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
d2y | |
dt2 |
=Ay2/3
admits as an integrating factor:
d2y | |
dt2 |
dy | |
dt |
=Ay2/3
dy | |
dt |
.
To integrate, note that both sides of the equation may be expressed as derivatives by going backwards with the chain rule:
d | \left( | |
dt |
1 | \left( | |
2 |
dy | |
dt |
\right)2\right)=
d | |
dt |
\left(A
3 | |
5 |
y5/3\right).
Therefore,
\left( | dy |
dt |
\right)2=
6A | |
5 |
y5/3+C0.
where
C0
This form may be more useful, depending on application. Performing a separation of variables will give
y(t) | |
\int | |
y(0) |
dy | ||||
|
This is an implicit solution which involves a nonelementary integral. This same method is used to solve the period of a simple pendulum.
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, say
M(x)
e\int
Note that it is not necessary to include the arbitrary constant in the integral, or absolute values in case the integral of
P(x)
|f(x)|=f(x)sgnf(x)
sgn
f(x)
ln|f(x)|
f(x)=0
To derive this, let
M(x)
M(x)
M(x)\underset{non-integrableexpression
M(x)y'+M(x)P(x)y
\underbrace{M(x)y'+M'(x)y}integrablederivative
Going from step 2 to step 3 requires that
M(x)P(x)=M'(x)
M(x)
P(x)
M(x)P(x)=M'(x)
P(x)=
M'(x) | |
M(x) |
\intP(x)dx=lnM(x)+c
M(x)=Ce\int
To verify, multiplying by
M(x)
M(x)y'+P(x)M(x)y=Q(x)M(x)
By applying the product rule in reverse, we see that the left-hand side can be expressed as a single derivative in
x
M(x)y'+P(x)M(x)y=M(x)y'+M'(x)y=
d | |
dx |
(M(x)y)
We use this fact to simplify our expression to
d | |
dx |
\left(M(x)y\right)=Q(x)M(x)
Integrating both sides with respect to
x
Ce\inty=\intQ(x)Ce\intdx
e\inty=\left(\intQ(x)e\intdx\right)+C
where
C
Moving the exponential to the right-hand side, the general solution to Ordinary Differential Equation is:
y=e-\int\left(\intQ(x)e\intdx\right)+Ce-
In the case of a homogeneous differential equation,
Q(x)=0
y=Ce-
for example, consider the differential equation
y'- | 2y |
x |
=0.
We can see that in this case
P(x)=
-2 | |
x |
| ||||||||||
M(x)=e |
| |||||||||||||
M(x)=e |
=e-2={\left(eln\right)}-2=x-2
M(x)= | 1 |
x2 |
.
Multiplying both sides by
M(x)
y' | |
x2 |
-
2y | |
x3 |
=0
The above equation can be rewritten as
d(x-2y) | |
dx |
=0
By integrating both sides with respect to x we obtain
x-2y=C
y=Cx2
The same result may be achieved using the following approach
y' | |
x2 |
-
2y | |
x3 |
=0
y'x3-2x2y | |
x5 |
=0
x(y'x2-2xy) | |
x5 |
=0
y'x2-2xy | |
x4 |
=0.
Reversing the quotient rule gives
\left( | y |
x2 |
\right)'=0
or
y | |
x2 |
=C,
or
y=Cx2.
where
C
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
M(x)
y'+p(x)y=h(x)
(M(x)y)'=M(x)h(x)
M(x)
y
M(x)=e\int
(M(x)y)''=M(x)\left(y''+2p(x)y'+\left(p(x)2+p'(x)\right)y\right)=M(x)h(x)
This implies that a second order equation must be exactly in the form
y''+2p(x)y'+\left(p(x)2+p'(x)\right)y=h(x)
For example, the differential equation
y''+2xy'+\left(x2+1\right)y=0
can be solved exactly with integrating factors. The appropriate
p(x)
y'
2p(x)=2x
p(x)=x
y
p(x)2+p'(x)=x2+1
e\int=
x2/2 | |
e |
x2/2 | |
e |
x2/2 | |
y''+2e |
x2/2 | |
p(x)y'+e |
\left(p(x)2+p'(x)\right)y=0
which can be rearranged to give
x2/2 | |
\left(e |
y\right)''=0
Integrating twice yields
x2/2 | |
e |
y=c1x+c2
Dividing by the integrating factor gives:
y= | c1x+c2 | |||
|
A slightly less obvious application of second order integrating factors involves the following differential equation:
y''+2\cot(x)y'-y=1
At first glance, this is clearly not in the form needed for second order integrating factors. We have a
2p(x)
y'
p(x)2+p'(x)
y
p(x)2+p'(x)=\cot2(x)-\csc2(x)
and from the Pythagorean identity relating cotangent and cosecant,
\cot2(x)-\csc2(x)=-1
so we actually do have the required term in front of
y
e\int=eln(\sin(x))=\sin(x)
Multiplying each term by
\sin(x)
\sin(x)y''+2\cot(x)\sin(x)y'-\sin(x)y=\sin(x)
which rearranged is
(\sin(x)y)''=\sin(x)
Integrating twice gives
\sin(x)y=-\sin(x)+c1x+c2
Finally, dividing by the integrating factor gives
y=c1x\csc(x)+c2\csc(x)-1
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
M(x)y
n
n
M(x)F\left(y,y',y'',\ldots,y(n)\right)
If an
n
F\left(y,y',y'',\ldots,y(n)\right)
n
h(x)M(x)
n
A third order usage of integrating factors gives
(M(x)y)'''=M(x)\left(y'''+3p(x)y''+\left(3p(x)2+3p'(x)\right)y'+\left(p(x)3+3p(x)p'(x)+p''(x)\right)y\right)
thus requiring our equation to be in the form
\left(y'''+3p(x)y''+(3p(x)2+3p'(x)\right)y'+\left(p(x)3+3p(x)p'(x)+p''(x)\right)y=h(x)
For example in the differential equation
y'''+3x2y''+\left(3x4+6x\right)y'+\left(x6+6x3+2\right)y=0
p(x)=x2
x3/3 | |
e |
x3/3 | |
\left(e |
y\right)'''=0
Integrating thrice and dividing by the integrating factor yields
y= |
| |||||||||
|