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The real topic is how to solve
inhomogeneous systems,

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but the subtext is what I wrote
on the board.

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I think you will see that
really thinking in terms of

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matrices makes certain things a
lot easier than they would be

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otherwise.
And I hope to give you a couple

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of examples of that today in
connection with solving systems

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of inhomogeneous equations.
Now, there is a little problem.

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We have to have a little bit of
theory ahead of time before

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that, which I thought rather
than interrupt the presentation

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as I try to talk about the
inhomogeneous systems it would

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be better to put a little theory
in the beginning.

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I think you will find it
harmless.

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And about half of it you know
already.

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The theory I am talking about
is, in general,

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the theory of the systems x
prime equal a x.

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I will just state it when n is
equal to two.

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A two-by-two system likely you
have had up until now.

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It is also true for end-by-end.
It is just a little more

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tedious to write out and to give
the definitions.

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Here is a little two-by-two
system.

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It is homogeneous.
There are no zeros.

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And it is not necessary to
assume this, but since the

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matrix is going to be constant
until the end of the term let's

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assume it in and not go for a
spurious generality.

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So constant matrices like you
will have on your homework.

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Now, there are two theorems,
or maybe three that I want you

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to know, that you need to know
in order to understand what is

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going on.
The first one,

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fortunately,
is already in your bloodstream,

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I hope.
Let's call it theorem A.

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It is simply the one that says
that the general solution to the

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system, that system I wrote on
the board, the two-by-two system

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is what you know it to be.
Namely, from all the examples

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that you have calculated.
It is a linear combination with

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arbitrary constants for the
coefficients of two solutions.

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In other words,
to solve it,

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to find the general solution
you put all your energy into

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finding two independent
solutions.

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And then, as soon as you found
them, the general one is gotten

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by combining those with
arbitrary constants.

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The only thing to specify is
what the x1 and the x2 are.

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"Where," I guess,
would be the right word to use.

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Where x1 and x2 are two
solutions, but neither must be a

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constant multiple of the other.
That is the only thing I want

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to stress, they have to be
independent.

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Or, as it is better to say,
linearly independent.

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Are two linearly independent
solutions.

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And department of fuller
explanation, i.e.,

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neither is a constant multiple
of the other.

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That is what it means to be
linearly independent.

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Now, this theorem I am not
going to prove.

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I am just going to say that the
proof is a lot like the one for

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second order equations.
It has an easy part and a hard

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part.
The easy part is to show that

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all of these guys are solutions.
And, in fact,

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that is almost self-evident by
looking at the equation.

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For example,
if x1 and x2,

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each of those solve that
equation so does their sum

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because, when you plug it in,
you differentiate the sum by

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differentiating each term and
adding.

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And here A times x1 plus x2 is
Ax1 plus Ax2.

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In other words,

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you are using the linearity and
the superposition principle.

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It is easy to show that all of
these, well, maybe I should

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actually write something down
instead of just talking.

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Easy that all these are
solutions.

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Every one of those guys,
regardless of what c1 and c2

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is, is a solution.
That is linearity,

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if I use that buzz word,
plus the superposition

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principle, that the sum of two
solutions is a solution.

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The hard thing is not to show
that these are solutions but to

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show that these are all the
solutions, that there are no

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other solutions.
No matter how you do that it is

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hard.
The hard thing is that there

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are no other solutions.
These are all.

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Now, you could sort of say,
well, it has two arbitrary

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constants in it.
That is sort of a rough and

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ready reason,
but it is not considered

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adequate by mathematicians.
And, in fact,

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I could go into a song and
dance as to just why it is

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inadequate.
But we have other things to do,

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bigger fish to fry,
as they say.

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Let's fry a fish.
No, we have another theorem

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first.
This one it is mostly the words

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that I am interested in.
Once again, we have our old

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friend the Wronskian back.
The Wronskian of what?

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Of two solutions.
It is the Wronskian of the

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solution x1 and x2.
They don't, by the way,

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have to be independent.
Just two solutions to the

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system.
And what is it?

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Hey, didn't we already have a
Wronskian?

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Yeah.
Forget about that one for the

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moment.
Postpone it for a minute.

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This is a determinant,
just like the old one way.

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This is going to be a great
lecture.

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(x1, x2).
Now what is this?

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x1 is a column vector,
right?

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x2 is a column vector.
Two things in it.

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Two things in it.
Together they make a square

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matrix.
And this means it is

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determinant.
It is the determinant of this.

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It is a determinant,
in other words,

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of a square matrix.
And that is what it is.

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I will change this equality.
To indicate it is a definition,

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I will put the colon there,
which is what you add,

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to indicate this is only equal
because I say so.

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It is a definition,
in other words.

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Now, there is a connection
between this and the earlier

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Wronskian which I,
unfortunately,

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cannot explain to you because
you are going to explain it to

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me.
I gave it to you as part one of

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your homework problem.
Make sure you do it.

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And, if you cannot remember
what the old Wronskian is,

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please look it up in the book.
Don't look it up in the

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solution to the problem.
If you do that you will learn

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something.
Then you will see how,

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in a certain sense,
this is a more general

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definition than I gave you
before.

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The one I gave you before is,
in a certain sense,

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a special case of it.
Now that is just the

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definition.
There is a theorem.

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And the theorem is going to
look just like the one we had

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for second order equations,
if you can remember back that

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far.
The theorem is that if these

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are two solutions there are only
two possibilities for the

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Wronskian.
So either or.

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Either the Wronskian is --
Now, the Wronskian,

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these are functions,
the column vectors are the

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solutions, so those are
functions of the variable t,

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so are these.
The Wronskian as a whole is a

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function of the independent
variable t after you have

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calculated out that determinant.
I will write it now this way to

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indicate that it s a function of
t.

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Either the Wronskian is --
One possibility is identically

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zero.
That is zero for all values of

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t, in other words.
And this happens if x1 and x2

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are not linearly independent.
Usually people just say

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dependent and hope they are
interpreted correctly.

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Are dependent.
But since I did not explain

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what dependent means,
I will say it.

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Not linearly independent.
I know that is horrible,

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but nobody has figured out
another way to say it.

