I want to solve a problem already asked on MSE, but I want to use Newton's notation and be careful on every assumption I make at every step, so this is not a duplicate.
I restate the problem :
Given an homogeneous differential equation $y' = f(x,y)$, rewrite it in polar coordinates and show it becomes separable.
Attempt :
first I can pass to polar coordinates with
$$x= r \cos \theta$$ $$y= r \sin \theta$$
Substituting in the r.h.s. we get
$$f(x,y) = f(r \cos \theta, r \sin \theta) $$
so when $r \neq 0$, since the d.e. is homogeneous ($f(x,y) = f(tx,ty)$ for every $t \neq 0$) I obtain
$$f(r \cos \theta, r \sin \theta) = f(\cos \theta, \sin \theta) = g(\theta) $$
The assumption that $r \neq 0 $ is fulfilled whenever the function $y(x)$ (which is supposed to exist) is defined on any open set, because in this case there exists an interval where $r$ is never $0$ which is what we want so I continue.
For the l.h.s. instead I suppose that a function $r(\theta)$ exists, so that I can differentiate with respect to $\theta$, then
$$x'(\theta) = r'(\theta) \cos \theta - r(\theta) \sin \theta$$ $$y'(\theta)= r'(\theta) \sin \theta + r(\theta) \cos \theta $$
Because the function $y(x)$ is supposed to exist then it exists also $y(x(\theta)) = y(r(\theta)\cos(\theta))$ then
$$[y(x(\theta))]' = y'(x(\theta)) [x(\theta)]' $$
I tried to follow the same path of the answer to the other question but (without the hidden help of Leibniz's notation) I don't know how to continue. Maybe there is more than meets the eye in the passages involving Leibniz's notation in the linked question. I don't know if the question is somewhat related to differential geometry but I didn't study this subject yet.
Can someone provide a proof in Newton's notation and which deals with every 'edge case' ?
Thanks a lot in advance
I think I just lost dependence of $y$ from $x$ trasforming in polar coordinates.
Since we suppose $y(x)$ exists and also $r(\theta)$ exists then I have $$y(x(\theta)) = r \sin \theta $$ from which follows that $$[y(x(\theta))]' = r' \sin\theta - r \cos(\theta)$$
fron here the proof is the same as in the other post