Two ways of solving this problem are shown here. You can choose whichever one makes more sense to you, or seems easiest.

#### Solution: Alternative 1

We will solve this in the same way as the previous section, 2nd Order Linear DEs.

A.E. `m^2+4m+4`

Solutions are a repeated root: `m_1=-2`, and `m_2=-2`

The response is critically damped, since the roots are equal.

`i(t)=(A+Bt)e^(-2t)`

`i(0)=2` implies `A=2`

So we can now write:

`i(t)=(2+Bt)e^(-2t)`

Differentiating gives:

`(di)/(dt)=(2+Bt)(-2e^(-2t))+Be^(-2t)`

`[(di)/(dt)]_(t=0)=4 `

This implies:

`(2)(-2)+B(1)=4 `

So `B=8`

Therefore `i(t)=(2+8t)e^(-2t)`

#### Solution: Alternative 2

Using the variables given in the damping theory above:

`L=1`, `R=4`, `1/C=4`, so `C=1/4`

`R^2=16`

`(4L)/C=4/(1"/"4)=16`

So `R^2=(4L)/C` and therefore we have critical damping.

Now `m_1=m_2=-R/(2L)=-4/2=-2`

The general solution is given by `i(t)=(A+Bt)e^(-Rt"/"2L`

So `i(t)=(A+Bt)e^(-4t"/"(2xx1))` `=(A+Bt)e^(-2t)`

This is the same solution we have using Alternative 1. The rest of the solution (finding *A* and *B*) will be identical.

#### Solution using Scientific Notebook

We need to set up the 2nd order DE with initial conditions as follows.

`(d^2i)/(dt^2)+4(di)/(dt)+4i=0`

`i(0)=2`

`i^'(0)=4`

[Then go to **Compute** menu, **Solve ODE...**, **Exact**]

The graph of our solution is:

Graph of `i(t)=(2+8t)e^(-2t)`, a critcally damped case.