Solution to Exact Differential Equation

Theorem
The first order ordinary differential equation‎:
 * $\displaystyle F = M \left({x, y}\right) + N \left({x, y}\right) \frac {dy} {dx} = 0$

is an exact differential equation‎ iff:
 * $\displaystyle \frac {\partial M} {\partial y} = \frac {\partial N} {\partial x}$

The general solution of such an equation‎ is:
 * $f \left({x, y}\right) = C$

where:
 * $\displaystyle \frac {\partial f} {\partial x} = M, \frac {\partial f} {\partial y} = N$

Necessary Condition

 * Suppose $F$ be exact.

Then by definition there exists a function whose second partial derivatives of $f$ exist and are continuous:
 * $f \left({x, y}\right)$

where:
 * $\displaystyle \frac {\partial f} {\partial x} = M, \frac {\partial f} {\partial y} = N$

Differentiating $M$ and $N$ partially WRT $y$ and $x$ respectively:


 * $\displaystyle \frac {\partial M} {\partial y} = \frac {\partial^2 f} {\partial x \partial y}, \frac {\partial N} {\partial x} = \frac {\partial^2 f} {\partial y \partial x}$

The mixed second partial derivatives are equal, so we have:
 * $\displaystyle \frac {\partial M} {\partial y} = \frac {\partial N} {\partial x}$

Sufficient Condition
Now suppose $F$ is such that:
 * $\displaystyle \frac {\partial M} {\partial y} = \frac {\partial N} {\partial x}$

Take the equation:
 * $\displaystyle \frac {\partial f} {\partial x} = M$

and integrate it WRT $x$:
 * $\displaystyle (1) \qquad f = \int M dx + g \left({y}\right)$

Note that the arbitrary constant here is an arbitrary function of $y$, since when differentiating partially WRT $x$ it disappears.

So, now we need to find such a $g \left({y}\right)$ so as to make $f$ as defined in $(1)$ satisfy $\displaystyle \frac {\partial f} {\partial y} = N$.

We differentiate $(1)$ WRT $y$ and equate it to $N$ to get:
 * $\displaystyle \frac {\partial} {\partial y} \int M dx + g' \left({y}\right) = N$

So:
 * $\displaystyle g' \left({y}\right) = N - \frac {\partial} {\partial y} \int M dx$

which we can integrate WRT $y$:
 * $\displaystyle g \left({y}\right) = \int \left({N - \frac {\partial} {\partial y} \int M dx}\right) dy$

which works as long as the integrand is a function of $y$ only.

This will happen if its derivative WRT $x$. We need to make sure of that, so we try it out:

Lo and behold, this is our initial condition.

The Solution
Then $F$ can be written in the form:
 * $\displaystyle \frac {\partial f} {\partial x} dx + \frac {\partial f} {\partial y} dy = 0$

and the solution follows.