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The task for today is to find
particular solutions.

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So, let me remind you where
we've gotten to.

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We're talking about the
second-order equation with

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constant coefficients,
which you can think of as

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00:00:21,000 --> 00:00:27,000
modeling springs,
or simple electrical circuits.

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But, what's different now is
that the right-hand side is an

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input which is not zero.
So, we are considering,

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I'm going to use x as your book
does, keeping to a neutral

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letter.
But, again, in the

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applications,
and in many of the applications

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at any rate, it wants to be t.
But, I make it x.

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So, the independent variable is
x, and the problem is,

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remember, that to find a
particular solution,

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and the reason why we want to
do that is then the general

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solution will be of the form y
equals that particular solution

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plus the complementary solution,
the general solution to the

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reduced equation,
which we can write this way.

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So, all the work depends upon
finding out what that yp is,

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and that's what we're going to
talk about today,

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or rather, talk about for two
weeks.

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But, the point is,
not all functions that you

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could write on the right-hand
side are equally interesting.

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There's one kind,
which is far more interesting,

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but more important in the
applications than all the

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others.
And, that's the one out of

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which, in fact,
as you will see later on this

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week and into next week,
an arbitrary function can be

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built out of these simple
functions.

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So, the important function is
on the right-hand side to be

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00:02:13,000 --> 00:02:19,000
able to solve it when it's a
simple exponential.

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But, if you allow me to make it
a complex exponential,

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so, here are the important
right-hand sides where we want,

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we want to be able to do it
when it's of the form,

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e to the ax.
In general, that will be,

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in most applications,
a is not a growing exponential,

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but a decaying exponential.
So, typically,

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a is negative.
But, it doesn't have to be.

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I'll put it in parentheses,
though, often.

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00:02:48,000 --> 00:02:54,000
That's not any assumption that
I'm going to make today.

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It's just culture.
But, we want to be able to do

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00:02:56,000 --> 00:03:02,000
it for sine omega x and cosine
omega x.

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00:02:59,000 --> 00:03:05,000
In other words,

43
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when the right-hand side is a
pure oscillation,

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that's another important type
of input both for electrical

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circuits, think alternating
current, or the spring systems.

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00:03:15,000 --> 00:03:21,000
That's a pure vibration is
you're imposing pure vibration

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00:03:19,000 --> 00:03:25,000
on the spring-mass-dashpot
system, and you want to see how

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it responds to that.
Or, you could put them

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00:03:27,000 --> 00:03:33,000
together, and make these
decaying oscillations.

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00:03:32,000 --> 00:03:38,000
So, we could also have
something like e to the ax times

51
00:03:36,000 --> 00:03:42,000
sine omega x,
or times cosine omega x.

52
00:03:41,000 --> 00:03:47,000
Now, the point is,

53
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all of these together are
really just special cases of one

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00:03:50,000 --> 00:03:56,000
general thing,
exponential,

55
00:03:52,000 --> 00:03:58,000
if you allow the exponent not
to be a real number,

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00:03:56,000 --> 00:04:02,000
but to be a complex number.
So, they're all special cases

57
00:04:02,000 --> 00:04:08,000
of e to the, I'll write it alpha
x,

58
00:04:07,000 --> 00:04:13,000
well, why don't we write it (a
plus i omega) x, right?

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If omega is zero,

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00:04:16,000 --> 00:04:22,000
then I've got this case.
If a is zero,

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I got this case separating into
its real and imaginary parts.

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00:04:26,000 --> 00:04:32,000
And, if neither is zero,
I have this case.

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00:04:32,000 --> 00:04:38,000
But, I don't want to keep
writing (a plus i omega) all the

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00:04:36,000 --> 00:04:42,000
time.
So, I'm going to write that

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00:04:38,000 --> 00:04:44,000
simply as e to the alpha x.

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00:04:41,000 --> 00:04:47,000
And, you understand that alpha
is a complex number now.

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It doesn't look like a real
number.

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00:04:48,000 --> 00:04:54,000
Okay, so the complex number.
So, the equation we are solving

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00:04:52,000 --> 00:04:58,000
is which one?
This pretty purple equation.

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00:04:55,000 --> 00:05:01,000
And, we are trying to find a
particular solution of it.

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00:04:59,000 --> 00:05:05,000
And, the special functions we
are going to use are these,

72
00:05:03,000 --> 00:05:09,000
well, this one in particular,
e to the alpha x.

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That's going to be our input.

