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 $y'$ of the unknown function and write the equation in the form

(1)

If there is a solution of (1) which may be expressed in the form $$f(x,\,y) = C$$ with $f$ having continuous partial derivatives in $R$ and with $C$ an arbitrary constant, then it's not difficult to see that such an $f$ satisfies the linear partial differential equation

(2)

It's straightforward 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 $F$ 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 $$\frac{f_x'}{X} = \frac{f_y'}{Y}.$$ If we denote the common value of these two ratios by $\mu(x,\,y) = \mu$ , then we have $$f_x' = \mu X,\quad f_y' = \mu Y.$$ This gives to the differential of the function $f$ the expression $$d\,f(x,\,y) = \mu(x,\,y)(X(x,\,y)\,dx+Y(x,\,y)\,dy).$$ We see that $\mu(x,\,y)$ is the integrating factor or Euler multiplicator of the given differential equation (1), i.e. the left hand 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 $f$ , 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 $f$ can be gotten from the line integral $$f(x,\,y) := \int_{P_0}^P [\mu(x,\,y)X(x,\,y)\,dx+\mu(x,\,y)Y(x,\,y)\,dy]$$ along any curve $\gamma$ connecting an arbitrarily chosen point $P_0 =(x_0,\,y_0)$ and the point $P = (x,\,y)$ in the region $R$ .

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

Example. In the differential equation $$(x^4+y^4)\,dx-xy^3\,dy = 0$$ we see that $X := x^4+y^4$ and $Y := -xy^3$ both define a homogeneous function 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 hand side of the equation $$\left(\frac{1}{x}+\frac{y^4}{x^5}\right)\,dx-\frac{y^3}{x^4}\,dy = 0$$ 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 $$f(x,\,y) := \int_1^x\left(\frac{1}{x}+\frac{0^4}{x^5}\right)\,dx -\int_0^y\frac{y^3\,dy}{x^4} = \ln|x|-\frac{y^4}{4x^4}.$$ So the general solution of the given differential equation is $$\ln|x|-\frac{y^4}{4x^4} = C.$$

Bibliography

1

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