The current in a circuit at time `t \ s` is given by the solution of the Differential Equation `(dI)/dt+4I=20`. Find the solution given that `I=2 \ A` when `t=0`, and find the time taken to reach half the steady state solution?

We have:

`(dI)/dt+4I=20 \ \ \ \ ...... [1]`

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

`dy/dx + P(x)y=Q(x)`

Then the integrating factor is given by;

`IF = e^(int P(x) dx)`
`" " = exp(int \ 4 \ dt)`
`" " = exp( 4t )`
`" " = e^(4t)`

And if we multiply the DE [1] by this Integrating Factor, `IF`, we will have a perfect product differential;

`(dI)/dt+4I=20`

`:. e^(4t)(dI)/dt+4I e^(4t)=20e^(4t)`

`:. d/dt {e^(4t) I} = 20e^(4t)`

Which we can directly integrate to get:

`int \ d/dt {e^(4t) I} \ dt = int \ 20e^(4t) \ dt`
`:. e^(4t) I = int \ 20e^(4t) \ dt`
`:. e^(4t) I = 5e^(4t) + C`
`:. I = 5 + Ce^(-4t)`

Applying the initial condition, `I(0)=2`, we get:

`2 = 5 + Ce^(0) => C=-3`

Thus the solution is:

`I(t) = 5-3e^(-4t)`

For the steady state solution we look at:

`lim_(t rarr oo) I(t) = lim_(t rarr oo) {5-3e^(-4t)}`
`" " = lim_(t rarr oo) 5-3lim_(t rarr oo)e^(-4t)`
`" " = 5-3lim_(t rarr oo)e^(-4t)`
`" " = 5`

So to find the time taken to reach half the steady state solution, we require the value of `t` such that:

`I(t)=5/2 => 5-3e^(-4t) = 5/2`

`:. 3e^(-4t) = 5/2`
`:. e^(-4t) = 5/6`
`:. -4t = ln(5/6)`

`:. t = -1/4ln(5/6)`
`" " = -1/4ln(5/6)`
`" " = 0.045580 ...`
`" " = 0.046` (2sf)