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That is one possibility,
or the opposite of this is

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never zero for any t value.
I mean a normal function is

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zero here and there,
and the rest of the time not

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zero.
Well, not this Wronskian.

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You only have two choices.
Either it is zero all the time

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or it is never zero.
It is like the function e to

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the t.
In other words,

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an exponential which is never
zero, always positive and never

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zero.
Or, it could be a constant.

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Anyway, it has to be a function
which is never zero.

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And this happens in the other
case, so this is --

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There is no place to write it.
This is the case if x1 and x2

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are independent,
by which I mean linearly

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independent.
It is just I didn't have room

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to write it.
That is pretty much the end of

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the theory.
And now, let's start in on the

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matrices.
The basic new matrix we are

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going to be talking about this
period and next one on Monday

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also is the way that most people
who work with systems actually

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look at the solutions to
systems, so it is important you

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learn this word and this way of
looking at it.

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What they do is look not at
each solution separately,

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as we have been doing up until
now.

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They put them all together in a
single matrix.

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And it is the properties of
that matrix that they study and

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try to do the calculations
using.

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And that matrix is called the
fundamental matrix for the

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system.

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Sometimes people don't bother
writing in the whole system.

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They just say it is a
fundamental matrix for A

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because, after all,
A is the only thing that is

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varying there.
Once you know A,

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you know what the system is.
So what is this guy?

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Well, it is a two-by-two
matrix.

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And it is the most harmless
thing.

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It is the precursor of the
Wronskian.

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It is what the Wronskian was
before the determinant was

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taken.
In other words,

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it is the matrix whose two
columns are those two solutions.

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The other question is what we
are going to call it.

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I kept trying everything and
settled on calling it capital X

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because I think that is the one
that guides you in the

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calculations the best.
This is definition two,

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so colon equality.
Notice I am not using vertical

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lines now because that would
mean a determinant.

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It is the matrix whose columns
are two independent solutions.

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Is that all?
Yeah.

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You just put them side-by-side.
Why?

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That will come out.
Why should one do this?

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Well, first of all,
in order not to interrupt the

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basic calculation that I want to
make with this during the

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period, it has two basic
properties that we are going to

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need during this period.
These are the properties.

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Just two.
And one is obvious and the

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other you will think,
I hope, is a little less

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familiar.
I think you will see there is

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nothing to it.
It is just a way of talking,

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really.
The first is the one that is

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already embedded in the theorem,
namely that the determinant of

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the fundamental matrix is not
zero for any t.

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Why?
Well, I just told you it

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wasn't.
This is the Wronskian.

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The Wronskian is never zero?
Why is it never zero?

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Well, because I said these
columns had to be independent

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solutions.
So this is not just not zero,

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00:14:44,000 --> 00:14:50,000
it is never zero.
It is not zero for any value of

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t.
That is good.

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00:14:50,000 --> 00:14:56,000
As you will see,
we are going to need that

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property.
But the other one is a little

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stranger.
The only thing I can say is,

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00:15:02,000 --> 00:15:08,000
get used to it.
Namely that X prime equals AX.

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Now, why is that strange?
That is not the same as this.

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This is a column vector.
That is a square matrix and

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this is a column vector.
This is not a column vector.

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This is a square matrix.
This is what is called a matrix

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differential equation where the
variable is not a single x or a

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column vector of a set of x's
like the x and the y.

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It is a whole matrix.

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Well, first of all,
I should say what is it saying?

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This is a two-by-two matrix.
When I multiply them I get a

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two-by-two matrix.
What is this?

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This is a two-by-two matrix,
every entry of which has been

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differentiated.
That is what it means to put

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that prime there.
To differentiate a matrix means

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nothing fancy.
It just means differentiate

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every entry.
It is just like to

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differentiate a vector (x,
y), to make a velocity vector

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00:16:20,000 --> 00:16:26,000
you differentiate the x and the
y.

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00:16:22,000 --> 00:16:28,000
Well, a column vector is a
special kind of matrix.

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00:16:26,000 --> 00:16:32,000
The definition applies to any
matrix.

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Well, why is that so?
I state it as a property,

234
00:16:33,000 --> 00:16:39,000
but I will continue it by
giving you, so to speak,

235
00:16:37,000 --> 00:16:43,000
the proof of it.
In fact, there is nothing in

236
00:16:40,000 --> 00:16:46,000
this.
It is nothing more than a

237
00:16:42,000 --> 00:16:48,000
little matrix calculation of the
most primitive kind.

238
00:16:46,000 --> 00:16:52,000
Namely, what does this mean?
Let's try to undo that.

239
00:16:50,000 --> 00:16:56,000
What does the left-hand side
really mean?

240
00:16:53,000 --> 00:16:59,000
Well, if that is what x means,
the left-hand side must mean

241
00:16:58,000 --> 00:17:04,000
the derivative of the first
column.

242
00:17:02,000 --> 00:17:08,000
That is its first column.
And the derivative of the

243
00:17:05,000 --> 00:17:11,000
second column.
That is what it means to

244
00:17:07,000 --> 00:17:13,000
differentiate the matrix X.
You differentiate each column

245
00:17:11,000 --> 00:17:17,000
separately.
And to differentiate the column

246
00:17:14,000 --> 00:17:20,000
you need to differentiate every
function in it.

247
00:17:17,000 --> 00:17:23,000
Well, what does the right-hand
side mean?

248
00:17:19,000 --> 00:17:25,000
Well, I am supposed to take A
and multiply that

249
00:17:23,000 --> 00:17:29,000
by [x1,x2].
Now, I don't know how to prove

250
00:17:26,000 --> 00:17:32,000
this, except ask you to think
about it.

251
00:17:30,000 --> 00:17:36,000
Or, I could write it all out
here.

252
00:17:34,000 --> 00:17:40,000
But think of this as a bing,
bing, bing, bing.

253
00:17:39,000 --> 00:17:45,000
And this is a bing,
bing.

254
00:17:42,000 --> 00:17:48,000
And this is a bong,
bong.

255
00:17:45,000 --> 00:17:51,000
How do I do the multiplication?
In other words,

256
00:17:50,000 --> 00:17:56,000
what is in the first column of
the matrix?

