Integrating factor

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.

Use in solving first order linear ordinary differential equations

Consider an ordinary differential equation of the form

$y'+P(x)y=Q(x)\quad \quad \quad (1)$

Consider a function $M(x)$. We multiply both sides of (1) by $M(x):$

$M(x)y'+M(x)P(x)y=M(x)Q(x).\quad \quad \quad (2)$

We want the left hand side to be in the form of the derivative of a product (see product rule), such that (2) can be written as

$(M(x)y)'=M(x)Q(x).\quad \quad \quad (3)$

The left hand side in (3) can now be integrated

$yM(x)=\int Q(x)M(x)\,dx,$

We can now solve for $y,$

$y={\frac {\int Q(x)M(x)\,dx}{M(x)}}.\,$

Applying the product rule to the left hand side of (3) and equating to the left hand side of (2)

$M'(x)y+M(x)y'=M(x)y'+M(x)P(x)y.\quad \quad \quad$

From which it is clear that $M(x)$ obeys the differential equation :

$M'(x)=M(x)P(x).\quad \quad \quad (4)\,$

${\frac {M'(x)}{M(x)}}=P(x).\quad \quad \quad (5)$

Keeping in mind that for:

$F(x)=e^{G(x)}.$

$F'(x)=G'(x)*e^{G(x)}.$

$F'(x)=G'(x)*F(x).$

We can see that when we let:

$P(x)=G'(x)$

We can solve (5):

$\int P(x)\,dx=G(x)$

$e^{\int P(x)\,dx}=e^{G(x)}$

$e^{\int P(x)\,dx}=M(x)$

$M(x)$ is called an integrating factor.

Example

Solve the differential equation

$y'-{\frac {2y}{x}}=0.$

We can see that in this case $P(x)={\frac {-2}{x}}$

$M(x)=e^{\int P(x)\,dx}$

$M(x)=e^{\int {\frac {-2}{x}}\,dx}=e^{-2\ln x}={(e^{\ln x})}^{-2}=x^{-2}$ (Note we do not need to include the integrating constant - we need only a solution, not the general solution)

$M(x)={\frac {1}{x^{2}}}.$

Multiplying both sides by $M(x)$ we obtain

${\frac {y'}{x^{2}}}-{\frac {2y}{x^{3}}}=0$

${\frac {y'x^{3}-2x^{2}y}{x^{5}}}=0$

${\frac {x(y'x^{2}-2xy)}{x^{5}}}=0$

${\frac {y'x^{2}-2xy}{x^{4}}}=0.$

Reversing the quotient rule gives

$\left({\frac {y}{x^{2}}}\right)'=0$

or

${\frac {y}{x^{2}}}=C\,$

which gives

$y\left(x\right)=Cx^{2}.$

General use

The term "integrating factor" is synonymous with the solution of first order linear equations. One should bear in mind, however, that an integrating factor is any expression that a differential equation is multiplied by to facilitate integration, and that it is by no means restricted to first order linear equations. For example, the nonlinear second order equation

${\frac {d^{2}y}{dt^{2}}}=Ay^{2/3}$

admits ${\tfrac {dy}{dt}}$ as an integrating factor:

${\frac {d^{2}y}{dt^{2}}}{\frac {dy}{dt}}=Ay^{2/3}{\frac {dy}{dt}}.$

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

${\frac {d}{dt}}\left({\frac {1}{2}}\left({\frac {dy}{dt}}\right)^{2}\right)={\frac {d}{dt}}\left(A{\frac {3}{5}}y^{5/3}\right).$

Therefore

$\left({\frac {dy}{dt}}\right)^{2}={\frac {6A}{5}}y^{5/3}+C_{0}.$

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

$\int {\frac {dy}{\sqrt {{\frac {6A}{5}}y^{5/3}+C_{0}}}}=t+C_{1};$

this is an implicit solution which involves a nonelementary integral. Though likely too obscure to be useful, this is a general solution. Also, because the previous equation is first order, it could be used for numeric solution in favor of the original equation.

References

• Munkhammar, Joakim, "Integrating Factor", MathWorld {{citation}}: Unknown parameter |link= ignored (help).

See also

• Method of variation of parameters
• Examples of differential equations
• Product rule
• Exact differential