<s>[INST] 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. 
 [/INST] 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.
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