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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

212
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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

217
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t.
That is good.

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As you will see,
we are going to need that

219
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property.
But the other one is a little

220
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stranger.
The only thing I can say is,

221
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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.

223
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This is a column vector.
That is a square matrix and

224
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this is a column vector.
This is not a column vector.

225
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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

227
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column vector of a set of x's
like the x and the y.

228
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It is a whole matrix.

229
00:15:40 --> 00:15:46

230
00:15:50 --> 00:15:56
Well, first of all,
I should say what is it saying?

231
00:15:53 --> 00:15:59
This is a two-by-two matrix.
When I multiply them I get a

232
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two-by-two matrix.
What is this?

233
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This is a two-by-two matrix,
every entry of which has been

234
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differentiated.
That is what it means to put

235
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that prime there.
To differentiate a matrix means

236
00:16:11 --> 00:16:17
nothing fancy.
It just means differentiate

237
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every entry.
It is just like to

238
00:16:16 --> 00:16:22
differentiate a vector (x,
y), to make a velocity vector

239
00:16:20 --> 00:16:26
you differentiate the x and the
y.

240
00:16:22 --> 00:16:28
Well, a column vector is a
special kind of matrix.

241
00:16:26 --> 00:16:32
The definition applies to any
matrix.

242
00:16:30 --> 00:16:36
Well, why is that so?
I state it as a property,

243
00:16:33 --> 00:16:39
but I will continue it by
giving you, so to speak,

244
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the proof of it.
In fact, there is nothing in

245
00:16:40 --> 00:16:46
this.
It is nothing more than a

246
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little matrix calculation of the
most primitive kind.

247
00:16:46 --> 00:16:52
Namely, what does this mean?
Let's try to undo that.

248
00:16:50 --> 00:16:56
What does the left-hand side
really mean?

249
00:16:53 --> 00:16:59
Well, if that is what x means,
the left-hand side must mean

250
00:16:58 --> 00:17:04
the derivative of the first
column.

251
00:17:02 --> 00:17:08
That is its first column.
And the derivative of the

252
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second column.
That is what it means to

253
00:17:07 --> 00:17:13
differentiate the matrix X.
You differentiate each column

254
00:17:11 --> 00:17:17
separately.
And to differentiate the column

255
00:17:14 --> 00:17:20
you need to differentiate every
function in it.

256
00:17:17 --> 00:17:23
Well, what does the right-hand
side mean?

257
00:17:19 --> 00:17:25
Well, I am supposed to take A
and multiply that 

258
00:17:23 --> 00:17:29
by [x1,x2].
Now, I don't know how to prove

259
00:17:26 --> 00:17:32
this, except ask you to think
about it.

260
00:17:30 --> 00:17:36
Or, I could write it all out
here.

261
00:17:34 --> 00:17:40
But think of this as a bing,
bing, bing, bing.

262
00:17:39 --> 00:17:45
And this is a bing,
bing.

263
00:17:42 --> 00:17:48
And this is a bong,
bong.

264
00:17:45 --> 00:17:51
How do I do the multiplication?
In other words,

265
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what is in the first column of
the matrix?

266
00:17:55 --> 00:18:01
Well, it is dah,
dah, and the lower thing is

267
00:18:01 --> 00:18:07
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.