Differential equation with nasty coefficients $ x^2(1-x)^2 y'' + (Ax + b)y = 0 $ I have encountered a differential equation on the form
$$ x^2(1-x)^2 y'' + (Ax + b)y = 0 $$
My math background is too limited to even know where to begin, so any help of solving the equation (if a solution exist?) would be greatly appreciated!

In fact it just belongs to an ODE of the form http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=262.
Let $y=x^p(1-x)^qu$ ,
Then $y'=x^p(1-x)^qu'+x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u$
$y''=x^p(1-x)^qu''+x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u=x^p(1-x)^qu''+2x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u$
$\therefore x^2(1-x)^2\left(x^p(1-x)^qu''+2x^p(1-x)^q\left(\dfrac{p}{x}-\dfrac{q}{1-x}\right)u'+x^p(1-x)^q\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u\right)+(Ax+b)x^p(1-x)^qu=0$
$u''+\left(\dfrac{2p}{x}-\dfrac{2q}{1-x}\right)u'+\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u+\dfrac{Ax+b}{x^2(1-x)^2}u=0$
$u''+\left(\dfrac{2p}{x}-\dfrac{2q}{1-x}\right)u'+\left(\dfrac{p(p-1)}{x^2}-\dfrac{2pq}{x(1-x)}+\dfrac{q(q-1)}{(1-x)^2}\right)u+\left(\dfrac{A+2b}{x}+\dfrac{b}{x^2}+\dfrac{A+2b}{1-x}+\dfrac{A+b}{(1-x)^2}\right)u=0$
$u''+\left(\dfrac{2p}{x}-\dfrac{2q}{1-x}\right)u'+\left(\dfrac{p(p-1)+b}{x^2}+\dfrac{A+2b-2pq}{x(1-x)}+\dfrac{q(q-1)+A+b}{(1-x)^2}\right)u=0$
Choose $p(p-1)+b=0$ and $q(q-1)+A+b=0$ , the ODE reduces to the Gaussian hypergeometric equation.
Note that the special case of $A=0$ the ODE reduces to the form of http://eqworld.ipmnet.ru/en/solutions/ode/ode0224.pdf.
