Initial Value Problem Differential Equation

[MathWarrior]

L\frac{dI}{dt}+RI=E
I(0) = I_{0}

Where E is a constant.


I know I need to separate the equation and integrate but I am not quite sure how given all the variables running around...
I don't see how the condition of I(0) = I_{0} helps in any way.

 

[genericusrnme]

Okay, you have the equation \frac{dI}{dt}=\frac{E}{L}, if L is a constant also then you should know what the solution to this differential equation is, solving for the initial condition then reduces to the problem of finding what the arbitrary constant should be!

Edit;
Whoops, just noticed that the first + isn't an equals sign :p
Still, it's just a first order ODE with constant coefficients which you should know how to solve.
If not I'm pretty sure there are 100+ videos on youtube that will teach you how!

good luck

 

[HallsofIvy]

"All the variables running around"? There are only two variables, the independent variable t and the dependent variable I. All The other letters are constants.

The "dI/dt" tells you that I is a function of t. If there were any other variables, that would have to be a partial derivative, \partial I/\partial t. But even if it were, partial derivatives are taken treating other variables as constants.

Of course, you could have a problem in which E, R, and L are functions of t but in that case you would have to be told what those functions are.

L\frac{dI}{dt}= E- R

\frac{L}{E- R}dI= dt

Integrating will give a "constant of integration" which you can use [/itex]I(0)= I_0[/itex]. Of course, your equation for I(t) will involve the constants E, L, R, and I_0.

 

[ehild]

This is a first order linear equation. First find the general solution of the homogeneous part LdI/dt+RI=0. It is of the form I=Ce^kt. Then find a particular solution of the inhomogeneous equation, and add to the general solution of the homogeneous one. The particular solution can be a constant in this case. So the solution will be of the form I=Ce^kt+B. Plug into the original equation and find the parameters k and B.

C is fitted to the initial condition: At t=0, I=Io.

ehild

 

[ehild]

Okay, you have the equation \frac{dI}{dt}=\frac{E}{L}

You omitted the term RI.
The equation is L\frac{dI}{dt}+RI=E


ehild

 

[ehild]

L\frac{dI}{dt}= E- R

The equation is L\frac{dI}{dt}= E- RI

ehild