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 t,
and y and its derivatives
satisfy this equation.
As you can see, it involves
y, y prime, and all the way
to its third derivative.
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 A*t.
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 read 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-order ordinary
differential equation of y
into a first-order
differential equation of u(t).
Although u(t) 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 pause
the video 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 -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_1*I,
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 read 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 x_1 is
equal to [1, 1, 1] transpose.
So we choose the
first 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 x_2 the eigenvector
corresponding
to the second
eigenvalue, is going
to be equal to 1, -1, and 1.
And x_3 is going
to be 4, -2, and 1.
Now we've got everything
we need in order
to create the general
solutions for u(t)
So we have eigenvalues, we have
the corresponding eigenvectors.
What should be u(t)?
The general solution for u(t)
is equal to some constant C_1
times e to the power
lambda_1*t-- so in this case,
e to the power t.
Then times the first
eigenvector, x_1.
Plus some other constant
C_2 times e to the power
lambda_2*t--
so e to the power
negative t-- times x_2.
That's the second eigenvector.
Then plus some other constant,
C_3 times e to the power
lambda_3t-- so negative
2t-- times x_3.
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 read
the last coordinate
of x_1, x_2, and x_3.
And you can see that
all of them are 1.
So y(t) is simply equal to C_1
e to the power lambda t plus C_2
e to the power negative t plus
C_3 e to the power negative 2t.
And the choice of C_1, C_2, and
C_3 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 A*t.
So let me first give you
the recipe to cook up
the exponential of A*t.
The exponential of A*t 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 inverse.
And you may ask what S is,
and what this matrix is.
So S is the matrix that has x_1,
x_2, and x_3 being its column
vectors.
So S is x_1, x_2, and x_3.
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 its diagonal entries
are given by e to the power
lambda_1*t-- so that's
e to the power t--
then e to the power
lambda_2*t-- negative t--
and e to the power
lambda_3*t-- negative 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
x_1 times e to
the power-- sorry.
The second column should be e
to the power negative t, x_2.
The third column should be e
to the power negative 2t, x_3.
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 is 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 negative 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 (1, 2) entry, so
you should put a negative sign
in the front.
Then the last spot 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 read out the first
column of exponential of A*t.
So the first column of
the exponential of A*t,
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 x_1.
Plus this times this, so that's
going to be minus 1/2 of e
to the power
negative t times x_2.
Then plus 1/3 of e to the
power negative 2t times x_3.
That's the first column
of the exponential A*t.
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 A*t.
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.
