# Solution to Exact Differential Equation

## Theorem

The first order ordinary differential equation:

- $F = \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$

is an exact differential equation if and only if:

- $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$

The general solution of such an equation is:

- $\map f {x, y} = C$

where:

- $\dfrac {\partial f} {\partial x} = M$
- $\dfrac {\partial f} {\partial y} = N$

## Proof

### Necessary Condition

Let $F$ be exact.

Then by definition there exists a function whose second partial derivatives of $f$ exist and are continuous:

- $\map f {x, y}$

where:

- $\dfrac {\partial f} {\partial x} = M$
- $\dfrac {\partial f} {\partial y} = N$

Differentiating $M$ and $N$ partially with respect to $y$ and $x$ respectively:

- $\dfrac {\partial M} {\partial y} = \dfrac {\partial^2 f} {\partial x \partial y}$
- $\dfrac {\partial N} {\partial x} = \dfrac {\partial^2 f} {\partial y \partial x}$

From Mixed Second Partial Derivatives are Equal:

- $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$

$\Box$

### Sufficient Condition

Let $F$ be such that:

- $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$

Take the equation:

- $\dfrac {\partial f} {\partial x} = M$

and integrate it with respect to $x$:

- $(1): \quad f = \ds \int M \rd x + \map g y$

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

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

Differentiating $(1)$ with respect to $y$ and equating it to $N$:

- $\ds \dfrac {\partial} {\partial y} \int M \rd x + \map {g'} y = N$

So:

- $\ds \map {g'} y = N - \dfrac {\partial} {\partial y} \int M \rd x$

which can be integrated with respect to $y$ thus:

- $\ds \map g y = \int \paren {N - \frac {\partial} {\partial y} \int M \rd x} \rd y$

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

This will happen if its derivative with respect to $x$ is equal to zero.

We need to make sure of that, so we try it out:

\(\ds \frac {\partial} {\partial x} \paren {N - \frac {\partial} {\partial y} \int M \rd x}\) | \(=\) | \(\ds \frac {\partial N} {\partial x} - \frac {\partial^2} {\partial x \partial y} \int M \rd x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {\partial N} {\partial x} - \frac {\partial^2} {\partial y \partial x} \int M \rd x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {\partial N} {\partial x} - \frac {\partial M} {\partial y}\) |

This is our initial condition.

$\Box$

## The Solution

Then $F$ can be written in the form:

- $\dfrac {\partial f} {\partial x} \rd x + \dfrac {\partial f} {\partial y} \rd y = 0$

and the solution follows.

$\blacksquare$

## Also see

- Results about
**examples of exact differential equations**can be found**here**.

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 18$: Basic Differential Equations and Solutions: $18.4$: Exact equation - 1972: George F. Simmons:
*Differential Equations*... (previous) ... (next): $\S 2.8$: Exact Equations - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next):**differential equation** - 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next):**differential equation**