LINAN CHEN: Hi.
Welcome back to recitation.
In the lecture, you've learned eigenvalues and eigenvectors
of a matrix.
One of the many important applications of them is
solving a higher order linear differential equation with
constant coefficients.
A typical example is like what I've written
on the board here.
y is a function of p, and y and its derivatives satisfy
this equation.
As you can see, it involves y, y prime, and all the way to
its third derivatives.
So our first goal is to solve this differential equation for
its general solution using the method of matrix.
So the very first thing that we should do is to find out
which matrix we should be working with.
So after that, we also want to say something about the
explanation of this matrix At.
We want to find out the first column of this matrix
exponential.
Why don't you hit the pause now, and try to write down
this matrix A by yourself.
But before you continue, make sure you come back to this
video and check with me you've got the correct A. I'll see
you in a while.
OK, let's work together to transform this problem into
linear algebra.
The idea is to put y double prime, y prime, and y together
as a vector.
And let me call this vector u.
So of course, vector u is also a function in t.
So this is vector u.
If this is u, what's going to be u prime?
OK, u prime is going to be, so I take derivative of every
coordinate here that's going to be y triple prime, y double
prime, and y prime.
So this is our u prime t.
And my goal is to write u prime as a matrix, call it A
times vector u itself.
So I want to put a matrix here.
And I want to create this matrix by incorporating this
differential equation.
If you move everything, except y triple prime to the
right-hand side of the equation, you can read y
triple prime is equal to negative 2
times y double prime.
So y triple prime is negative 2 times y double
prime plus y prime.
That's plus 1 times y prime plus 2y.
That's 2y, right?
That gives you the first row.
Then look at the second coordinate,
this y double prime.
y double prime is simply itself.
So you rate y double prime is equal to 1, y double
prime then 0 0.
That's the second row.
Well same thing happens to the last row. y
prime is again itself.
So that's 0 1 and 0.
That is our matrix A. Did you get the right answer?
So we have transformed this equation, this third other
ordinary differential equation of y into a first other
differential equation of ut.
Although ut is a vector, but if we can solve this equation
for u, we have all the information we need for y.
So let's plan on solving this equation.
In order to solve this equation, we will need the
eigenvalues and eigenvectors of this matrix A. Again, this
is a good practice for you.
Why don't you possibly do again, and try to complete
this problem on your own.
When you're ready, I'm going to come back and show
you how I did it.
Let's finish up everything together.
So as we said, we need the eigenvalues and eigenvectors
of matrix A, and that involves computing the determinant of
the following matrix.
So I want to compute the determinant of A minus lambda
times the identity matrix I.
Let's write it out.
That's the determinant of -2 minus lambda 1 2, 1 negative
lambda 0, and 0 1 negative lambda.
So we need the determinant of this three by three matrix.
Do it in your favorite way.
You can either use the big summation formula, or you can
do by cofactor along any row or any column.
The correct answer should be this is equal to 1 minus
lambda times 1 plus lambda times 2 plus lambda.
And this polynomial has three roots.
1, -1, and negative -2.
These are the eigenvalues we're looking for.
So let me write it here.
Lambda 1 is equal to 1.
Lambda 2 is equal to -1.
And lambda 3 is equal to -2.
So now what we need is the eigenvector corresponding to
each eigenvalue.
Let's take lambda 1 for example.
The eigenvector of A corresponding to lambda 1 is
in the null space of the matrix A minus lambda 1I, so
in this case it's A minus I. So it's in the null space of
this matrix.
In other words, we are looking for a vector, let's call it a
b c, a column vector a b and c, such that this matrix
multiplying a b c gives me 0.
So if you write it out that's going to be A minus I is -3.
1, 2, 1, -1, 0, 0, 1, -1 times a b c is equal to 0.
OK.
Could we choose constants a b c such that
this is always true?
Well if you read the last row, so the last dot product, it
says that b has to be equal to c.
And if you rate the second row it says that a has
to be equal to b.
Which means a is equal to b is equal to c.
And if this relation is true, the first product is always
going to be 0.
So that simply means we can choose the first eigenvector,
the eigenvector corresponding to lambda 1, to be x1 is equal
to 1, 1, 1, transpose.
So we choose the first time eigenvector to be the column
vector with all the coordinates being 1.
And you can do the same thing to lambda 2 and lambda 3.
But please allow me to skip the computation here.
I'm going to write out the answer for you.
So x2, the eigenvector corresponding to the second
eigenvalue is going to be equal to 1, -1, and 1.
And x3 is going to be 4, -2, and 1.
Now we've got everything we need in order to create the
general solutions for ut.
