00:01
All right, let's have some fun with some differential equation type problems.
00:06
The first has to do with a differential equation where we're told that the solution can be found by this particular taylor series.
00:17
Notice that the taylor series is about x equals one.
00:22
All right, so then we look at our differential equation.
00:25
We can divide everything by the coefficient of y double prime.
00:30
So that means we'll divide by x minus three.
00:34
And we're looking for singular points, basically bad x's that make the denominator go to zero.
00:39
We see right away that we don't want x to be three.
00:42
That's a bad point for us.
00:44
So to find that lower bound at the radius of convergence, we just find the distance between where we're doing the taylor and our singular point.
00:53
So we get absolute value of negative two, which is two.
00:58
So therefore, for our radius of convergence is two.
01:01
That's our lower bound of the radius of convergence.
01:04
All right, excellent.
01:06
So let me clear the screen and we will do the next part.
01:12
All right, so here we have a function with step function parts and our goal is to find the laplace transform.
01:21
So basically find the laplace transform.
01:31
All right, first thing we notice is that we can combine.
01:34
We have a couple terms with our sempleast transform.
01:36
Function that starts at t equals one.
01:39
So let's go ahead and just do some combining.
01:42
So that will give us t plus you step function starting at one.
01:49
And let's see what we have.
01:50
We have minus t and we have a two minus t.
01:54
So that becomes t plus our step function at um let's see a minus two t plus.
02:05
2 minus 2 t plus 2.
02:07
I can go ahead and factor out a 2.
02:11
So that will give me step function.
02:13
And what's left behind then is i'm actually going to factor out a minus 2 because i want it nice and clean inside.
02:19
So if i factor out a minus 2, then i'll get t minus 1.
02:25
Excellent.
02:25
Okay.
02:26
So that is perfect because i have the right form now where i can do the low loss transform.
02:33
So that worked out very nicely.
02:35
Okay, perfect.
02:36
Okay, so now we are going to do the loeplaus transform.
02:39
So f of s.
02:41
The laplace transform of t is just one over s squared.
02:48
And there's laplace tables where you can look at and you can see how to do this.
02:53
The trick is now with the step function, so i have minus two.
02:57
And basically with the step function, i'm going to get a term e to the s.
03:05
That's actually sorry, e to the minus s.
03:08
So this term is e to the minus s...