Find solution to the differential equation : $(2x+1)dx+\frac{x^2-y}{x}dy=0$

I was wondering how to solve this DE.
I have tried finding integral factor $\mu$ but failed to.
My try :
$$M=(2x+1) \quad,\quad N(x-\frac{y}{x})$$ $$M_y=0 \quad \ne \quad N_x=1+\frac{y}{x^2}$$ So I have checked if $\frac {M_y-N_x}{N} =f(x)$ or $\frac {N_x-My}{M}=g(y)$ but neither works.

I don't know how to proceed and find integral factor or perhaps solve it in another way.
Symbolab/Wolfram/Mathdf doesn't show solution for this DE.
Thank you in advance.

Solution 1:

Your equation is $$y'=\frac{2x^2+x}{y-x^2}.$$ A solution to this equation is an first integral for the system: $$x'=y-x^2$$ $$y'=2x^2+x$$ This system has equilibrium points at $p_0=(0,0)$ and $p_1=(\frac{-1}{2},\frac{1}{4})$. The jacobian of this vector field in $p_1$ is $$ \begin{bmatrix} 1 & 1\\ -1 & 0\\ \end{bmatrix} $$ wich has eigenvalues $\lambda_1=\frac{1+\sqrt{3}}{2}$ and $\lambda_2=\frac{1-\sqrt{3}}{2}$. The equation $$m\lambda_1+n\lambda_2=0$$ has no integer solutions. So, by theorem 5.2 from Goriely's "Integrability and Nonintegrability of Dynamical Systems", there are no analytic solution around $p_1$. That is: your equation has no analytic solutions. Interesting enough is that the 'similar' equation $$y'=\frac{2x^2+4x}{y-x^2}.$$ Has analytic solutions.