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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
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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.
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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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you differentiate the x and the
y.
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Well, a column vector is a
special kind of matrix.
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The definition applies to any
matrix.
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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,
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the proof of it.
In fact, there is nothing in
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this.
It is nothing more than a
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little matrix calculation of the
most primitive kind.
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Namely, what does this mean?
Let's try to undo that.
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What does the left-hand side
really mean?
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Well, if that is what x means,
the left-hand side must mean
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the derivative of the first
column.
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That is its first column.
And the derivative of the
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second column.
That is what it means to
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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.
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Well, what does the right-hand
side mean?
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Well, I am supposed to take A
and multiply that
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by [x1,x2].
Now, I don't know how to prove
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this, except ask you to think
about it.
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Or, I could write it all out
here.
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But think of this as a bing,
bing, bing, bing.
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And this is a bing,
bing.
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And this is a bong,
bong.
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How do I do the multiplication?
In other words,
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what is in the first column of
the matrix?
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Well, it is dah,
dah, and the lower thing is
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dah, dah.
In other words,
268
00:18:03 --> 00:18:09
it is A times x1.
269
00:18:08 --> 00:18:14
270
00:18:15 --> 00:18:21
Shut your eyes and visualize
it.
271
00:18:17 --> 00:18:23
Got it?
Dah, dah is the top entry,
272
00:18:19 --> 00:18:25
and dah, dah is the bottom
entry.
273
00:18:22 --> 00:18:28
It is what you get by
multiplying A by the column
274
00:18:25 --> 00:18:31
vector x1.
And the same way the other guy
275
00:18:28 --> 00:18:34
is --
-- what you get by multiplying
276
00:18:33 --> 00:18:39
A by the column vector x2.
This is just matrix
277
00:18:37 --> 00:18:43
multiplication.
That is the law of matrix
278
00:18:41 --> 00:18:47
multiplication.
That is how you multiply
279
00:18:45 --> 00:18:51
matrices.
Well, good, but where does this
280
00:18:49 --> 00:18:55
get us?
What does it mean for those two
281
00:18:53 --> 00:18:59
guys to be equal?
That is going to happen,
282
00:18:57 --> 00:19:03
if and only if x1 prime is
equal to A x1.
283
00:19:02 --> 00:19:08
This guy equals that guy.
And similarly for the x2's.
284
00:19:09 --> 00:19:15
285
00:19:14 --> 00:19:20
The end result is that this
matrix, saying that the
286
00:19:18 --> 00:19:24
fundamental matrix satisfies
this matrix differential
287
00:19:22 --> 00:19:28
equation is only a way of
saying, in one breath,
288
00:19:26 --> 00:19:32
that its two columns are both
solutions to the original
289
00:19:30 --> 00:19:36
system.
It is, so to speak,
290
00:19:33 --> 00:19:39
an efficient way of turning
these two equations into a
291
00:19:38 --> 00:19:44
single equation by making a
matrix.
292
00:19:41 --> 00:19:47
I guess it is time,
finally, to come to the topic
293
00:19:46 --> 00:19:52
of the lecture.
I said the thing the matrices
294
00:19:50 --> 00:19:56
were going to be used for is
solving inhomogeneous systems,
295
00:19:55 --> 00:20:01
so let's take a look at those.
I thought I would give you an
296
00:20:00 --> 00:20:06
example.
Inhomogeneous systems.
297
00:20:05 --> 00:20:11
298
00:20:10 --> 00:20:16
Well, what is one going to look
like?
299
00:20:12 --> 00:20:18
So far what we have done is,
up until now has been solving,
300
00:20:17 --> 00:20:23
we spent essentially two weeks
solving and plotting the
301
00:20:21 --> 00:20:27
solutions to homogeneous
systems.
302
00:20:23 --> 00:20:29
There was nothing over there.
And homogeneous systems,
303
00:20:27 --> 00:20:33
in fact, with constant
coefficients.
304
00:20:31 --> 00:20:37
Stuff that looked like that
that we abbreviated with
305
00:20:34 --> 00:20:40
matrices.
Now, to make the system
306
00:20:36 --> 00:20:42
inhomogeneous what I do is add
the extra term on the right-hand
307
00:20:41 --> 00:20:47
side, which is some function of
t.
308
00:20:43 --> 00:20:49
Except, I will have to have two
functions of t because I have
309
00:20:47 --> 00:20:53
two equations.
