Solve $x^2(xdx+ydy)+2y(xdy-ydx)=0$ Solve $x^2(xdx+ydy)+2y(xdy-ydx)=0$
My Attempts 
$$x^2(xdx+ydy)+2y(xdy-ydx)=0$$
$$\dfrac {xdx+ydy}{xdy-ydx}=-\dfrac {2y}{x}$$
Put $x=r\cos (\theta)$ and $y=r\sin (\theta)$
So, $r^2=x^2+y^2$ and $\tan (\theta)=\dfrac {y}{x}$
Now,
$$x^2+y^2=r^2$$
Differentiating both sides,
$$2xdx+2ydy=2rdr$$
$$xdx+ydy=rdr$$
 A: $$\dfrac {xdx+ydy}{xdy-ydx}=-\dfrac {2y}{x}$$
Duvide both sides by $({x^2+y^2})$:
$$\dfrac {xdx+ydy}{x^2+y^2}=-\dfrac {2y}{x}\left ( \dfrac {xdy-ydx}{x^2+y^2} \right )$$
$$\dfrac {xdx+ydy}{x^2+y^2}=-\dfrac {2y}{x}\left ( d(\arctan (\frac {y}{x})) \right )$$
Susbtitute $\dfrac {y}{x}=z$
$$\frac 12\dfrac {d(x^2+y^2)}{x^2+y^2}=-2zd(\arctan z)$$
$$\frac 12\dfrac {d(x^2+y^2)}{x^2+y^2}=-\dfrac {2zdz}{z^2+1}$$
Integrate.

Edit I didn't pay attention that the right side of the DE was wrong so:
$$\frac 12\dfrac {d(x^2+y^2)}{x^2+y^2}=-2\frac {y}{\color{red}{x^2}}d(\arctan \frac {y}{x})$$
$$\frac 12\dfrac {dr^2}{r^2}=-2\frac {r\sin \theta}{r^2 \cos^2 \theta}d\theta$$
More simply:
$${dr}=-2\frac {\sin \theta}{\cos^2 \theta}d\theta$$
Integrate.
A: Starting from
$$x^2(xdx+ydy)+2y(xdy-ydx)=0$$
an alternative method is to multiply all terms by $\dfrac{2}{x^2}$ so that
$$2(xdx+ydy)+{4y}\left(\frac{xdy-ydx}{x^2}\right)=0$$
which is equivalent to
$$d(x^2+y^2)+4yd\left(\frac{y}{x}\right)=0$$
where we can then convert to polar coordinates. Letting $r^2=x^2+y^2,\tan(\theta)=\dfrac{y}{x},y=r\sin(\theta)$
$$d(r^2)+4r\sin(\theta)d(\tan(\theta))=0$$
hence
$$2r\,dr+4r\sin(\theta)\sec^2(\theta)\,d\theta=0$$
or
$$dr=-2\sec(\theta)\tan(\theta)\,d\theta$$
which is a separable equation and can be solved through integration.
A: Hint:
Differentiate both sides of:
$$\tan \theta = \dfrac{y}{x}$$
This gives:
$$\sec^2 \theta d\theta = \dfrac{xdy-ydx}{x^2}$$
Since $x^2 = r^2\cos^2\theta$, we have:
$$r^2d\theta = xdy-ydx$$
