WEBVTT
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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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it is never zero.
It is not zero for any value of
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t.
That is good.
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As you will see,
we are going to need that
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property.
But the other one is a little
00:14:57.000 --> 00:15:03.000
stranger.
The only thing I can say is,
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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.
00:15:20.000 --> 00:15:26.000
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
00:15:32.000 --> 00:15:38.000
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?
00:15:53.000 --> 00:15:59.000
This is a two-by-two matrix.
When I multiply them I get a
00:15:58.000 --> 00:16:04.000
two-by-two matrix.
What is this?
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This is a two-by-two matrix,
every entry of which has been
00:16:04.000 --> 00:16:10.000
differentiated.
That is what it means to put
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that prime there.
To differentiate a matrix means
00:16:11.000 --> 00:16:17.000
nothing fancy.
It just means differentiate
00:16:14.000 --> 00:16:20.000
every entry.
It is just like to
00:16:16.000 --> 00:16:22.000
differentiate a vector (x,
y), to make a velocity vector
00:16:20.000 --> 00:16:26.000
you differentiate the x and the
y.
00:16:22.000 --> 00:16:28.000
Well, a column vector is a
special kind of matrix.
00:16:26.000 --> 00:16:32.000
The definition applies to any
matrix.
00:16:30.000 --> 00:16:36.000
Well, why is that so?
I state it as a property,
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but I will continue it by
giving you, so to speak,
00:16:37.000 --> 00:16:43.000
the proof of it.
In fact, there is nothing in
00:16:40.000 --> 00:16:46.000
this.
It is nothing more than a
00:16:42.000 --> 00:16:48.000
little matrix calculation of the
most primitive kind.
00:16:46.000 --> 00:16:52.000
Namely, what does this mean?
Let's try to undo that.
00:16:50.000 --> 00:16:56.000
What does the left-hand side
really mean?
00:16:53.000 --> 00:16:59.000
Well, if that is what x means,
the left-hand side must mean
00:16:58.000 --> 00:17:04.000
the derivative of the first
column.
00:17:02.000 --> 00:17:08.000
That is its first column.
And the derivative of the
00:17:05.000 --> 00:17:11.000
second column.
That is what it means to
00:17:07.000 --> 00:17:13.000
differentiate the matrix X.
You differentiate each column
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separately.
And to differentiate the column
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you need to differentiate every
function in it.
00:17:17.000 --> 00:17:23.000
Well, what does the right-hand
side mean?
00:17:19.000 --> 00:17:25.000
Well, I am supposed to take A
and multiply that
00:17:23.000 --> 00:17:29.000
by [x1,x2].
Now, I don't know how to prove
00:17:26.000 --> 00:17:32.000
this, except ask you to think
about it.
00:17:30.000 --> 00:17:36.000
Or, I could write it all out
here.
00:17:34.000 --> 00:17:40.000
But think of this as a bing,
bing, bing, bing.
00:17:39.000 --> 00:17:45.000
And this is a bing,
bing.
00:17:42.000 --> 00:17:48.000
And this is a bong,
bong.
00:17:45.000 --> 00:17:51.000
How do I do the multiplication?
In other words,
00:17:50.000 --> 00:17:56.000
what is in the first column of
the matrix?
00:17:55.000 --> 00:18:01.000
Well, it is dah,
dah, and the lower thing is
00:18:01.000 --> 00:18:07.000
dah, dah.
In other words,
00:18:03.000 --> 00:18:09.000
it is A times x1.
00:18:15.000 --> 00:18:21.000
Shut your eyes and visualize
it.
00:18:17.000 --> 00:18:23.000
Got it?
Dah, dah is the top entry,
00:18:19.000 --> 00:18:25.000
and dah, dah is the bottom
entry.
00:18:22.000 --> 00:18:28.000
It is what you get by
multiplying A by the column
00:18:25.000 --> 00:18:31.000
vector x1.
And the same way the other guy
00:18:28.000 --> 00:18:34.000
is --
-- what you get by multiplying
00:18:33.000 --> 00:18:39.000
A by the column vector x2.
This is just matrix
00:18:37.000 --> 00:18:43.000
multiplication.
That is the law of matrix
00:18:41.000 --> 00:18:47.000
multiplication.
That is how you multiply
00:18:45.000 --> 00:18:51.000
matrices.
Well, good, but where does this
00:18:49.000 --> 00:18:55.000
get us?
What does it mean for those two
00:18:53.000 --> 00:18:59.000
guys to be equal?
That is going to happen,
00:18:57.000 --> 00:19:03.000
if and only if x1 prime is
equal to A x1.
00:19:02.000 --> 00:19:08.000
This guy equals that guy.
And similarly for the x2's.
