Let $K$ be a field of characteristic $0$, let $A = K[[t\_1, \ldots, t\_n]]$ be a power series ring over $K$, and let $V$ be a free $A$-module. Let $\nabla \colon V \rightarrow V \otimes\_A \Omega^1\_{A/K} = \bigoplus\_{k=1}^n V \ dt\_k$ be an integrable connection, so $\nabla^2 = 0$. I want to show that there always exist solutions $s \in V$ to the equation $\nabla(s) = 0$ subject to any initial condition $s(0) = \alpha$ for $\alpha \in V(0) = V \otimes\_A A/\mathfrak{m}\_A$. 


More concretely, letting $e\_1, \ldots, e\_r$ be a basis for $V$, I have a system of $n \times r$ linear first-order homogeneous partial differential equations
$$
\frac{\partial}{\partial t\_k} s\_i = \sum\_j M^j\_{ik} s\_j
$$
For some $M\_{ik}^j \in A$. Or, writing $M\_k$ for the matrix $(M\_{i,k}^j)\_{i,j}$, 
$$
\frac{\partial}{\partial t\_k} s = M\_k s
$$


How can I show that integrability implies there are always solutions? Concretely, integrability is exactly the statement that for all $i,j,k, \ell$, we have
$$
\frac{\partial}{\partial t\_\ell} M\_{k} - \frac{\partial}{\partial t\_k} M\_{\ell} = M\_k M\_\ell - M\_\ell M\_k
$$ 



Reference request - existence of formal solutions for integrable connections
I hope you'll allow one or two slight adjustments to your question:


* I think you are interested in $(t\_1, \dots, t\_n)$-adically continuous connections rather than arbitrary connections (for which the desired solution principle would not generally hold).
* Judging from how you use "$r$," I think you intend $V$ to be a finite-rank free $A$-module (rather than an arbitrary free $A$-module).


You will find a very nice treatment of your question thus modified in Proposition 8.9 of Nicholas Katz's paper "Nilpotent connections and the monodromy theorem: Applications of a result of Turrittin."


Here is the statement of that proposition:



> 
> Let $K$ be a field of characteristic zero, $K[[t\_1, \dots, t\_n]]$ the
>  ring of formal power series over $K$ in $n$ variables. Let $M$ be a
>  finitely generated module over $K[[t\_1, \dots, t\_n]]$, given with an
>  integrable connection $\nabla$ (for the continuous $K$-derivations of
>  $K[[t\_1, \dots, t\_n]]$ to itself). Then $M^\nabla$, the $K$-space of
>  horizontal elements of $M$, is finite-dimensional over $K$, and the
>  pair $(M, \nabla)$ is isomorphic to the pair
>  $(M^\nabla\otimes\_K K[[t\_1, \dots, t\_n]], 1\otimes d)$
>  where $d$ denotes the ``identical'' connection on $K[[t\_1, \dots, t\_n]]$.
> 
> 
> 


Note that in this proposition, one can even assume only that the module $M$ with integrable connection is finitely generated (rather than free of finite rank). As a consequence of fact that $M^\nabla \otimes\_K K[[t\_1, \dots, t\_n]] \to M$ is an isomorphism, one can then conclude that $M$ is necessarily free.