Now it is inhomogeneous.
310
00:20:50 --> 00:20:56
And what makes it inhomogeneous
is the fact that these are not
311
00:20:54 --> 00:21:00
zero anymore.
There is something there.
312
00:20:57 --> 00:21:03
Functions of t are there.
These are given functions of t
313
00:21:02 --> 00:21:08
like exponentials,
polynomials,
314
00:21:04 --> 00:21:10
the usual stuff you have on the
right-hand side of the
315
00:21:08 --> 00:21:14
differential equation.
What is confusing here is that
316
00:21:11 --> 00:21:17
when we studied second order
equations it was homogeneous if
317
00:21:15 --> 00:21:21
the right-hand side was zero,
and if there was something else
318
00:21:20 --> 00:21:26
there it was inhomogeneous.
Unfortunately,
319
00:21:23 --> 00:21:29
I have stuck this stuff on the
right-hand side so it is not
320
00:21:27 --> 00:21:33
quite so clear anymore.
It has got to look like that,
321
00:21:32 --> 00:21:38
in other words.
How would the matrix
322
00:21:34 --> 00:21:40
abbreviation look?
Well, the left-hand side is x
323
00:21:37 --> 00:21:43
prime.
The homogenous part is ax,
324
00:21:40 --> 00:21:46
just as it has always been.
The only extra part is those
325
00:21:43 --> 00:21:49
functions r.
And this is a column vector,
326
00:21:46 --> 00:21:52
after the multiplication this
is a column vector,
327
00:21:50 --> 00:21:56
what is left is column vector.
Now, explicitly it is a
328
00:21:53 --> 00:21:59
function of t,
given by explicit functions of
329
00:21:56 --> 00:22:02
t, again, like exponentials.
Or, they could be fancy
330
00:22:02 --> 00:22:08
functions.
That is the thing we are trying
331
00:22:06 --> 00:22:12
to solve.
Why don't I put it up in green?
332
00:22:10 --> 00:22:16
Our new and better and improved
system.
333
00:22:13 --> 00:22:19
Think back to what we did when
we studied inhomogeneous
334
00:22:18 --> 00:22:24
equations.
We are not talking about
335
00:22:22 --> 00:22:28
systems but just a single
equation.
336
00:22:25 --> 00:22:31
What we did was the main
theorem --
337
00:22:30 --> 00:22:36
I guess there are going to be
three theorems today,
338
00:22:36 --> 00:22:42
not just two.
Theorem C.
339
00:22:39 --> 00:22:45
Is that right?
Yes, A, B.
340
00:22:42 --> 00:22:48
We are up to C.
Theorem C says that the general
341
00:22:48 --> 00:22:54
solution, that is,
the general solution to the
342
00:22:54 --> 00:23:00
system, is equal to the
complimentary function,
343
00:23:00 --> 00:23:06
which is the general solution
to x prime equals Ax,
344
00:23:06 --> 00:23:12
--
-- the homogeneous equation,
345
00:23:11 --> 00:23:17
in other words,
plus, what am I going to call
346
00:23:15 --> 00:23:21
it?
(x)p, right you are,
347
00:23:17 --> 00:23:23
a particular solution.
But the principle is the same
348
00:23:22 --> 00:23:28
and is proved exactly the same
way.
349
00:23:25 --> 00:23:31
It is just linearity and
superposition.
350
00:23:30 --> 00:23:36
351
00:23:35 --> 00:23:41
The linearity of the original
system and the superposition
352
00:23:40 --> 00:23:46
principle.
The essence is that to solve
353
00:23:43 --> 00:23:49
this inhomogeneous system,
what we have to do is find a
354
00:23:48 --> 00:23:54
particular solution.
This part I already know how to
355
00:23:52 --> 00:23:58
do.
We have been doing that for two
356
00:23:55 --> 00:24:01
weeks.
The new thing is to find this.
357
00:24:00 --> 00:24:06
358
00:24:05 --> 00:24:11
Now, if you remember back
before spring break,
359
00:24:08 --> 00:24:14
most of the work in solving the
second order equation was in
360
00:24:12 --> 00:24:18
finding that particular
solution.
361
00:24:15 --> 00:24:21
You quickly enough learned how
to solve the homogeneous
362
00:24:19 --> 00:24:25
equation, but there was no real
general method for finding this.
