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...
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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