Hint on solving Riccati differential equation Hint on solving the following ODE
$$y' + a x^{2} y + x y^{2} = a$$
where $a$ is a real number.
 A: Riccati equations arise as  natural equation solutions.
Please refer to page 26 LP Eisenhart and page 37 DJ Struik and look them through. 
Cross Ratios of four particular integrals is constant.
A: Assume $a\neq0$ for the key case:
Hint:
Approach $1$:
Let $y=\dfrac{u'}{xu}$ ,
Then $y'=\dfrac{u''}{xu}-\dfrac{u'}{x^2u}-\dfrac{(u')^2}{xu^2}$
$\therefore\dfrac{u''}{xu}-\dfrac{u'}{x^2u}-\dfrac{(u')^2}{xu^2}+\dfrac{axu'}{u}+\dfrac{(u')^2}{xu^2}=a$
$\dfrac{u''}{xu}-\dfrac{u'}{x^2u}+\dfrac{axu'}{u}-a=0$
$xu''+(ax^3-1)u'-ax^2u=0$
We can make comparison with another approach, as all Riccati equations with dependent variable $y$ have an interesting property that the substitution $y=\dfrac{1}{u}$ will brings the Riccati equations to the Riccati equations again, but the coefficients are rearranged.
Approach $2$:
Let $y=\dfrac{1}{u}$ ,
Then $y'=-\dfrac{u'}{u^2}$
$\therefore-\dfrac{u'}{u^2}+\dfrac{ax^2}{u}+\dfrac{x}{u^2}=a$
$u'+au^2-ax^2u-x=0$
Let $u=\dfrac{v'}{av}$ ,
Then $u'=\dfrac{v''}{av}-\dfrac{(v')^2}{av^2}$
$\therefore\dfrac{v''}{av}-\dfrac{(v')^2}{av^2}+\dfrac{(v')^2}{av^2}-\dfrac{x^2v'}{v}-x=0$
$\dfrac{v''}{av}-\dfrac{x^2v'}{v}-x=0$
$v''-ax^2v'-axv=0$
