Question: I'm trying to understand how to solve this differential equation: 

$ [z^2(1-z)\dfrac{d^2}{dz} - z^2 \dfrac{d}{dz} - \lambda] f(z) = 0 $ 

I know the solution is related to the hypergeometric function $\_2F^1$, but as I recall from many sources: this functions satisfies another differential equation: 

$ [z(1-z)\dfrac{d^2}{dz} + (c - (ab + 1)z) \dfrac{d}{dz} - ab] f(z) = 0 $ 

with $a,b,c \in \mathcal{R} $

I've tried to transform it in this form: 

$ \dfrac{d}{dz} [(z-1)f'(z)] + \dfrac{\lambda}{z^2} f(z) = 0 $ 

or use quadratic transformation or other properties of $\_2F^1$, but I failed. 
Any suggestions?

Thanks in advance for the attention.

Answer: 
Let $\alpha$ be a root of $\alpha^2-\alpha-\lambda=0$. The change of the dependent
variable $f(z)=z^\alpha w(z)$ reduces to the hypergeometric equation in $w$:
$$z(1-z)w''+(2\alpha(1-z)-z)w'-\alpha^2 w=0.$$
The transformation $x = (z-2)/z$ takes your differential equation to 
$$ (x^2-1) f'' + 2 x f' - \lambda f = 0$$

which is a Gegenbauer differential equation. Its solutions can be written using Legendre P and Q functions:

$$f \left( x \right) =c\_1 \,{\it LegendreP} \left( {\frac {1}{2}
\sqrt {1+4\,\lambda}}-{\frac{1}{2}},x \right) +c\_2 \,{\it 
LegendreQ} \left( {\frac {1}{2}\sqrt {1+4\,\lambda}}-{\frac{1}{2}},x
 \right)
$$