257
00:17:55,000 --> 00:18:01,000
Well, it is dah,
dah, and the lower thing is

258
00:18:01,000 --> 00:18:07,000
dah, dah.
In other words,

259
00:18:03,000 --> 00:18:09,000
it is A times x1.

260
00:18:15,000 --> 00:18:21,000
Shut your eyes and visualize
it.

261
00:18:17,000 --> 00:18:23,000
Got it?
Dah, dah is the top entry,

262
00:18:19,000 --> 00:18:25,000
and dah, dah is the bottom
entry.

263
00:18:22,000 --> 00:18:28,000
It is what you get by
multiplying A by the column

264
00:18:25,000 --> 00:18:31,000
vector x1.
And the same way the other guy

265
00:18:28,000 --> 00:18:34,000
is --
-- what you get by multiplying

266
00:18:33,000 --> 00:18:39,000
A by the column vector x2.
This is just matrix

267
00:18:37,000 --> 00:18:43,000
multiplication.
That is the law of matrix

268
00:18:41,000 --> 00:18:47,000
multiplication.
That is how you multiply

269
00:18:45,000 --> 00:18:51,000
matrices.
Well, good, but where does this

270
00:18:49,000 --> 00:18:55,000
get us?
What does it mean for those two

271
00:18:53,000 --> 00:18:59,000
guys to be equal?
That is going to happen,

272
00:18:57,000 --> 00:19:03,000
if and only if x1 prime is
equal to A x1.

273
00:19:02,000 --> 00:19:08,000
This guy equals that guy.
And similarly for the x2's.

274
00:19:14,000 --> 00:19:20,000
The end result is that this
matrix, saying that the

275
00:19:18,000 --> 00:19:24,000
fundamental matrix satisfies
this matrix differential

276
00:19:22,000 --> 00:19:28,000
equation is only a way of
saying, in one breath,

277
00:19:26,000 --> 00:19:32,000
that its two columns are both
solutions to the original

278
00:19:30,000 --> 00:19:36,000
system.
It is, so to speak,

279
00:19:33,000 --> 00:19:39,000
an efficient way of turning
these two equations into a

280
00:19:38,000 --> 00:19:44,000
single equation by making a
matrix.

281
00:19:41,000 --> 00:19:47,000
I guess it is time,
finally, to come to the topic

282
00:19:46,000 --> 00:19:52,000
of the lecture.
I said the thing the matrices

283
00:19:50,000 --> 00:19:56,000
were going to be used for is
solving inhomogeneous systems,

284
00:19:55,000 --> 00:20:01,000
so let's take a look at those.
I thought I would give you an

285
00:20:00,000 --> 00:20:06,000
example.
Inhomogeneous systems.

286
00:20:10,000 --> 00:20:16,000
Well, what is one going to look
like?

287
00:20:12,000 --> 00:20:18,000
So far what we have done is,
up until now has been solving,

288
00:20:17,000 --> 00:20:23,000
we spent essentially two weeks
solving and plotting the

289
00:20:21,000 --> 00:20:27,000
solutions to homogeneous
systems.

290
00:20:23,000 --> 00:20:29,000
There was nothing over there.
And homogeneous systems,

291
00:20:27,000 --> 00:20:33,000
in fact, with constant
coefficients.

292
00:20:31,000 --> 00:20:37,000
Stuff that looked like that
that we abbreviated with

293
00:20:34,000 --> 00:20:40,000
matrices.
Now, to make the system

294
00:20:36,000 --> 00:20:42,000
inhomogeneous what I do is add
the extra term on the right-hand

295
00:20:41,000 --> 00:20:47,000
side, which is some function of
t.

296
00:20:43,000 --> 00:20:49,000
Except, I will have to have two
functions of t because I have

297
00:20:47,000 --> 00:20:53,000
two equations.
Now it is inhomogeneous.

298
00:20:50,000 --> 00:20:56,000
And what makes it inhomogeneous
is the fact that these are not

299
00:20:54,000 --> 00:21:00,000
zero anymore.
There is something there.

300
00:20:57,000 --> 00:21:03,000
Functions of t are there.
These are given functions of t

301
00:21:02,000 --> 00:21:08,000
like exponentials,
polynomials,

302
00:21:04,000 --> 00:21:10,000
the usual stuff you have on the
right-hand side of the

303
00:21:08,000 --> 00:21:14,000
differential equation.
What is confusing here is that

304
00:21:11,000 --> 00:21:17,000
when we studied second order
equations it was homogeneous if

305
00:21:15,000 --> 00:21:21,000
the right-hand side was zero,
and if there was something else

306
00:21:20,000 --> 00:21:26,000
there it was inhomogeneous.
Unfortunately,

307
00:21:23,000 --> 00:21:29,000
I have stuck this stuff on the
right-hand side so it is not

308
00:21:27,000 --> 00:21:33,000
quite so clear anymore.
It has got to look like that,

309
00:21:32,000 --> 00:21:38,000
in other words.
How would the matrix

310
00:21:34,000 --> 00:21:40,000
abbreviation look?
Well, the left-hand side is x

311
00:21:37,000 --> 00:21:43,000
prime.
The homogenous part is ax,

312
00:21:40,000 --> 00:21:46,000
just as it has always been.
The only extra part is those

313
00:21:43,000 --> 00:21:49,000
functions r.
And this is a column vector,

314
00:21:46,000 --> 00:21:52,000
after the multiplication this
is a column vector,

315
00:21:50,000 --> 00:21:56,000
what is left is column vector.
Now, explicitly it is a

316
00:21:53,000 --> 00:21:59,000
function of t,
given by explicit functions of

317
00:21:56,000 --> 00:22:02,000
t, again, like exponentials.
Or, they could be fancy

318
00:22:02,000 --> 00:22:08,000
functions.
That is the thing we are trying

319
00:22:06,000 --> 00:22:12,000
to solve.
Why don't I put it up in green?

320
00:22:10,000 --> 00:22:16,000
Our new and better and improved
system.

