All right in this video, I'm going to do one case of how to solve a linear system of differential equations.
I'm using the eigenvalue, eigenvector method.
The three cases would be one case is you have complex eigenvalues,
another case is where you have two real values, and then another case would be if you had repeated real values.
But I'm going to just do
the case where you get complex eigenvalues in this example. And also in all my examples, probably the
matrix of constant coefficients
which is labeled A here will be 2 x 2. So the solution is a linear combination of two equations.
So let's go ahead and get started.
I have the problem written here. Since I already said we're using the eigenvalue and eigenvector method,
we're going to first find the eigenvalues, and I already said that in this case the values will be complex.
Let's go ahead and start solving for the two eigenvalues.
We have the equation lambda squared minus T lambda plus D equals zero to
solve for what lambda is. In this case, the T is the
trace of the matrix A and D stands for the determinant of the matrix. The trace is
the sum of this diagonal. In this case it's going to be 0. The determinant is the multiplication of this. So that's
-16 minus the multiplication of the other diagonal. That's minus -32 which gives 16.
The equation is lambda squared minus 0 lambda plus 16 equals to 0. If we solve for lambda,
we get -0 plus or minus the square root of
0 squared minus 4 times 16 all over 2. That's going to give you 0 plus or minus
4i. So we have solved
for the first and the second eigenvalues, which is lambda.
Next we can use another equation to find the eigenvectors.
The equation is the matrix A minus the eigenvalues times the identity times
your
vectors. In this case it would be v1 v2 equals to 0.
We're going to do the first case, lambda 1, which is 0 plus
4i. I usually just do the first one as a positive
and you don't have to do the second one because the
answer will just be the conjugate of
the first one you do. So let's go ahead and plug everything in. We have that A is
-4 minus 8, 4, 4, minus--the identity matrix if you don't know that's just
ones on the diagonal--so if we have a 2 x 2 it's gonna look like that. If we multiply
lambda times the identity, we get--and we're just doing the first lambda--
we get 0 plus 4i or just 4i
0, 0, and then 4i times v1 v2. This is
multiplied--or this is in brackets first and then it's all multiplied by v1 v2 equal to 0. Let's go ahead and
subtract across matrices.
We get -4 minus 4i.
That's -4 minus 0. So -4 and then 0 minus--or
0i minus 4i is
-4i. -8 minus 0 is -8, 4 minus 0 is 4, and 4
minus 4i is 4 minus 4i, times v1 v2 equal to 0.
The next step that you want to do is find the
solution to v1 and v2. When you do matrix multiplication you can get two
equations equal to 0 to solve for v1 or v2.
You can just, for these complex cases--
most of the times you can just use the first row, right? But for complex cases, if you think it's going to give you weird
values that are not simple or whole numbers, for example, then you can look at the second row. For me it just seems easier
because you know you're gonna divide by four to use the second row instead of the first one.
We'll just ignore the first row and just use the second equation to find v1 and v2. When you multiply the
row and the column, you get 4v1 plus 4 minus 4i
v2 equal to zero. And then from here, usually again what I do is just plug in 1 for either v1 or v2
and then
see what that gives me for the other value. Since this has the complex case and this just has the 4,
I'll plug in 1 to v2.
I'm going to say v2 is 1 and when that's 1 we have -4 plus 4i
equal to 4
v1. I just plugged in 1 and then moved this part to the right-hand side of the equation. Then we're going to solve for v1.
So v1 is equal to -1 plus
i. When v2 is 1 the second row gives us that v1 will be -1 plus i.
Now we have the eigenvectors--or the eigenvector--and we have the eigenvalue for the first
lambda 1 that we found. And like I said, this means that they--we already know the solution
because the second case would just be the conjugate, so the
second answer is lambda 2 is 0 minus 4i and the eigenvector for that would be -1 minus i
1 because that's just the conjugate of both the vector and the eigenvalue.
If we could stop there at just solving for the imaginary
solution, which we can't because the question says solve for real solutions,
we would have that y of t is equal to c1 e to the 0 plus 4i
-1 plus i1
plus c2 e to the 0 minus 4i
-1 minus i, 1. We can use this to go one step further and find the real solutions.
What you're going to need to do--well first of all you can simplify because e to the 0 is just 1. So you
can really just act like that's not there, and you can act like that's not there. Then we're going to use Euler's formula
to get rid of the imaginary values. Euler's formula
is that e to the i t is equal to cosine t plus i sine t. In this case,
we have e to the 4i. So t is 4.
Sorry this should be t's. We have the coefficient as 4, so the
coefficient goes with the t's. What I mean by that is e to the 4it is equal to
cosine 4t plus i sine 4t.
All right, and then we can plug in--where we see e to the 4it--we can plug in this right-hand side.
Let's go ahead and do that next.
We're really not going to worry about the conjugated second part because
we're going to basically separate real and imaginary values and use that as the linear combination
for the two solutions.
What I mean by that is you don't really have to worry about the second part and I'll show you why and it'll all make
sense later. So let's go ahead and plug in Euler's formula.
We have c1--and then we're going to plug in here--times cosine 4t plus i sine
4t times -1 plus i times 1. And then the second part don't worry about it.
What we're going to do next is we're going to multiply this part by the vectors.
That would give you c1 times negative cosine 4t. That's that part and then you're basically just foiling, right?
So it's plus i
cosine 4t minus i sine
4t plus i squared sine
4t. That's the first part. Then you also have to do the second part of the vector. That would be cosine
4t plus i sine
4t. And then again
you're not worrying about the second part. Next what we're going to do is we're going to separate
here the one where you have--where you had to foil, where you're multiplying two binomials--you have to
separate the
imaginary and the real part. So i squared, i is the square root of -1 obviously,
so i squared is -1. This is actually a real value. We're going to separate it.
Really that matrix is negative cosine
4t minus sine
4t plus i cosine 4t minus i sine
4t and then the bottom is just the same. Our final solution is going to be us
separating the real values from the imaginary values
and then you can just get rid of the i.
What I mean by that is that the final solution is that y of t is equal to c1
and then e to the 0 was just 1. So it's times the vector.
Let's do the real parts first. The real part here is--let me write this out again. So that's cosine
4t plus i sine
4t. If we're separating just the real parts, we're going to get negative cosine 4t minus sine
4t and then the real part here is cosine 4t.
That's the first vector and the second vector will be c2 times the imaginary part, which we're just
including the real valued part of the imaginary part if that makes any sense. It's going to be cosine
4t minus sine
4t and the second value is sine
4t and that's the final solution.
The reason I said you could ignore the
conjugate part was because you can actually separate the real and imaginary parts from just the first
lambda that you do to get two linearly independent solutions
to your system. That works as a final answer and hopefully that makes sense.
This is basically the same steps that you would do for the case where you have two real
eigenvalues. However, there's just the extra step of after you find the imaginary solution, you apply Euler's formula
and then you just have to simplify and separate the real and imaginary parts.