`Ri+L(di)/(dt)=V`

`4i+2(di)/(dt)=10 sin 5t`

`4I+2(sI-i(0))=10 5/(s^2+25)`

Assume `i_0=i(0)=0`.

`4I+2sI=50/(s^2+25)`

`(2+s)I=25/(s^2+25)`

So

`I=25/((s+2)(s^2+25))` `=A/(s+2)+(Bs+C)/(s^2+25)`

`25 = A(s^2+ 25) + (Bs + C)(s + 2)`

We need to solve for A, B and C.

First, let `s = −2` and this gives

`25 = 29A`

Thus `A = 25/29`

Next, we equate coefficients of `s^2`:

`0=A+B` gives `B=-25/29`

Equating coefficients of `s`:

`0=2B+C` gives `C=50/29`

So

`I=25/((s+2)(s^2+25))`

`=25/(29(s+2))-25/29 s/(s^2+25)` `+50/29 1/(s^2+25)`

`=25/29(1/(s+2)-s/(s^2+25)+` `{:2/5 5/(s^2+25))`

So we have

`i=25/29(e^(-2t)-cos 5t+2/5sin 5t)\ "A"`

Here is the graph of `i(t)`:

1234560.20.40.60.811.2-0.2-0.4-0.6-0.8-1-1.2ti(t)Open image in a new page

Graph of `i(t)=25/29(e^(-2t)-cos 5t+2/5sin 5t)`.