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!
 A: 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.
A: The solution to this ODE  is obtained by applying a Moebius transformation to hypergeometric functions and then by multiplying the result by another function. See case (D) in my answer to Looking for closed form solutions to linear ordinary differential equations with time dependent coefficients. .
