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.
