General solution for the motion of a 1D particle with drag

Question

Consider a particle of mass $m$, with equation of motion $$m\ddot{x}=-U'(x)-f(\dot{x})\dot{x}.$$ I am trying to find an equation of the form $v=g(x)$ so that we can reduce this to an integral of the form $\int\frac{dx}{g(x)}=t$ and thus have the solution. The work-energy theorem gives $$\frac{1}{2}m\dot{x}(t)^2+U(x(t))=\frac{1}{2}mu^2+U(x_0)-\int_0^t f(\dot{x}(t'))\dot{x}(t')^2dt',$$ where $u$ and $x_0$ are the initial velocity and position, respectively. However, this result is not very useful because to do the integral, we need to know $\dot{x}(t)$, i.e., already have solved the problem we are trying to solve! So, we cannot use the same approach we use with conservative systems. So, what can we do instead? How can we find a constant of motion that would allow us to reduce this problem to doing an integral?

Answer 1

For what it is worth, any autonomous 1D 2nd-order ODE $\ddot{x}=a(x,\dot{x})$ can be written as an autonomous 2D 1st-order ODE $$\begin{align}\dot{x}~=~&v,\cr \dot{v}~=~&a(x,v),\end{align}$$ which in principle has locally a conservative Hamiltonian formulation. The Hamiltonian $H(x,v)$ is an integral of motion, cf. e.g. this Phys.SE post.