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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...

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

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