This question is a followup to this one, which received a response but which was not quite answered. I am hoping that rephrasing and simplifying it will help elicit more comprehensive feedback. My apologies if this is bad form (i.e. if it's too close to the original). Also, the genesis of this question (a research topic) is a post I made on Math Overflow, but I think this particular aspect of the problem is too elementary for that site (please correct me if I'm wrong).
This is a bit like turning the method of characteristics on its head. Let me take the case $n=3$: Say I have two smooth vector fields $X,Y$ on $P$, the strictly positive orthant of $R^3$, which at each point of P are linearly independent, thus span a two-dimensional subspace of $R^3$. Further suppose that their Lie bracket $\mathbf{[}X,Y\mathbf{]} = 0$ everywhere, so that by Frobenius there is a foliation of P by two-dimensional integral surfaces. Pick one of those leafs, call if $L$. Is it then true that there is a smooth function $u(x,y)$ on $R^2_{++}$ with the property that $L$ is the graph of $u$? So, as I say, I'm asking (I think) if under these conditions there is a quasi-linear partial differential equation in $u$ such that if I solve it by the method of characteristics the surface generated by all the integral curves ends up being $L$.
Note that implicit in my question is that $L$ is globally a graph, although responses about local results are welcome.
Edit: Jack Lee kindly alerted me to the fact that I left out an assumption, which in fact is that the normal vector to the plane spanned by $X,Y$ at each point of P is strictly positive.
Edit: In case it helps I can give a concrete example of the vector fields I'm working with. $X^1 = (-m(x_1,x_2,x_3),1, 0), \; X^2 = (-n(x_1,x_2,x_3),0,1)$. Here the smooth functions $m$ and $n$ are strictly positive everywhere on $M$ and bounded away from zero and $\infty$ on every compact subset of $M$. I also simply impose the condition $[X^1,X^2] = mn_1 - nm_1 -n_2 +m_3 = 0$. Note that the normal field is $N_{(x_1,x_2,x_3)} = (1,m(x_1,x_2,x_3),n(x_1,x_2,x_3)) \gg 0$.
Further clarification: Assumptions on m are, for each fixed $x_2,x_3$ that $\lim_{x \rightarrow \infty}m(x,x_2,x_3) = 0, \lim_{x \rightarrow 0}m(x,x_2,x_3) = \infty$. A similar assumption holds for $n$; i.e. for each fixed $x_1,x_3$ $\lim_{x \rightarrow \infty}n(x_1,x,x_3) = 0, \lim_{x \rightarrow 0}n(x_1,x,x_3) = \infty$.