In mathematics, an integrating factor is a function that is chosen to facilitate the solving of a given equation involving differentials. It is commonly used to solve ordinary differential equations, but is also used within multivariable calculus when multiplying through by an integrating factor allows an inexact differential to be made into an exact differential (which can then be integrated to give a scalar field). This is especially useful in thermodynamics where temperature becomes the integrating factor that makes entropy an exact differential.

An integrating factor is any expression that a differential equation is multiplied by to facilitate integration. For example, the nonlinear second order equation

 

admits   as an integrating factor:

 

To integrate, note that both sides of the equation may be expressed as derivatives by going backwards with the chain rule:

 

Therefore,

 

where   is a constant.

This form may be more useful, depending on application. Performing a separation of variables will give

 

This is an implicit solution which involves a nonelementary integral. This same method is used to solve the period of a simple pendulum.

Solving first order linear ordinary differential equations

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Integrating factors are useful for solving ordinary differential equations that can be expressed in the form

 

The basic idea is to find some function, say  , called the "integrating factor", which we can multiply through our differential equation in order to bring the left-hand side under a common derivative. For the canonical first-order linear differential equation shown above, the integrating factor is  .

Note that it is not necessary to include the arbitrary constant in the integral, or absolute values in case the integral of   involves a logarithm. Firstly, we only need one integrating factor to solve the equation, not all possible ones; secondly, such constants and absolute values will cancel out even if included. For absolute values, this can be seen by writing  , where   refers to the sign function, which will be constant on an interval if   is continuous. As   is undefined when  , and a logarithm in the antiderivative only appears when the original function involved a logarithm or a reciprocal (neither of which are defined for 0), such an interval will be the interval of validity of our solution.

To derive this, let   be the integrating factor of a first order linear differential equation such that multiplication by   transforms a non-integrable expression into an integrable derivative, then:

  1.  
  2.  
  3.  

Going from step 2 to step 3 requires that  , which is a separable differential equation, whose solution yields   in terms of  :

  1.  
  2.  
  3.  
  4.  

To verify, multiplying by   gives

 

By applying the product rule in reverse, we see that the left-hand side can be expressed as a single derivative in  

 

We use this fact to simplify our expression to

 

Integrating both sides with respect to  

 
 

where   is a constant.

Moving the exponential to the right-hand side, the general solution to Ordinary Differential Equation is:

 

In the case of a homogeneous differential equation,   and the general solution to Ordinary Differential Equation is:

 .

for example, consider the differential equation

 

We can see that in this case  

 
 
 

Multiplying both sides by   we obtain

 

The above equation can be rewritten as

 

By integrating both sides with respect to x we obtain

 

or

 

The same result may be achieved using the following approach

 
 
 
 

Reversing the quotient rule gives

 

or

 

or

 

where   is a constant.

Solving second order linear ordinary differential equations

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The method of integrating factors for first order equations can be naturally extended to second order equations as well. The main goal in solving first order equations was to find an integrating factor   such that multiplying   by it would yield  , after which subsequent integration and division by   would yield  . For second order linear differential equations, if we want   to work as an integrating factor, then

 

This implies that a second order equation must be exactly in the form   for the integrating factor to be usable.

Example 1

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For example, the differential equation

 

can be solved exactly with integrating factors. The appropriate  can be deduced by examining the   term. In this case,  , so  . After examining the   term, we see that we do in fact have  , so we will multiply all terms by the integrating factor  . This gives us

 

which can be rearranged to give

 

Integrating twice yields

 

Dividing by the integrating factor gives:

 

Example 2

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A slightly less obvious application of second order integrating factors involves the following differential equation:

 

At first glance, this is clearly not in the form needed for second order integrating factors. We have a   term in front of   but no   in front of  . However,

 

and from the Pythagorean identity relating cotangent and cosecant,

 

so we actually do have the required term in front of   and can use integrating factors.

 

Multiplying each term by   gives

 

which rearranged is

 

Integrating twice gives

 

Finally, dividing by the integrating factor gives

 

Solving nth order linear differential equations

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Integrating factors can be extended to any order, though the form of the equation needed to apply them gets more and more specific as order increases, making them less useful for orders 3 and above. The general idea is to differentiate the function     times for an  th order differential equation and combine like terms. This will yield an equation in the form

 

If an  th order equation matches the form   that is gotten after differentiating   times, one can multiply all terms by the integrating factor and integrate     times, dividing by the integrating factor on both sides to achieve the final result.

Example

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A third order usage of integrating factors gives

 

thus requiring our equation to be in the form

 

For example in the differential equation

 

we have  , so our integrating factor is  . Rearranging gives

 

Integrating thrice and dividing by the integrating factor yields

 

See also

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References

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  • Munkhammar, Joakim, "Integrating Factor", MathWorld.