Nonlinear Ordinary Differential Equation - $x$'s and $y$'s all over the place, integer coefficients I have come across the following differential equation:
$$\dfrac{dy}{dx}=\dfrac{y^{2}+125x^{2}+42x^{3}+12xy-4x^{2}y}{xy}$$
It quite possibly has no closed-form solution but I would appreciate any insights if possible. I hope I put the equation in properly.
 A: The first things you should attempt are to see if it is seperable or exact.  It's not seperable (you can't get all the x's with the dx and all the y's with the dy).   So, to see if it's exactly, we put it into standard form ($Mdx+NdY=C$)  and see if $M_y=N_x$  So, in standard form, we have $(y^2+125x^2+42x^3+12xy-4x^2y)dx+(-xy)dy=0$   So,  $M_y=2y+12x-4x^2$,  $N_x=-y$,  we are not exact.   Then one would try to come up with an integrating factor, some function $\mu(x,y)$ such that if we multiplied through in standard form we would be exact...unfortunately, not only is this not always possible, there's often no solution. 
Beyond that, there are techniques to examine linearization and equilibrium, but I'm just taking that course now myself :).
A: $\dfrac{dy}{dx}=\dfrac{y^2+125x^2+42x^3+12xy-4x^2y}{xy}$
$y\dfrac{dy}{dx}=\dfrac{y^2}{x}+(12-4x)y+42x^2+125x$
This belongs to an Abel equation of the second kind.
Let $y=xu$ ,
Then $\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore xu\left(x\dfrac{du}{dx}+u\right)=\dfrac{x^2u^2}{x}+(12-4x)xu+42x^2+125x$
$x^2u\dfrac{du}{dx}+xu^2=xu^2+(12-4x)xu+42x^2+125x$
$x^2u\dfrac{du}{dx}=(12-4x)xu+42x^2+125x$
$u\dfrac{du}{dx}=\left(\dfrac{12}{x}-4\right)u+42+\dfrac{125}{x}$
