How might we find $\sigma$? How does one solve a "differential equation" for $\sigma$ of the form 
$$ 
\sigma(v)w_i(v)={\partial \over \partial v_j}\left[\sigma(v)A_{ij}(v)\right]
\quad i=1,\dots,n.
$$ 
where the summation convention applies.
$w,v$ are an $n$-D vectors, $\sigma$ is a scalar function, $A$ is an invertible $n\times n$ matrix?
Perhaps there is a general solution form? References (links) for the treatment of such an equation is also appreciated.
Thank you.
Added: 
In light of drak's suggestion, here is a bit more
Some thoughts:
It might be friendlier to change "variables" to $A\sigma$?
Is there a more familiar expression for the index notation ${\partial \over \partial v_j}M_{ij}(v)$ such as one in  terms of $\nabla$? It would seem to me that it is taking the divergence of each row of the matrix $M$.
Some more thoughts: since the function $\sigma$ appears on both sides of the equation, it is likely that it is an exponential.
A simplified version: What if we suppose that $A$ is a constant matrix? 
 A: $\def\w{{\bf w}}
\def\A{{\bf A}}
\def\B{{\bf B}}
\def\v{{\bf v}}
\def\u{{\bf u}}
\def\grad{\nabla}
\def\darg{{\overleftarrow \nabla}}
\def\t{\tau}$There is a notation used in physics that can handle these sorts of operations without indices.
The differential equation takes the form 
$$\begin{equation*}
(\A \sigma)\darg = \w \sigma,\tag{1}
\end{equation*}$$
where $(\B\darg)_{i} = \frac{\partial}{\partial v_j} B_{ij}$. 
Then 
$$(\A\darg + \A\grad)\sigma = \w\sigma,$$
so $\A\grad\sigma = (\w - \A\darg)\sigma$, or
$$\begin{equation*}
\frac{1}{\sigma} \grad\sigma = \A^{-1}(\w - \A\darg).\tag{2}
\end{equation*}$$
This is the equation given by @Mercy in the comments.  
A natural ansatz is $\sigma = e^\t$, since $e^{-\t}\grad e^\t = \grad \t$.
Thus, we must solve
$$\grad \t = \A^{-1}(\w - \A\darg),$$
to which we can apply the gradient theorem.
We find
$$\t(\v) - \t(\v_0) = \int_{\v_0}^{\v} d\u^T\, 
\A^{-1}(\u)\left(\w(\u) - \A(\u)\darg_\u\right).$$
Therefore,
\begin{equation*}
\sigma(\v) = \sigma(\v_0)\exp
\int_{\v_0}^{\v} d\u^T\, \A^{-1}(\u)\left(\w(\u) - \A(\u)\darg_\u\right).\tag{3}
\end{equation*}
Let's make sure we can unwind this expression.
It is shorthand for
$$\sigma(\v) = \sigma(\v_0) \exp
\int_{\v_0}^{\v} d u_i\,
(A^{-1}(\u))_{ij}\left(w_{j}(\u) - \frac{\partial}{\partial u_k} A_{jk}(\u)\right).$$
Note that the exponent is a scalar.
It is the line integral of the vector field
$\A^{-1}(\w - \A\darg)$.
Special case
Suppose $\A$ and $\w$ are constant and $\v_0 = 0$. 
The solution is then 
$$\sigma(\v) = \sigma(0) \exp \left(\v^T\A^{-1}\w\right),$$ 
which satisfies the differential equation (1) since 
$(\A \sigma)\darg = \A\grad \sigma = \A \A^{-1}\w \sigma = \w\sigma$. 
(We choose the gradient to be a column vector so 
$\grad(\v^T\A^{-1}\w) = \A^{-1}\w$.)
