Polar Coordinates tranformation for Linear Homogeneous Differential Equations (1st order)

A differential equation of the form

$M(x, y)dx + N(x, y)dy = 0 \tag{1}$

is homogeneous of degree $n$ provided that there is an $n \in \Bbb Z$ such that

$M(tx, ty) = t^nM(x, y) \tag{2}$

and

$N(tx, ty) = t^nN(x, y); \tag{3}$

if (1) is such an equation, setting

$x = r\cos \theta \tag{4}$

and

$y = r\sin \theta, \tag{5}$

we find

$M(r\cos \theta, r\sin \theta)dx + N(r\cos \theta, r\sin \theta)dy = 0 \tag{6}$

becomes

$r^nM(\cos \theta, \sin \theta)dx + r^nN(\cos \theta, \sin \theta) dy = 0. \tag{7}$

Now (4) and (5) yield

$dx = dr\cos \theta - rd\theta \sin \theta \tag{8}$

and

$dy = dr \sin \theta + rd\theta \cos \theta; \tag{9}$

inserting these two equations into (7) we obtain

$r^nM(\cos \theta, \sin \theta)(dr\cos \theta - rd\theta \sin \theta)$ $+ r^nN(\cos \theta, \sin \theta) (dr \sin \theta + rd\theta \cos \theta) = 0. \tag{10}$

We gather like terms (in $dr$ and $d\theta$):

$r^n(M(\cos \theta, \sin \theta)\cos \theta + N(\cos \theta, \sin \theta)\sin \theta)dr$ $- r^{n + 1}(M(\cos \theta, \sin \theta)\sin \theta - N(\cos\theta, \sin \theta)\cos \theta) d\theta = 0. \tag{11}$

Some minor algebraic fiddling yields (note we can cancel $r^n$ as long as $r \ne 0$, where polars are in any event undefined):

$\dfrac{dr}{r} = \dfrac{M(\cos \theta, \sin \theta)\sin \theta - N(\cos\theta, \sin \theta)\cos \theta}{M(\cos \theta, \sin \theta)\cos \theta + N(\cos \theta, \sin \theta)\sin \theta} d\theta, \tag{12}$

and voila!!! seperated variables.

Now try integrating it.