363
00:24:24 --> 00:24:30
We had an exponential input
theorem with some modifications
364
00:24:28 --> 00:24:34
to it.
We took a week's detour in
365
00:24:32 --> 00:24:38
Fourier series to see how to do
it for periodic functions or
366
00:24:37 --> 00:24:43
functions defined on finite
intervals.
367
00:24:40 --> 00:24:46
There were other techniques
which I did not get around to
368
00:24:45 --> 00:24:51
showing you, techniques
involving the so-called method
369
00:24:50 --> 00:24:56
of undetermined coefficients.
Although, some of you peaked in
370
00:24:55 --> 00:25:01
your book and learned it from
there.
371
00:25:00 --> 00:25:06
But the work is in finding
(x)p.
372
00:25:03 --> 00:25:09
The miracle that occurs here,
by contrast,
373
00:25:07 --> 00:25:13
is that it turns out to be easy
to find (x)p.
374
00:25:11 --> 00:25:17
And easy in this further sense
that I do not have to restrict
375
00:25:17 --> 00:25:23
the kind of function I use.
For example,
376
00:25:21 --> 00:25:27
the second homework problem I
have given you,
377
00:25:26 --> 00:25:32
the second part two homework
problem.
378
00:25:31 --> 00:25:37
379
00:25:36 --> 00:25:42
You will see how to use
systems.
380
00:25:38 --> 00:25:44
For example,
to solve this simple equation,
381
00:25:42 --> 00:25:48
I will write it out for you,
consider that equation,
382
00:25:46 --> 00:25:52
tangent t.
What technique will you apply
383
00:25:50 --> 00:25:56
to solve that?
In other words,
384
00:25:52 --> 00:25:58
suppose you wanted to find a
particular solution to that.
385
00:25:57 --> 00:26:03
The right-hand side is not an
exponential.
386
00:26:00 --> 00:26:06
It is not a polynomial.
It is not like sine or cosine
387
00:26:06 --> 00:26:12
of bt.
I could use the Laplace
388
00:26:10 --> 00:26:16
transform.
No, because you don't know how
389
00:26:14 --> 00:26:20
to take the Laplace transform of
tangent t.
390
00:26:19 --> 00:26:25
Neither, for that matter,
do I.
391
00:26:21 --> 00:26:27
Fourier series.
Not a good choice for a
392
00:26:25 --> 00:26:31
function that goes to infinity
at pi over two.
393
00:26:30 --> 00:26:36
394
00:26:35 --> 00:26:41
So you cannot do this until you
do your homework.
395
00:26:39 --> 00:26:45
Now you will be able to do it.
In other words,
396
00:26:44 --> 00:26:50
one of the big things is not
only will I give you a formula
397
00:26:50 --> 00:26:56
for the Xp but that formula will
work even for tangent t,
398
00:26:56 --> 00:27:02
any function at all.
Well, I thought I would try to
399
00:27:00 --> 00:27:06
put a little meat on the bones
of the inhomogeneous systems by
400
00:27:04 --> 00:27:10
actually giving you a physical
problem so we would actually be
401
00:27:08 --> 00:27:14
able to solve a physical problem
instead of just demonstrate a
402
00:27:12 --> 00:27:18
solution method.
Here is a mixing problem.
403
00:27:16 --> 00:27:22
404
00:27:23 --> 00:27:29
Just to illustrate what makes a
system of equations
405
00:27:27 --> 00:27:33
inhomogeneous,
here at two ugly tanks.
406
00:27:31 --> 00:27:37
I am not going to draw these
carefully, but they are both 1
407
00:27:37 --> 00:27:43
liter.
And they are connected by
408
00:27:40 --> 00:27:46
pipes.
And I won't bother opening
409
00:27:43 --> 00:27:49
holes in them.
There is a pipe with fluids
410
00:27:47 --> 00:27:53
flowing back there and this
direction it is flowing this
411
00:27:52 --> 00:27:58
way, but that is not the end.
The end is there is stuff
412
00:28:00 --> 00:28:06
coming in to both of them.
And I think I will just make it
413
00:28:08 --> 00:28:14
coming out of this one.
There is something realistic.
414
00:28:15 --> 00:28:21
The numbers 2,
3, 2.
415
00:28:18 --> 00:28:24
Let's start there and see what
the others have to be.
