Stable equilibrium given force If a particle moves under the influence of a resistive force proportianal to velocity and a potential $U$,      
$$F(x,\dot x)=-b\dot x-\frac {\partial U}{\partial x}$$
Where b>0 and $U(x)=(x^2-a^2)^2$
My thoughts were to make $F=0$, which would result in:
$$\dot x= \frac 1b(4x^3-4a^2x)$$
This means that the points of equilibrium are a function of time?
EDIT::
So I set up as suggested, using $F=m\ddot x$ to get
$$0=m\ddot x+b\dot x+4x(x^2-a^2)$$
I am trying to solve for $x(t)$.  I know how to solve the equation that has only x, but I do not know how to solve for the x^3 term.  I am curious how to combine these to get x(t)
 A: The stable point of equilibrium is at $x=0,x=\pm a$. This becomes obvious when you realize that for these kinds of problems with linear friction, you can actually ignore the friction term when computing the equilibrium state of the system. 
Why? If the system is at equilibrium, then it is both at rest ($\dot{x}=0$) and has no net force acting on it ($F=0$). Combining these two statements yields
$$0=F=-\frac{\partial U}{\partial x}=4 x \left(x^2-a^2\right)$$
which implies
$$x\in\{-a,0,a\}.$$
A: You can determine equilibrium points by one of two approaches:


*

*Use Newton second law, which means replace $F$ by $m\ddot x$. The equation becomes a 2nd order ODE. Solve it for $x$.

*Using your formulation by setting $F = 0$ you obtained a 1st order ODE. Solve it for $x$ as function of time. At the time when the velocity becomes zero, the force is already zero and the potential gradient is zero as well. So the particle doesn't experience any force and has zero momentum. The value of $x$ at that time is the equilibrium position.
