Integrating factor

Use in solving first order linear ordinary differential equations

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

y ′ + P ( x ) y = Q ( x )

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

M ( x ) = e ∫ s 0 x P ( s ) d s

In order to derive this, let M ( x ) be the integrating factor of a first order, linear differential equation such that multiplication by M ( x ) transforms a partial derivative into a total derivative, then:

( 1 ) M ( x ) ( y ′ + P ( x ) y ⏟ ) partial derivative ( 2 ) M ( x ) y ′ + M ( x ) P ( x ) y ( 3 ) M ( x ) y ′ + M ′ ( x ) y ⏟ total derivative

Going from step 2 to step 3 requires that M ( x ) P ( x ) = M ′ ( x ) , which is a separable differential equation, whose solution yields M ( x ) in terms of P ( x ) :

( 4 ) M ( x ) P ( x ) = M ′ ( x ) ( 5 ) P ( x ) = M ′ ( x ) M ( x ) ( 6 ) ∫ s 0 x P ( s ) d s = ln ⁡ M ( x ) ( 7 ) e ∫ s 0 x P ( s ) d s = M ( x )

To verify see that multiplying through by M ( x ) gives

y ′ e ∫ s 0 x P ( s ) d s + P ( x ) y e ∫ s 0 x P ( s ) d s = Q ( x ) e ∫ s 0 x P ( s ) d s

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

y ′ e ∫ s 0 x P ( s ) d s + P ( x ) y e ∫ s 0 x P ( s ) d s = d d x ( y e ∫ s 0 x P ( s ) d s )

We use this fact to simplify our expression to

d d x ( y e ∫ s 0 x P ( s ) d s ) = Q ( x ) e ∫ s 0 x P ( s ) d s

We then integrate both sides with respect to x , firstly by renaming x to t , obtaining

y e ∫ s 0 x P ( s ) d s = ∫ t 0 x Q ( t ) e ∫ s 0 t P ( s ) d s d t + C

Finally, we can move the exponential to the right-hand side to find a general solution to our ODE:

y = e − ∫ s 0 x P ( s ) d s ∫ t 0 x Q ( t ) e ∫ s 0 t P ( s ) d s d t + C e − ∫ s 0 x P ( s ) d s

In the case of a homogeneous differential equation, in which Q ( x ) = 0 , we find that

y = C e ∫ s 0 x P ( s ) d s

where C is a constant.

Example

Solve the differential equation

y ′ − 2 y x = 0.

We can see that in this case P ( x ) = − 2 x

M ( x ) = e ∫ 1 x P ( s ) d s (Note we do not need to keep a general s 0 - we need only a solution, not the general solution) M ( x ) = e ∫ 1 x − 2 s d s = e − 2 ln ⁡ x = ( e ln ⁡ x ) − 2 = x − 2 M ( x ) = 1 x 2 .

Multiplying both sides by M ( x ) we obtain

y ′ x 2 − 2 y x 3 = 0 y ′ x 3 − 2 x 2 y x 5 = 0 x ( y ′ x 2 − 2 x y ) x 5 = 0 y ′ x 2 − 2 x y x 4 = 0.

Reversing the quotient rule gives

( y x 2 ) ′ = 0

or

y x 2 = C

which gives

y ( x ) = C x 2 .

General use

An integrating factor is any expression that a differential equation is multiplied by to facilitate integration and is not restricted to first order linear equations. For example, the nonlinear second order equation

d 2 y d t 2 = A y 2 / 3

admits d y d t as an integrating factor:

d 2 y d t 2 d y d t = A y 2 / 3 d y d t .

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

d d t ( 1 2 ( d y d t ) 2 ) = d d t ( A 3 5 y 5 / 3 ) .

Therefore

( d y d t ) 2 = 6 A 5 y 5 / 3 + C 0 .

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

∫ y ( 0 ) y ( t ) d y 6 A 5 y 5 / 3 + C 0 = t

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