416
00:28:25 --> 00:28:31
So these are flow rates.
One liter tanks.
417
00:28:31 --> 00:28:37
The flow rates are in,
let's say, liters per hour.
418
00:28:38 --> 00:28:44
And I have some dissolved
substance in,
419
00:28:42 --> 00:28:48
so here is going to be x salt
in there and the same chemical
420
00:28:50 --> 00:28:56
in there, whatever it is.
x is the amount of salt,
421
00:28:56 --> 00:29:02
let's say, in tank one.
And y, the same thing in tank
422
00:29:02 --> 00:29:08
two.
Now, if you have stuff flowing
423
00:29:06 --> 00:29:12
unequally this way,
you must have balance.
424
00:29:09 --> 00:29:15
You have to make sure that
neither tank is getting emptied
425
00:29:15 --> 00:29:21
or bursting and exploding.
What is flowing in?
426
00:29:19 --> 00:29:25
What is x?
Three is going out,
427
00:29:22 --> 00:29:28
two is coming in,
so this has to be one in order
428
00:29:27 --> 00:29:33
that tank x stay full and not
explode.
429
00:29:32 --> 00:29:38
And how about y?
How much is going out?
430
00:29:35 --> 00:29:41
Two there and two here.
Four is going out,
431
00:29:39 --> 00:29:45
three is coming in.
This also has to be one.
432
00:29:43 --> 00:29:49
Those are just the flow rates
of water or the liquid that is
433
00:29:48 --> 00:29:54
coming in.
Now, the only thing I am going
434
00:29:52 --> 00:29:58
to specify is the concentration
of what is coming in.
435
00:29:58 --> 00:30:04
Here the concentration is 5 e
to the minus t.
436
00:30:02 --> 00:30:08
And that is what makes the
problem inhomogeneous.
437
00:30:07 --> 00:30:13
Here the concentration is going
to be zero.
438
00:30:10 --> 00:30:16
In other words,
pure water is flowing in here
439
00:30:14 --> 00:30:20
to create the liquid balance.
Here, on the other hand,
440
00:30:18 --> 00:30:24
salt solution is flowing in but
with a steadily declining
441
00:30:23 --> 00:30:29
concentration.
So, what is the system?
442
00:30:28 --> 00:30:34
Well, you have set it up
exactly the way you did when you
443
00:30:34 --> 00:30:40
studied first order equations.
It is inflow minus outflow.
444
00:30:41 --> 00:30:47
What is the outflow?
The outflow is all in this
445
00:30:46 --> 00:30:52
pipe.
The flow rates are liters per
446
00:30:50 --> 00:30:56
hour.
Three liters per hour flowing
447
00:30:54 --> 00:31:00
out.
How much salt does that
448
00:30:57 --> 00:31:03
represent?
It is negative three times the
449
00:31:02 --> 00:31:08
concentration of salt.
But the concentration,
450
00:31:06 --> 00:31:12
notice, equals x divided
by one.
451
00:31:10 --> 00:31:16
In other words,
x represents both the
452
00:31:13 --> 00:31:19
concentration and the amount.
So I don't have to distinguish.
453
00:31:18 --> 00:31:24
If I had made it two liter
tanks then I would have had to
454
00:31:23 --> 00:31:29
divide this by two.
I am cheating,
455
00:31:26 --> 00:31:32
but it is enough already.
x prime equals minus 3x.
456
00:31:32 --> 00:31:38
That is what is going out.
What is coming in?
457
00:31:36 --> 00:31:42
Well, 2y is coming in.
Concentration here.
458
00:31:39 --> 00:31:45
What is coming in?
Is it y 2 liter?
459
00:31:43 --> 00:31:49
Plus what is coming in from the
outside.
460
00:31:46 --> 00:31:52
We have to add that in,
and that will be plus 5 e to
461
00:31:51 --> 00:31:57
the negative t.
How about y?
462
00:31:54 --> 00:32:00
y prime is changing.
What comes in from x?
463
00:32:00 --> 00:32:06
That is 3x.
What goes out?
464
00:32:01 --> 00:32:07
Well, two is leaving here and
two is leaving here.
465
00:32:05 --> 00:32:11
It doesn't matter that they are
going out through separate
466
00:32:10 --> 00:32:16
pipes.
They are both going out.
467
00:32:12 --> 00:32:18
It is minus 4,
2 and 2.
