How to solve the following differential equation: $(2x^3-x^2-y)dx-(2x^2y-x)dy=0$ I'd like your help with this differential equation:
$(2x^3-x^2-y)dx-(2x^2y-x)dy=0$
It seems not difficult, but I have no idea about it
Thanks a lot for help)

Seek a total derivative $df=0$ by multiplying by an Ansatz $x^ay^b$ viz.$$f_x=2x^{a+3}y^b-x^{a+2}y^b-x^ay^{b+1},\,f_y=x^{a+1}y^b-2x^{a+2}y^{b+1}$$so$$0=\frac{f_{xy}-f_{yx}}{x^{a-1}y^{b-1}}=2bx^4-bx^3-(a+b+2)xy+2(a+2)x^2y^2.$$Equating coefficients gives $a=-2,\,b=0$, so$$df=(2x-1-y/x^2)dx+(1/x-2y)dy\implies f=x^2-x+y/x-y^2-C.$$The constraint therefore reduces to $x^2-x+y/x-y^2=C$.
