A specific linear differential equation on $\mathbb{C}-\{0,1\}$ whose monodromy group represents the fundamental group of $\mathbb{C}-\{0,1\}$ 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?
 A: 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 
$$z(z-1)y''+(1-2z)w'-w/4=0.$$
A: Yes. they exist. The problem breaks up into two steps:
(1) Embed the free group on two generators into $GL_n(\mathbb{C})$; see here or here among many sources.
(2) Find a connection on $\mathbb{CP}^1 \setminus \{ 0,1,\infty \}$ whose monodromy is the embedding from step (1).
Step (2) is always possible, by the Riemann-Hilbert correspondence, but that result is usually highly non-constructive. I write this answer to advertise that step (2) is not bad for $2 \times 2$ matrices.

Let $\theta$ be $\cos^{-1}(1/3)$. Let $A$ and $B$ in $SO(3)$ be rotation by $\theta$ around the $x$ and $z$-axes. It is standard (see here) that $A$ and $B$ generate a free group.
We'd rather work with the preimages of $A$ and $B$ in the double cover $SU(2) \to SO(3)$. These are
$$X := \begin{pmatrix} e^{i \theta/2} & 0 \\ 0 & e^{-i \theta/2} \end{pmatrix} \qquad Y:=\begin{pmatrix} \cos(\theta/2) & \sin(\theta/2) \\ - \sin(\theta/2) & \cos(\theta/2) \end{pmatrix}.$$
Since $A$ and $B$ generate a free group, so do the preimages $X$ and $Y$.
The great thing about $2 \times 2$ matrices is that $X$ and $Y$ are determined up to simultaneous conjugacy by $Tr(X)$, $Tr(Y)$, $Tr(XY)$, $\det(X)$ and $\det(Y)$. In this case, these are $\cos(\theta/2)$, $\cos(\theta/2)$, $\cos(\phi/2)$, where $\phi$ is an angle I could compute explicitly if I had to, and $1$, $1$. So it is enough to write down an ODE where the monodromy is all determinant $1$, and where the traces of loops around $0$, $1$ and $\infty$ are $\cos(\theta/2)$ and $\cos(\theta/2)$, $\cos(\phi/2)$ respectively.
To this end, we take our ODE to be $v'(t) = \frac{1}{(2 \pi i) t} Pv + \frac{1}{(2 \pi i)(t-1)} Qv$, where $P$ and $Q$ are $2 \times 2$ constant matrices with eigenvalues $\pm i \theta/2$, $\pm i \theta/2$ such that $P+Q$ has eigenvalues $\pm i \phi/2$. This is not hard to achieve, taking $P = \left( \begin{smallmatrix} i \theta/2 & 0 \\ 0 & -i \theta/2 \end{smallmatrix} \right)$ and $Q = S P S^{-1}$ for $S$ of the form $\left( \begin{smallmatrix} e^{i \alpha} & 0 \\ 0 & e^{-i\alpha} \end{smallmatrix} \right)$ and solving numerically for what $\alpha$ gives $P+Q$ the correct determinant.

That said, this answer is kind of wrong headed. One rarely wants to give a specific embedding of the free group into $SU(2)$; one just wants to know that almost any two generators do it. And, while this method will give you an explicit connection, in general one uses the Riemann-Hilbert correspondence non-constructively and notes that almost any connection will have no relations between the monodromy matrices.
