# Initial value problem with lots of constants

#### find_the_fun

##### Active member
$$\displaystyle L \frac{di}{dt}+Ri=E$$ 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?

#### MarkFL

Staff member
Re: initial value problem with lots of constants

$$\displaystyle \mu(t)=e^{\frac{R}{L}\int\,dt}=e^{\frac{R}{L}t}$$
$$\displaystyle e^{\frac{R}{L}t}\frac{dI}{dt}+\frac{R}{L}e^{\frac{R}{L}t}I=\frac{E}{L}e^{\frac{R}{L}t}$$
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).