solve $x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$ Help with this excercise.. :)
$$x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$$
the book says it is an exact differential equation, but how?

$$x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$$
$$x(x^2+y^2)^{-1/2}dx+y(x^2+2y^2)dy=0$$
$M=x(x^2+y^2)^{-1/2}$
$N=y(x^2+2y^2)$
$$\frac{\partial M}{\partial y}=-\frac{xy}{(x^2+y^2)^{3/2}}$$
$$\frac{\partial N}{\partial x}=2xy$$
I cant find the integrating factor,, :(

Obviously $x(x^2+y^2)^{-1/2}+yy\prime(x^2+2y^2)=0$ isn't an exact differential equation.
So, there is a typo. Moreover the equation is certainly simple. The simplest guess is :
$$x(x^2+y^2)+yy\prime(x^2+2y^2)=0$$
$$x(x^2+y^2)dx+y(x^2+2y^2)dy=0$$
If it is the exact differential of $F(x,y)$ then :
$\begin{cases}
\frac{\partial F}{\partial x}=x(x^2+y^2) \quad\to\quad F=\frac{x^4}{4}+\frac{x^2y^2}{2}+f(y)\\
\frac{\partial F}{\partial y}=y(x^2+2y^2) \quad\to\quad F=\frac{x^2y^2}{2}+2\frac{y^4}{4}+g(x)
\end{cases}$
which implies $f(y)=\frac{y^4}{2}$ and $g(x)=\frac{x^4}{4}$
$F=\frac{x^4}{4}+\frac{x^2y^2}{2}+\frac{y^4}{2}=\text{constant}\quad$ because $dF=0$
$$x^4+2x^2y^2+2y^4=c$$
Solving it for $y$ leads to :
$$y(x)=\pm\sqrt{\frac{-x^2\pm \sqrt{2c-x^4}}{2}}$$
