Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 33

Given ,
L (dI)/(dt) + RI = E
=> L I' + RI = E
now dividing with L on both sides we get
=> (LI')/L +(R/L)I=(E/L)
=>I' +(R/L)I=(E/L)-----(1)
which is a linear differential equation of first order
Solve the differential equation for the current given a constant voltage E_0,
so E = E_0.
So , Re-writing the equation (1) as,
(1) => I' +(R/L)I=(E_0/L) -----(2)
On comparing the above equation with the general linear differential equation we get as follows
y' +py=q ---- (3) -is the general linear differential equation form.
so on comparing the equations (2) and (3)
we get,
p= (R/L) and q= (E_0/L)
so , now
let us find the integrating factor (I.F)= e^(int p dt)

so now ,I.F = e^(int (R/L) dt)
= e^((R/L)int (1) dt)
= e^((R/L)(t)) =e^(((Rt)/L))
So , now the general solution for linear differential equation is
I * (I.F) = int (I.F) q dt +c
=>I*(e^(((Rt)/L))) = int (e^(((Rt)/L))) (E_0/L) dt +c
=>Ie^((Rt)/L) = E_0/L int e^((Rt)/L) dt +c -----(4)
Now let us evaluate the part
int e^((Rt)/L) dt
this is of the form
int e^(at) dt and so we know it is equal to
= (e^(at))/a
so , now ,
int e^((Rt)/L) dt
where a= R/L
so ,
int e^((Rt)/L) dt = e^((Rt)/L)/(R/L)
now substituting in the equation (4) we get ,
I*(e^(((Rt)/L))) = (E_0/L)(e^(((Rt)/L)))/(R/L) +c
I = ((E_0/L)(e^(((Rt)/L)))/(R/L)+c) /((e^(((Rt)/L))))
I = ((E_0/L)(e^(((Rt)/L)))/(R/L)) /((e^(((Rt)/L))))+c((e^(((-Rt)/L))))
upon cancelling L and e^((Rt)/L) , we get
= E_0/R +c((e^(((-Rt)/L))))
so ,
I = E_0/R +ce^((-Rt)/L) is the solution