Show that differential equation has integrating factor $\eta (x, y) = \nu (x^2 + y^2)$ and find it

Question

I have differential equation $$x + x^4 + 2x^2 y^2 + y^4 + yy' = 0$$

The problem is to find integrating factor $\eta (x, y) = \nu (x^2 + y^2)$ for such big equation :(

Does anybody have any ideas?

Answer 1

$$x + \color {red}{x^4 + 2x^2 y^2 + y^4} + yy' = 0$$ $$ \color{red}{(x^2+ y^2 )^2 }+x+ yy' = 0$$ $$ (x^2+ y^2 )^2 +\dfrac 12(x^2)'+ \dfrac 12 (y^2)' = 0$$ $$ (x^2+ y^2 )^2 +\dfrac 12(x^2 +y^2)' = 0$$ $$1=- \dfrac 12 \dfrac {(x^2 +y^2)'}{(x^2+ y^2 )^2 }$$ $$ \left(\dfrac {1}{x^2+ y^2 }\right)'=2$$ Integrate.

Answer 2

$$\frac{d}{dx}(x^2 + y^2) = 2x + 2yy'$$

Let $v = x^2 + y^2$

The above equation becomes

$$v^2 + \frac{1}{2}\frac{dv}{dx} = 0$$