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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:

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

which leads to

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

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?

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

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  • $\begingroup$ Thank you! I solved it easily after I corrected this mistake! :) $\endgroup$ – John Mar 7 at 17:19

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