00:01
So we're having zero initial conditions, and we're wanting to use classical methods to find the solution.
00:08
So based on this, if we're looking at the first one, we see that the way this can be written is such that we start off with the fact that we need to find mu, which is going to be e to the integral of p of t d t, or p of t is the, what precedes the x term.
00:37
So that's 5.
00:38
So this is going to be e to the 5t.
00:42
Then we know that x of t is going to be 1 over mu, so 1 over e to the 5t times the integral of mu of t.
00:53
So that's e to the 5t times our q of t, which is 2 cosine 3t d t.
01:02
So that right there would be our x of t function.
01:05
And then we're going to have to do integration by parts to solve these in particular.
01:11
Then if we look at, say, c, we'd have a characteristic equation.
01:18
So that's going to be s squared plus 6s plus 20 equals, since we have zero initial conditions, is equal to zero.
01:29
And then with that we see that what multiplies to 20 and adds to 6...