# What is the geometric interpretation behind the method of exact differential equations?

Given an equation in the form $M(x)dx + N(y)dy = 0$ we test that the partial derivative of $M$ with respect to $y$ is equal to the partial derivative of $N$ with respect to $x$. If they are equal, then the equation is exact. What is the geometric interpretation of this?

Further more to solve the equation we may integrate $M(x) dx$ or $N(y)dy$, whichever we like better, and then add a constant as a function in terms of the constant variable and solve this.

e.g. If $f(x) = 3x^2$ then $F(x) = x^3 + g(y)$.

After we have our integral we set its partial differential with respect to the other variable our other given derivative and solve for $g(y)$. I have done the entire homework assignment correctly, but I have no clue why I am doing these steps. What is the geometric interpretation behind this method, and how does it work?

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Great question. The idea is that $(M(x), N(y))$ defines a vector field, and the condition you're checking is equivalent (on $\mathbb{R}^2$) to the vector field being conservative, i.e. being the gradient of some scalar function $p$ called the potential. Common physical examples of conservative vector fields include gravitational and electric fields, where $p$ is the gravitational or electric potential.

Geometrically, being conservative is equivalent to the curl vanishing. It is also equivalent to the condition that line integrals between two points depend only on the beginning and end points and not only on the path chosen. (The connection between this and the curl is Green's theorem.)

The differential equation $M(x) \, dx + N(y) \, dy = 0$ is then equivalent to the condition that $p$ is a constant, and since this is not a differential equation it is a much easier condition to work with. The analogous one-variable statement is that $M(x) \, dx = 0$ is equivalent to $\int M(x) \, dx = \text{const}$. Geometrically, the solutions to $M(x) \, dx + N(y) \, dy = 0$ are therefore the level curves of the potential, which are always orthogonal to its gradient. The most well-known example of this is probably the diagram of the electric field and the level curves of the electrostatic potential around a dipole. This is one way to interpret the expression $M(x) \, dx + N(y) \, dy = 0$; it is precisely equivalent to the "dot product" of $(M(x), N(y)$ and $(dx, dy)$ being zero, where you should think of $(dx, dy)$ as being an infinitesimal displacement along a level curve.

(For those in the know, I am ignoring the distinction between vector fields and 1-forms and also the distinction between closed forms and exact forms.)

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At the end of the first paragraph, what were you meaning to say? "I am being slightly biased?" –  PEV Jan 17 '11 at 3:55
@Trevor: whoops. I meant to say what I said at the end. –  Qiaochu Yuan Jan 17 '11 at 4:00

You might want to look up the wiki article on exact differentials and inexact differentials. The famous physical quantity which you cannot write as an exact differential is heat.

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