Local systems as solution of PDE I want to understand the correspondence between locally constant sheaves and vector bundles with flat connection on a manifold $X$.
Given a local system $\mathcal L$, it is clear how to define a vector bundle $E$ with locally constant transition functions. This allows us to define a connection $\nabla$  on the sheaf of smooth sections of $L$, i.e. on  $\mathcal L \otimes \mathcal C^\infty_X$ it by demanding that $\nabla$ vanishes on  $\mathcal L$, i.e. locally $\nabla = d$.
This connection is clearly flat, as $d^2=0$, and $\mathcal L = \ker \nabla$.
But what about the other direction. Assume we have a vector bundle $E$ which admits a flat connection $\nabla$. I read, that then $\ker \nabla$ defines a local system $\mathcal L$, that is we consider the sheaf of solutions to the differential equation $\nabla s=0$ 
How do I show that $\mathcal L$ is locally constant?
For any $x_0\in X$, we can find a neighbourhood $U$, where $\nabla = d + A$ for some matrix $A$ of $1$-forms.
Thus there, a section $s$ we have$$s=(f_1,\dots, f_n) \in \ker \nabla ~~~~~~\Leftrightarrow ~~~~~ ~ df_i +\sum_j a_{ij} f_j=0, ~~\forall i.$$
Assuming that for any initial value $s(x_0)=(f_1(x_0),\dots,f_n(x_0))\in \mathbb C^n$, there exists a unique solution to the above differential equation, we get an isomorphism $\mathcal L(U) \to \mathbb C^n$, which is consistent with restrictions.
But how can I show this? In the texts I read, it was always said that it follows from Frobenius theorem and then the proof was finished without expanding on it. So here is my question:

Why does the differential equation $$df_i +\sum_j a_{ij} f_j=0, ~~\forall i$$ have a unique solution for any initial value
  $s(x_0)=(f_1(x_0),\dots,f_n(x_0))\in \mathbb C^n$.
  Why does it follow
  from Frobenius theorem and how do I use flatness of the bundle?

 A: The short proof is:

$\ker\nabla$, the choice horizontal space, is a distribution on $E$.
  It is involutive by flatness of $\nabla$ (flat horizontal lifts preserves Lie brackets), so it can be integrated by Frobenius theorem.

It seems that you are not very comfortable with this, so as an alternative, let's try to do it without Frobenius.
Since this is local, assume your base is $(-1,1)^n$, and let's state the flatness condition as $[\nabla_i,\nabla_j]=0$ (where $\nabla_i=\nabla_{\partial/\partial x^i}$).  We have a the starting value $f_{(0)}\in E_0$ at $0$.


*

*Integrating $\nabla_1f=0$ along $x^1$-direction is no problem, gets a horizontal section $f_{(1)}$ on $(-1,1)\times\{0\}^{n-1}$.

*Next, for each $x^1$, we integrate $\nabla_2f=0$ along the $x^2$-direction with initial value $f_{(1)}(x^1)$ at $(x^1,0,\dots,0)$.  This gives $f_{(2)}$ that is horizontal in $x^2$-direction, but for $x^1$-direction?  Away from $x^2=0$ we haven't guarantee $\nabla_1f_{(2)}=0$.  For that, the flatness condition comes to rescue.  Since $[\nabla_1,\nabla_2]=0$, we have $\nabla_2\nabla_1f_{(2)}=\nabla_1\nabla_2f_{(2)}=0$, i.e. $\nabla_1f_{(2)}$ is independent of $x^2$.  So $\nabla_1f_{(2)}=0$ at all points $(x^1,x^2,0,\dots,0)$.

*Similarly $x^3,\dots,x^n$.  QED
