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In Exercises 33 and $34,$ use the differential equation for electric circuits given by $$L \frac{d I}{d t}+R I+E$$In this equation, $I$ is the current, $R$ is the resistance, $L$ is the inductance, and $E$ is the electromotive force (voltage).Solve the differential equation for the current given a constant voltage $E_{0}$.

$i(t)=\frac{E_{0}}{R}\left(1-e^{-\frac{R t}{L}}\right)$

Calculus 2 / BC

Chapter 6

Differential Equations

Section 4

First-Order Linear Differential Equations

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Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

33:32

00:45

lectric Circuits In Exerci…

03:12

Use the differential equat…

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06:16

In Exercises $35-38,$ cons…

04:31

The current $i$ in an elec…

So we have the differential equation. Oh, t I d t plus our I equal to e c rule he here e zero is the voltage and it's constant. Are is the resistance and it's also constant. And l it's they induct er's That also is constant. So ah, we can assume that l is different from zero. Otherwise, we will not have any differential equation since this whole term will be zero. So we're gonna send that That's not happened. Then we can rewrite the differential equation House T I DT plus are over ill. I equal toe e zero over and now we will have that are over. L is going to be the function p Off T and the zero over l is going to be the function cure this case. Both are constant. So now what we compute, we can compute their integrate integrating factor That is a function you off t equal toe e to the integral off p effects. Pfft. This case that is our over l DT and this is going to be equal to a toe. There are over l t. And then the solution of the torrential a question it's going to be I off t able to one over the integrating factor toe are lt times deigned to grow off cure 50. That is zero over l times integrating factor plus some integration. Constancy. No, this is going to be ableto e to the minus are over. Lt times, remember that this value here and these will you hear are constant. So here we will get he zero over l times l over our he tow the are over. Lt. Plus here and here we see a lot of consolation this one, counselor with this one. And when we make the product off this too, they cancel out each other. So we will get zero over our times e. So they are over. Oh, sorry. These to cancel out plus c e to the minus are over lt. And that's the solution off the differential equation.

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