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Initial value problem with lots of constants


Active member
Feb 1, 2012
\(\displaystyle L \frac{di}{dt}+Ri=E\) and we're given \(\displaystyle i(0)=i_o\) \(\displaystyle I,R,E,i_o\) are constants.

So I rewrite equation as \(\displaystyle \frac{di}{dt}+\frac{R}{L}i=\frac{E}{L}\) therefore \(\displaystyle P(i)=\frac{R}{L}\)

let \(\displaystyle \mu(x)=e^{\int \frac{R}{L}dt}=e^{\frac{tr}{L}+C}\)
multiply equation by integrating factor to get
\(\displaystyle e^{\frac{tR}{L}} \frac{di}{dt}+e^{\frac{tr}{L}} \frac{Ri}{L}=e^{\frac{tr}{L}}\frac{E}{L}\)
\(\displaystyle \frac{d}{dt}[\mu(x)i]=e^{\frac{tR}{L}}\frac{E}{L}i\)

I think I've done something wrong because the above statement is not true. Also, every question I've seen the e^some-integral involves ln so the e's go away. Is this always the case?


Staff member
Feb 24, 2012
Re: initial value problem with lots of constants

Your integrating factor is:

\(\displaystyle \mu(t)=e^{\frac{R}{L}\int\,dt}=e^{\frac{R}{L}t}\)

You don't need a constant of integration, since it would be divided out anyway. So, your ODE becomes:

\(\displaystyle e^{\frac{R}{L}t}\frac{dI}{dt}+\frac{R}{L}e^{\frac{R}{L}t}I=\frac{E}{L}e^{\frac{R}{L}t}\)

Note: I have changed $i$ to $I$ so as to not confuse current with the imaginary constant.

\(\displaystyle \frac{d}{dt}\left(Ie^{\frac{R}{L}t} \right)=\frac{E}{L}e^{\frac{R}{L}t}\)

Now, integrate with respect to $t$, then use the initial value to determine the parameter (constant of integration).