About $(x^2+y^2)dy-y^2dx=0$ differential equation Let us show $(x^2+y^2)dy-y^2dx=0$ differantial equation is homogeneous then solve the equation.
Firstly I did show we can also write $\frac{y^2}{x^2+y^2}=\frac{dy}{dx}$ equation depending on the first equation. Let $v=y/x$. Then we get $\frac{v^2}{v^2+1}=\frac{dy}{dx}x+v$ equation. If we go on and integrate that equation, we will get $\int\frac{(v^2+1)dv}{v^3-v^2+v}+\int\frac{dx}{x}=C$ equation. I failed to integrate the left term, what to do about it?
By the way, I am a bit bad at English so if you don't understand something, tell me to explain it.

First of all you have $y=vx$ and $\frac{dy}{dx}=x\frac{dv}{dx}+v$, so we need to solve
$$\frac{v^2}{1+v^2}=x\frac{dv}{dx}+v$$
$$\frac{v^2-v(1+v^2)}{1+v^2}=x\frac{dv}{dx}$$
or
$$\frac{1+v^2}{-v^3+v^2-v}dv=\frac{1}{x}dx.$$
The right is easy to integrate; the left you should be able to tackle with partial fraction:
$$\frac{1+v^2}{v(-v^2+v-1)}=-\frac{1}{v^2-v+1}-\frac{1}{v}.$$
The only difficulty is the first term. Note that
$$\frac{1}{v^2-v+1}=\frac{1}{(v-1/2)^2+3/4}.$$
Use the fact that $\frac{d}{dx}\arctan\frac{x}{r}=\frac{r}{r^2+x^2}$. Then what is the derivative of $\arctan\frac{x-1/2}{\sqrt 3/2}$?
