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

$$\left(2x+ 1-\frac{y^2}{x^2}\right)dx+ \frac{2y}{x}dy= 0$$
$$\left(2x+1\right)dx+\dfrac {(-y^2dx +{2xy}dy)}{x^2}= 0$$
$$\left(2x+1\right)dx+\dfrac {(-y^2dx +{x}dy^2)}{x^2}= 0$$
$$\left(2x+1\right)dx+d \left (\dfrac {y^2}{x}\right)= 0$$
Integrate.
