How to take this exterior derivative of the expression $du - \sum_i p_i dx_i$? I am reading the wikipedia page about applying the method of characteristics in the fully nonlinear case. We have the fully nonlinear equation
$$ \tag{1} F(x_1, \cdots, x_n , u, p_1, \cdots, p_n) = 0,$$
here 
$$\tag{2} p_i = \frac{\partial u}{\partial x_i}$$
is the partial derivative of $u$ with respect to $x_i$. 
In the method of characteristics, we wish to reduce the PDE to a family of ODE. Let assume that $u$ is a solution to (1). 
$$s\mapsto (x_1(s), \cdots, x_n(s), u(s), p_1(s), \cdots, p_n(s))$$
be a curve so that (1) is satisfied for all $s$. Then it is claimed that the following holds: 
\begin{equation}
\begin{split}
\sum_i (F_{x_i} +F_up_i)\dot x_i + \sum_i F_{p_i}\dot p_i &=0\\
\dot u - \sum_i p_i\dot x_i &=0\\
\sum_i ( \dot x_i dp_i - \dot p_i dx_i) &= 0.
\end{split}
\end{equation}
I can see that the first two equations follow from taking total derivative with respect to $s$ of (1) and the expression $u(s) = u(x_1(s),\cdots, x_n(s))$. In the wiki page, it is claimed that 

... the third follows by taking an exterior derivative of the relation $du - \sum p_i dx_i = 0$. 

Unfortunately, I fail to see how the third equation are derived using exterior derivative. Could you give me the steps so I could check my work, please?
P.S.: I’ve been struggling a lot lately on Frobenius theorem and systems of total differential equations. If you could explain that, I would deeply appreciate it.
 A: Indeed, the last equality is a consequence of taking exterior derivative, but the wikipedia could have been more precise. 
First, think of $\alpha = du - \sum_i p_i dx_i$ as a one form on $\mathbb R^{2n+1}$, the exterior derivative is 
$$ d\alpha= \sum_i dp_i \wedge dx_i.$$
Now, for any function $u : \mathbb R^n \to \mathbb R$, we define the mapping 
$$ I_u : \mathbb R^n\to \mathbb R^{2n+1}, \ \ \ I_u(x) = \left(x_1, \cdots, x_n, u(x), \frac{\partial u}{\partial x_1}, \cdots, \frac{\partial u}{\partial x_n} \right).$$
Then we have $I_u^* \alpha = 0$ since 
$$ I_u^* \alpha = I_u^* (du-\sum_i p_i dx_i) = du(x) - \sum_i\frac{\partial u}{\partial x_i} dx_i = du-du = 0.$$
(So it is really the (usual) abuse of notations: $\alpha$ is not a zero form, by "$du - \sum_i p_i dx_i = 0$" it really means $I_u^* \alpha = 0$). Now since pullback commutes with exterior derivative, 
$$0=d(I_u^*\alpha) = I_u^* (d\alpha) = I_u^* \left( \sum_i dp_i \wedge dx_i\right).$$
So the two form $d\alpha$ also restrict to zero under the map $I_u$. In particular, the tangent vector $X=(\dot x(s), \dot u(s), \dot p(s))$ also satisfies 
$$\sum_i dp_i \wedge dx_i (X, \cdot)= 0\Rightarrow \sum_i \left( \dot p_i dx_i - \dot x_i dp_i\right) = 0,$$
which is the last equation. Note that this last equation (thus, this answer) has completely nothing to do with the equation $F$. 