468
00:32:14 --> 00:32:20
How about the inhomogeneous
term?
469
00:32:16 --> 00:32:22
There is one coming in,
but there is no salt in it.
470
00:32:20 --> 00:32:26
Therefore, that is not
changing.
471
00:32:23 --> 00:32:29
What is coming through that
pipe is necessary for the liquid
472
00:32:27 --> 00:32:33
balance.
But it has no effect
473
00:32:31 --> 00:32:37
whatsoever.
I will put a zero here but,
474
00:32:35 --> 00:32:41
of course, you don't have to
put that in.
475
00:32:38 --> 00:32:44
This is now an inhomogeneous
system.
476
00:32:42 --> 00:32:48
In other words,
the system is x prime equals
477
00:32:46 --> 00:32:52
this matrix, negative 3,
the same sort of stuff we
478
00:32:50 --> 00:32:56
always had, plus the
inhomogeneous term which is the
479
00:32:55 --> 00:33:01
column vector 5 e to the minus t
and zero.
480
00:33:00 --> 00:33:06
It is the presence of this term
481
00:33:04 --> 00:33:10
that makes this system
inhomogeneous.
482
00:33:06 --> 00:33:12
And what that corresponds to is
this little closed system being
483
00:33:11 --> 00:33:17
attacked from the outside by
these external pipes which are
484
00:33:15 --> 00:33:21
bringing salt in.
Without those,
485
00:33:17 --> 00:33:23
of course the balance would be
all wrong.
486
00:33:20 --> 00:33:26
I would have to change this to
three and cut that out,
487
00:33:24 --> 00:33:30
I guess.
But then, it would be just a
488
00:33:27 --> 00:33:33
simple homogenous system.
It is these pipes that make it
489
00:33:32 --> 00:33:38
inhomogeneous.
Now, I should start to solve
490
00:33:35 --> 00:33:41
that.
I did this just to illustrate
491
00:33:38 --> 00:33:44
where a system might come from.
Before I solve that,
492
00:33:42 --> 00:33:48
what I want to do is,
of course, is solve it in
493
00:33:45 --> 00:33:51
general.
In other words,
494
00:33:47 --> 00:33:53
how do you solve this in
general?
495
00:33:49 --> 00:33:55
Because I promised you that you
would be able to do in general,
496
00:33:54 --> 00:34:00
regardless of what sort of
functions were in the r of t,
497
00:33:58 --> 00:34:04
that column vector.
So let's do it.
498
00:34:03 --> 00:34:09
499
00:34:10 --> 00:34:16
First of all,
you have to learn the name of
500
00:34:15 --> 00:34:21
the method.
This method is for solving x
501
00:34:20 --> 00:34:26
prime equals Ax.
It is a method for finding a
502
00:34:26 --> 00:34:32
particular solution.
503
00:34:30 --> 00:34:36
504
00:34:36 --> 00:34:42
Of course, to actually solve it
then you have to add the
505
00:34:39 --> 00:34:45
complimentary function.
We are looking for a particular
506
00:34:43 --> 00:34:49
solution for this system.
Now, the whole cleverness of
507
00:34:47 --> 00:34:53
the method, which I think was
discovered a couple hundred
508
00:34:51 --> 00:34:57
years ago by,
I think, Lagrange,
509
00:34:53 --> 00:34:59
I am not sure.
The method is called variation
510
00:34:57 --> 00:35:03
of parameters.
I am giving you that so that
511
00:35:00 --> 00:35:06
when you forget you will be able
to look it up and be indexes to
512
00:35:05 --> 00:35:11
some advanced engineering
mathematics book or something,
513
00:35:08 --> 00:35:14
whatever is on your shelf.
But, if course,
514
00:35:11 --> 00:35:17
you won't remember the name
either so maybe this won't work.
515
00:35:15 --> 00:35:21
Variation of parameters,
I will explain to you why it is
516
00:35:19 --> 00:35:25
called that.
All the cleverness is in the
517
00:35:22 --> 00:35:28
very first line.
If you could remember the very
518
00:35:25 --> 00:35:31
first line then I trust you to
do the rest yourself.
519
00:35:30 --> 00:35:36
I don't know any motivation for
this first step,
520
00:35:34 --> 00:35:40
but mathematics is supposed to
be mysterious anyway.
521
00:35:40 --> 00:35:46
It keeps me eating.
