Show that $$y(z) = z^{1-c} F(a+1-c, b+1-c, 2-c, z)$$ satisfies the differential equation $$z(1-z)\frac{d^{2}y}{dz^{2}} + (c-(a+b+1)z)\frac{dy}{dz} - aby = 0.$$
I know that:
$F(a,b,c,z) = \sum_{n=0}^{\infty} \frac{(a)_{n}(b)_{n}}{n!(c)_{n}} z^{n}$
$(a)_{n} = a(a+1) \dots (a+n-1)$ if $n \geq 1$ and $1$ if $n=0$
$(a)_{n} = \frac{\Gamma{(a+n)}}{\Gamma{(a)}}$ and $\Gamma{(n+1)} = n!$
I think I should somehow rewrite $F(a+1-c, b+1-c, 2-c, z)$ with the things I know and then differentiate $y$ twice to $z$.