# Homogeneous differential equation in polar coordinates

Prove that a homogeneous first-order differential equation of the form $$y' = f(x, y)$$ is separable when written in polar coordinates.

I can solve this without polar coordinates straightforwardly, but the Apostol textbook requires to show this in polar coordinates. I tried:

$$y' = f(x, y)\quad \Rightarrow \quad (r\sin\theta)' = f(r\cos\theta, r\sin\theta)$$

$$\frac{dr}{d\theta}\sin\theta + r\cos\theta = f(1, \tan\theta)$$

since $$f(r\cos\theta, r\sin\theta) = f(1, r\sin\theta/r\cos\theta)$$. I expect that $$r$$ as a function of $$\theta$$ can be grouped somehow on one side with its derivative, but cannot figure how.

How do I proceed from here to show that this equation is separable in polar coordinates? Or it has to be done differently?

• Divide by $\sin(\theta)$ – EnlightenedFunky Mar 7 at 15:48
• Then you'll have $r' +P(\theta)r = f(1,tan(\theta))/\sin(\theta)$ Then use Integrating factor – EnlightenedFunky Mar 7 at 15:50

You wrote $$y'=(r\sin\theta)' = \frac{dr}{d\theta}\sin\theta + r\cos\theta$$ This is false because the mistake is that the meaning of prime is not the same in $$y'$$ and in $$(r\sin\theta)'$$ $$\frac{dy}{dx}\neq\frac{d}{d\theta}(r\sin\theta)$$ One have to compute $$\frac{d}{dx}(r\sin\theta)$$ . $$\begin{cases} dx=\cos\theta dr-r\sin\theta d\theta\\ dy=\sin\theta dr+r\cos\theta d\theta \end{cases}$$ $$\frac{dy}{dx}=\frac{\sin\theta dr+r\cos\theta d\theta}{\cos\theta dr-r\sin\theta d\theta}$$ $$\frac{dy}{dx}=\frac{\frac{1}{r} \frac{dr}{d\theta}\tan\theta+1}{\frac{1}{r} \frac{dr}{d\theta}-\tan\theta}$$ I suppose that you can continue up to the separable equation.