It says, look for a solution
522
00:35:45 --> 00:35:51
and there will be one of the
following form.
523
00:35:49 --> 00:35:55
Now, it will look exactly like
--
524
00:35:54 --> 00:36:00
525
00:36:00 --> 00:36:06
Look carefully because it is
going to be gone in a moment.
526
00:36:04 --> 00:36:10
It will look exactly like this.
But, of course,
527
00:36:08 --> 00:36:14
it cannot be this because this
solves the homogeneous system.
528
00:36:13 --> 00:36:19
If I plug this in with these as
constants it cannot possibly be
529
00:36:18 --> 00:36:24
a particular solution to this
because it will stop there and
530
00:36:23 --> 00:36:29
satisfy that with r equals zero.
The whole trick is you think of
531
00:36:29 --> 00:36:35
these are parameters which are
now variable.
532
00:36:32 --> 00:36:38
Constants that are varying.
That is why it is called
533
00:36:35 --> 00:36:41
variation of parameters.
You think of these,
534
00:36:38 --> 00:36:44
in other words,
as functions of t.
535
00:36:41 --> 00:36:47
We are going to look for a
solution which has the form,
536
00:36:45 --> 00:36:51
since they are functions of t,
I don't want to call them c1
537
00:36:49 --> 00:36:55
and c2 anymore.
I will call them v because that
538
00:36:52 --> 00:36:58
is what most people call them,
v or u, sometimes.
539
00:36:57 --> 00:37:03
540
00:37:02 --> 00:37:08
The method says look for a
solution of that form.
541
00:37:06 --> 00:37:12
The variation parameters,
these are the parameters that
542
00:37:10 --> 00:37:16
are now varying instead of being
constants.
543
00:37:14 --> 00:37:20
Now, if you take it in that
form and start trying to
544
00:37:18 --> 00:37:24
substitute into the equation you
are going to get a mess.
545
00:37:23 --> 00:37:29
I think I was wrong in saying I
could trust you from this point
546
00:37:28 --> 00:37:34
on.
I will take the first step from
547
00:37:32 --> 00:37:38
you, and then I could trust you
to do the rest after that first
548
00:37:38 --> 00:37:44
step.
The first step is to change the
549
00:37:42 --> 00:37:48
way this looks by using the
fundamental matrix.
550
00:37:46 --> 00:37:52
Remember what the fundamental
matrix was?
551
00:37:50 --> 00:37:56
Its entries were the two
columns of solutions.
552
00:37:54 --> 00:38:00
These are solutions to the
homogeneous system.
553
00:38:00 --> 00:38:06
And I am going to write it
using the fundamental matrix as,
554
00:38:05 --> 00:38:11
now thinks about it.
The fundamental matrix has
555
00:38:09 --> 00:38:15
columns x1 and x2.
Your instinct might be using
556
00:38:13 --> 00:38:19
matrix multiplication to put the
v1 and the v2 here,
557
00:38:18 --> 00:38:24
but that won't work.
You have to put them here.
558
00:38:23 --> 00:38:29
559
00:38:30 --> 00:38:36
This says the same thing as
that.
560
00:38:33 --> 00:38:39
Let's just take a second out to
calculate.
561
00:38:37 --> 00:38:43
The x is going to look like
(x1, y1).
562
00:38:40 --> 00:38:46
That is my first solution.
My second solution,
563
00:38:44 --> 00:38:50
here is the fundamental matrix,
is (x2, y2).
564
00:38:49 --> 00:38:55
And I am multiplying this on
the right by (v1,
565
00:38:53 --> 00:38:59
v2).
Does it come out right?
566
00:38:56 --> 00:39:02
Look.
What is it?
567
00:38:59 --> 00:39:05
The top is x1 v1 plus x2 v2.
568
00:39:02 --> 00:39:08
The top, x1 v1 plus x2 v2.
569
00:39:06 --> 00:39:12
It is in the wrong order,
but multiplication is
570
00:39:10 --> 00:39:16
commutative, fortunately.
And the same way the bottom
571
00:39:14 --> 00:39:20
thing will be v1 y1 plus v2 y2.
572
00:39:18 --> 00:39:24
If I had written it on the
other side instead,
573
00:39:22 --> 00:39:28
which is tempting because the
v's occur on the left here,
574
00:39:27 --> 00:39:33
that won't work.