00:19:14.000 --> 00:19:20.000
The end result is that this
matrix, saying that the
00:19:18.000 --> 00:19:24.000
fundamental matrix satisfies
this matrix differential
00:19:22.000 --> 00:19:28.000
equation is only a way of
saying, in one breath,
00:19:26.000 --> 00:19:32.000
that its two columns are both
solutions to the original
00:19:30.000 --> 00:19:36.000
system.
It is, so to speak,
00:19:33.000 --> 00:19:39.000
an efficient way of turning
these two equations into a
00:19:38.000 --> 00:19:44.000
single equation by making a
matrix.
00:19:41.000 --> 00:19:47.000
I guess it is time,
finally, to come to the topic
00:19:46.000 --> 00:19:52.000
of the lecture.
I said the thing the matrices
00:19:50.000 --> 00:19:56.000
were going to be used for is
solving inhomogeneous systems,
00:19:55.000 --> 00:20:01.000
so let's take a look at those.
I thought I would give you an
00:20:00.000 --> 00:20:06.000
example.
Inhomogeneous systems.
00:20:10.000 --> 00:20:16.000
Well, what is one going to look
like?
00:20:12.000 --> 00:20:18.000
So far what we have done is,
up until now has been solving,
00:20:17.000 --> 00:20:23.000
we spent essentially two weeks
solving and plotting the
00:20:21.000 --> 00:20:27.000
solutions to homogeneous
systems.
00:20:23.000 --> 00:20:29.000
There was nothing over there.
And homogeneous systems,
00:20:27.000 --> 00:20:33.000
in fact, with constant
coefficients.
00:20:31.000 --> 00:20:37.000
Stuff that looked like that
that we abbreviated with
00:20:34.000 --> 00:20:40.000
matrices.
Now, to make the system
00:20:36.000 --> 00:20:42.000
inhomogeneous what I do is add
the extra term on the right-hand
00:20:41.000 --> 00:20:47.000
side, which is some function of
t.
00:20:43.000 --> 00:20:49.000
Except, I will have to have two
functions of t because I have
00:20:47.000 --> 00:20:53.000
two equations.
Now it is inhomogeneous.
00:20:50.000 --> 00:20:56.000
And what makes it inhomogeneous
is the fact that these are not
00:20:54.000 --> 00:21:00.000
zero anymore.
There is something there.
00:20:57.000 --> 00:21:03.000
Functions of t are there.
These are given functions of t
00:21:02.000 --> 00:21:08.000
like exponentials,
polynomials,
00:21:04.000 --> 00:21:10.000
the usual stuff you have on the
right-hand side of the
00:21:08.000 --> 00:21:14.000
differential equation.
What is confusing here is that
00:21:11.000 --> 00:21:17.000
when we studied second order
equations it was homogeneous if
00:21:15.000 --> 00:21:21.000
the right-hand side was zero,
and if there was something else
00:21:20.000 --> 00:21:26.000
there it was inhomogeneous.
Unfortunately,
00:21:23.000 --> 00:21:29.000
I have stuck this stuff on the
right-hand side so it is not
00:21:27.000 --> 00:21:33.000
quite so clear anymore.
It has got to look like that,
00:21:32.000 --> 00:21:38.000
in other words.
How would the matrix
00:21:34.000 --> 00:21:40.000
abbreviation look?
Well, the left-hand side is x
00:21:37.000 --> 00:21:43.000
prime.
The homogenous part is ax,
00:21:40.000 --> 00:21:46.000
just as it has always been.
The only extra part is those
00:21:43.000 --> 00:21:49.000
functions r.
And this is a column vector,
00:21:46.000 --> 00:21:52.000
after the multiplication this
is a column vector,
00:21:50.000 --> 00:21:56.000
what is left is column vector.
Now, explicitly it is a
00:21:53.000 --> 00:21:59.000
function of t,
given by explicit functions of
00:21:56.000 --> 00:22:02.000
t, again, like exponentials.
Or, they could be fancy
00:22:02.000 --> 00:22:08.000
functions.
That is the thing we are trying
00:22:06.000 --> 00:22:12.000
to solve.
Why don't I put it up in green?
00:22:10.000 --> 00:22:16.000
Our new and better and improved
system.
00:22:13.000 --> 00:22:19.000
Think back to what we did when
we studied inhomogeneous
00:22:18.000 --> 00:22:24.000
equations.
We are not talking about
00:22:22.000 --> 00:22:28.000
systems but just a single
equation.
00:22:25.000 --> 00:22:31.000
What we did was the main
theorem --
00:22:30.000 --> 00:22:36.000
I guess there are going to be
three theorems today,
00:22:36.000 --> 00:22:42.000
not just two.
Theorem C.
00:22:39.000 --> 00:22:45.000
Is that right?
Yes, A, B.
00:22:42.000 --> 00:22:48.000
We are up to C.
