Two ways of solving this problem are shown here. You can choose whichever one makes more sense to you, or seems easiest.
We will solve this in the same way as the previous section, 2nd Order Linear DEs.
Solutions are a repeated root: `m_1=-2`, and `m_2=-2`
The response is critically damped, since the roots are equal.
`i(0)=2` implies `A=2`
So we can now write:
Using the variables given in the damping theory above:
`L=1`, `R=4`, `1/C=4`, so `C=1/4`
So `R^2=(4L)/C` and therefore we have critical damping.
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.
We need to set up the 2nd order DE with initial conditions as follows.
[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.