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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
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(1) |
We assume that the functions  and  have continuous partial derivatives in a region  of
 .
If there is a solution of (1) which may be expressed in the form
with  having continuous partial derivatives in  and with  an arbitrary constant, then it's not difficult to see that such an  satisfies the linear partial differential equation
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(2) |
Conversely, every non-constant solution  of (2) gives also a solution
 of (1). Thus, solving (1) and solving (2) are equivalent tasks.
It's straightforward to show that if
 is a non-constant solution of the equation (2), then all solutions of this equation are
 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
 is any solution of (1). Then (2) implies the proportion equation
If we denote the common value of these two ratios by
 , then we have
This gives to the differential of the function  the expression
We see that
 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
 , to an exact differential.
Conversely, any integrating factor  of (1), i.e. such that
 is the differential of some function  , is easily seen to determine the solutions of the form
 of (1). Altogether, solving the differential equation (1) is equivalent with finding an integrating factor of the equation.
When an integrating factor  of (1) is available, the solution function  can be gotten from the line integral
along any curve  connecting an arbitrarily chosen point
 and the point
 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
 is an integrating factor.
Example. In the differential equation
we see that
 and
 both define a homogeneous function of degree 4. Thus we have the integrating factor
 , and the left hand side of the equation
is an exact differential. We can integrate it along the broken line, first from  to  and then still to  , obtaining
So the general solution of the given differential equation is
- 1
- E. LINDELÖF: Differentiali- ja integralilasku III 1. Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1935).
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![$\displaystyle f(x,\,y) := \int_{P_0}^P [\mu(x,\,y)X(x,\,y)\,dx+\mu(x,\,y)Y(x,\,y)\,dy]$](http://images.planetmath.org:8080/cache/objects/8710/l2h/img32.png "$\displaystyle f(x,\,y) := \int_{P_0}^P [\mu(x,\,y)X(x,\,y)\,dx+\mu(x,\,y)Y(x,\,y)\,dy]$")