Theorem C says that the general
00:22:48.000 --> 00:22:54.000
solution, that is,
the general solution to the
00:22:54.000 --> 00:23:00.000
system, is equal to the
complimentary function,
00:23:00.000 --> 00:23:06.000
which is the general solution
to x prime equals Ax,
00:23:06.000 --> 00:23:12.000
--
-- the homogeneous equation,
00:23:11.000 --> 00:23:17.000
in other words,
plus, what am I going to call
00:23:15.000 --> 00:23:21.000
it?
(x)p, right you are,
00:23:17.000 --> 00:23:23.000
a particular solution.
But the principle is the same
00:23:22.000 --> 00:23:28.000
and is proved exactly the same
way.
00:23:25.000 --> 00:23:31.000
It is just linearity and
superposition.
00:23:35.000 --> 00:23:41.000
The linearity of the original
system and the superposition
00:23:40.000 --> 00:23:46.000
principle.
The essence is that to solve
00:23:43.000 --> 00:23:49.000
this inhomogeneous system,
what we have to do is find a
00:23:48.000 --> 00:23:54.000
particular solution.
This part I already know how to
00:23:52.000 --> 00:23:58.000
do.
We have been doing that for two
00:23:55.000 --> 00:24:01.000
weeks.
The new thing is to find this.
00:24:05.000 --> 00:24:11.000
Now, if you remember back
before spring break,
00:24:08.000 --> 00:24:14.000
most of the work in solving the
second order equation was in
00:24:12.000 --> 00:24:18.000
finding that particular
solution.
00:24:15.000 --> 00:24:21.000
You quickly enough learned how
to solve the homogeneous
00:24:19.000 --> 00:24:25.000
equation, but there was no real
general method for finding this.
00:24:24.000 --> 00:24:30.000
We had an exponential input
theorem with some modifications
00:24:28.000 --> 00:24:34.000
to it.
We took a week's detour in
00:24:32.000 --> 00:24:38.000
Fourier series to see how to do
it for periodic functions or
00:24:37.000 --> 00:24:43.000
functions defined on finite
intervals.
00:24:40.000 --> 00:24:46.000
There were other techniques
which I did not get around to
00:24:45.000 --> 00:24:51.000
showing you, techniques
involving the so-called method
00:24:50.000 --> 00:24:56.000
of undetermined coefficients.
Although, some of you peaked in
00:24:55.000 --> 00:25:01.000
your book and learned it from
there.
00:25:00.000 --> 00:25:06.000
But the work is in finding
(x)p.
00:25:03.000 --> 00:25:09.000
The miracle that occurs here,
by contrast,
00:25:07.000 --> 00:25:13.000
is that it turns out to be easy
to find (x)p.
00:25:11.000 --> 00:25:17.000
And easy in this further sense
that I do not have to restrict
00:25:17.000 --> 00:25:23.000
the kind of function I use.
For example,
00:25:21.000 --> 00:25:27.000
the second homework problem I
have given you,
00:25:26.000 --> 00:25:32.000
the second part two homework
problem.
00:25:36.000 --> 00:25:42.000
You will see how to use
systems.
00:25:38.000 --> 00:25:44.000
For example,
to solve this simple equation,
00:25:42.000 --> 00:25:48.000
I will write it out for you,
consider that equation,
00:25:46.000 --> 00:25:52.000
tangent t.
What technique will you apply
00:25:50.000 --> 00:25:56.000
to solve that?
In other words,
00:25:52.000 --> 00:25:58.000
suppose you wanted to find a
particular solution to that.
00:25:57.000 --> 00:26:03.000
The right-hand side is not an
exponential.
00:26:00.000 --> 00:26:06.000
It is not a polynomial.
It is not like sine or cosine
00:26:06.000 --> 00:26:12.000
of bt.
I could use the Laplace
00:26:10.000 --> 00:26:16.000
transform.
No, because you don't know how
00:26:14.000 --> 00:26:20.000
to take the Laplace transform of
tangent t.
00:26:19.000 --> 00:26:25.000
Neither, for that matter,
do I.
00:26:21.000 --> 00:26:27.000
Fourier series.
Not a good choice for a
00:26:25.000 --> 00:26:31.000
function that goes to infinity
at pi over two.
00:26:35.000 --> 00:26:41.000
So you cannot do this until you
do your homework.
00:26:39.000 --> 00:26:45.000
Now you will be able to do it.
In other words,
00:26:44.000 --> 00:26:50.000
one of the big things is not
only will I give you a formula
00:26:50.000 --> 00:26:56.000
for the Xp but that formula will
work even for tangent t,
00:26:56.000 --> 00:27:02.000
any function at all.
