# Use the given differential equation for electrical circuits given by `L dI/dt + RI = E ` In this equation, I is the current, R is the resistance, L is the inductance, and E is the electromotive force (voltage). Solve the differential equation for the current given a constant voltage E_0

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