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Newton's second law for a nonrelativistic particle of mass $m$ in 1D, reads, $$F \bigg(x, \frac{\mathrm{d}x}{\mathrm{d} t}, t \bigg)=m \frac{\mathrm{d}^2 x}{\mathrm{d} t^2}$$, where $F$ is the net force function. Now, if we assume that $F$ is time-independent and that $v=\frac{\mathrm{d} x}{\mathrm{d} t}$, can be written as a function of position $x$, $v=v(x)$, and using the chain rule we find the equation $$mv\frac{\mathrm{d} v}{\mathrm{d} x}=F(x,v).$$

After some manipulation, we arrive at the differential form $mv\mathrm{d}v-F(x,v)\mathrm{d}x=0$. In order for this differential form to be exact, it needs to satisfy the equation $$\frac{\partial}{\partial x} (mv)=-\frac{\partial F}{\partial v}=0$$, which shows that this differential can only be exact iff $F$ does not depend on velocity. In order to include forces that depend on velocity, one needs to multiply the differential from by an integrating factor $\lambda=\lambda(x,v)$, to obtain $\lambda mv\mathrm{d}v-\lambda F(x,v)\mathrm{d}x=0$. Let $A$ be the general conserved quantity for 1D motion. Then $A$ must satisfy $$\frac{\partial A}{\partial v}=\lambda mv, \frac{\partial A}{\partial x}=-\lambda F(x,v).$$

Now, here comes my question. How can one obtain an explicit formula for $A$ in terms of integrals of $F$ and $mv$. What will be in general the physical interpretation of $A$ and the integrating factor $\lambda$?

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For what it's worth, one may show that a 2nd-order ODE $$\ddot{x}~=~f(x,\dot{x})$$ without explicit time-dependence, or equivalently, a pair of 1st-order ODEs of the form $$ \dot{x}~=~v, \qquad \dot{v}~=~f(x,v),$$ always has a local Hamiltonian formulation, cf. e.g. this Phys.SE post. The Hamiltonian is a conserved quantity.

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It is all essentially about the energy conservation. When forces do not depend on velocity, they are conservative, and characterized by the potential energy, so the total energy is conserved. When the forces do depend on the velocity, these may be either forces like Lorentz force that do not change the total energy or forces like viscous friction, when the energy is not conserved and conserved quantity does not exist (there is no proof in the question of the existence of a non-trivial integrating factor).

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  • $\begingroup$ As far as I know, it is a theorem in differential equations that a non-trivial integrating factor always exists, it may not be easy to find, but it is always there. $\endgroup$
    – Don Al
    Jul 16, 2020 at 7:57
  • $\begingroup$ @DonAl You are probably right. But your question is how to find integrating factor solving an arbitrary equation in Newtonian mechanics, whereas you acknowledge yourself that it is difficult. $\endgroup$
    – Roger V.
    Jul 16, 2020 at 8:04

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