Well, I thought I would try to
00:27:00.000 --> 00:27:06.000
put a little meat on the bones
of the inhomogeneous systems by
00:27:04.000 --> 00:27:10.000
actually giving you a physical
problem so we would actually be
00:27:08.000 --> 00:27:14.000
able to solve a physical problem
instead of just demonstrate a
00:27:12.000 --> 00:27:18.000
solution method.
Here is a mixing problem.
00:27:23.000 --> 00:27:29.000
Just to illustrate what makes a
system of equations
00:27:27.000 --> 00:27:33.000
inhomogeneous,
here at two ugly tanks.
00:27:31.000 --> 00:27:37.000
I am not going to draw these
carefully, but they are both 1
00:27:37.000 --> 00:27:43.000
liter.
And they are connected by
00:27:40.000 --> 00:27:46.000
pipes.
And I won't bother opening
00:27:43.000 --> 00:27:49.000
holes in them.
There is a pipe with fluids
00:27:47.000 --> 00:27:53.000
flowing back there and this
direction it is flowing this
00:27:52.000 --> 00:27:58.000
way, but that is not the end.
The end is there is stuff
00:28:00.000 --> 00:28:06.000
coming in to both of them.
And I think I will just make it
00:28:08.000 --> 00:28:14.000
coming out of this one.
There is something realistic.
00:28:15.000 --> 00:28:21.000
The numbers 2,
3, 2.
00:28:18.000 --> 00:28:24.000
Let's start there and see what
the others have to be.
00:28:25.000 --> 00:28:31.000
So these are flow rates.
One liter tanks.
00:28:31.000 --> 00:28:37.000
The flow rates are in,
let's say, liters per hour.
00:28:38.000 --> 00:28:44.000
And I have some dissolved
substance in,
00:28:42.000 --> 00:28:48.000
so here is going to be x salt
in there and the same chemical
00:28:50.000 --> 00:28:56.000
in there, whatever it is.
x is the amount of salt,
00:28:56.000 --> 00:29:02.000
let's say, in tank one.
And y, the same thing in tank
00:29:02.000 --> 00:29:08.000
two.
Now, if you have stuff flowing
00:29:06.000 --> 00:29:12.000
unequally this way,
you must have balance.
00:29:09.000 --> 00:29:15.000
You have to make sure that
neither tank is getting emptied
00:29:15.000 --> 00:29:21.000
or bursting and exploding.
What is flowing in?
00:29:19.000 --> 00:29:25.000
What is x?
Three is going out,
00:29:22.000 --> 00:29:28.000
two is coming in,
so this has to be one in order
00:29:27.000 --> 00:29:33.000
that tank x stay full and not
explode.
00:29:32.000 --> 00:29:38.000
And how about y?
How much is going out?
00:29:35.000 --> 00:29:41.000
Two there and two here.
Four is going out,
00:29:39.000 --> 00:29:45.000
three is coming in.
This also has to be one.
00:29:43.000 --> 00:29:49.000
Those are just the flow rates
of water or the liquid that is
00:29:48.000 --> 00:29:54.000
coming in.
Now, the only thing I am going
00:29:52.000 --> 00:29:58.000
to specify is the concentration
of what is coming in.
00:29:58.000 --> 00:30:04.000
Here the concentration is 5 e
to the minus t.
00:30:02.000 --> 00:30:08.000
And that is what makes the
problem inhomogeneous.
00:30:07.000 --> 00:30:13.000
Here the concentration is going
to be zero.
00:30:10.000 --> 00:30:16.000
In other words,
pure water is flowing in here
00:30:14.000 --> 00:30:20.000
to create the liquid balance.
Here, on the other hand,
00:30:18.000 --> 00:30:24.000
salt solution is flowing in but
with a steadily declining
00:30:23.000 --> 00:30:29.000
concentration.
So, what is the system?
00:30:28.000 --> 00:30:34.000
Well, you have set it up
exactly the way you did when you
00:30:34.000 --> 00:30:40.000
studied first order equations.
It is inflow minus outflow.
00:30:41.000 --> 00:30:47.000
What is the outflow?
The outflow is all in this
00:30:46.000 --> 00:30:52.000
pipe.
The flow rates are liters per
00:30:50.000 --> 00:30:56.000
hour.
Three liters per hour flowing
00:30:54.000 --> 00:31:00.000
out.
How much salt does that
00:30:57.000 --> 00:31:03.000
represent?
It is negative three times the
00:31:02.000 --> 00:31:08.000
concentration of salt.
But the concentration,
00:31:06.000 --> 00:31:12.000
notice, equals x divided
by one.
00:31:10.000 --> 00:31:16.000
In other words,
x represents both the
00:31:13.000 --> 00:31:19.000
concentration and the amount.
So I don't have to distinguish.
00:31:18.000 --> 00:31:24.000
If I had made it two liter
tanks then I would have had to
00:31:23.000 --> 00:31:29.000
divide this by two.
