$r'=f(p)/r^4, p'=g(p)/r^4$ What is the method to analytically solve this system of non-linear differential equations? I'm working on a physical system in polar coordinates and I'm a bit lost on how to start solving a problem like this. The system has two dependent variables: the distance between two particles (r) and the angle between two particles ($\phi$). These evolve with time and I'm looking for an expression that explicitly shows r and $\phi$ only as functions of time and the initial conditions.
The system is described by two non-linear differential equations:
$r’ = {dr \over dt} = f(\phi)/r^4$
$\phi’ = {d\phi \over dt} = g(\phi)/r^4$
Where $f(\phi)$ and $g(\phi)$ are integratable/differentiable trigonometric functions dependent only on $\phi$.
What is the best way to go about solving this system of equations?
 A: Off the top of my head, here's what I would try:
from the equations
$r' = \dfrac{dr}{dt} = \dfrac {f(\phi)}{r^4} \tag{1}$
and
$\phi' = \dfrac{d\phi}{dt} = \dfrac{g(\phi)}{r^4}, \tag{2}$
we obtain upon division of (1) by (2)
$\dfrac{dr}{d\phi} = \dfrac{\dfrac{f(\phi)}{r^4}}{\dfrac{g(\phi)}{r^4}} = \dfrac{f(\phi)}{g(\phi)}, \tag{3}$
which in fact may be integrated to give $r$ as a function of $\phi$:
$r - r_0 = \displaystyle  \int_{r_0}^r ds = \displaystyle \int_{\phi_0}^{\phi}\dfrac{f(w)}{g(w)}dw = \Phi(\phi), \tag{4}$
whence
$r(\phi) = \Phi(\phi) + r_0. \tag{5}$
We may then substitute (5) into (2):
$\dfrac{d\phi}{dt} = \dfrac{g(\phi)}{r^4(\phi)} = \dfrac{g(\phi)}{(\Phi(\phi) + r_0)^4}, \tag{6}$
which, though nasty in appearance, can in principle be integrated to give $t$ as a function of $\phi$:
$t - t_0 = \displaystyle \int_{\phi_0}^\phi \dfrac{(\Phi(w) + r_0)^4}{g(w)}dw =\Psi(\phi); \tag{7}$
in principle, as long as $dt/d\phi \ne 0$, we may invert (7) to obtain $\phi$ as a function of $t$, and insert such $\phi(t)$ in (1):
$\dfrac{dr}{dt} = \dfrac{f(\phi(t))}{r^4}, \tag{8}$
which we can then solve for $r$ in terms or $t$ as well; indeed, we have:
$r^4 \dfrac{dr}{dt} = f(\phi(t)), \tag{9}$
whence
$\dfrac{1}{5}(r^5 - r_0^5) = \displaystyle \int_{t_0}^t f(\phi(s))ds, \tag{10}$
$r^5(t) = 5 \displaystyle \int_{t_0}^t f(\phi(s))ds + r_0^5,  \tag{11}$
leading to the somewhat arcane expression
$r(t) = \sqrt[5]{5\displaystyle \int_{t_0}^t f(\phi(s))ds + r_0^5} = (5\displaystyle \int_{t_0}^t f(\phi(s))ds + r_0^5)^{1/5}. \tag{12}$
In principle this works, but is probably going to be impossible in closed form for most $f(\phi)$, $g(\phi)$.  My advice:  use a computer.
