Solution to Exact Differential Equation

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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

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