I am cheating,
00:31:26.000 --> 00:31:32.000
but it is enough already.
x prime equals minus 3x.
00:31:32.000 --> 00:31:38.000
That is what is going out.
What is coming in?
00:31:36.000 --> 00:31:42.000
Well, 2y is coming in.
Concentration here.
00:31:39.000 --> 00:31:45.000
What is coming in?
Is it y 2 liter?
00:31:43.000 --> 00:31:49.000
Plus what is coming in from the
outside.
00:31:46.000 --> 00:31:52.000
We have to add that in,
and that will be plus 5 e to
00:31:51.000 --> 00:31:57.000
the negative t.
How about y?
00:31:54.000 --> 00:32:00.000
y prime is changing.
What comes in from x?
00:32:00.000 --> 00:32:06.000
That is 3x.
What goes out?
00:32:01.000 --> 00:32:07.000
Well, two is leaving here and
two is leaving here.
00:32:05.000 --> 00:32:11.000
It doesn't matter that they are
going out through separate
00:32:10.000 --> 00:32:16.000
pipes.
They are both going out.
00:32:12.000 --> 00:32:18.000
It is minus 4,
2 and 2.
00:32:14.000 --> 00:32:20.000
How about the inhomogeneous
term?
00:32:16.000 --> 00:32:22.000
There is one coming in,
but there is no salt in it.
00:32:20.000 --> 00:32:26.000
Therefore, that is not
changing.
00:32:23.000 --> 00:32:29.000
What is coming through that
pipe is necessary for the liquid
00:32:27.000 --> 00:32:33.000
balance.
But it has no effect
00:32:31.000 --> 00:32:37.000
whatsoever.
I will put a zero here but,
00:32:35.000 --> 00:32:41.000
of course, you don't have to
put that in.
00:32:38.000 --> 00:32:44.000
This is now an inhomogeneous
system.
00:32:42.000 --> 00:32:48.000
In other words,
the system is x prime equals
00:32:46.000 --> 00:32:52.000
this matrix, negative 3,
the same sort of stuff we
00:32:50.000 --> 00:32:56.000
always had, plus the
inhomogeneous term which is the
00:32:55.000 --> 00:33:01.000
column vector 5 e to the minus t
and zero.
00:33:00.000 --> 00:33:06.000
It is the presence of this term
00:33:04.000 --> 00:33:10.000
that makes this system
inhomogeneous.
00:33:06.000 --> 00:33:12.000
And what that corresponds to is
this little closed system being
00:33:11.000 --> 00:33:17.000
attacked from the outside by
these external pipes which are
00:33:15.000 --> 00:33:21.000
bringing salt in.
Without those,
00:33:17.000 --> 00:33:23.000
of course the balance would be
all wrong.
00:33:20.000 --> 00:33:26.000
I would have to change this to
three and cut that out,
00:33:24.000 --> 00:33:30.000
I guess.
But then, it would be just a
00:33:27.000 --> 00:33:33.000
simple homogenous system.
It is these pipes that make it
00:33:32.000 --> 00:33:38.000
inhomogeneous.
Now, I should start to solve
00:33:35.000 --> 00:33:41.000
that.
I did this just to illustrate
00:33:38.000 --> 00:33:44.000
where a system might come from.
Before I solve that,
00:33:42.000 --> 00:33:48.000
what I want to do is,
of course, is solve it in
00:33:45.000 --> 00:33:51.000
general.
In other words,
00:33:47.000 --> 00:33:53.000
how do you solve this in
general?
00:33:49.000 --> 00:33:55.000
Because I promised you that you
would be able to do in general,
00:33:54.000 --> 00:34:00.000
regardless of what sort of
functions were in the r of t,
00:33:58.000 --> 00:34:04.000
that column vector.
So let's do it.
00:34:10.000 --> 00:34:16.000
First of all,
you have to learn the name of
00:34:15.000 --> 00:34:21.000
the method.
This method is for solving x
00:34:20.000 --> 00:34:26.000
prime equals Ax.
It is a method for finding a
00:34:26.000 --> 00:34:32.000
particular solution.
00:34:36.000 --> 00:34:42.000
Of course, to actually solve it
then you have to add the
00:34:39.000 --> 00:34:45.000
complimentary function.
We are looking for a particular
00:34:43.000 --> 00:34:49.000
solution for this system.
Now, the whole cleverness of
00:34:47.000 --> 00:34:53.000
the method, which I think was
discovered a couple hundred
00:34:51.000 --> 00:34:57.000
years ago by,
I think, Lagrange,
00:34:53.000 --> 00:34:59.000
I am not sure.
The method is called variation
00:34:57.000 --> 00:35:03.000
of parameters.