321
00:22:13,000 --> 00:22:19,000
Think back to what we did when
we studied inhomogeneous

322
00:22:18,000 --> 00:22:24,000
equations.
We are not talking about

323
00:22:22,000 --> 00:22:28,000
systems but just a single
equation.

324
00:22:25,000 --> 00:22:31,000
What we did was the main
theorem --

325
00:22:30,000 --> 00:22:36,000
I guess there are going to be
three theorems today,

326
00:22:36,000 --> 00:22:42,000
not just two.
Theorem C.

327
00:22:39,000 --> 00:22:45,000
Is that right?
Yes, A, B.

328
00:22:42,000 --> 00:22:48,000
We are up to C.
Theorem C says that the general

329
00:22:48,000 --> 00:22:54,000
solution, that is,
the general solution to the

330
00:22:54,000 --> 00:23:00,000
system, is equal to the
complimentary function,

331
00:23:00,000 --> 00:23:06,000
which is the general solution
to x prime equals Ax,

332
00:23:06,000 --> 00:23:12,000
--
-- the homogeneous equation,

333
00:23:11,000 --> 00:23:17,000
in other words,
plus, what am I going to call

334
00:23:15,000 --> 00:23:21,000
it?
(x)p, right you are,

335
00:23:17,000 --> 00:23:23,000
a particular solution.
But the principle is the same

336
00:23:22,000 --> 00:23:28,000
and is proved exactly the same
way.

337
00:23:25,000 --> 00:23:31,000
It is just linearity and
superposition.

338
00:23:35,000 --> 00:23:41,000
The linearity of the original
system and the superposition

339
00:23:40,000 --> 00:23:46,000
principle.
The essence is that to solve

340
00:23:43,000 --> 00:23:49,000
this inhomogeneous system,
what we have to do is find a

341
00:23:48,000 --> 00:23:54,000
particular solution.
This part I already know how to

342
00:23:52,000 --> 00:23:58,000
do.
We have been doing that for two

343
00:23:55,000 --> 00:24:01,000
weeks.
The new thing is to find this.

344
00:24:05,000 --> 00:24:11,000
Now, if you remember back
before spring break,

345
00:24:08,000 --> 00:24:14,000
most of the work in solving the
second order equation was in

346
00:24:12,000 --> 00:24:18,000
finding that particular
solution.

347
00:24:15,000 --> 00:24:21,000
You quickly enough learned how
to solve the homogeneous

348
00:24:19,000 --> 00:24:25,000
equation, but there was no real
general method for finding this.

349
00:24:24,000 --> 00:24:30,000
We had an exponential input
theorem with some modifications

350
00:24:28,000 --> 00:24:34,000
to it.
We took a week's detour in

351
00:24:32,000 --> 00:24:38,000
Fourier series to see how to do
it for periodic functions or

352
00:24:37,000 --> 00:24:43,000
functions defined on finite
intervals.

353
00:24:40,000 --> 00:24:46,000
There were other techniques
which I did not get around to

354
00:24:45,000 --> 00:24:51,000
showing you, techniques
involving the so-called method

355
00:24:50,000 --> 00:24:56,000
of undetermined coefficients.
Although, some of you peaked in

356
00:24:55,000 --> 00:25:01,000
your book and learned it from
there.

357
00:25:00,000 --> 00:25:06,000
But the work is in finding
(x)p.

358
00:25:03,000 --> 00:25:09,000
The miracle that occurs here,
by contrast,

359
00:25:07,000 --> 00:25:13,000
is that it turns out to be easy
to find (x)p.

360
00:25:11,000 --> 00:25:17,000
And easy in this further sense
that I do not have to restrict

361
00:25:17,000 --> 00:25:23,000
the kind of function I use.
For example,

362
00:25:21,000 --> 00:25:27,000
the second homework problem I
have given you,

363
00:25:26,000 --> 00:25:32,000
the second part two homework
problem.

364
00:25:36,000 --> 00:25:42,000
You will see how to use
systems.

365
00:25:38,000 --> 00:25:44,000
For example,
to solve this simple equation,

366
00:25:42,000 --> 00:25:48,000
I will write it out for you,
consider that equation,

367
00:25:46,000 --> 00:25:52,000
tangent t.
What technique will you apply

368
00:25:50,000 --> 00:25:56,000
to solve that?
In other words,

369
00:25:52,000 --> 00:25:58,000
suppose you wanted to find a
particular solution to that.

370
00:25:57,000 --> 00:26:03,000
The right-hand side is not an
exponential.

371
00:26:00,000 --> 00:26:06,000
It is not a polynomial.
It is not like sine or cosine

372
00:26:06,000 --> 00:26:12,000
of bt.
I could use the Laplace

373
00:26:10,000 --> 00:26:16,000
transform.
No, because you don't know how

374
00:26:14,000 --> 00:26:20,000
to take the Laplace transform of
tangent t.

375
00:26:19,000 --> 00:26:25,000
Neither, for that matter,
do I.

376
00:26:21,000 --> 00:26:27,000
Fourier series.
Not a good choice for a

377
00:26:25,000 --> 00:26:31,000
function that goes to infinity
at pi over two.

378
00:26:35,000 --> 00:26:41,000
So you cannot do this until you
do your homework.

379
00:26:39,000 --> 00:26:45,000
Now you will be able to do it.
In other words,

380
00:26:44,000 --> 00:26:50,000
one of the big things is not
only will I give you a formula

381
00:26:50,000 --> 00:26:56,000
for the Xp but that formula will
work even for tangent t,

382
00:26:56,000 --> 00:27:02,000
any function at all.
Well, I thought I would try to

383
00:27:00,000 --> 00:27:06,000
put a little meat on the bones
of the inhomogeneous systems by

384
00:27:04,000 --> 00:27:10,000
actually giving you a physical
problem so we would actually be

385
00:27:08,000 --> 00:27:14,000
able to solve a physical problem
instead of just demonstrate a

386
00:27:12,000 --> 00:27:18,000
solution method.
Here is a mixing problem.

387
00:27:23,000 --> 00:27:29,000
Just to illustrate what makes a
system of equations

388
00:27:27,000 --> 00:27:33,000
inhomogeneous,
here at two ugly tanks.

