method of integrating factors

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

Let us consider a differential equation solved for the derivative $y^{\prime }$ of the unknown function and write the equation in the form

$X(x,y)dx+Y(x,y)dy=\mathrm{\hspace{0.33em}0}.$(1)

We assume that the functions $X$ and $Y$ have continuous partial derivatives in a region $R$ of ${ℝ}^2$.

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

$X{\displaystyle \frac{\partial f}{\partial y}}-Y{\displaystyle \frac{\partial f}{\partial x}}=\mathrm{\hspace{0.33em}0}.$(2)

Conversely, every non-constant solution $f$ of (2) gives also a solution  $f(x,y)=C$  of (1).  Thus, solving (1) and solving (2) are equivalent (http://planetmath.org/Equivalent3) tasks.

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

$$\frac{f_x^{\prime }}{X}=\frac{f_y^{\prime }}{Y}.$$

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

$$f_x^{\prime }=\mu X,f_y^{\prime }=\mu Y.$$

This gives to the differential of the function $f$ the expression

$$df(x,y)=\mu (x,y)(X(x,y)dx+Y(x,y)dy).$$

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 (http://planetmath.org/ExactDifferentialForm).

Conversely, any integrating factor $\mu$ of (1), i.e. such that $\mu Xdx+\mu Ydy$  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 (http://planetmath.org/HomogeneousFunction): then the expression ${\displaystyle \frac{1}{xX+yY}}$ is an integrating factor.

Example.  In the differential equation

$$(x^4+y^4)dx-x{y}^{3}dy=\mathrm{\hspace{0.33em}0}$$

we see that  $X=:x^4+y^4$  and  $Y=:-x{y}^3$  both define a homogeneous function of degree (http://planetmath.org/HomogeneousFunction) 4.  Thus we have the integrating factor  $\mu =:{\displaystyle \frac{1}{x^5+x{y}^4-x{y}^4}}={\displaystyle \frac{1}{x^5}}$,  and the left hand side of the equation

$$\left(\frac{1}{x}+\frac{y^4}{x^5}\right)dx-\frac{y^3}{x^4}dy=\mathrm{\hspace{0.33em}0}$$

is an exact differential.  We can integrate it along the broken line, first from  $(1,\mathrm{\hspace{0.17em}0})$  to  $(x,\mathrm{\hspace{0.17em}0})$  and then still to  $(x,y)$,  obtaining

$$f(x,y)=:{\int }_1^x(\frac{1}{x}+\frac{0^4}{x^5})dx-{\int }_0^y\frac{y^{3}dy}{x^4}=\ln |x|-\frac{y^4}{4x^4}.$$

So the general solution of the given differential equation is

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