Existence of Integrating Factor

From ProofWiki
Jump to: navigation, search


Let the first order ordinary differential equation:

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

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 $\mu \left({x, y}\right)$ such that:

$\mu \left({x, y}\right) \left({M \left({x, y}\right) + N \left({x, y}\right) \dfrac {\mathrm d y} {\mathrm d x}}\right) = 0$

is an exact differential equation.

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


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

$(2): \quad M \left({x, y}\right) \mathrm d x + N \left({x, y}\right) \mathrm d y = 0$

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

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

where $C$ is some constant.

We can eliminate $C$ by differentiating:

$\dfrac {\partial f} {\partial x} \, \mathrm d x + \dfrac {\partial f} {\partial y} \, \mathrm d y = 0$

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

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

and so:

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

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


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

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

$\mu M \, \mathrm d x + \mu N \, \mathrm d y = 0$


$\dfrac {\partial f} {\partial x} \, \mathrm d x + \dfrac {\partial f} {\partial y} \, \mathrm d y = 0$

which is exact.

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


Also see