389
00:27:31,000 --> 00:27:37,000
I am not going to draw these
carefully, but they are both 1

390
00:27:37,000 --> 00:27:43,000
liter.
And they are connected by

391
00:27:40,000 --> 00:27:46,000
pipes.
And I won't bother opening

392
00:27:43,000 --> 00:27:49,000
holes in them.
There is a pipe with fluids

393
00:27:47,000 --> 00:27:53,000
flowing back there and this
direction it is flowing this

394
00:27:52,000 --> 00:27:58,000
way, but that is not the end.
The end is there is stuff

395
00:28:00,000 --> 00:28:06,000
coming in to both of them.
And I think I will just make it

396
00:28:08,000 --> 00:28:14,000
coming out of this one.
There is something realistic.

397
00:28:15,000 --> 00:28:21,000
The numbers 2,
3, 2.

398
00:28:18,000 --> 00:28:24,000
Let's start there and see what
the others have to be.

399
00:28:25,000 --> 00:28:31,000
So these are flow rates.
One liter tanks.

400
00:28:31,000 --> 00:28:37,000
The flow rates are in,
let's say, liters per hour.

401
00:28:38,000 --> 00:28:44,000
And I have some dissolved
substance in,

402
00:28:42,000 --> 00:28:48,000
so here is going to be x salt
in there and the same chemical

403
00:28:50,000 --> 00:28:56,000
in there, whatever it is.
x is the amount of salt,

404
00:28:56,000 --> 00:29:02,000
let's say, in tank one.
And y, the same thing in tank

405
00:29:02,000 --> 00:29:08,000
two.
Now, if you have stuff flowing

406
00:29:06,000 --> 00:29:12,000
unequally this way,
you must have balance.

407
00:29:09,000 --> 00:29:15,000
You have to make sure that
neither tank is getting emptied

408
00:29:15,000 --> 00:29:21,000
or bursting and exploding.
What is flowing in?

409
00:29:19,000 --> 00:29:25,000
What is x?
Three is going out,

410
00:29:22,000 --> 00:29:28,000
two is coming in,
so this has to be one in order

411
00:29:27,000 --> 00:29:33,000
that tank x stay full and not
explode.

412
00:29:32,000 --> 00:29:38,000
And how about y?
How much is going out?

413
00:29:35,000 --> 00:29:41,000
Two there and two here.
Four is going out,

414
00:29:39,000 --> 00:29:45,000
three is coming in.
This also has to be one.

415
00:29:43,000 --> 00:29:49,000
Those are just the flow rates
of water or the liquid that is

416
00:29:48,000 --> 00:29:54,000
coming in.
Now, the only thing I am going

417
00:29:52,000 --> 00:29:58,000
to specify is the concentration
of what is coming in.

418
00:29:58,000 --> 00:30:04,000
Here the concentration is 5 e
to the minus t.

419
00:30:02,000 --> 00:30:08,000
And that is what makes the
problem inhomogeneous.

420
00:30:07,000 --> 00:30:13,000
Here the concentration is going
to be zero.

421
00:30:10,000 --> 00:30:16,000
In other words,
pure water is flowing in here

422
00:30:14,000 --> 00:30:20,000
to create the liquid balance.
Here, on the other hand,

423
00:30:18,000 --> 00:30:24,000
salt solution is flowing in but
with a steadily declining

424
00:30:23,000 --> 00:30:29,000
concentration.
So, what is the system?

425
00:30:28,000 --> 00:30:34,000
Well, you have set it up
exactly the way you did when you

426
00:30:34,000 --> 00:30:40,000
studied first order equations.
It is inflow minus outflow.

427
00:30:41,000 --> 00:30:47,000
What is the outflow?
The outflow is all in this

428
00:30:46,000 --> 00:30:52,000
pipe.
The flow rates are liters per

429
00:30:50,000 --> 00:30:56,000
hour.
Three liters per hour flowing

430
00:30:54,000 --> 00:31:00,000
out.
How much salt does that

431
00:30:57,000 --> 00:31:03,000
represent?
It is negative three times the

432
00:31:02,000 --> 00:31:08,000
concentration of salt.
But the concentration,

433
00:31:06,000 --> 00:31:12,000
notice, equals x divided
by one.

434
00:31:10,000 --> 00:31:16,000
In other words,
x represents both the

435
00:31:13,000 --> 00:31:19,000
concentration and the amount.
So I don't have to distinguish.

436
00:31:18,000 --> 00:31:24,000
If I had made it two liter
tanks then I would have had to

437
00:31:23,000 --> 00:31:29,000
divide this by two.
I am cheating,

438
00:31:26,000 --> 00:31:32,000
but it is enough already.
x prime equals minus 3x.

439
00:31:32,000 --> 00:31:38,000
That is what is going out.
What is coming in?

440
00:31:36,000 --> 00:31:42,000
Well, 2y is coming in.
Concentration here.

441
00:31:39,000 --> 00:31:45,000
What is coming in?
Is it y 2 liter?

442
00:31:43,000 --> 00:31:49,000
Plus what is coming in from the
outside.

443
00:31:46,000 --> 00:31:52,000
We have to add that in,
and that will be plus 5 e to

444
00:31:51,000 --> 00:31:57,000
the negative t.
How about y?

445
00:31:54,000 --> 00:32:00,000
y prime is changing.
What comes in from x?

446
00:32:00,000 --> 00:32:06,000
That is 3x.
What goes out?

447
00:32:01,000 --> 00:32:07,000
Well, two is leaving here and
two is leaving here.

448
00:32:05,000 --> 00:32:11,000
It doesn't matter that they are
going out through separate

449
00:32:10,000 --> 00:32:16,000
pipes.
They are both going out.

450
00:32:12,000 --> 00:32:18,000
It is minus 4,
2 and 2.

451
00:32:14,000 --> 00:32:20,000
How about the inhomogeneous
term?

452
00:32:16,000 --> 00:32:22,000
There is one coming in,
but there is no salt in it.

453
00:32:20,000 --> 00:32:26,000
Therefore, that is not
changing.

