00:02
In this question, we want to find the charge on the capacitor in an rlc series circuit when we're given a bunch of parameters.
00:09
Now, i've already set up the differential equation in order to solve this.
00:15
It's based on the blue text above, and i simply just multiplied everything by two to get rid of the fraction in front of the second derivative term.
00:25
Now, in order to do this equation, we're going to need two parts.
00:28
We're going to need the homogeneous solution and the particular solution.
00:34
We know that it's going to be of the form.
00:37
In order to do that, we're going to consider our differential equation, d squared q d t squared plus 20 dq d t plus 200 q equals 0, which translates into the characteristic equation s squared plus 20s plus 200 equals 0.
01:07
Solving this equation by using either the quadratic formula or something, you can use any method you want, but what you'll get is you'll get the solutions, s would be equal to negative 10 plus or minus j10, where j is equal to the square root of negative 1.
01:35
Our equation would take the form of the homogeneous solution.
01:40
It would take the form c1, e to negative 10t, cosine 10, plus c2, e to negative 10 t, sine 10, t, plus c2, e to negative 10 t, t, sine of 10.
01:52
But now we would need the particular solution.
02:00
Where does one get the particular solution for this? well, let's see.
02:11
Since we have a constant, we're going to guess that our constant, that our particular solution is of the form, qp of t is equal to some constant a, then qp dot is equal to qp double dot is equal to zero.
02:54
Now we're going to plug this into our different original differential equation...