First order differential equation, please help?

Is it really so difficult to type in "V_s= Ri_1+ L di_1/dt+ E" that you need to take a picture of the page, upload it, and require others to open an unknown website in order to help you?

That is a separable first order differential equation. It can be written as L di_1/dt= V_s- E- Ri_1. That can be separate into (L/(V_s- E- Ri_1))di_1= dt and you can integrate both sides.
To integrate the left, let u= V_s- E- Ri_1 so that du= -R di_1, di_1= (1/R)du and the equation becomes (L/R)(1/u)du= dt. Can you integrate that?
 
Here is a portion of the text that the original poster wants us to go find:

The load current for mode 1 can be found from

. . . . .\(\displaystyle V_s\, =\, Ri_1\, +\, L\dfrac{di_1}{dt}\, +\, E\)

which with the initial load current i1 (t = 0) = I1 gives the load current as

. . . . .\(\displaystyle i_1\, =\, I_1 e^{\left(-tR\,/\,L\right)}\, +\, \dfrac{V_s\, -\, E}{R}\, \left(1\, -\, e^{\left(-tR\,/\,L\right)}\right)\)


The poster's work is displayed in a small fuzzy graphic. I think the poster's steps are as follows:

\(\displaystyle V_s\, =\, R_{i_1}\, +\, L\, \dfrac{di_1}{dt}\, +\, E\)

In laplace

\(\displaystyle V_s\, =\, RI(S)\, +\, E\, + \, L\left[SI(S)\, -\, I_1\right]\)

\(\displaystyle V_s\, =\, RI(S)\, +\, LSI(S)\, +\, E\, -\, -\, LI_1\)

\(\displaystyle I(S)\left[R\, +\, LS\right]\, =\, V_S\, -\, E\, +\, LI_1\)

\(\displaystyle LIS\, \left[S\, +\, \dfrac{R}{L}\right]\)

\(\displaystyle LIS\, =\, \dfrac{V_S\, -\, E}{\left(S\, +\, \dfrac{R}{L}\right)}\, +\, \dfrac{LI_1}{\left(S\, +\, \dfrac{R}{L}\right)}\)

\(\displaystyle I(S)\, =\, \dfrac{V_S\, -\, E}{L \left(S\, +\, \dfrac{R}{L}\right)}\, +\, \dfrac{LI_1}{L \left(S\, +\, \dfrac{R}{L}\right)}\)

\(\displaystyle i(t)\, =\, \left(\dfrac{V_S\, -\, E}{L}\right)\, e^{(-R/L)t}\)
 
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