How to solve the differential equation $y' = \frac{x+y}{x-y}$? Solve the following differential equation:

$y' = \frac{x+y}{x-y}$

Someone please help me start this problem. This does not look like a regular first-order differential equation in the form  $y' + 2xy = x$. Thank you.
 A: Hint: It's homogenous of degree 1 which suggests the substitution 
$$
y=xv\implies y'=xv'+v
$$
and 
$$
xv'+v=\frac{x+xv}{x-xv}=\frac{1+v}{1-v}\stackrel{\text{algebra}}{\implies}
\frac{1-v}{v^2+1}\mathrm dv=\frac{1}{x}\mathrm dx\implies \int \frac{1-v}{v^2+1}\,\mathrm dv=\ln x+c\\
\implies \int\frac{1}{v^2+1}\,\mathrm dv-\int \frac{v}{v^2+1}\,\mathrm dv=\ln x+c\\
\implies\arctan(v)-\frac{1}{2}\ln (v^2+1)=\ln x+c\\
\implies\arctan\left(\frac{y}{x}\right)-\frac{1}{2}\ln \left(\frac{y^2}{x^2}+1\right)=\ln x+c
$$
and I am afraid we will have to be satisfied with the implicit solution.
edit: Following @Artem's suggestion we can rewrite this curve in polar coordinates and it simplifies nicely, although the caveat is it makes $y$'s role difficult to 
see
$$
\arctan\left(\frac{y}{x}\right)-\frac{1}{2}\ln \left(\frac{y^2}{x^2}+1\right)=\ln x+c\\
\implies \theta-\ln \left(\sqrt{\frac{r^2}{r^2\cos^2 \theta}}\right)-\ln r\cos \theta=c\\
\implies \theta+\ln (\cos \theta)-\ln (r\cos \theta)=c\\
\implies \theta+\ln\left(\frac{1}{r}\right)=c\\
\implies \theta-c=-\ln\left(\frac{1}{r}\right)=\ln r\\
\implies Ae^{\theta}=r
$$