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00:05:13,000 --> 00:05:19,000
Now, it turns out this is
amazingly easy to do because

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00:05:19,000 --> 00:05:25,000
it's an exponential because I
write it in exponential form.

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The idea is simply to use a
rule which, in fact,

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00:05:32,000 --> 00:05:38,000
you know already,
the rule of substitution.

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So, I'm going to write the
equation in the form,

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00:05:43,000 --> 00:05:49,000
so, there it is.
It's y double prime plus A y

80
00:05:48,000 --> 00:05:54,000
prime plus B y equals f of x.

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00:05:55,000 --> 00:06:01,000
But, I'm going to think of the
left-hand side as the polynomial

82
00:06:02,000 --> 00:06:08,000
operator, AD plus B.
A and B are constants,

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00:06:07,000 --> 00:06:13,000
applied to y equals f of x.

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00:06:10,000 --> 00:06:16,000
That's the way I write the
thing.

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00:06:13,000 --> 00:06:19,000
And, this part,
I'm going to think of in the

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00:06:16,000 --> 00:06:22,000
form.
This is p of D,

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00:06:18,000 --> 00:06:24,000
a polynomial in D.
In fact, it's a simple

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00:06:21,000 --> 00:06:27,000
quadratic polynomial.
But, most of what I'm going to

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00:06:25,000 --> 00:06:31,000
say today would apply equally
well if we were a higher order

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00:06:29,000 --> 00:06:35,000
polynomial, a polynomial of
higher degree.

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00:06:34,000 --> 00:06:40,000
And, just to reinforce the
idea, I've given you one problem

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00:06:38,000 --> 00:06:44,000
in your problem set when p is a
polynomial of higher degree.

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00:06:42,000 --> 00:06:48,000
I should say,
the notes are written for

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general polynomials,
not just for quadratic ones.

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I'm simplifying it by leaving
it, today, I'll do what's in the

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notes, but I'll do it in the
quadratic case to save a little

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00:06:56,000 --> 00:07:02,000
time, and because that's the one
you will be most concerned with

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00:07:00,000 --> 00:07:06,000
in the problems.
All right, so p of D y equals f

99
00:07:05,000 --> 00:07:11,000
of x.
And now, there are just a

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00:07:09,000 --> 00:07:15,000
couple of basic formulas that
we're going to use all the time.

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00:07:15,000 --> 00:07:21,000
The first is that if you apply
p of D to a complex

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00:07:19,000 --> 00:07:25,000
exponential, or a real one,
it doesn't matter,

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00:07:23,000 --> 00:07:29,000
the answer is you get just what
you started with,

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00:07:27,000 --> 00:07:33,000
with D substituted by alpha.
So, it's p of alpha.

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00:07:32,000 --> 00:07:38,000
In other words,
put an alpha wherever you saw a

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00:07:36,000 --> 00:07:42,000
D in the polynomial.
And, what is this?

107
00:07:40,000 --> 00:07:46,000
Well, this is now just an
ordinary complex number,

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and multiply that by what you
started with,

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e to the alpha x.

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So, that's a basic formula.
It's called in the notes the

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substitution rule because the
heart of it is,

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you substitute for the D,
you substitute alpha.

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Now, this hardly requires
proof.

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00:08:02,000 --> 00:08:08,000
But, let's prove it just so you
see, to reinforce things and

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00:08:07,000 --> 00:08:13,000
make things go a little more
slowly to make sure you are on

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board all the time.
How would I prove that?

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Well, just calculate it out,
what in fact is (D squared plus

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AD plus B) times e to the alpha
x.

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Well, it's D squared e to the
alpha x by linearity plus AD

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00:08:30,000 --> 00:08:36,000
times e to the alpha x plus B
times e to the alpha x.

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00:08:40,000 --> 00:08:46,000
Well, what are these?
What's the derivative of e to

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the alpha x?
It's just alpha times e to the

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alpha x.
What's a second derivative?

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Well, if you remember from the
exam, you can do tenth

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derivatives now.
So, the second derivative is

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easy.
It's alpha squared times e to

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the alpha x.

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In other words,
this law, what I'm saying

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really is that this law is
obviously, quote unquote,

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"true."
Okay, I'm not even going to put

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it in quotes.
It's obviously true for the

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operator, D, and the operator D
squared.

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In other words,
D of e to the alpha x equals

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alpha times e to the alpha x.

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D squared times e to the alpha

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x equals alpha squared times e
to the alpha x.