So we have eigenvalues, we have the corresponding
eigenvectors.
What should be ut?
The general solution for ut is equal to some constants, c1,
times e to the power lambda 1t.
So in this case e to the power t.
Then times the first eigenvector, x1.
Plus some other constant c2 times e to the
power lambda 2t.
So e to the power negative t times x2.
That's the second eigenvector.
Then plus some other constant, c3, times e to the
power lambda 3t.
So negative 2t times x3.
That gives you the general solution for u.
As we just said, if you know what u is, you have all the
information you need for y.
Just in case you're curious about what y is, you can just
rate the last coordinate of x1, x2, and x3.
And you can see that all of them are 1.
So yt is simply equal to c1 e to the power lambda t plus c2
e to the power negative t plus c3 e to the power negative 2t.
And the choice of c1, c2, and c3 is completely arbitrary.
So that completes the first part of this question.
In the second part, we want to say something about the
exponential of At.
So let me first give you the recipe to cook up the
exponential of At.
The exponential of At is equal to the
product of three matrices.
So you usually we denote them by S times e to the power
capital lambda t times S invert.
And you may ask what S is, and what this matrix is.
So S is the matrix that has x1, x2, and x3 being its
column vectors.
So S is x1, x2, and x3.
Let me copy it down here.
So that's 1, 1, 1, 1 -1, 1, 4, -2, 1.
And the matrix in the middle, e to the power lambda t is a
diagonal matrix.
So e to the power lambda t, it's a diagonal matrix, and
it's diagonal and trace are given by e to the power lambda
1t, so that's e to the power t, then e to the power lambda
2t negative t, and e to the power lambda 3t makes it 2t.
0 everywhere else.
So that's e to the power lambda t.
Then the exponential of this At is given by the product of
these three matrices.
It looks a bit complicated because it involves the
inverse of S. But luckily, we only want the first column of
the result.
So if we consider this product, we can see the
product of the first two matrices is relatively easy,
because this is a diagonal matrix, and we know that S is
given by these columns.
So the result of the product of these two is simply
multiplying the columns of S by these coefficients
respectively.
So you expect to get e to the power lambda t, x1, times e to
the power--
sorry.
The second column should be e to the power negative t, x2.
The third column should be e to the power negative 2t, x3.
And here, what we should put is S inverse.
But we don't need everything from S inverse, because as we
just said, we only need the first column of this result.
And the first column of this product is going to be given
by linear combinations of these columns, and the
coefficients are going to be given by the
first column S inverse.
So our goal should be just to get the
first column of S inverse.
Then what if the first column of S inverse?
Well the formula for S inverse is S inverse is going to be
the reciprocal of the determinant of S, so 1 over
determinant of S times the transpose of a matrix C. This
matrix C, the entries of this matrix C are given by
cofactors of matrix S. And then you take transpose, you
divide everything by the determinant of S. The result
will be S inverse.
And we only need the first column of this matrix.
Let's try to write the first column out.
Well again, do it in your favorite way to compute the
determinant of S. The result should be 1 over 6.
So the determinant of S is 6.
Then what is the first column of C transpose?
Well we can read it from here.
This spot, the 1, 1 spot, should be the cofactor
of this spot here.
That make this 1, minus negative 2, which is 1, so we
put 1 here.
Now this spot will be the cofactor of this entry here.
so that's 1 minus negative 2, that's 3.
But this is 12 entry, so you should put a
negative sign in the front.
Then the last stop should be the cofactor of this entry
here, which is 1 minus negative 1, that's 2.
Something else here.
Two warnings.
First, don't forget this transpose sign.
Second, don't forget this negative sign.
We've got the first column of S inverse, and
that's all we need.
So we put it here.
That's 1 over 6, -1/2 and 1/3.
That's good enough for me.
Now I can write up the first column of exponential of At.
So the first column of the exponential of At, I'm going
to write it here.
That's going to be equal to the linear combination of
these columns.
So that's 1/6 of the first column, that's e to the power
t over 6 times x1.
Plus this times this, so that's going to be minus 1/2
of e to the power negative t times x2.
The plus 1/3 of e to the power negative 2t times x3.
That's the first column of the exponential At.
And then with the other two columns.
That's the answer.
If you want more practice, you can certainly complete this S
inverse, and then you can also complete the
exponential of At.
But I will leave the rest to you.
OK, I hope this example shows you that linear algebra can be
a powerful tool in solving higher order ordinary
differential equations with constant coefficients.
And we have demonstrated the standard procedure to do it,
and we also practiced how to calculate the
exponential of a matrix.
Thanks for watching, and see you next time.