Integrating Factor

Suppose that $M(x,y)dx + N(x,y)dy = 0$ is not exact differential equation. If

$\displaystyle \mu(x,y)M(x,y)dx + \mu(x,y)N(x,y)dy = 0$

is exact differential equation provided $\mu(x,y)\not \equiv 0$ , then $\mu(x,y)$ is called the integrating factor.

In general, an integrating factor is not unique. For example, a function $x^{-2}$ and $y$ are the integrating factor of the following differential equation

$\displaystyle 2xdx + \frac{x^{2}}{y}dy = 0 .$

Now the question is how to find a integratin factor. Suppose that

$\displaystyle \mu(x,y)M(x,y)dx + \mu(x,y)N(x,y)dy = 0$

is exact. Then

$\displaystyle \frac{\partial (\mu M)}{\partial y} = \frac{\partial (\mu N)}{\partial x}$

Rewrite this to get

$\displaystyle N\frac{\partial \mu }{\partial x} - M\frac{\partial \mu }{\partial y} = \mu(\frac{\partial M}{\partial y} - \frac{\partial N }{\partial x})$

This partial differential equation is hard to solve. Thus we concentrate on the special cases.

For $\mu(x,y) = \mu(x)$ , $M\frac{\partial \mu }{\partial y} = 0$ . Thus,

$\displaystyle N\frac{\partial \mu}{\partial x} = \mu(\frac{\partial M}{\partial y} - \frac{\partial N }{\partial x})$

Then

$\displaystyle \frac{1}{\mu}d\mu = \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N }{\partial x})dx.$

Note that if

$\displaystyle \frac{1}{N}(\frac{\partial M}{\partial y} - \frac{\partial N }{\partial x})$

is a function of $x$

only, then the integrating factor is given by

$\displaystyle \mu(x) = \exp\{\int\frac{1}{N}[M_{y} - N_{x}]dx\}$

For $\mu(x,y) = \mu(y)$ , $N\frac{\partial \mu }{\partial x} = 0$ . Thus,

$\displaystyle -M\frac{\partial \mu}{\partial y} = \mu(\frac{\partial M}{\partial y} - \frac{\partial N }{\partial x})$

Then

$\displaystyle \frac{1}{\mu}d\mu = -\frac{1}{M}(\frac{\partial M}{\partial y} - \frac{\partial N }{\partial x})dy.$

Note that if

$\displaystyle \frac{1}{M}(\frac{\partial M}{\partial y} - \frac{\partial N }{\partial x})$

is a function of $y$

only, then the integrating factor is given by

$\displaystyle \mu(y) = \exp\{-\int\frac{1}{M}[M_{y} - N_{x}]dy\}$

Note that since we are looking for one integrating factor, we ignore the constant.

Example 1..14 Find the general solution of $(2x^{2} - y)dx + (x^{2}y + x)dy = 0$ .

SOLUTION Note that $M_{y} = -1, N_{x} = 2xy + 1$ . Thus, this differential equation is not exact. Then we calculate $({1}/{N})[M_{y} - N_{x}]$ .

$\displaystyle \frac{1}{N}[M_{y} - N_{x}] = \frac{1}{x^{2}y + x}[-1 - (2xy + 1)] = -\frac{2(xy+1)}{x(xy + 1)} = -\frac{2}{x}.$

This is a function of $x$

only. Then the integrating factor is

$\displaystyle \mu = \exp(-\int\frac{2}{x}dx) = \exp(-2\log{\vert x\vert}) = \frac{1}{x^{2}}$

Now multiply this to the original differential equation to obtain

$\displaystyle (2 - \frac{y}{x^{2}})dx + (y + \frac{1}{x})dy = 0 .$

This is exact differential equation and we solve by the grouping method.

$\displaystyle 2dx - (\frac{y}{x^2}dx - \frac{1}{x}dy) + ydy = 0$

Thus the general solution is

$\displaystyle 2x + \frac{y}{x} + \frac{y^{2}}{2} = c\ensuremath{\ \blacksquare}$

Example 1..15 Find the general solution of $(4xy^{2} + 6y)dx + (5x^{2}y + 8x)dy = 0$ .

SOLUTION Since $M_{y} = 8xy + 6, N_{x} = 10xy + 8$ , the given differential equation is not exact. Now $M_{y} - N_{x} = -2(xy + 1)$ . Thus neither $(1/N)[M_{y} - N_{x}]$ nor $(1/M)[M_{y} - N_{x}]$ is a function of $x$ only or $y$ only. Then we must find an integrating factor by the different method. $M_{y}$ and $N_{x}$ are polynomials. Then we let $\mu$ be $x^{m}y^{n}$ . If $x^{m}y^{n}$ is an integrating factor, then

$\displaystyle (4x^{m+1}y^{n+2} + 6x^{m}y^{n+1})dx + (5x^{m+2}y^{n+1} + 8x^{m+1}y^{n})dy = 0$

must be an exact differential equation.

$\displaystyle \frac{\partial M}{\partial y}$	$\displaystyle =$	$\displaystyle 4(n+2)x^{m+1}y^{n+1} + 6(n+1)x^{m}y^{n}$
	$\displaystyle =$	$\displaystyle 5(m+2)x^{m+1}y^{n+1} + 8(m+1)x^{m}y^{n} = \frac{\partial N}{\partial x}$