Space of solutions to a system of first-order PDEs I would like to know what is known (both explanations and references) about the spaces of smooth solutions to linear systems of PDEs of the following form: 
Let $g_{1},...,g_{n}$ be smooth functions on $\mathbb{R}^{n}$ with the integrability condition $\partial{g_{i}}/\partial{x^{j}}=\partial{g_{j}}/\partial{x^{i}}$ and consider the space of smooth functions $f$ on $\mathbb{R}^{n}$ satisfying $\partial{f}/\partial{x^{i}}=fg_{i}$ for all $i$. 
Similarly for the $g_{i}$ and $f$ being holomorphic on $\mathbb{C}^{n}$, and replacing $\mathbb{R}^{n}, \mathbb{C}^{n}$ with open contractible subsets.
My hope is that the answer is there is a unique solution, up to scaling.
 A: You can rewrite the system of PDEs as $\vec{\nabla} \ln f = \vec{g}$. If the vector field $\vec{g}$ has a potential function (which I believe is true if and only if its Jacobian matrix is symmetric, but I don't recall the source of this fact off-hand), then we may denote the potential as $G$ and solve the system as $f=\exp G$, which is unique up to rescalings. (There may be a negative sign in the exponential depending on what definition of potential you're using.) On the other hand, if $\vec{g}$ doesn't have a potential then it isn't a gradient field and hence no $f$ exists.
A: anon's answer is almost complete, here I'm filling important gaps. 
As anon pointed your system is $\frac{\partial}{\partial x_i} \ln(f)= g_i$ then by Froebenius theorem, the PDE system has a solution if and only if the $\nabla g$ is a symmetric matrix, which is true given your assumption that $\frac{\partial}{\partial x_i} g_j= \frac{\partial}{\partial x_j} g_j$. 
Moreover, given initial data, Frobenius theorem also gives you a unique local solution for you system, see page 92 of this paper (http://nyjm.albany.edu/j/2004/10-6.pdf‎) for the Frobenius theorem in a PDE/local coordinates/classical version. Perhaps with your additional assumptions about the domain, you maybe able to extent it to a global solution. 