What will I get?
575
00:39:31 --> 00:39:37
I will get v1 x1 plus v2 y1,
576
00:39:35 --> 00:39:41
which is not at all what I
want.
577
00:39:37 --> 00:39:43
You must put it on the right.
But this is a very important
578
00:39:42 --> 00:39:48
thing.
This is going to plague us on
579
00:39:45 --> 00:39:51
Monday, too.
It must be written on the right
580
00:39:49 --> 00:39:55
and not on the left as a column
vector.
581
00:39:52 --> 00:39:58
The rest of the program is very
simple.
582
00:39:55 --> 00:40:01
I will write it out as a
program.
583
00:40:00 --> 00:40:06
Substitute into the system,
into that, in other words,
584
00:40:04 --> 00:40:10
and see what v has to be.
That is what we are looking
585
00:40:08 --> 00:40:14
for.
We know what the x1 and x2 are.
586
00:40:11 --> 00:40:17
It is a question of what those
coefficients are.
587
00:40:14 --> 00:40:20
And see what v is.
Let's do it.
588
00:40:18 --> 00:40:24
589
00:40:32 --> 00:40:38
Let's substitute.
Let's see.
590
00:40:35 --> 00:40:41
The system is x prime equals Ax
plus r.
591
00:40:41 --> 00:40:47
I want to put in (x)p,
this proposed particular
592
00:40:46 --> 00:40:52
solution.
And it is a fundamental matrix,
593
00:40:50 --> 00:40:56
and the v is unknown.
How do I differentiate the
594
00:40:56 --> 00:41:02
product of two matrices?
You differentiate the product
595
00:41:02 --> 00:41:08
of two matrices using the
product rule that you learned
596
00:41:07 --> 00:41:13
the first day of 18.01.
Trust me.
597
00:41:10 --> 00:41:16
Let's do it.
I am going to substitute in.
598
00:41:14 --> 00:41:20
In other words,
here is my (x)p,
599
00:41:17 --> 00:41:23
(x)p, and I am going to write
in what that is.
600
00:41:21 --> 00:41:27
The left-hand side is the
derivative of,
601
00:41:25 --> 00:41:31
X prime times v,
plus X times the derivative of
602
00:41:29 --> 00:41:35
v.
Notice that one of these is a
603
00:41:34 --> 00:41:40
column vector and the other is a
square matrix.
604
00:41:37 --> 00:41:43
That is perfectly Okay.
Any two matrices which are the
605
00:41:39 --> 00:41:45
rate shape so you can multiply
them together,
606
00:41:42 --> 00:41:48
if you want to differentiate
their product,
607
00:41:44 --> 00:41:50
in other words,
if the entries are functions of
608
00:41:46 --> 00:41:52
t it is the product rule.
The derivative of this times
609
00:41:49 --> 00:41:55
time plus that times the
derivative of this.
610
00:41:52 --> 00:41:58
You have to keep them in the
right order.
611
00:41:54 --> 00:42:00
You are not allowed to shuffle
them around carelessly.
612
00:41:57 --> 00:42:03
So that is that.
What is it equal to?
613
00:42:00 --> 00:42:06
Well, the right-hand side is A.
And now I substitute just (x)p
614
00:42:08 --> 00:42:14
in, so that is X times v plus r.
Is this progress?
615
00:42:14 --> 00:42:20
What is v?
It looks like a mess but it is
616
00:42:20 --> 00:42:26
not.
Why not?
617
00:42:22 --> 00:42:28
It is because this is not any
old matrix X.
618
00:42:29 --> 00:42:35
This is a matrix whose columns
are solutions to the system.
619
00:42:34 --> 00:42:40
And what does that do?
That means X prime satisfies
620
00:42:39 --> 00:42:45
that matrix differential
equation.
621
00:42:42 --> 00:42:48
X prime is the same as Ax.
622
00:42:45 --> 00:42:51
And, by a little miracle,
the v is tagging along in both
623
00:42:50 --> 00:42:56
cases.
This cancels that and now there
624
00:42:54 --> 00:43:00
is very little left.
The conclusion,
625
00:42:58 --> 00:43:04
therefore, is that Xv is equal
to r.
626
00:43:02 --> 00:43:08
What is v?
It is v that we are looking
627
00:43:05 --> 00:43:11
for, right?
You have to solve a matrix
628
00:43:09 --> 00:43:15
equation, now.
