Hypergeometric Function Differential Equation Is there some nice obvious way to see that the hypergeometric function $$_2F_1(a,b;c:z) = \sum_{i=0}^\infty \tfrac{(a)_n(b)_n}{(c)_n}\tfrac{z^n}{n!}$$
should satisfy the differential equation
$$z(1-z)\tfrac{d^2u}{dz^2} + [c-(a+b+1)]\tfrac{du}{dz}-abu=0?$$
I can't get it to work out by directly differentiating & it's driving me crazy - can it be done directly or does it require some nice identity? Thanks
 A: Please feel free to simplify!
We can apply the ``Method of Coefficients'';{http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.231.6616&rep=rep1&type=pdf}{Method of Coefficients}
. 
We are looking for a recursion between terms; and then using the
Method of Coefficients to translate it into a differential equation.
Some excerpts from Method of Coefficients
$[x^{n}]x^{k}f(x)=[x^{n-k}]f(t)$ ;
$[x^{n}]\frac{f(x)}{x^{k}}=[x^{n+k}]f(x)$
$[x^{n}]n\cdot f(x)=[x^{n-1}]f'(x)=[x^{n}]f'(x)\cdot x$
Now writing down the relationship between successive coefficients
of 
$_{2}F_{1}(a,b;c;x)={\displaystyle \sum_{n=0}^{\infty}}\frac{\left(a\right)_{n}\left(b\right)_{n}}{\left(c\right)_{n}}\frac{x^{n}}{n!}$
$[x^{n+1}]f(x)=\frac{\left(a+n\right)\cdot\left(b+n\right)}{\left(c+n\right)}\cdot\frac{n!}{\left(n+1\right)!}\cdot[x^{n}]f(x)=\frac{\left(a+n\right)\cdot\left(b+n\right)}{\left(c+1\right)}\cdot\frac{1}{\left(n+1\right)}\cdot[x^{n}]f(x)$
$\left(n+1\right)\cdot\left(c+n\right)\cdot[x^{n+1}]f(x)-\left(a+n\right)\cdot\left(b+n\right)\cdot[x^{n}]f(x)=0$
$\left(c+n\right)\cdot[x^{n}]f'(x)-\left(a+n\right)\cdot b\cdot[x^{n}]f(x)-(a+n)\cdot[x^{n}]x\cdot f'(x)=0$
$[x^{n}]x\cdot f''(x)+c\cdot[x^{n}]f'(x)-b\cdot[x^{n}]x\cdot f'(x)-a\cdot b[x^{n}]f(x)-a\cdot[x^{n}]x\cdot f'(x)-[x^{n}]x^{2}\cdot f''(x)-[x^{n}]f'(x)=0$
$[x^{n}]\left(x-x^{2}\right)\cdot f''(x)+\left(c-\left(b+a+1\right)\cdot x\right)\cdot f'(x)-a\cdot b\cdot f(x)=0$
QED
