8 Solving equations via the integrating factor

Having found the integrating factor for a linear equation we now proceed to solve the equation.

Returning to the differential equation, written in standard form:

$\frac{di}{dt}+\frac{R}{L}i=\frac{E}{L}cos\omega t$

for which the integrating factor is

${\text{e}}^{Rt∕L}$

we multiply the equation by the integrating factor to obtain

${\text{e}}^{Rt∕L}\frac{di}{dt}+\frac{R}{L}{\text{e}}^{Rt∕L}i=\frac{E}{L}{\text{e}}^{Rt∕L}cos\omega t$

At this stage the left-hand side of this equation can always be simplified as follows:

$\frac{d}{dt}\left({\text{e}}^{Rt∕L}i\right)=\frac{E}{L}{\text{e}}^{Rt∕L}cos\omega t.$

Now this is in the form of an exact differential equation and so we can integrate both sides to obtain the solution:

${\text{e}}^{Rt∕L}i=\frac{E}{L}\int \; <!--nolimits\; -->{\text{e}}^{Rt∕L}cos\omega tdt.$

All that remains is to complete the integral on the right-hand side. Using the method of integration by parts we find

$\int \; <!--nolimits\; -->{\text{e}}^{Rt∕L}cos\omega tdt=\frac{L}{L^2{\omega }^2+R^2}\left[\omega Lsin\omega t+Rcos\omega t\right]{\text{e}}^{Rt∕L}$

Hence

${\text{e}}^{Rt∕L}i=\frac{E}{L^2{\omega }^2+R^2}\left[\omega Lsin\omega t+Rcos\omega t\right]{\text{e}}^{Rt∕L}+C.$

Finally

$i=\frac{E}{L^2{\omega }^2+R^2}\left[\omega Lsin\omega t+Rcos\omega t\right]+C{\text{e}}^{-Rt∕L}.$

is the solution to the original differential equation (1). Note that, as we should expect for the solution to a first order differential equation, it contains a single arbitrary constant $C$ .

Task!

Using the integrating factors found earlier in the Task on pages 22-23, find the general solutions to the differential equations

1. $x^2\frac{dy}{dx}+2x^{2}y=x^2{\text{e}}^{-2x}$
2. $t^2\frac{di}{dt}+2t^{2}i=t^2{\text{e}}^{-2t}$
3. $\frac{dy}{dx}-\left(tanx\right)y=1.$

Answer