# Procedure for finding integrals of motion

In some examples I have read that if you want to find the integral of motion for some equation of motion, say on the form $\ddot{x}+ax=0$ for some constant $a$, you multiply the EOM by $$\dot{x}=q(x) \implies \ddot{x} = \frac{dq}{dt} = \frac{dq}{dx}\frac{dx}{dt}.$$ You then separate $q$ and $x$ and integrate both sides. If you then rearrange to get the integration constant (constant of motion I presume(?)) alone on either side you find some new equation on the form (in this case) $$\frac{\dot{x}^2}{2}+a\frac{x^2}{2} = C.$$

Assuming this is correct, is this a general procedure for finding integrals of motion for explicitly time independent EOMs? Is the multiplication of $\dot{x}$ just a mathematical trick or is there a physical interpretation of it?

• As much as I know, I presume its just a mathematical trick to solve differential equations of this type. – SchrodingersCat Nov 28 '15 at 14:42
• It's not really solving the ODE though, it's just finding a relation between the generalized coordinate and its derivative(s). – theorem Nov 28 '15 at 14:57
• You can solve the ODE from here on if you want by simple integration methods or even from the 2nd order differential equation. – SchrodingersCat Nov 28 '15 at 14:58
• Yes, of course. In any case, the solution to the ODE is not what I am interested in :) – theorem Nov 28 '15 at 15:00
• Then what else do you want to know? :-) – SchrodingersCat Nov 28 '15 at 15:00

Separation of variables works on the following three cases:

1. Acceleration is a function of time only $\ddot{x} = a(t)$ with solution the direct integration of time $$v(t) = \int a(t)\,{\rm d}t \\ x(t) = \int v(t)\,{\rm d}t$$
2. Acceleration is a function of speed only $\ddot{x} = a(\dot{x})$ with solution $$\left.\frac{{\rm d}t}{{\rm d}\dot{x}} = \frac{1}{a(\dot{x})} \right\} t(\dot{x}) = \int {\rm d}t = \int \frac{1}{a(\dot{x})} \,{\rm d}\dot{x} \\ \left. \dot{x} \frac{{\rm d}t}{{\rm d}\dot{x}} = \frac{{\rm d}x}{{\rm d}\dot{x}}= \frac{\dot{x}}{a(\dot{x})} \right\} x(\dot{x}) = \int {\rm d}x = \int \frac{\dot{x}}{a(\dot{x})} \,{\rm d}\dot{x}$$
3. Acceleration is a function of distance only $\ddot{x} = a(x)$ with solution $$\left. \frac{\ddot{x}}{\dot{x}} = \frac{{\rm d}\dot{x}}{{\rm d}t} \frac{{\rm d}t}{{\rm d}x} = \frac{{\rm d}\dot{x}}{{\rm d}x} = \frac{a(x)}{\dot{x}} \right\} \frac{1}{2}\dot{x}^2= \int \dot{x}\, {\rm d}\dot{x} = \int a(x) \,{\rm d}x$$ $$t = \int \frac{1}{\dot{x}(x)}\,{\rm d}x$$

These are all mathematical "tricks". Some have physical interpretation. For Case 3 you can see equation $$\frac{1}{2} m \dot{x}^2 = {\rm Work}$$