Making an ODE exact, when formula's of exactness do not provide a solution I have
$$\left(2x+ 1-\frac{y^2}{x^2}\right)dx+ \frac{2y}{x}dy= 0$$
which is not exact.  However, it could be made exact by using the formula for integrating factors. I have two ways,
With $$M=\left(2x+ 1-\frac{y^2}{x^2}\right)$$ and $$N=\frac{2y}{x}$$ we can form the following integrating factors:
$$\phi(x)=\frac{N_x-M_y}{M}=-\frac{\frac{4y}{x^2}}{2x+1-\frac{y^2}{x^2}}$$
or
$$\psi(x)=\frac{M_y-N_x}{N}=\frac{ (x^3 - x^2 + y^2)}{yx^2}$$
Then multiplying these in, should give an exact form of the ODE. But neither of the two make the ODE exact. Are there other formulas one can use?
Thanks

Ok, this differential form is indeed exact. The integrating factor is 1, and the first integral is
$$
I=\dfrac{x^3+x^2+y^2}{x}
$$
