In this case, `R^2 = 10^2 = 100` and `(4L)/C = (4xx1)/0.0025 = 1600`, so `R^2<(4L)/C`.

We have an example of **Case 3, Under-damped**, from Section 8.

Once again, using `L(di)/(dt)+Ri+1/Cq=E` we can write:

`(di)/(dt)+10i+400q=E\ \ \ (1)`

where `E` is a constant.

Differentiating gives:

`(d^2i)/(dt^2)+10(di)/(dt)+400i=0`

We are given:

`i(0)=0`

`i'(0)=0.1`

The auxiliary equation this time is: `m^2+10m+400=0`

Solution is: `m_1=-5+19.365j,` `\ m_2=-5-19.365j`

So since it is the under-damped case, we have

`i=e^(-5t)(A cos 19.365 t+` `{:B sin 19.365t)`

Since `i(0)=0`, we substitute and get:

`0=A`

So

`i=e^(-5t)(B sin 19.365t)\ \ \ (2)`

Differentiating (2) gives:

`(di)/(dt)=e^(-5t)(19.365B cos 19.365t) +` `{: (B sin 19.365t)(-5e^(-5t))

` =e^(-5t)(19.365B cos 19.365t -` `{:5B sin 19.365t)

Since `i'(0)=0.1`, we substitute and obtain:

`0.1=(19.365B)`

So

`B=0.1/19.365 = 0.00516`

So we conclude:

`i=0.00516 e^(-5t)(sin 19.365t)`

Here is the graph of the solution for the natural response case:

Graph of `i(t)=0.00516 e^{-5.0t} sin 19.36\ t`.

Notice the signal decays quickly.