00:01
Okay, so in part a, we want to solve this given system of linear equations.
00:08
We want to solve the system of differential equations by first finding the eigenvalues, and then we'll find the corresponding eigenvectors, and that will give us our general solution.
00:20
So i already have written down these differentials in matrix form.
00:26
So then we're going to then find the characteristic polynomial, which then will allow us to find the eigenvalues.
00:33
So let's go ahead and start with that.
00:35
So to get the characteristic polynomial, then a .k .a.
00:41
Eigenvalues, we have to find the determinant of our matrix a, which is the matrix 1423, and subtract eigenvalue times identity matrix, which will be the following.
01:01
So find the determinant of 1 minus lambda, 2, and 3 minus 7.
01:10
Lambda.
01:12
So the final determinant, you essentially cross multiply your a and d elements and subtract from your b and c elements.
01:21
They want to solve for lambda.
01:22
But just multiplying this out here, finally determinant gives us this lambda squared minus 4 lambda minus 5.
01:36
So now we have to solve this characteristic polynomial.
01:40
So the characteristic polynomial is always set to zero because we want to find appropriate eigenvectors from these eigenvalues.
01:52
So if you factor this out, we get our two eigenvalues, get lambda minus five times lambda plus one.
02:02
So this tells us that our first eigenvalue will be five, and our second eigenvalue will be negative one.
02:09
So now we have the eigenvalues.
02:11
Let's write those in red.
02:13
So eigenvalue 1, so lambda 1 is 5.
02:19
And lambda 2 is equal to negative 1.
02:26
So now we want to solve for the eigenvectors from these corresponding eigenvalues.
02:31
Right, and we're going to solve for these eigenvectors with the following technique.
02:45
Okay, so we're going to do lambda 1 first.
02:49
So since we're solving for eigenvectors, we want to solve the homogeneous equation because we don't want our eigenvectors just be 0 -0, as that's trivial, and not what we're looking for here.
03:01
To solve these differential equations.
03:03
So we want to solve the homogeneous equation to get the eigenvectors for each of these eigenvalues.
03:11
So let's do lambda 1 first.
03:17
Lambda 1 equals 5.
03:20
So this means our a matrix will change.
03:24
So 1 minus 5 is negative 4.
03:27
We have 4 still 2, 3 minus 5 negative 2.
03:34
Then we want to multiply by two vectors, by two vectors such that you get zero, zero.
03:52
Now we don't want, we don't want the homogeneous solution.
03:56
We don't want our vectors to be zero zero because that doesn't really show anything when we go to find our general solution for our two differential equations.
04:04
So we're going to multiply out and then solve for each of the components of this first vector.
04:10
So this means we get minus 4, 801 plus 4, 8 ,2, 2801, 2801, we get minus 2.
04:36
And these both equals 0.
04:38
So essentially what we're going to do is solve for one of these, and just kind of make our own vector that looks the nicest.
04:46
It doesn't matter which one we choose or what we choose, because all we're dealing with are scaled forms.
04:55
Of our base vector that we're going to find.
04:58
So if we solve for this, we say that 801 is equal to 802, right? because if you subtract the minus, if you subtract 4, 82 over, you get this relationship because the force will cancel out.
05:13
And you'll get the same relationship if you do the same thing here, right? because you'll add this over and get that 2801 is equal to 2802.
05:21
So 801 is equal to a2.
05:23
So this means our first eigenvector, our first eigenvector has this setup, where essentially the components are the same.
05:42
So all we need to do is choose a value for this component.
05:47
Let's say a to 2 equals 1.
05:50
This means our eigenvector for the eigenvalue 5 is 1 -1.
05:55
So we're done with that one.
05:58
So now we have to do lambda 2, which is equal to negative 1.
06:03
Again, same process.
06:05
Find the non -homogeneous solution...