Integrating factor

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In mathematics, an integrating factor is a function that is chosen to facilitate the solving of a given ordinary differential equation.

1 Technique
2 Example
3 General use
4 See also

Technique

Consider an ordinary differential equation of the form

$y'+a(x)y = b(x)\quad\quad\quad (1)$

where $y = y (x)$ is an unknown function of $x$ , and $a (x)$ and $b (x)$ are given functions.

The integrating factor method works by turning the left hand side into the form of the derivative of a product.

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

$M(x)y' + M(x)a(x)y = M(x)b(x).\quad\quad\quad (2)$

We want the left hand side to be in the form of the derivative of a product (see product rule). In fact, if we assume this the left hand side can be rearranged as

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

The left hand side in (3) can now be integrated much more easily by means of the fundamental theorem of calculus,

$y(x) M(x) = \int b(x) M(x)\,dx + C,$

where $C$ is a constant (see arbitrary constant of integration). We can now solve for $y (x),$

$y(x) = \frac{\int b(x) M(x)\, dx + C}{M(x)}.\,$

However, to explicitly solve for $y (x)$ we need to find an expression for $M (x).$

Rewrite (3) using the product rule.

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

Identify terms in (2) and it's clear that $M (x)$ obeys the differential equation :

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

To get $M (x)$ , divide both sides by $M (x):$

$\frac{M'(x)}{M(x)}-a(x) = 0.\quad\quad\quad (5)$

Equation (5) is now in the form of a logarithmic derivative. Solving (5) gives

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

We see that multiplying by $M (x)$ and the property $M'(x) = a (x) M (x)$ were essential in solving this differential equation. $M (x)$ is called an integrating factor.

Example

Solve the differential equation

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

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

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

$M(x)=e^{\int \frac{-2}{x}\,dx} = e^{-2 \ln x} = {(e^{\ln x})}^{-2} = x^{-2}$ (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

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

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

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

which gives

y (x) = C x 2 .

General use

The term "integrating factor" is almost universally bound to 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

$\frac{d^2 y}{d t^2} = A y^{2/3}$

admits $\tfrac{d y}{d t}$ as an integrating factor:

$\frac{d^2 y}{d t^2} \frac{d y}{d t} = A y^{2/3} \frac{d y}{d t}.$

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

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

Therefore

$\left(\frac{d y}{d t}\right)^2 = \frac{6 A}{5} y^{5/3} + C_0$

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

$\int \frac{d y}{\sqrt{\frac{6 A}{5} y^{5/3} + C_0}} = t + C_1;$

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

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