TITLE: Solving $P(x,y)dx + Q(x,y)dy =0$: interpretation in terms of forms
QUESTION [7 upvotes]: I asked a similar question here which I will formulate more sharply: 
When we write a differential equation as $P(x,y)dx + Q(x,y)dy = 0$, what is the interpretation in terms of differential forms? 
(I suppose the language of differential forms is the proper one to understand it.) Suppose we can separate into $\alpha(x)dx + \beta(y)dy = 0$. We integrate to find a relation between $x$ and $y$. What is the interpretation of this action in terms of differential forms? At first I thought we were flowing along the vector field $\alpha(x)dx + \beta(y)dy$, and the relation between $x$ and $y$ describes the flow lines. But then I realized $\alpha(x)dx + \beta(y)dy$ is a covector field, not a vector field, so this interpretation is not correct.
When we integrate the right-hand side and get a constant, what is the justification of that in terms of forms?

REPLY [2 votes]: Such a differential equation is called an Exact Differential Equation. The equation is solved by finding a multiplicative "integrating" factor $R$ such that the following form is exact:
$$
               R(Qdx+Pdy) = QRdx+PRy=0.
$$
In other words,
$$
                   QRdx + PRdy = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y} dx = 0.
$$
for some function $f$. That's what it has to do with forms. Equivalently,
$$
    \frac{\partial f}{\partial x}=QR,\;\;\; \frac{\partial f}{\partial y}=PR.
$$
In such a case, the requirement $Qdx+Pdy=0$ means $df=0$ or, equivalently, $f$ is constant (at least locally). Then, by the implicit function theorem
$f(x,y)=C$ can be solved for $x=x(y)$ or $y=y(x)$ in a neighborhood of $(x_{0},y_{0})$ for which $f(x_{0},y_{0})=0$ if $\frac{\partial f}{\partial x}(x_{0},y_{0})\ne 0$ or if $\frac{\partial f}{\partial y}(x_{0},y_{0})\ne 0$. If such $R$ exists, then it factors out of the equation when considering the ratios of derivatives of $f$. If you can solve for $y=y(x)$, then $f(x,y(x))=C$ and $y(x_{0})=y_{0}$, which gives
$$
           \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}=0,
$$
or
$$
       \frac{dy}{dx} = -\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}} = -\frac{QR}{PR}=-\frac{Q}{P}.
$$
Similarly, if $x=x(y)$ satisfies $f(x(y),y)=C$ and $x(y_{0})=x_{0}$, then
$$
           \frac{dx}{dy} = -\frac{P}{Q}.
$$
That is how one arrives at (and justifies) the final form of equation
$$
        Q+P\frac{dy}{dx}=0\;\;\mbox{ or } \;\; Q\frac{dx}{dy}+P=0.
$$
The process generalizes to more variables through the notion of exactness. One is then able to solve for one variable in terms of the others by setting $f=0$.
Here is a nice introduction to relate geometry, exactness, and this ODE:
http://tutorial.math.lamar.edu/classes/de/exact.aspx