I am giving you that so that
00:35:00.000 --> 00:35:06.000
when you forget you will be able
to look it up and be indexes to
00:35:05.000 --> 00:35:11.000
some advanced engineering
mathematics book or something,
00:35:08.000 --> 00:35:14.000
whatever is on your shelf.
But, if course,
00:35:11.000 --> 00:35:17.000
you won't remember the name
either so maybe this won't work.
00:35:15.000 --> 00:35:21.000
Variation of parameters,
I will explain to you why it is
00:35:19.000 --> 00:35:25.000
called that.
All the cleverness is in the
00:35:22.000 --> 00:35:28.000
very first line.
If you could remember the very
00:35:25.000 --> 00:35:31.000
first line then I trust you to
do the rest yourself.
00:35:30.000 --> 00:35:36.000
I don't know any motivation for
this first step,
00:35:34.000 --> 00:35:40.000
but mathematics is supposed to
be mysterious anyway.
00:35:40.000 --> 00:35:46.000
It keeps me eating.
It says, look for a solution
00:35:45.000 --> 00:35:51.000
and there will be one of the
following form.
00:35:49.000 --> 00:35:55.000
Now, it will look exactly like
--
00:36:00.000 --> 00:36:06.000
Look carefully because it is
going to be gone in a moment.
00:36:04.000 --> 00:36:10.000
It will look exactly like this.
But, of course,
00:36:08.000 --> 00:36:14.000
it cannot be this because this
solves the homogeneous system.
00:36:13.000 --> 00:36:19.000
If I plug this in with these as
constants it cannot possibly be
00:36:18.000 --> 00:36:24.000
a particular solution to this
because it will stop there and
00:36:23.000 --> 00:36:29.000
satisfy that with r equals zero.
The whole trick is you think of
00:36:29.000 --> 00:36:35.000
these are parameters which are
now variable.
00:36:32.000 --> 00:36:38.000
Constants that are varying.
That is why it is called
00:36:35.000 --> 00:36:41.000
variation of parameters.
You think of these,
00:36:38.000 --> 00:36:44.000
in other words,
as functions of t.
00:36:41.000 --> 00:36:47.000
We are going to look for a
solution which has the form,
00:36:45.000 --> 00:36:51.000
since they are functions of t,
I don't want to call them c1
00:36:49.000 --> 00:36:55.000
and c2 anymore.
I will call them v because that
00:36:52.000 --> 00:36:58.000
is what most people call them,
v or u, sometimes.
00:37:02.000 --> 00:37:08.000
The method says look for a
solution of that form.
00:37:06.000 --> 00:37:12.000
The variation parameters,
these are the parameters that
00:37:10.000 --> 00:37:16.000
are now varying instead of being
constants.
00:37:14.000 --> 00:37:20.000
Now, if you take it in that
form and start trying to
00:37:18.000 --> 00:37:24.000
substitute into the equation you
are going to get a mess.
00:37:23.000 --> 00:37:29.000
I think I was wrong in saying I
could trust you from this point
00:37:28.000 --> 00:37:34.000
on.
I will take the first step from
00:37:32.000 --> 00:37:38.000
you, and then I could trust you
to do the rest after that first
00:37:38.000 --> 00:37:44.000
step.
The first step is to change the
00:37:42.000 --> 00:37:48.000
way this looks by using the
fundamental matrix.
00:37:46.000 --> 00:37:52.000
Remember what the fundamental
matrix was?
00:37:50.000 --> 00:37:56.000
Its entries were the two
columns of solutions.
00:37:54.000 --> 00:38:00.000
These are solutions to the
homogeneous system.
00:38:00.000 --> 00:38:06.000
And I am going to write it
using the fundamental matrix as,
00:38:05.000 --> 00:38:11.000
now thinks about it.
The fundamental matrix has
00:38:09.000 --> 00:38:15.000
columns x1 and x2.
Your instinct might be using
00:38:13.000 --> 00:38:19.000
matrix multiplication to put the
v1 and the v2 here,
00:38:18.000 --> 00:38:24.000
but that won't work.
You have to put them here.
00:38:30.000 --> 00:38:36.000
This says the same thing as
that.
00:38:33.000 --> 00:38:39.000
Let's just take a second out to
calculate.
00:38:37.000 --> 00:38:43.000
The x is going to look like
(x1, y1).
00:38:40.000 --> 00:38:46.000
That is my first solution.
My second solution,
00:38:44.000 --> 00:38:50.000
here is the fundamental matrix,
is (x2, y2).
00:38:49.000 --> 00:38:55.000
And I am multiplying this on
the right by (v1,
00:38:53.000 --> 00:38:59.000
v2).
Does it come out right?
00:38:56.000 --> 00:39:02.000
Look.
What is it?
00:38:59.000 --> 00:39:05.000
The top is x1 v1 plus x2 v2.
