MHB Initial value problem with lots of constants

[find_the_fun]

[math]L \frac{di}{dt}+Ri=E[/math] and we're given [math]i(0)=i_o[/math] [math]I,R,E,i_o[/math] are constants.

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

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

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?

 

[MarkFL]

Re: initial value problem with lots of constants

Your integrating factor is:

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

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

$$e^{\frac{R}{L}t}\frac{dI}{dt}+\frac{R}{L}e^{\frac{R}{L}t}I=\frac{E}{L}e^{\frac{R}{L}t}$$

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

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

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