Solve Differential Equation: $y' = \frac{\sqrt{x^2+y^2}-x}{y}$ Solve Differential Equation: $y' = \frac{\sqrt{x^2+y^2}-x}{y}$
This looks like a problem that would do well with a conversion to polar coordinates ($r^2 = x^2 + y^2$ and $x = r\cos(\theta)$ $y = r\sin(\theta)$). However, I am confused on how to change $y' = \frac{dy}{dx}$ to $\frac{dr}{d\theta}$, in order to do separation of variables. Could someone demonstrate a solution using this approach?
 A: $x = r \cos\theta, y = r \sin\theta$
$dx = \cos\theta \ dr - r \sin\theta \ d\theta$
$dy = \sin\theta \ dr + r \cos\theta \ d\theta$
Given equation is $dy = \frac{\sqrt{x^2+y^2}-x}{y} dx$
So we get,
$r \sin\theta \ (\sin\theta \ dr + r \cos\theta \ d\theta) = r \ (1 - \cos\theta) \ (\cos\theta \ dr - r \sin\theta \ d\theta)$
On simplifying we get,
$\displaystyle \frac{1}{r} \ dr = - \frac{\sin\theta}{1-\cos\theta} \ d\theta$
To integrate, simply observe that $1 - \cos\theta = t \implies \sin\theta \ d\theta = dt$
So both sides are simple integration.
A: hint
Instead of looking for a cartesian solution of the form  $ y=y(x) $, you want polar solution as $ r=r(\theta) $ with
$$x=r(\theta)\cos(\theta)\text{ and } y=r(\theta)\sin(\theta)$$
So, differentiation gives
$$dx=\Bigl(r'\cos(\theta)-r(\theta)\sin(\theta)\Bigr)d\theta$$
$$dy=\Bigl(r'\sin(\theta)+r(\theta)\cos(\theta)\Bigr)d\theta$$
thus, the equation becomes
$$\frac{dy}{dx}=\frac{r'\sin(\theta)+r\cos(\theta)}{r'\cos(\theta)-r\sin(\theta)}$$
$$=\frac{r-r\cos(\theta)}{r\sin(\theta)}$$
$$=\tan(\frac{\theta}{2})$$
which simplifies to
$$\frac{r'}{r}=-\frac{\cos(\frac{\theta}{2})}{\sin(\frac{\theta}{2})}$$
Yielding to the solution
$$\ln(\frac{r}{\lambda})=-2\ln(|\sin(\frac{\theta}{2})|)$$
and
$$\boxed{r=\frac{2\lambda}{1-\cos(\theta)}}$$
A: $$y'=f(x,y)$$
$$\dfrac {dy}{dx}=f(x,y)$$
Apply the chain rule:
$$\dfrac {dy}{d\theta}\dfrac {d\theta}{dx}=f(x,y)$$
$$\dfrac {dy}{d\theta}=g(r,\theta)\dfrac {dx}{d\theta}$$
Nowit's easy to compute the derivatives.
$$\dfrac {dy}{d\theta}=\dfrac d {d\theta}(r(\theta)\sin (\theta))$$
$$\dfrac {dx}{d\theta}=\dfrac d {d\theta}(r(\theta)\cos (\theta))$$
