# method of integrating factors

Let us consider a differential equation solved for the derivative  $y^{\prime}$ 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 $X$ and $Y$ have continuous   partial derivatives  in a region $R$ of $\mathbb{R}^{2}$.

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

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

 $f_{x}^{\prime}\;=\;\mu X,\qquad f_{y}^{\prime}\;=\;\mu Y.$

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

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 (http://planetmath.org/HomogeneousFunction): then the expression $\displaystyle\frac{1}{xX+yY}$ is an integrating factor.

Example.  In the differential equation

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

we see that  $X=:x^{4}+y^{4}$  and  $Y=:-xy^{3}$  both define a homogeneous function of degree (http://planetmath.org/HomogeneousFunction) 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

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

## References

• 1 E. Lindelöf: Differentiali- ja integralilasku III 1.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1935).
Title method of integrating factors MethodOfIntegratingFactors 2013-03-22 16:31:48 2013-03-22 16:31:48 pahio (2872) pahio (2872) 21 pahio (2872) Topic msc 35-00 msc 34-00 ErnstLindelof integrating factor Euler multiplicator