454
00:32:23,000 --> 00:32:29,000
What is coming through that
pipe is necessary for the liquid

455
00:32:27,000 --> 00:32:33,000
balance.
But it has no effect

456
00:32:31,000 --> 00:32:37,000
whatsoever.
I will put a zero here but,

457
00:32:35,000 --> 00:32:41,000
of course, you don't have to
put that in.

458
00:32:38,000 --> 00:32:44,000
This is now an inhomogeneous
system.

459
00:32:42,000 --> 00:32:48,000
In other words,
the system is x prime equals

460
00:32:46,000 --> 00:32:52,000
this matrix, negative 3,
the same sort of stuff we

461
00:32:50,000 --> 00:32:56,000
always had, plus the
inhomogeneous term which is the

462
00:32:55,000 --> 00:33:01,000
column vector 5 e to the minus t
and zero.

463
00:33:00,000 --> 00:33:06,000
It is the presence of this term

464
00:33:04,000 --> 00:33:10,000
that makes this system
inhomogeneous.

465
00:33:06,000 --> 00:33:12,000
And what that corresponds to is
this little closed system being

466
00:33:11,000 --> 00:33:17,000
attacked from the outside by
these external pipes which are

467
00:33:15,000 --> 00:33:21,000
bringing salt in.
Without those,

468
00:33:17,000 --> 00:33:23,000
of course the balance would be
all wrong.

469
00:33:20,000 --> 00:33:26,000
I would have to change this to
three and cut that out,

470
00:33:24,000 --> 00:33:30,000
I guess.
But then, it would be just a

471
00:33:27,000 --> 00:33:33,000
simple homogenous system.
It is these pipes that make it

472
00:33:32,000 --> 00:33:38,000
inhomogeneous.
Now, I should start to solve

473
00:33:35,000 --> 00:33:41,000
that.
I did this just to illustrate

474
00:33:38,000 --> 00:33:44,000
where a system might come from.
Before I solve that,

475
00:33:42,000 --> 00:33:48,000
what I want to do is,
of course, is solve it in

476
00:33:45,000 --> 00:33:51,000
general.
In other words,

477
00:33:47,000 --> 00:33:53,000
how do you solve this in
general?

478
00:33:49,000 --> 00:33:55,000
Because I promised you that you
would be able to do in general,

479
00:33:54,000 --> 00:34:00,000
regardless of what sort of
functions were in the r of t,

480
00:33:58,000 --> 00:34:04,000
that column vector.
So let's do it.

481
00:34:10,000 --> 00:34:16,000
First of all,
you have to learn the name of

482
00:34:15,000 --> 00:34:21,000
the method.
This method is for solving x

483
00:34:20,000 --> 00:34:26,000
prime equals Ax.
It is a method for finding a

484
00:34:26,000 --> 00:34:32,000
particular solution.

485
00:34:36,000 --> 00:34:42,000
Of course, to actually solve it
then you have to add the

486
00:34:39,000 --> 00:34:45,000
complimentary function.
We are looking for a particular

487
00:34:43,000 --> 00:34:49,000
solution for this system.
Now, the whole cleverness of

488
00:34:47,000 --> 00:34:53,000
the method, which I think was
discovered a couple hundred

489
00:34:51,000 --> 00:34:57,000
years ago by,
I think, Lagrange,

490
00:34:53,000 --> 00:34:59,000
I am not sure.
The method is called variation

491
00:34:57,000 --> 00:35:03,000
of parameters.
I am giving you that so that

492
00:35:00,000 --> 00:35:06,000
when you forget you will be able
to look it up and be indexes to

493
00:35:05,000 --> 00:35:11,000
some advanced engineering
mathematics book or something,

494
00:35:08,000 --> 00:35:14,000
whatever is on your shelf.
But, if course,

495
00:35:11,000 --> 00:35:17,000
you won't remember the name
either so maybe this won't work.

496
00:35:15,000 --> 00:35:21,000
Variation of parameters,
I will explain to you why it is

497
00:35:19,000 --> 00:35:25,000
called that.
All the cleverness is in the

498
00:35:22,000 --> 00:35:28,000
very first line.
If you could remember the very

499
00:35:25,000 --> 00:35:31,000
first line then I trust you to
do the rest yourself.

500
00:35:30,000 --> 00:35:36,000
I don't know any motivation for
this first step,

501
00:35:34,000 --> 00:35:40,000
but mathematics is supposed to
be mysterious anyway.

502
00:35:40,000 --> 00:35:46,000
It keeps me eating.
It says, look for a solution

503
00:35:45,000 --> 00:35:51,000
and there will be one of the
following form.

504
00:35:49,000 --> 00:35:55,000
Now, it will look exactly like
--

505
00:36:00,000 --> 00:36:06,000
Look carefully because it is
going to be gone in a moment.

506
00:36:04,000 --> 00:36:10,000
It will look exactly like this.
But, of course,

507
00:36:08,000 --> 00:36:14,000
it cannot be this because this
solves the homogeneous system.

508
00:36:13,000 --> 00:36:19,000
If I plug this in with these as
constants it cannot possibly be

509
00:36:18,000 --> 00:36:24,000
a particular solution to this
because it will stop there and

510
00:36:23,000 --> 00:36:29,000
satisfy that with r equals zero.
The whole trick is you think of

511
00:36:29,000 --> 00:36:35,000
these are parameters which are
now variable.

512
00:36:32,000 --> 00:36:38,000
Constants that are varying.
That is why it is called

513
00:36:35,000 --> 00:36:41,000
variation of parameters.
You think of these,

514
00:36:38,000 --> 00:36:44,000
in other words,
as functions of t.

515
00:36:41,000 --> 00:36:47,000
We are going to look for a
solution which has the form,

516
00:36:45,000 --> 00:36:51,000
since they are functions of t,
I don't want to call them c1

517
00:36:49,000 --> 00:36:55,000
and c2 anymore.
I will call them v because that

518
00:36:52,000 --> 00:36:58,000
is what most people call them,
v or u, sometimes.

519
00:37:02,000 --> 00:37:08,000
The method says look for a
solution of that form.

520
00:37:06,000 --> 00:37:12,000
The variation parameters,
these are the parameters that

521
00:37:10,000 --> 00:37:16,000
are now varying instead of being
constants.

