PROFESSOR: Welcome back.
So in this session, we're going to use the matrix
method to solve this linear system of differential
equations.
These are x dot equals 6x plus 5y, and y dot equals x plus 2y.
So why don't you take a few minutes
to write down the system in matrix form
and go through the matrix method to solve it.
And I'll be right back.
Welcome back.
So let's write down this system in matrix form.
You would have a vector with entries x and y prime equals
a matrix with entries 6, 5; 1, 2 multiplying the column vector
[x, y].
So now, we did big part of the work.
The matrix method tells us that we
need to find the eigenvalues of this matrix
to be able to basically diagonalize it and seek
eigenvectors so that then we can just read off the solutions
and write the solution of the system
as a linear combination of the eigenvectors that we found.
So let's look for the eigenvalues first.
The eigenvalues would be computed
by seeking the determinant of this matrix in this form:
6 minus lambda, 5; 1, 2 minus lambda.
We're going to have an equation on lambda, solve for lambda,
and the solutions will be our eigenvalues.
So the determinant would be 6 minus lambda multiplying 2
minus lambda minus 5, 1 dot 5, equals to 0.
So here, the lambda that lambda gives us a lambda squared.
We have minus 6*lambda minus 2*lambda,
which would be minus 8*lambda.
And then, we would have 2 dot 6, which is 12, minus 5,
which gives us 7.
So quadratic equation in lambda, and you
can factorize it and find the solutions, which is lambda_1
equals to 1, lambda_2 equals to 7.
So we're done with the first part.
These are our eigenvalues.
They're not repeated.
They're just completely different and real valued.
So now, we're going to look at the eigenvectors associated
to each eigenvalue.
So first eigenvector would be associated with lambda_1 equals
to 1.
So we would be solving this system.
We would be solving this system with a new matrix, 6 minus 1.
I'm going to spell out this one so that-- 2 minus 1.
So this is just our lambda, multiplying an unknown vector
with components a_1 and a_2, equals to zero vector.
And basically here, the unknowns are a_1 and a_2.
So this is simply 5, 5; 1, and 1, [a1, a2] equals to 0, 0.
So as you saw before, here, basically, we
can read off the equation as being
5a_1 plus 5a_2 equals to 0 and another one which
is a_1 plus a_2 equals to 0.
They're the same equations.
So really, we just have a_1 plus a_2 equals to 0.
And so our vector v_1 could be picked
to just have component 1, for example, a_1 equals to 1.
And its second component would just be minus 1.
That would be one pick for our v_1.
We could normalize this vector if you wanted to.
I'm just going to keep it like this for now.
So if we look now for the second eigenvector corresponding
to the second eigenvalue of 7, I would
be looking for the components of these vectors
by doing a similar solving for the same thing.
And I'm going to spell it out again so that you see where
the terms are coming from.
It's just 6 minus the value of my lambda...
[0, 0].
So here, we have 6 minus 7, which is 1, 5.
And then, we have 1 and 2 minus 7, which is minus 5.
So really, what do we have is an equation minus 1
plus 5a_2 in both cases.
So we can pick a value for a_1 or a_2
and write down a vector v_2, in for example the form of a_1
equals to-- let's pick a_2 equals to 1.
And we would have a_1 equals to 5, for example.
Again, if you wanted an orthonormal basis formed
by your v_1, v_2, you would just normalize these two vectors.
So here, basically, we can then rewrite the solution
to the original system as being linear combinations
of-- so I'm just going to write it in vector form.
The first vector 1-- I'll keep it in v_1, v_2,
that way you see it.
And then, I'll go into the component.
We'd have v_1 exponential of the value of lambda
we found that corresponds to v_1.
So it would be 1 dot t.
And then, v_2 exponential of the lambda
value that corresponds to v_2.
And then, basically, we just have
constants of integration here.
And so the solution to this problem
would be linear combination of the vectors
by the basis of our eigenvectors and multiplied
by the exponentials assigned a value of the eigenvalues
that we found when we looked for the eigenvalues
of the matrix of the system.
So here, just know that like for the 1D problem
that we saw before, we're building
a solution based on linear combination of lucky guesses
that we used.
And in the one equation case, we used a guess of e to lambda*t
in 1D.
Here, in this case, we had a guess of a vector v
and the form of lambda*t that we used.
And then, basically, we just solved for the lambdas,
and solved for the v's, and did a linear combination
of all the solutions.
Like we did before in the 1D case, we solved the lambda.
We had different values of lambda.
We did a linear combinations of the exponentials.
So that ends this problem.
And here, the key is just to go through the method
of diagonalizing your matrix.
Basically, it's finding the eigenvalues,
and then computing the eigenvectors associated
with that, and writing your solutions
in terms of a linear combination of the solution that you found.