Stuck at Homogeneos Differential Equation example

Question

I'm stuck on a problem: $$(x^2 + y^2)\,dy + 2y^2\, dx = 0$$

Here is the part I've solved so far:

Can you help me with the rest?

Answer 1

With constants $A$, $B$ and $C$ let $$\frac{u^2+1}{u(u^2+2u+1)}=\frac{A}{u}+\frac{B}{u+1}+\frac{C}{(u+1)^2}$$ so $A=1$, $B=0$ and $C=-2$ $$\frac{u^2+1}{u(u^2+2u+1)}=\frac{1}{u}+\frac{-2}{(u+1)^2}$$ and $$-\int\frac{1}{u}+\frac{-2}{(u+1)^2}=\int\frac{1}{x}dx$$ $$-\ln u-2\frac{1}{u+1}=\ln x+C$$ after simplify, the general solution is $$\color{blue}{\boxed{\ln y+\frac{2x}{x+y}=C}}$$

Answer 2

Hint: the rest part of the integral is $$\\ \int { \frac { { u }^{ 2 }+2u+1-2u }{ u{ \left( u+1 \right) }^{ 2 } } du } =\int { \frac { { \left( u+1 \right) }^{ 2 }-2u }{ u{ \left( u+1 \right) }^{ 2 } } du= } \int { \frac { du }{ u } -2\int { \frac { du }{ { \left( u+1 \right) }^{ 2 } } } } =\\ =\int { \frac { du }{ u } -2\int { \frac { d\left( u+1 \right) }{ { \left( u+1 \right) }^{ 2 } } } } =\ln { u+\frac { 2 }{ u+1 } } $$