# Solve the general initial value problem modeling the LR circuit, frac{dl}{dt}+ RI=E, I(0)= I_{0}, where E is a constant source of emf.

Solve the general initial value problem modeling the LR circuit, $\frac{dl}{dt}+RI=E,I\left(0\right)={I}_{0}$, where E is a constant source of emf.
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Corben Pittman

Step 1

Let our equation be $L{I}^{\prime }\left(t\right)+RI\left(t\right)=E\dots \left(\cdot \right)$ where are I - moving charge L - inductor R - resistor E - const First, divide both sides of equation with L

$LI{\left(t\right)}^{\prime }+RI\left(t\right)=\frac{E}{:}L$
${I}^{\prime }\left(t\right)+\frac{R}{L}I\left(t\right)=\frac{E}{L}$

We have first order linear differential equation. To solve her, we must find integration factor $\mu \left(t\right)$. First, let's define function a(t) $a\left(t\right)=\frac{R}{L}$

We will get integration factor using next formula

$\mu \left(t\right)={e}^{\int a\left(t\right)dt}={e}^{\int \frac{R}{L}dt}={e}^{\frac{R}{L}t}$

Now, multiply both sides of our equation with integration factor

${I}^{\prime }\left(t\right)+\frac{R}{L}I\left(t\right)=\frac{\frac{E}{L}}{{e}^{\frac{R}{L}t}}$
${e}^{\frac{R}{L}t}{I}^{\prime }\left(t\right)+{e}^{\frac{R}{L}t}\frac{1}{RC}I\left(t\right)={e}^{\frac{R}{L}t}\frac{E}{L}$

Step 2

${e}^{\frac{R}{L}t}{I}^{\prime }\left(t\right)+{e}^{\frac{R}{L}t}\frac{R}{L}I\left(t\right)={e}^{\frac{R}{L}t}\frac{R}{L}$
$=\frac{d}{dt}\left({e}^{\frac{R}{L}t}I\left(t\right)\right)$
$\frac{d}{dt}\left({e}^{\frac{R}{L}t}I\left(t\right)\right)={e}^{\frac{R}{L}t}\frac{E}{L}$

Integrate both sides of equation

$\int \frac{d}{dt}\left({e}^{\frac{R}{L}t}I\left(t\right)\right)dt=\int {e}^{\frac{R}{L}t}\frac{E}{L}dt$