# Existence of Integrating Factor

## Theorem

$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$

be such that $M$ and $N$ are real functions of two variables which are not homogeneous functions of the same degree.

Suppose also that:

$\dfrac {\partial M} {\partial y} \ne \dfrac {\partial N} {\partial x}$

that is, $(1)$ is not exact.

Finally, suppose that $(1)$ has a general solution.

Then it is always possible to find an integrating factor $\map \mu {x, y}$ such that:

$\map \mu {x, y} \paren {\map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} } = 0$

Hence it is possible to find that solution by Solution to Exact Differential Equation.

## Proof

Let us for ease of manipulation express $(1)$ in the form of differentials:

$(2): \quad \map M {x, y} \rd x + \map N {x, y} \rd y = 0$

Suppose that $(2)$ has a general solution:

$(3): \quad \map f {x, y} = C$

where $C$ is some constant.

We can eliminate $C$ by differentiating:

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

It follows from $(2)$ and $(3)$ that:

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

and so:

$(4): \quad \dfrac {\partial f / \partial x} M = \dfrac {\partial f / \partial y} N$

Let this common ratio in $(4)$ be denoted $\map \mu {x, y}$.

Then:

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

So, if we multiply $(2)$ by $\mu$, we get:

$\mu M \rd x + \mu N \rd y = 0$

or:

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

which is exact.

So, if $(2)$ has a general solution, it has at least one integrating factor $\map \mu {x, y}$.

$\blacksquare$