TITLE: A specific linear differential equation on $\mathbb{C}-\{0,1\}$ whose monodromy group represents the fundamental group of $\mathbb{C}-\{0,1\}$
QUESTION [8 upvotes]: Is there a linear differential equation on $\mathbb{C}$ with singularities at $0$ and $1$ whose monodromy group represents the fundamental group of $\mathbb{C}-\{0,1\}$? If so, can someone give a specific example?

REPLY [13 votes]: If the word "represents" in the question means "isomorphic to", one can take
the hypergeometric equation: $w''+Qw=0$ with
$$Q(z)=\frac{1}{4z^2}+\frac{1}{4(z-1)^2}-\frac{1}{4z(z-1)}.$$
See Hurwitz-Courant, Ch7, or (with much more detail) Caratheodory, vol. II.
The ratio $f=w_1/w_2$ of two linearly independent solutions of this equation
maps the upper half-plane onto a triangle with all angles $0$. One can normalize
(choose particular solutions $w_1,w_2$) so that this triangle is inscribed in the unit disk. Then the well-known picture (see any course of complex variable) shows that $f$ is the inverse to the universal covering of $C\backslash\{0,1\}$ by the unit
disk. Which implies that the monodromy group is the fundamental group of $C\backslash\{0,1\}$.
EDIT. Written in the standard hypergometric form, the equation is 
$$z(z-1)y''+(1-2z)w'-w/4=0.$$