00:39:02.000 --> 00:39:08.000
The top, x1 v1 plus x2 v2.
00:39:06.000 --> 00:39:12.000
It is in the wrong order,
but multiplication is
00:39:10.000 --> 00:39:16.000
commutative, fortunately.
And the same way the bottom
00:39:14.000 --> 00:39:20.000
thing will be v1 y1 plus v2 y2.
00:39:18.000 --> 00:39:24.000
If I had written it on the
other side instead,
00:39:22.000 --> 00:39:28.000
which is tempting because the
v's occur on the left here,
00:39:27.000 --> 00:39:33.000
that won't work.
What will I get?
00:39:31.000 --> 00:39:37.000
I will get v1 x1 plus v2 y1,
00:39:35.000 --> 00:39:41.000
which is not at all what I
want.
00:39:37.000 --> 00:39:43.000
You must put it on the right.
But this is a very important
00:39:42.000 --> 00:39:48.000
thing.
This is going to plague us on
00:39:45.000 --> 00:39:51.000
Monday, too.
It must be written on the right
00:39:49.000 --> 00:39:55.000
and not on the left as a column
vector.
00:39:52.000 --> 00:39:58.000
The rest of the program is very
simple.
00:39:55.000 --> 00:40:01.000
I will write it out as a
program.
00:40:00.000 --> 00:40:06.000
Substitute into the system,
into that, in other words,
00:40:04.000 --> 00:40:10.000
and see what v has to be.
That is what we are looking
00:40:08.000 --> 00:40:14.000
for.
We know what the x1 and x2 are.
00:40:11.000 --> 00:40:17.000
It is a question of what those
coefficients are.
00:40:14.000 --> 00:40:20.000
And see what v is.
Let's do it.
00:40:32.000 --> 00:40:38.000
Let's substitute.
Let's see.
00:40:35.000 --> 00:40:41.000
The system is x prime equals Ax
plus r.
00:40:41.000 --> 00:40:47.000
I want to put in (x)p,
this proposed particular
00:40:46.000 --> 00:40:52.000
solution.
And it is a fundamental matrix,
00:40:50.000 --> 00:40:56.000
and the v is unknown.
How do I differentiate the
00:40:56.000 --> 00:41:02.000
product of two matrices?
You differentiate the product
00:41:02.000 --> 00:41:08.000
of two matrices using the
product rule that you learned
00:41:07.000 --> 00:41:13.000
the first day of 18.01.
Trust me.
00:41:10.000 --> 00:41:16.000
Let's do it.
I am going to substitute in.
00:41:14.000 --> 00:41:20.000
In other words,
here is my (x)p,
00:41:17.000 --> 00:41:23.000
(x)p, and I am going to write
in what that is.
00:41:21.000 --> 00:41:27.000
The left-hand side is the
derivative of,
00:41:25.000 --> 00:41:31.000
X prime times v,
plus X times the derivative of
00:41:29.000 --> 00:41:35.000
v.
Notice that one of these is a
00:41:34.000 --> 00:41:40.000
column vector and the other is a
square matrix.
00:41:37.000 --> 00:41:43.000
That is perfectly Okay.
Any two matrices which are the
00:41:39.000 --> 00:41:45.000
rate shape so you can multiply
them together,
00:41:42.000 --> 00:41:48.000
if you want to differentiate
their product,
00:41:44.000 --> 00:41:50.000
in other words,
if the entries are functions of
00:41:46.000 --> 00:41:52.000
t it is the product rule.
The derivative of this times
00:41:49.000 --> 00:41:55.000
time plus that times the
derivative of this.
00:41:52.000 --> 00:41:58.000
You have to keep them in the
right order.
00:41:54.000 --> 00:42:00.000
You are not allowed to shuffle
them around carelessly.
00:41:57.000 --> 00:42:03.000
So that is that.
What is it equal to?
00:42:00.000 --> 00:42:06.000
Well, the right-hand side is A.
And now I substitute just (x)p
00:42:08.000 --> 00:42:14.000
in, so that is X times v plus r.
Is this progress?
00:42:14.000 --> 00:42:20.000
What is v?
It looks like a mess but it is
00:42:20.000 --> 00:42:26.000
not.
Why not?
00:42:22.000 --> 00:42:28.000
It is because this is not any
old matrix X.
00:42:29.000 --> 00:42:35.000
This is a matrix whose columns
are solutions to the system.
00:42:34.000 --> 00:42:40.000
And what does that do?
That means X prime satisfies
00:42:39.000 --> 00:42:45.000
that matrix differential
equation.
00:42:42.000 --> 00:42:48.000
X prime is the same as Ax.
00:42:45.000 --> 00:42:51.000
And, by a little miracle,
the v is tagging along in both
00:42:50.000 --> 00:42:56.000
cases.
