What is the "differential form" of an ODE? I'm reading an introductory textbook on Ordinary Differential Equations and the author does something I think needs some justification.
Here it goes:
Consider an ODE $y' = f(x,y)$, so we can write $f(x,y)=\frac{-M(x,y)}{N(x,y)}$.
Therefore $y'=\frac{-M(x,y)}{N(x,y)}$, which we can write in "differential form" as $M(x,y)dx+N(x,y)dy=0$.
Question: What is exactly the "differential form" of an ODE? Is it just a purely formal reasoning or there is some elementary context in which we can interpret the $dx,dy$ in the above equation?
Thank you.
 A: A differential form is an expression $\omega = a\,dx+b\,dy$ where $dx,dy$ are linear functionals on the tangent space. That is, if $v=(v_1,v_2)$ is a direction, then $dx(v)=v_1$ and $dy(v)=v_2$. The equation $ω=0$ describes a line $0=ω(v)=av_1+bv_2$ through the origin in the direction $(-b,a)$ or $(b,-a)$. No specific orientation is implied.
In that sense, $$0=M(x,y)dx+N(x,y)dy$$
has in every point $(x,y)=(x,y(x))$ of a trajectory of the ODE the direction $(1,y'(x))$$ as solution.
In general, $0=M(x,y)dx+N(x,y)dy$ describes a direction field in the $(x,y)$ plane, a direction in each point $(x,y)$, such that any solution of the ODE is tangent in every point to the direction at this point, and any curve tangent to the direction field is a re-parametrization of a solution curve.
If one can find a function $F(x,y)$ such that one can identify $\frac∂{∂x}F(x,y)=M(x,y)$ and $\frac∂{∂y}F(x,y)=M(x,y)$, then the level curves of $F$ are tangent to the direction field. This means that any solution of the ODE follows one level set. One then also says that $F$ is a first integral.
$M(x,y)dx+N(x,y)dy$ may not allow directly a first integral, but a rescaling with a so called "integrating factor" may overcome that, i.e., there may be some $\phi(x,y)$ such that 
$$(ϕ(x,y) M(x,y))dx+(ϕ(x,y) N(x,y))dy$$
is integrable.
A: There are a few interpretations. A geometrically simple interpretation that I am fond of is that this says that the curves of the ODE have tangents that are orthogonal to the direction field $(M,N)$. One way to see this is to consider parametrizing the curve by an auxiliary variable $t$ and then formally "divide the differential form of the ODE on both sides by $dt$". One can also see it from the usual form of the ODE, since $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ by the chain rule. 
Either way you get $M \frac{dx}{dt} + N \frac{dy}{dt} = 0$ which is exactly the orthogonality statement I mentioned. However, this orthogonality is purely geometric, it does not depend on any particular choice of parametrization.
One can also represent the situation in the differential form framework as LutzL has already suggested. IMO this is a bit unhelpful for an elementary ODE student, however, because there is some differential geometry that has to be set up before this approach is fully defined.