522
00:37:14,000 --> 00:37:20,000
Now, if you take it in that
form and start trying to

523
00:37:18,000 --> 00:37:24,000
substitute into the equation you
are going to get a mess.

524
00:37:23,000 --> 00:37:29,000
I think I was wrong in saying I
could trust you from this point

525
00:37:28,000 --> 00:37:34,000
on.
I will take the first step from

526
00:37:32,000 --> 00:37:38,000
you, and then I could trust you
to do the rest after that first

527
00:37:38,000 --> 00:37:44,000
step.
The first step is to change the

528
00:37:42,000 --> 00:37:48,000
way this looks by using the
fundamental matrix.

529
00:37:46,000 --> 00:37:52,000
Remember what the fundamental
matrix was?

530
00:37:50,000 --> 00:37:56,000
Its entries were the two
columns of solutions.

531
00:37:54,000 --> 00:38:00,000
These are solutions to the
homogeneous system.

532
00:38:00,000 --> 00:38:06,000
And I am going to write it
using the fundamental matrix as,

533
00:38:05,000 --> 00:38:11,000
now thinks about it.
The fundamental matrix has

534
00:38:09,000 --> 00:38:15,000
columns x1 and x2.
Your instinct might be using

535
00:38:13,000 --> 00:38:19,000
matrix multiplication to put the
v1 and the v2 here,

536
00:38:18,000 --> 00:38:24,000
but that won't work.
You have to put them here.

537
00:38:30,000 --> 00:38:36,000
This says the same thing as
that.

538
00:38:33,000 --> 00:38:39,000
Let's just take a second out to
calculate.

539
00:38:37,000 --> 00:38:43,000
The x is going to look like
(x1, y1).

540
00:38:40,000 --> 00:38:46,000
That is my first solution.
My second solution,

541
00:38:44,000 --> 00:38:50,000
here is the fundamental matrix,
is (x2, y2).

542
00:38:49,000 --> 00:38:55,000
And I am multiplying this on
the right by (v1,

543
00:38:53,000 --> 00:38:59,000
v2).
Does it come out right?

544
00:38:56,000 --> 00:39:02,000
Look.
What is it?

545
00:38:59,000 --> 00:39:05,000
The top is x1 v1 plus x2 v2.

546
00:39:02,000 --> 00:39:08,000
The top, x1 v1 plus x2 v2.

547
00:39:06,000 --> 00:39:12,000
It is in the wrong order,
but multiplication is

548
00:39:10,000 --> 00:39:16,000
commutative, fortunately.
And the same way the bottom

549
00:39:14,000 --> 00:39:20,000
thing will be v1 y1 plus v2 y2.

550
00:39:18,000 --> 00:39:24,000
If I had written it on the
other side instead,

551
00:39:22,000 --> 00:39:28,000
which is tempting because the
v's occur on the left here,

552
00:39:27,000 --> 00:39:33,000
that won't work.
What will I get?

553
00:39:31,000 --> 00:39:37,000
I will get v1 x1 plus v2 y1,

554
00:39:35,000 --> 00:39:41,000
which is not at all what I
want.

555
00:39:37,000 --> 00:39:43,000
You must put it on the right.
But this is a very important

556
00:39:42,000 --> 00:39:48,000
thing.
This is going to plague us on

557
00:39:45,000 --> 00:39:51,000
Monday, too.
It must be written on the right

558
00:39:49,000 --> 00:39:55,000
and not on the left as a column
vector.

559
00:39:52,000 --> 00:39:58,000
The rest of the program is very
simple.

560
00:39:55,000 --> 00:40:01,000
I will write it out as a
program.

561
00:40:00,000 --> 00:40:06,000
Substitute into the system,
into that, in other words,

562
00:40:04,000 --> 00:40:10,000
and see what v has to be.
That is what we are looking

563
00:40:08,000 --> 00:40:14,000
for.
We know what the x1 and x2 are.

564
00:40:11,000 --> 00:40:17,000
It is a question of what those
coefficients are.

565
00:40:14,000 --> 00:40:20,000
And see what v is.
Let's do it.

566
00:40:32,000 --> 00:40:38,000
Let's substitute.
Let's see.

567
00:40:35,000 --> 00:40:41,000
The system is x prime equals Ax
plus r.

568
00:40:41,000 --> 00:40:47,000
I want to put in (x)p,
this proposed particular

569
00:40:46,000 --> 00:40:52,000
solution.
And it is a fundamental matrix,

570
00:40:50,000 --> 00:40:56,000
and the v is unknown.
How do I differentiate the

571
00:40:56,000 --> 00:41:02,000
product of two matrices?
You differentiate the product

572
00:41:02,000 --> 00:41:08,000
of two matrices using the
product rule that you learned

573
00:41:07,000 --> 00:41:13,000
the first day of 18.01.
Trust me.

574
00:41:10,000 --> 00:41:16,000
Let's do it.
I am going to substitute in.

575
00:41:14,000 --> 00:41:20,000
In other words,
here is my (x)p,

576
00:41:17,000 --> 00:41:23,000
(x)p, and I am going to write
in what that is.

577
00:41:21,000 --> 00:41:27,000
The left-hand side is the
derivative of,

578
00:41:25,000 --> 00:41:31,000
X prime times v,
plus X times the derivative of

579
00:41:29,000 --> 00:41:35,000
v.
Notice that one of these is a

580
00:41:34,000 --> 00:41:40,000
column vector and the other is a
square matrix.

581
00:41:37,000 --> 00:41:43,000
That is perfectly Okay.
Any two matrices which are the

582
00:41:39,000 --> 00:41:45,000
rate shape so you can multiply
them together,

583
00:41:42,000 --> 00:41:48,000
if you want to differentiate
their product,

584
00:41:44,000 --> 00:41:50,000
in other words,
if the entries are functions of

585
00:41:46,000 --> 00:41:52,000
t it is the product rule.
The derivative of this times

586
00:41:49,000 --> 00:41:55,000
time plus that times the
derivative of this.

587
00:41:52,000 --> 00:41:58,000
You have to keep them in the
right order.

588
00:41:54,000 --> 00:42:00,000
You are not allowed to shuffle
them around carelessly.

