Proving $\ddot x = w-2x+x^2$, where $w\ge0$, conserves energy Take $\ddot x = w-2x+x^2$, where $w\ge0$.
Show that the evolution of $x$ conserves a form of the energy and identify the potential function, sketching the phase portrait for $w=0$.
I am not too sure what is meant by the question. I know that to prove energy is conserved, you need to show that the derivative equals $0$. If we take $V = w-2x+x^2$, then $\dot V = -2\dot x+2x\dot x = 2\dot x(x-1) = 0$, but how do we prove this equals $0$, if this is what we need to show?
I also thought it may be helpful to rewrite the equation as the system:
$\dot x=y$, $\dot y = w-2x+x^2$, which would give $\dot V = 2y(x-1)$.
Any help would be hugely appreciated, thank you!
 A: The expression $F(x) = w-2x+x^2$ is not the energy, but the "force" exerted on the particle (assuming mass equal to one), depending on its position in the $x$ axis (remember Newton's second law).
By definition, potential energy is equal to $V(x) = - \int F(x)dx = \int(2x-x^2-w)dx = x^2 - \dfrac{x^3}{3} - wx$ (the constant of integration does not matter).
Now, on the other hand, if $v = \dfrac{dx}{dt}$ is the velocity of the particle, the equation says that:
$$\frac{dv}{dt} = -\frac{dV}{dx
}$$
$$\frac{dv}{dt}dx = -dV$$
$$\frac{dx}{dt}dv = -dV$$
$$vdv = -dV$$
$$vdv + dV = 0$$
$$d\left(\frac{v^2}{2} + V\right) = 0$$
Which implies that $\frac{v^2}{2} + V(x)$, which is the total energy (kinetic plus potential) is constant along the movement. This is the standard derivation of conservation of energy, and is valid whenever the force is a function of position only (not velocity or time).
A: In physics courses we tend to use the following trick: multiply $\ddot x = w-2x+x^2$ by $\dot x$ to obtain
$$ \dot x \ddot x = (w-2x+x^2)\dot x \ \ \Rightarrow \ \ \frac{d}{dt}\left( \frac{1}{2}\dot x^2 \right) = \frac{d}{dt} \left( wx-x^2+\frac{1}{3}x^3 \right)$$
Consequently,
$$ \frac{d}{dt}\left( \frac{1}{2}\dot x^2 - wx+x^2-\frac{1}{3}x^3 \right) = 0 $$
from which we find
$$ E = \frac{1}{2}\dot x^2 - wx+x^2-\frac{1}{3}x^3 $$
is a constant of the motion. Call it energy if you like.
