We use the following equation from before:

`Ri+L(di)/(dt)+1/Cinti\ dt=E`

and obtain:

`20i+0.05(di)/(dt)+1/10^(-4)inti\ dt` `=100 cos 200t`

After multiplying throughout by 20, we have:

`400i+(di)/(dt)+20xx10^4intidt` `=2000\ cos\ 200t`

Taking Laplace transform and using the fact that i(0) = 0:

`{:(400I+sI-i(0)+200000I/s=2000 s/(s^2+200^2)),(400sI+s^2I+200000I=2000 s^2/(s^2+200^2)),((s^2+400s+200000)I=2000 s^2/(s^2+200^2)),(I=2000 s^2/((s^2+400s+200000)(s^2+200^2))):}`

Using Scientific Notebook to find the partial fractions:

`2000 s^2/((s^2+400s+200000)(s^2+200^2))`

`=(2000-s)/(s^2+400s+200000)+(-400+s)/(s^2+40000)`

So

`I=-(s-2000)/(s^2+400s+200000)+(s-400)/(s^2+200^2)`

`=-(s-2000)/(s^2+400s+40000+160000)+(s-400)/(s^2+200^2)`

`=-(s-2000)/((s+200)^2+400^2)+(s-400)/(s^2+200^2)`

`=(s+200)/((s+200)^2+400^2)+2200/((s+200)^2+400^2)` `+s/(s^2+200^2)-400/(s^2+200^2)`

`=-(s+200)/((s+200)^2+400^2)` `+2200/400 400/((s+200)^2+400^2)` `+s/(s^2+200^2)-2 200/(s^2+200^2)`

So

`i=-e^(-200t)cos 400t` `+11/2e^(-200t)sin 400t` `+cos\ 200t` `- 2\ sin\ 200t`

NOTE: Scientific Notebook can do all this for us very easily. In one step, we have:

`ccL^(-1){2000 s^2/((s^2+400s+200000)(s^2+200^2))}`

`=-e^(-200t)cos\ 400t+11/2e^(-200t)sin\ 400t+cos\ 200t- 2\ sin\ 200t`

Transient part: `-e^(-200t)cos\ 400t+11/2e^(-200t)sin\ 400t`

Steady state part: `cos\ 200t-2\ sin\ 200t`