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 $Ecos\omega t$ , where $E$ and $\omega $ are constants, is
$\phantom{\rule{2em}{0ex}}L\phantom{\rule{1em}{0ex}}\frac{di}{dt}+Ri=Ecos\omega t$ (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 $\frac{dy}{dx}$ is 1. That is, it takes the form:
$\phantom{\rule{2em}{0ex}}\frac{dy}{dx}+f\left(x\right)\phantom{\rule{1em}{0ex}}y=g\left(x\right)$ (2)
in which $f\left(x\right)$ and $g\left(x\right)$ are functions of $x$ .
Comparing (1) and (2), $x$ is replaced by $t$ and $y$ by $i$ to produce $\frac{di}{dt}+f\left(t\right)\phantom{\rule{1em}{0ex}}i=g\left(t\right)$ . The function $f\left(t\right)$ 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 $L$ :
$\phantom{\rule{2em}{0ex}}\frac{di}{dt}+\frac{R}{L}i=\frac{E}{L}cos\omega t.$
Step 2 Integrate the coefficient of the dependent variable (that is, $f\left(t\right)=R\u2215L$ ) with respect to the independent variable (that is, $t$ ), and ignoring the constant of integration
$\phantom{\rule{2em}{0ex}}\int \frac{R}{L}\phantom{\rule{1em}{0ex}}dt=\frac{R}{L}\phantom{\rule{1em}{0ex}}t.$
Step 3 Take the exponential of the function obtained in Step 2.
This is the integrating factor (I.F.)
$\phantom{\rule{2em}{0ex}}\text{I.F.}={\text{e}}^{Rt\u2215L}.$
This leads to the following Key Point on integrating factors:
Key Point 4
The linear differential equation (written in standard form):
Task!
Find the integrating factors for the equations
 $x\phantom{\rule{1em}{0ex}}\frac{dy}{dx}+2x\phantom{\rule{1em}{0ex}}y=x{\text{e}}^{2x}$
 $t\phantom{\rule{1em}{0ex}}\frac{di}{dt}+2t\phantom{\rule{0.3em}{0ex}}i=t{\text{e}}^{2t}$
 $\frac{dy}{dx}\left(tanx\right)y=1.$

Step 1
Divide by
$x$
to obtain
$\frac{dy}{dx}+2y={\text{e}}^{2x}$
Step 2 The coefficient of the independent variable is 2 hence $\int 2\phantom{\rule{1em}{0ex}}dx=2x$
Step 3 $\text{I.F.}={\text{e}}^{2x}$
 The only difference from (1) is that $i$ replaces $y$ and $t$ replaces $x$ . Hence $\text{I.F.}={\text{e}}^{2t}$ .

Step 1
This is already in the standard form.
Step 2 $\int tanx\phantom{\rule{1em}{0ex}}dx=\int \frac{sinx}{cosx}\phantom{\rule{1em}{0ex}}dx=lncosx.$
Step 3 $\text{I.F.}={\text{e}}^{lncosx}=cosx$