00:01
Here i'm going to start by looking at the method of variation of parameters, which is a way to find a particular solution of a second -order differential equation with non -constant functions in front of some of the parts as is shown.
00:20
There's a p of x and a q of x.
00:22
And the particular solution involves finding the solution if there is an f -of -x driving the right -hand.
00:30
Side.
00:31
So this is the non -homogeneous case.
00:35
And it seems a little convoluted.
00:38
We're going to use this method for a very simple system that we could use much more basic tools, but just to demonstrate that it works.
00:47
So in the method of variation of parameters, the first thing you do is you find two independent solutions of the homogeneous case.
01:00
So find homogenous.
01:01
So find homogeous.
01:01
Genius solutions y1 and y2, however that works.
01:18
The next thing is you want to calculate what's called the ronskin, which is a determinant of a matrix.
01:27
It's a very easy thing to calculate.
01:31
So the matrix involves the two functions, the two independent functions that you have found, and their first derivatives.
01:40
So we'll call that w, and it is equal to y1, y2 prime, minus y2, y1 prime.
01:54
And why this is important, there's some theoretical work that underpins all this by plugging in a linear combination of the two solutions multiplied by arbitrary functions.
02:10
And this is kind of the mechanics that comes out of that.
02:13
We'll see an example of a similar logic as we work through the example.
02:19
But then what you do is you find two parts to the y particular.
02:28
You basically combine the ront skin and the two solutions in a mechanistic way.
02:39
So you take the integral of y2 times f of x, dx divide by the ron skin, and you add.
02:51
To it, sort of a complementary thing.
02:54
So you can kind of see you're doing very similar math to what you're doing in the ronskin to begin with, sort of integrating it, if you will, the ronskin is this.
03:16
So what you're basically doing is you're beating the function on the right -hand side, the non -homogeneous function against the two solutions that you know already work, and you're extracting a new solution from that.
03:35
So let's see how this works.
03:36
We're going to take as our example, the equation y double prime minus 3y prime plus 2y equals 5e to the minus x.
03:51
Now this is a second order differential equation linear with constant coefficients.
03:57
So let's quickly find a solution, but then we're going to take kind of a circuitous route.
04:04
And really the reason for doing this is to convince us that this technique can work with a more sophisticated system.
04:12
But usually for this type of system, you start with y equals e to the rt.
04:19
You take the first derivative, the second derivative, and you put all that in and turn this back into what's called the characteristic equation, which in this case is r squared minus 3y plus 2 equals 0.
04:38
So this is the homogeneous solution.
04:41
And that has two roots r minus 2 times r minus 1 equals 0.
04:48
So the two roots are e, two roots are 2 and 1.
04:56
And so y1, we can say, is e to the, and we shouldn't have t in there, we should have x, sorry, my physics background's coming back to haunt me.
05:10
We're always doing time stuff in there.
05:13
So y1 is equal to e to the x, and y2 is equal to e to the 2x.
05:24
Okay, a more circuitous route is to just find y1 however you can, and you can find y2 by assuming it is some function will call it, i think they call it v of x times y1.
05:49
Put that back into the differential equation.
05:53
This is a more powerful technique.
05:56
So we're going to put it back in.
05:59
Can we solve for v of x? and that's kind of the nature of what you're doing with the variation.
06:10
Of parameters.
06:11
You're kind of assuming your particular solution is some functions multiplying y1 and another function multiplying y2, and the ronski in business comes out of the work with the differential equation.
06:26
So let's see how this works.
06:28
Y2 prime is equal to v -v prime, y1 plus v double prime, not v plus v.
06:46
Y1 prime.
06:49
Okay.
06:50
And since y is equal to e to the x, this is just v prime plus v, e to the x.
07:04
Y2 double prime is equal to v double prime plus v, e to the x.
07:18
Okay, so we're doing a product rule.
07:20
We took the of the things in the parentheses, and then plus v prime plus v.
07:27
E to the x.
07:28
The derivative of e to the x just gives itself back.
07:32
And so y double prime is equal to v double prime plus 2 v.
07:39
V prime plus ve to the x.
07:45
Okay, and now we're going to put those back into the differential equation, and not too surprisingly, it will just multiply.
08:01
Multiply these functions by constants, and we can cancel out the e to the x.
08:08
So y double prime, y2 double prime just comes in as v double prime plus 2v prime plus v, and then we have minus 3 times v prime plus v, and then we have plus 2 times v, all equal to 0.
08:32
And if we gather up like terms, what do we get v double prime minus v prime? so 2 minus 1 plus 0 is equal to 0.
08:59
So this tells us that v satisfies the differential equation.
09:04
V double prime is equal to v prime.
09:07
And we could think about the first derivative of a function being equal to the function that is satisfied by v equals e to the x.
09:21
So our second solution is, let's see, our first solution times this or e to the 2x.
09:42
So this kind of gives an idea of how you can build up a different solution by multiplying by solutions that you already have.
09:53
It gives you that idea.
09:55
And so now we are going to try the variation of parameters method, which at the heart of it you're starting out.
10:06
Okay, and we'll just say heart of this is writing down why particular is some arbitrary function.
10:16
We'll call it u times y1 plus v times y2 and working through the consequences of that...