Finding the integrating factor for linear ODEs

7 Finding the integrating factor for linear ODEs

The differential equation governing the current $i$ in a circuit with inductance $L$ and resistance $R$ in series subject to a constant applied electromotive force $E cos ω t$ , where $E$ and $ω$ are constants, is

$L \frac{d i}{d t} + R i = E cos ω t$ (1)

This is an example of a linear differential equation in which $i$ is the dependent variable and $t$ is the independent variable. The general standard form of a linear first order differential equation is normally written with ‘ $y$ ’ as the dependent variable and with ‘ $x$ ’ as the independent variable and arranged so that the coefficient of $\frac{d y}{d x}$ is 1. That is, it takes the form:

$\frac{d y}{d x} + f (x) y = g (x)$ (2)

in which $f (x)$ and $g (x)$ are functions of $x$ .

Comparing (1) and (2), $x$ is replaced by $t$ and $y$ by $i$ to produce $\frac{d i}{d t} + f (t) i = g (t)$ . The function $f (t)$ is the coefficient of the dependent variable in the differential equation. We shall describe the method of finding the integrating factor for (1) and then generalise it to a linear differential equation written in standard form.

Step 1 Write the differential equation in standard form i.e. with the coefficient of the derivative equal to 1. Here we need to divide through by $L$ :

$\frac{d i}{d t} + \frac{R}{L} i = \frac{E}{L} cos ω t .$

Step 2 Integrate the coefficient of the dependent variable (that is, $f (t) = R ∕ L$ ) with respect to the independent variable (that is, $t$ ), and ignoring the constant of integration

$\int \frac{R}{L} d t = \frac{R}{L} t .$

Step 3 Take the exponential of the function obtained in Step 2.

This is the integrating factor (I.F.)

$I.F. = e^{R t ∕ L} .$

This leads to the following Key Point on integrating factors:

Key Point 4

The linear differential equation (written in standard form):

\frac{d y}{d x} + f (x) y = g (x) has an integrating factor I.F. = exp [\int f (x) d x]

Task!

Find the integrating factors for the equations

$x \frac{d y}{d x} + 2 x y = x e^{- 2 x}$
$t \frac{d i}{d t} + 2 t i = t e^{- 2 t}$
$\frac{d y}{d x} - (tan x) y = 1.$

Step 1 Divide by $x$ to obtain $\frac{d y}{d x} + 2 y = e^{- 2 x}$
Step 2 The coefficient of the independent variable is 2 hence $\int 2 d x = 2 x$

Step 3 $I.F. = e^{2 x}$
The only difference from (1) is that $i$ replaces $y$ and $t$ replaces $x$ . Hence $I.F. = e^{2 t}$ .
Step 1 This is already in the standard form.
Step 2 $\int - tan x d x = \int \frac{- sin x}{cos x} d x = ln cos x .$

Step 3 $I.F. = e^{ln cos x} = cos x$

7 Finding the integrating factor for linear ODEs

Key Point 4

Task!

Answer