Exact Differential Equation Geometry In a variety of contexts, I have noticed hints of a strong connection between exact differential equations and machinery from multivariable calculus.
From another question, I have gathered that the geometry of finding the general solution to an exact differential equation $M(x,\ y)\ dx + N(x,\ y)\ dy = 0$ includes finding the potential function of $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$, that is, the surface in $\mathbb{R^3}$ whose gradient is the vector field $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$.  What is missing from this description?  For example, we can write the LHS of a not-necessarily-exact $M(x,\ y)\ dx + N(x,\ y)\ dy = 0$ as $\begin{bmatrix}M(x,\ y) \\ N(x,\ y) \end{bmatrix} \cdot \begin{bmatrix}dx \\ dy\end{bmatrix}$, which expresses the differential of work done by $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$, and thus shows promise for extending the above geometric picture.  In fact, since work is a path function, I suspect the Wikipedia quote

In mathematics, an integrating factor is a function that is chosen to
  facilitate the solving of a given equation involving differentials. It
  is commonly used to solve ordinary differential equations, but is also
  used within multivariable calculus when multiplying through by an
  integrating factor allows an inexact differential to be made into an
  exact differential (which can then be integrated to give a scalar
  field). This is especially useful in thermodynamics where temperature
  becomes the integrating factor that makes entropy an exact
  differential.

is misleading, in that the ordinary differential equations and multivariable calculus uses are actually identical, which if correct, might allow geometry from other path functions (such as the referenced thermodynamics example) to be leveraged.
A big part of what I'm trying to remedy is that it's not clear to me why finding the potential function of $\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}$ should solve the ODE.  I understand the algebraic rationale that the solution technique undoes the multivariable chain rule (divide through by $dx$ then integrate both sides), but is the geometric picture just an afterthought, or does it offer its own independent rationale for the solution technique?
 A: Interesting question, althought I am not exactly sure what kind of answer you want. If this answer is not fully satisfactory, please leave a comment.
Why finding the potential function solves the differential equation. Suppose we have a potential function $\phi$ for $(M(x,y),N(x,y))$. Then all the curves $\phi(x,y)=c$ with $c$ a constant form a solution of the ODE. Indeed, if we take the total derivative of $\phi(x,y)=c$ we get
$$
  \phi_x(x,y) dx + \phi_y(x,y) dy = 0, \\
  M(x,y) dx + N(x,y) dy = 0.
$$ 
This is a rather formal explanation.
The physics picture. The differential $Mdx + N dy$ can indeed be regarded as the infinitesimal amount of work done by a field $\vec F=(M(x,y),N(x,y))$. This picture can help you understand intuitively why $F(x,y)=c$ solves the ODE $Mdx + Ndy =0$.
Note that a potential in physics is a scalar function $\phi$ such that $-\nabla \phi =\vec{F}=(M,N)$; one adds a minus sign. If such a potential exists, we say that the vector field is exact or, more commonly in physics, conservative. Well-known examples of conservative vector fields are the gravitational force field and electric field in electrostatics.
Conservative vector fields are characterized by the fact that the line integral of them only depend on the end points. Physically this means that the work done by the fields does not depend on the path, only on the beginning and end position. So if $(M(x,y),N(x,y))=\vec{F} = -\nabla \phi$, then
$$
  \int_C M(x,y)dx + N(x,y)dy = \int_C \vec{F} = -(\phi(p_2) - \phi(p_1))
$$
where $C$ is a curve going from $p_1$ to $p_2$. This line integral can be interpreted as the work done by the field $\vec F$ along the path $C$. This line integral (i.e. the total amount of work) is zero iff the potential of the endpoints are the same.
Back to the ODE.
Now let us look back at the ODE $M(x,y)dx + N(x,y)dy = 0$. Physically this equation means that the infinitesimal work should be zero. The solution curves $\phi(x,y)=c$ are the equipotential curves. These curves are exactly the paths along which the work is zero for every infinitesimal step. 
