Derivation of Lagrange-Charpit Equations I am working through the derivation of the Lagrange-Charpit equations presented in this Wikipedia article: http://en.wikipedia.org/wiki/Method_of_characteristics#Fully_nonlinear_case

I am interested in the "fully" nonlinear case. Where we have:

$$F(x_1,...,x_n,u,p_1,...,p_n)=0  $$
And $$ p_i=\frac{\partial u}{\partial x_i} $$

I am fine with the derivation up until the point where they say that (Also, $\dot{x_i}=dx_i/ds$):

$$\sum_i(\dot{x}_idp_i-\dot{p_i}dx_i)=0 $$

Follows from taking the exterior derivative of:

$$ du-\sum_ip_idx_i=0  $$

From the little I know about exterior derivatives, it seems like doing this would give (for the two dimensional case):

$$0=\left(\frac{\partial p_2}{\partial x_1}-\frac{\partial p_1}{\partial x_2}\right)dx_1\wedge dx_2 $$

Because $ddu=0$ and the anti-symmetry of the wedge product.  I don't see how this result leads to the one given.  Could someone help me out? I feel like there is something very simple that I am missing.
$ \newcommand{\pd}[2]{ \frac{\partial #1}{\partial #2} } $ 
Attending to the positive answer to my request from the OP, consider the non-linear 1st order PDE:

$$F(x_i,u,p_i) = 0, \quad i = 1,\ldots, n, \quad p_i = \pd{u}{x_i}, \quad u = u(x_1,\ldots,x_n). \tag{1}$$

Assume $F \in \mathcal{C}^1_{x_i}$ and compute the partial derivative of eq. $(1)$ with respect to $x_i$:

$$\pd{F}{x_i} + \pd{F}{u}\pd{u}{x_i} + \pd{F}{p_i}\pd{p_i}{x_i} = 0,$$ 

or equivalently:

$$\pd{F}{x_i} + p_i \pd{F}{u} + \pd{F}{p_i}\pd{p_i}{x_i} = 0,$$ 

which is a quasilinear PDE for $p_i$ which can be readily solved leading the set of equations known as Lagrange-Charpit equations:


  $$ \color{blue}{\large{\frac{\mathrm{d}x_i}{\pd{F}{p_i}} = -\frac{\mathrm{d}p_i}{\pd{F}{x_i}+p_i \pd{F}{u}} = \frac{\mathrm{d}u}{\sum_ip_i \pd{F}{p_i}}}}   $$


where the last equality comes from the fact $\sum_i p_i \mathrm{d}x_i = \mathrm{d}u. $

Hope this helps.

Cheers!
let the general pde be f(x,y,z,p,q)......1
since z depends on (y)and(x) then,
dz=dz/dx.dx+dzdy.dy
dz=pdx+qdy...............................2
let another relation be Q(x,y,z,p,q)......3
on solving 1 and 3 we get the value of p and q
to determine Q we differentiate w.r.t x and y.
then,eliminate dp/dx