This cancels that and now there
00:42:54.000 --> 00:43:00.000
is very little left.
The conclusion,
00:42:58.000 --> 00:43:04.000
therefore, is that Xv is equal
to r.
00:43:02.000 --> 00:43:08.000
What is v?
It is v that we are looking
00:43:05.000 --> 00:43:11.000
for, right?
You have to solve a matrix
00:43:09.000 --> 00:43:15.000
equation, now.
This is a square matrix so you
00:43:13.000 --> 00:43:19.000
have to do it by inverting the
matrix.
00:43:16.000 --> 00:43:22.000
You don't just sloppily divide.
You multiply on which side by
00:43:21.000 --> 00:43:27.000
what matrix?
Choice of left or right.
00:43:25.000 --> 00:43:31.000
You multiply by the inverse
matrix on the left or on the
00:43:29.000 --> 00:43:35.000
right?
It has to be on the left.
00:43:35.000 --> 00:43:41.000
Multiply both sides of the
equation by X inverse on the
00:43:42.000 --> 00:43:48.000
left, and then you will get v is
equal to X inverse r.
00:43:50.000 --> 00:43:56.000
How do I know the X inverse
00:43:55.000 --> 00:44:01.000
exists?
Does X inverse exist?
00:44:00.000 --> 00:44:06.000
For a matrix inverse to exist,
the matrix's determinant must
00:44:05.000 --> 00:44:11.000
be not zero.
Why is the determinant of this
00:44:10.000 --> 00:44:16.000
not zero?
Because its columns are
00:44:13.000 --> 00:44:19.000
independent solutions.
00:44:22.000 --> 00:44:28.000
Of course this is not right.
I forgot the prime here.
00:44:31.000 --> 00:44:37.000
I am not failing this course
after all.
00:44:34.000 --> 00:44:40.000
v prime equals that.
00:44:43.000 --> 00:44:49.000
This is done by differentiating
each entry in the column vector.
00:44:47.000 --> 00:44:53.000
And, therefore,
we should integrate it.
00:44:50.000 --> 00:44:56.000
It will be the integral,
just the ordinary
00:44:53.000 --> 00:44:59.000
anti-derivative of x inverse
times r.
00:44:57.000 --> 00:45:03.000
This is a column vector.
The entries are functions of t.
00:45:02.000 --> 00:45:08.000
You simply integrate each of
those functions in turn.
00:45:06.000 --> 00:45:12.000
So integrate each entry.
00:45:12.000 --> 00:45:18.000
There is my v.
Sorry, you cannot tell the v's
00:45:19.000 --> 00:45:25.000
from the r's here.
And so, finally,
00:45:25.000 --> 00:45:31.000
the particular solution is (x)p
is equal to --
00:45:34.000 --> 00:45:40.000
It is really not bad at all.
It is equal to X times v.
00:45:37.000 --> 00:45:43.000
It's equal to X times the
integral of X inverse r dt.
00:45:40.000 --> 00:45:46.000
Now, actually,
00:45:43.000 --> 00:45:49.000
there is not much work to doing
that.
00:45:45.000 --> 00:45:51.000
Once you have solved the
homogeneous system and gotten
00:45:49.000 --> 00:45:55.000
the fundamental matrix,
taking the inverse of a
00:45:52.000 --> 00:45:58.000
two-by-two matrix is almost
trivial.
00:45:54.000 --> 00:46:00.000
You flip those two and you
change the signs of these two
00:45:57.000 --> 00:46:03.000
and you divide by the
determinant.
00:46:01.000 --> 00:46:07.000
You multiply it by r.
And the hard part is if you can
00:46:04.000 --> 00:46:10.000
do the integration.
If not, you just leave the
00:46:07.000 --> 00:46:13.000
integral sign the way you have
learned to do in this silly
00:46:11.000 --> 00:46:17.000
course and you still have the
answer.
00:46:14.000 --> 00:46:20.000
What about the arbitrary
constant of integration?
00:46:17.000 --> 00:46:23.000
The answer is you don't need to
put it in.
00:46:20.000 --> 00:46:26.000
Just find one particular
solution.
00:46:22.000 --> 00:46:28.000
It is good enough.
You don't have to put in the
00:46:25.000 --> 00:46:31.000
arbitrary constants of
integration.
00:46:29.000 --> 00:46:35.000
Because they are already in the
complimentary function here.
00:46:33.000 --> 00:46:39.000
Therefore, you don't have to
add them.
00:46:35.000 --> 00:46:41.000
I am sorry I didn't get a
chance to actually solve that.
00:46:39.000 --> 00:46:45.000
I will have to let it go.
The recitations will do it on
00:46:42.000 --> 00:46:48.000
Tuesday, will solve that
particular problem,
00:46:45.000 --> 00:46:51.000
which means you will,
in effect.