3-1$$ how do we go about finding the stability of the various states of equilibria?

I am aware that, given a potential $V$ for the system, we can determine the equilibria by taking the first derivative with respect to time and then analysing the sign of the second derivative to determine whether each state is stable or unstable. Here is my attempt at the above question:

Clearly the RHS of our ODE is zero only if $x=1$ (we're dealing with real numbers only here). So to determine if this is a stable or unstable position of equilibrium, I would take the derivative of the RHS with respect to $x$, giving $3x^2$. At $x=1$, this is clearly positive. Therefore this is a position of unstable equilibrium.

Is the above correct? In particular, I know that $V^{\prime\prime}>0$ implies stability, but would it be the opposite that implies stability of a force ie. $(m\ddot{x})^{\prime\prime}<0$? (My thinking behind this was that $F=-V^\prime$, but this would surely be assuming that $F$ is conservative, which may not be the case?)

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$$ m \dot x\ddot x = x^3\dot x -\dot x \Rightarrow \frac{1}{2}\frac{d}{dt}(\dot x)^2-\frac{1}{4}\frac{d}{dt}x^4+\dot x = 0 $$

then integrating

$$ \frac{1}{2}(\dot x)^2-1/4x^4+x = C $$

This show us that no closed curve is allowed in the set $\dot x \times x$ hence no stable equilibrium points exist for this dynamic system.

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