This is a square matrix so you
629
00:43:13 --> 00:43:19
have to do it by inverting the
matrix.
630
00:43:16 --> 00:43:22
You don't just sloppily divide.
You multiply on which side by
631
00:43:21 --> 00:43:27
what matrix?
Choice of left or right.
632
00:43:25 --> 00:43:31
You multiply by the inverse
matrix on the left or on the
633
00:43:29 --> 00:43:35
right?
It has to be on the left.
634
00:43:35 --> 00:43:41
Multiply both sides of the
equation by X inverse on the
635
00:43:42 --> 00:43:48
left, and then you will get v is
equal to X inverse r.
636
00:43:50 --> 00:43:56
How do I know the X inverse
637
00:43:55 --> 00:44:01
exists?
Does X inverse exist?
638
00:44:00 --> 00:44:06
For a matrix inverse to exist,
the matrix's determinant must
639
00:44:05 --> 00:44:11
be not zero.
Why is the determinant of this
640
00:44:10 --> 00:44:16
not zero?
Because its columns are
641
00:44:13 --> 00:44:19
independent solutions.
642
00:44:16 --> 00:44:22
643
00:44:22 --> 00:44:28
Of course this is not right.
I forgot the prime here.
644
00:44:26 --> 00:44:32
645
00:44:31 --> 00:44:37
I am not failing this course
after all.
646
00:44:34 --> 00:44:40
v prime equals that.
647
00:44:38 --> 00:44:44
648
00:44:43 --> 00:44:49
This is done by differentiating
each entry in the column vector.
649
00:44:47 --> 00:44:53
And, therefore,
we should integrate it.
650
00:44:50 --> 00:44:56
It will be the integral,
just the ordinary
651
00:44:53 --> 00:44:59
anti-derivative of x inverse
times r.
652
00:44:57 --> 00:45:03
This is a column vector.
The entries are functions of t.
653
00:45:02 --> 00:45:08
You simply integrate each of
those functions in turn.
654
00:45:06 --> 00:45:12
So integrate each entry.
655
00:45:08 --> 00:45:14
656
00:45:12 --> 00:45:18
There is my v.
Sorry, you cannot tell the v's
657
00:45:19 --> 00:45:25
from the r's here.
And so, finally,
658
00:45:25 --> 00:45:31
the particular solution is (x)p
is equal to --
659
00:45:34 --> 00:45:40
It is really not bad at all.
It is equal to X times v.
660
00:45:37 --> 00:45:43
It's equal to X times the
integral of X inverse r dt.
661
00:45:40 --> 00:45:46
Now, actually,
662
00:45:43 --> 00:45:49
there is not much work to doing
that.
663
00:45:45 --> 00:45:51
Once you have solved the
homogeneous system and gotten
664
00:45:49 --> 00:45:55
the fundamental matrix,
taking the inverse of a
665
00:45:52 --> 00:45:58
two-by-two matrix is almost
trivial.
666
00:45:54 --> 00:46:00
You flip those two and you
change the signs of these two
667
00:45:57 --> 00:46:03
and you divide by the
determinant.
668
00:46:01 --> 00:46:07
You multiply it by r.
And the hard part is if you can
669
00:46:04 --> 00:46:10
do the integration.
If not, you just leave the
670
00:46:07 --> 00:46:13
integral sign the way you have
learned to do in this silly
671
00:46:11 --> 00:46:17
course and you still have the
answer.
672
00:46:14 --> 00:46:20
What about the arbitrary
constant of integration?
673
00:46:17 --> 00:46:23
The answer is you don't need to
put it in.
674
00:46:20 --> 00:46:26
Just find one particular
solution.
675
00:46:22 --> 00:46:28
It is good enough.
You don't have to put in the
676
00:46:25 --> 00:46:31
arbitrary constants of
integration.
677
00:46:29 --> 00:46:35
Because they are already in the
complimentary function here.
678
00:46:33 --> 00:46:39
Therefore, you don't have to
add them.
679
00:46:35 --> 00:46:41
I am sorry I didn't get a
chance to actually solve that.
680
00:46:39 --> 00:46:45
I will have to let it go.
The recitations will do it on
681
00:46:42 --> 00:46:48
Tuesday, will solve that
particular problem,
682
00:46:45 --> 00:46:51
which means you will,
in effect.