Awkward terms in Differential equation Find the solution of the equation:
$(x^2+y^2+x)d x-(2x^2+2y^2-y)dy=0$
As a standard approach, I brought $dy/dx$ to the RHS of the equation. Then, I added and subtracted $x^2+y^2-y$ to the numerator of the LHS to find some symmetric relation between the numerator and denominator. Perfecting squares of the numerator or the denominator does not help either. Please advice. 

The equation can be written as
$$xdx +ydy + (x^2+y^2)(dx-2dy) = 0$$
Away from $(x,y) = (0,0)$ we can divide by $x^2+y^2$ to get
$$\frac{x}{x^2+y^2} dx + \frac{y}{x^2+y^2} dy + (dx-2dy) = 0,$$
which is the same as 
$$d(\frac{1}{2}\ln({x^2+y^2)} + x-2y) = 0. $$
That leads to $\frac{1}{2}\ln{(x^2+y^2)} +x-2y = k$, as in @almagest's comment.
