Extended Riccati Equation Could you help me find the general solution to the ODE bellow?
$y'(x)=y(x)/x+(7x+0.5/x). y(x)^2+(6x+1) y(x)^3$
This is somewhat an extention to the Riccati equation. I have found two particular solutions $y(x)=0$ and $y(x)=-0.5/x$, but is it possible to derive the general solution? Thanks.
 A: Hint:
Let $y=\dfrac{1}{u}$ ,
Then $y'=-\dfrac{u'}{u^2}$
$\therefore-\dfrac{u'}{u^2}=\dfrac{1}{xu}+\left(7x+\dfrac{1}{2x}\right)\dfrac{1}{u^2}+\dfrac{6x+1}{u^3}$
$uu'=-\dfrac{u^2}{x}-\left(7x+\dfrac{1}{2x}\right)u-6x-1$
This belongs to an Abel equation of the second kind.
Let $u=\dfrac{v}{x}$ ,
Then $u'=\dfrac{v'}{x}-\dfrac{v}{x^2}$
$\therefore\dfrac{v}{x}\left(\dfrac{v'}{x}-\dfrac{v}{x^2}\right)=-\dfrac{v^2}{x^3}-\left(7x+\dfrac{1}{2x}\right)\dfrac{v}{x}-6x-1$
$\dfrac{vv'}{x^2}-\dfrac{v^2}{x^3}=-\dfrac{v^2}{x^3}-\left(7+\dfrac{1}{2x^2}\right)v-6x-1$
$\dfrac{vv'}{x^2}=-\left(7+\dfrac{1}{2x^2}\right)v-6x-1$
$v\dfrac{dv}{dx}=-\left(7x^2+\dfrac{1}{2}\right)v-6x^3-x^2$
Let $v=-\dfrac{w}{2}$ ,
Then $\dfrac{dv}{dx}=-\dfrac{1}{2}\dfrac{dw}{dx}$
$\therefore-\dfrac{w}{2}\left(-\dfrac{1}{2}\dfrac{dw}{dx}\right)=-\left(7x^2+\dfrac{1}{2}\right)\left(-\dfrac{w}{2}\right)-6x^3-x^2$
$\dfrac{w}{4}\dfrac{dw}{dx}=\left(7x^2+\dfrac{1}{2}\right)\dfrac{w}{2}-6x^3-x^2$
$w\dfrac{dw}{dx}=(14x^2+1)w-24x^3-4x^2$
