PlanetMath: method of integrating factors

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method of integrating factors

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The method of integrating factors is in principle a means for solving ordinary differential equations of first order. It has not great practical significance, but is theoretically important.

Let us consider a differential equation solved for the derivative of the unknown function and write the equation in the form

$\displaystyle X(x,\,y)\,dx+Y(x,\,y)\,dy = 0.$

(1)

We assume that the functions

and

have continuous partial derivatives in a region

of $\mathbb{R}^2$ .

If there is a solution of (1) which may be expressed in the form

$\displaystyle f(x,\,y) = C$

with

having continuous partial derivatives in

and with

an arbitrary constant, then it's easy to see that such an

satisfies the linear partial differential equation

$\displaystyle X\frac{\partial f}{\partial y}-Y\frac{\partial f}{\partial x} = 0.$

(2)

Conversely, every non-constant solution

of (2) gives also a solution $f(x,\,y) = C$ of (1). Thus, solving (1) and solving (2) are equivalent tasks.

It's not difficult to show that if $f_0(x,\,y)$ is a non-constant solution of the equation (2), then all solutions of this equation are $F(f_0(x,\,y))$ where is a freely chosen function with (mostly) continuous derivative.

The connection of the equations (1) and (2) may be presented also in another form. Suppose that $f(x,\,y) = C$ is any solution of (1). Then (2) implies the proportion equation

$\displaystyle \frac{f_x'}{X} = \frac{f_y'}{Y}.$

If we denote the common value of these two ratios by $\mu(x,\,y) = \mu$ , then we have

$\displaystyle f_x' = \mu X,\quad f_y' = \mu Y.$

This gives to the differential of the function

the expression

$\displaystyle d\,f(x,\,y) = \mu(x,\,y)(X(x,\,y)\,dx+Y(x,\,y)\,dy).$

We see that $\mu(x,\,y)$ is the integrating factor of the given differential equation (1), i.e. the left side of (1) turns, when multiplied by $\mu(x,\,y)$ , to an exact differential.

Conversely, any integrating factor $\mu$ of (1), i.e. such that $\mu X\,dx+\mu Y\,dy$ is the differential of some function , is easily seen to determine the solutions of the form $f(x,\,y) = C$ of (1). Altogether, solving the differential equation (1) is equivalent with finding an integrating factor of the equation.

When an integrating factor $\mu$ of (1) is available, the solution function can be gotten from the line integral

$\displaystyle f(x,\,y) := \int_{P_0}^P [\mu(x,\,y)X(x,\,y)\,dx+\mu(x,\,y)Y(x,\,y)\,dy]$

along any curve $\gamma$ connecting an arbitrarily chosen point $P_0 =(x_0,\,y_0)$ and the point $P = (x,\,y)$ in the region

Note. In general, it's very hard to find a suitable integrating factor. One special case where such can be found, is that and are homogeneous functions of same degree: then the expression $\displaystyle\frac{1}{xX+yY}$ is an integrating factor.

Example. In the differential equation

$\displaystyle (x^4+y^4)\,dx-xy^3\,dy = 0$

we see that

and

define homogeneous functions of degree 4. Thus we have the integrating factor $\displaystyle\mu := \frac{1}{x^5+xy^4-xy^4} = \frac{1}{x^5}$ , and the left side of the equation

$\displaystyle \left(\frac{1}{x}+\frac{y^4}{x^5}\right)\,dx-\frac{y^3}{x^4}\,dy = 0$

is an exact differential. We can integrate it along the broken line, first from $(1,\,0)$ to $(x,\,0)$ and then still to $(x,\,y)$ , obtaining

$\displaystyle f(x,\,y) := \int_1^x\left(\frac{1}{x}+\frac{0^4}{x^5}\right)\,dx -\int_0^y\frac{y^3\,dy}{x^4} = \ln\vert x\vert-\frac{y^4}{4x^4}.$

So the general solution of the given differential equation is

$\displaystyle \ln\vert x\vert-\frac{y^4}{4x^4} = C.$

Bibliography

1: E. LINDELÖF: Differentiali- ja integralilasku III 1. Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1935).

"method of integrating factors" is owned by pahio.

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