589
00:41:57,000 --> 00:42:03,000
So that is that.
What is it equal to?

590
00:42:00,000 --> 00:42:06,000
Well, the right-hand side is A.
And now I substitute just (x)p

591
00:42:08,000 --> 00:42:14,000
in, so that is X times v plus r.
Is this progress?

592
00:42:14,000 --> 00:42:20,000
What is v?
It looks like a mess but it is

593
00:42:20,000 --> 00:42:26,000
not.
Why not?

594
00:42:22,000 --> 00:42:28,000
It is because this is not any
old matrix X.

595
00:42:29,000 --> 00:42:35,000
This is a matrix whose columns
are solutions to the system.

596
00:42:34,000 --> 00:42:40,000
And what does that do?
That means X prime satisfies

597
00:42:39,000 --> 00:42:45,000
that matrix differential
equation.

598
00:42:42,000 --> 00:42:48,000
X prime is the same as Ax.

599
00:42:45,000 --> 00:42:51,000
And, by a little miracle,
the v is tagging along in both

600
00:42:50,000 --> 00:42:56,000
cases.
This cancels that and now there

601
00:42:54,000 --> 00:43:00,000
is very little left.
The conclusion,

602
00:42:58,000 --> 00:43:04,000
therefore, is that Xv is equal
to r.

603
00:43:02,000 --> 00:43:08,000
What is v?
It is v that we are looking

604
00:43:05,000 --> 00:43:11,000
for, right?
You have to solve a matrix

605
00:43:09,000 --> 00:43:15,000
equation, now.
This is a square matrix so you

606
00:43:13,000 --> 00:43:19,000
have to do it by inverting the
matrix.

607
00:43:16,000 --> 00:43:22,000
You don't just sloppily divide.
You multiply on which side by

608
00:43:21,000 --> 00:43:27,000
what matrix?
Choice of left or right.

609
00:43:25,000 --> 00:43:31,000
You multiply by the inverse
matrix on the left or on the

610
00:43:29,000 --> 00:43:35,000
right?
It has to be on the left.

611
00:43:35,000 --> 00:43:41,000
Multiply both sides of the
equation by X inverse on the

612
00:43:42,000 --> 00:43:48,000
left, and then you will get v is
equal to X inverse r.

613
00:43:50,000 --> 00:43:56,000
How do I know the X inverse

614
00:43:55,000 --> 00:44:01,000
exists?
Does X inverse exist?

615
00:44:00,000 --> 00:44:06,000
For a matrix inverse to exist,
the matrix's determinant must

616
00:44:05,000 --> 00:44:11,000
be not zero.
Why is the determinant of this

617
00:44:10,000 --> 00:44:16,000
not zero?
Because its columns are

618
00:44:13,000 --> 00:44:19,000
independent solutions.

619
00:44:22,000 --> 00:44:28,000
Of course this is not right.
I forgot the prime here.

620
00:44:31,000 --> 00:44:37,000
I am not failing this course
after all.

621
00:44:34,000 --> 00:44:40,000
v prime equals that.

622
00:44:43,000 --> 00:44:49,000
This is done by differentiating
each entry in the column vector.

623
00:44:47,000 --> 00:44:53,000
And, therefore,
we should integrate it.

624
00:44:50,000 --> 00:44:56,000
It will be the integral,
just the ordinary

625
00:44:53,000 --> 00:44:59,000
anti-derivative of x inverse
times r.

626
00:44:57,000 --> 00:45:03,000
This is a column vector.
The entries are functions of t.

627
00:45:02,000 --> 00:45:08,000
You simply integrate each of
those functions in turn.

628
00:45:06,000 --> 00:45:12,000
So integrate each entry.

629
00:45:12,000 --> 00:45:18,000
There is my v.
Sorry, you cannot tell the v's

630
00:45:19,000 --> 00:45:25,000
from the r's here.
And so, finally,

631
00:45:25,000 --> 00:45:31,000
the particular solution is (x)p
is equal to --

632
00:45:34,000 --> 00:45:40,000
It is really not bad at all.
It is equal to X times v.

633
00:45:37,000 --> 00:45:43,000
It's equal to X times the
integral of X inverse r dt.

634
00:45:40,000 --> 00:45:46,000
Now, actually,

635
00:45:43,000 --> 00:45:49,000
there is not much work to doing
that.

636
00:45:45,000 --> 00:45:51,000
Once you have solved the
homogeneous system and gotten

637
00:45:49,000 --> 00:45:55,000
the fundamental matrix,
taking the inverse of a

638
00:45:52,000 --> 00:45:58,000
two-by-two matrix is almost
trivial.

639
00:45:54,000 --> 00:46:00,000
You flip those two and you
change the signs of these two

640
00:45:57,000 --> 00:46:03,000
and you divide by the
determinant.

641
00:46:01,000 --> 00:46:07,000
You multiply it by r.
And the hard part is if you can

642
00:46:04,000 --> 00:46:10,000
do the integration.
If not, you just leave the

643
00:46:07,000 --> 00:46:13,000
integral sign the way you have
learned to do in this silly

644
00:46:11,000 --> 00:46:17,000
course and you still have the
answer.

645
00:46:14,000 --> 00:46:20,000
What about the arbitrary
constant of integration?

646
00:46:17,000 --> 00:46:23,000
The answer is you don't need to
put it in.

647
00:46:20,000 --> 00:46:26,000
Just find one particular
solution.

648
00:46:22,000 --> 00:46:28,000
It is good enough.
You don't have to put in the

649
00:46:25,000 --> 00:46:31,000
arbitrary constants of
integration.

650
00:46:29,000 --> 00:46:35,000
Because they are already in the
complimentary function here.

651
00:46:33,000 --> 00:46:39,000
Therefore, you don't have to
add them.

652
00:46:35,000 --> 00:46:41,000
I am sorry I didn't get a
chance to actually solve that.

653
00:46:39,000 --> 00:46:45,000
I will have to let it go.
The recitations will do it on

654
00:46:42,000 --> 00:46:48,000
Tuesday, will solve that
particular problem,

655
00:46:45,000 --> 00:46:51,000
which means you will,
in effect.