Show second-order system $x''(t)=x(t)^3-x(t)$ is conservative Could someone please take me through the process of showing that the second-order system $\frac{d^2x}{dt^2} = x^3 -x$ is conservative?
 A: As the equation doesn't have first derivatives and it's in one variable, you always can write the equation in the form:
$\dfrac{1}{2}\left(\dfrac{dv}{dt}\right)^2=\dfrac{x^4}{4}-\dfrac{x^2}{2}+C$ (with $v=\dfrac{dx}{dt}$), meaning that $\dfrac{1}{2}\left(\dfrac{dv}{dt}\right)^2-\left(\dfrac{x^4}{4}-\dfrac{x^2}{2}\right)=C$, so is, the sum of the two terms is constant for any solution. We interpret the first term of the resulting equation as the kinetic energy of the system, the second term as the potential energy and the third, the constant $C$ as the total energy.
We can get the equation for the energy this way. We have $\dfrac{d^2x}{dt^2} = x^3 -x$, now,
$v\dfrac{dv}{dt}=\dfrac{dx}{dt}\left(\dfrac{x^4}{4}-\dfrac{x^2}{2}\right)$ and integrating $vdv=(\dfrac{x^4}{4}-\dfrac{x^2}{2})dx$ leads to the desired result. In general, as the problem is in one dimension, for equations of the form $\dfrac{d^2x}{dt^2} = F(x)$ always exist a function $V(x)$ such that $F(x)=V'(x)$ and hence, a term interpretable as the potential energy of the system. For equations of the form $\dfrac{d^2x}{dt^2} +A\dfrac{dx}{dt}= F(x)$ is not possible to find such an equation.
