Integrating Factor

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

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

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

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

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

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

is exact. Then

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

Rewrite this to get

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

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,

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

Then

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

Note that if

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

is a function of $x$ only, then the integrating factor is given by

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

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

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

Then

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

Note that if

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

is a function of $y$ only, then the integrating factor is given by

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

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 $({1}/{N})[M_{y} - N_{x}]$.

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

This is a function of $x$ only. Then the integrating factor is

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

Now multiply this to the original differential equation to obtain

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

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

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

Thus the general solution is

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

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

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

must be an exact differential equation.

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

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

Compare these equations to get

$$4(n+2) = 5(m+2), \ 6(n+1) = 8(m+1)$$

Solving the system of equations to get $n = 3, m = 2$. Thus

$$(4x^{3}y^{5} + 6x^{2}y^{4})dx + (5x^{4}y^{4} + 8x^{3}y^{3})dy = 0$$

is an exact differential equation, and the general solution is

$$x^{4}y^{5} + 2x^{3}y^{4} = c\ensuremath{\ \blacksquare}$$