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And, therefore,

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it's true for linear
combinations of these as well by

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linearity.
So, therefore,

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also true for p of D.
And, in fact,

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so if you calculate it out,
what is it?

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This is alpha squared e to the
alpha x plus alpha e to the

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alpha x times the

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coefficient plus b times e to
the alpha x.

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00:10:00,000 --> 00:10:06,000
So, it's in fact exactly this.
It's e to the alpha x times

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00:10:04,000 --> 00:10:10,000
(alpha squared plus A alpha plus
B).

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00:10:10,000 --> 00:10:16,000
Now, how are we going to use

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this?
Well, the idea is very simple.

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Remember, we're trying to solve
this, I should have some

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consistent notation for these
equations.

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Purple, I think,
will be the right thing here.

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You are solving purple
equations.

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The formulas which will solve
them will be orange formulas,

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and we will see what we need as
we go along.

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So, I would like to just
formulate it,

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this solution,
the particular solution now.

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I'm going to call it a theorem.
It's really too simple to be a

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theorem.
On the other hand,

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00:10:57,000 --> 00:11:03,000
it's too important not to be a
theorem.

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So, let's call it,
as I called it in the notes,

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00:11:06,000 --> 00:11:12,000
the exponential input theorem,
which says it all.

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Theorem says it's important.
Exponential input means it's

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00:11:17,000 --> 00:11:23,000
taking f of x to be an
exponential.

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00:11:21,000 --> 00:11:27,000
It's an exponential input,
and the theorem tells you what

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the response is.
So, for that equation,

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I'm not going to recopy the
equation for the purple

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00:11:40,000 --> 00:11:46,000
equation, adequately indicated
this way.

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00:11:46,000 --> 00:11:52,000
There.
Now try to take notes.

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For the purple equation,
a solution is e to the alpha x.

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Somewhere I neglected to say

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that f of x, all right,
so for the purple equals e to

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00:12:09,000 --> 00:12:15,000
the alpha x, how about that?

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That equation,
y double prime plus a y prime

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00:12:17,000 --> 00:12:23,000
plus b y equals e to
the alpha x.

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00:12:22,000 --> 00:12:28,000
So, here's the exponential

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input.
The solution is e to the alpha

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00:12:28,000 --> 00:12:34,000
x divided by p of alpha.

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00:12:34,000 --> 00:12:40,000
Now, that's a very useful
formula.

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00:12:36,000 --> 00:12:42,000
In fact, Haynes Miller,
who also teaches this course,

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00:12:40,000 --> 00:12:46,000
in his notes calls of the most
important theorem in the course.

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Well, I don't have to totally
agree with him,

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but it's certainly important.
It's probably the most

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00:12:53,000 --> 00:12:59,000
important theorem for these two
weeks, anyway.

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00:12:57,000 --> 00:13:03,000
But, you will have others as
well.

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00:12:59,000 --> 00:13:05,000
Okay, so that's a theorem.
The theorem is going green.

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00:13:03,000 --> 00:13:09,000
You can tell what they are by
their color code.

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00:13:09,000 --> 00:13:15,000
Well, in other words,
what I've done is simply write

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00:13:12,000 --> 00:13:18,000
down the solution for you,
write down the particular

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00:13:16,000 --> 00:13:22,000
solution.
But let's verify it in general.

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00:13:19,000 --> 00:13:25,000
So, the proof would be what?
Well, I have to substitute it

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00:13:24,000 --> 00:13:30,000
into the equation.
So, the equation is p of D

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00:13:27,000 --> 00:13:33,000
applied to y is equal to alpha
x.

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00:13:31,000 --> 00:13:37,000
And, I want to know,
when I substitute that

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00:13:34,000 --> 00:13:40,000
expression in,
is it the case that when I plug

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00:13:37,000 --> 00:13:43,000
it in, that the right-hand side,
I calculate it out,

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00:13:41,000 --> 00:13:47,000
apply p of D to it.
Is it the case that I get e to

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00:13:46,000 --> 00:13:52,000
the alpha x on the right?

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00:13:49,000 --> 00:13:55,000
Well, all you have to do is do
it.

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00:13:51,000 --> 00:13:57,000
What is p of D applied to e to
the alpha x divided by p of

200
00:13:56,000 --> 00:14:02,000
alpha?

201
00:14:00,000 --> 00:14:06,000
Well, p of D applied to e to
the alpha x is p of alpha times

202
00:14:03,000 --> 00:14:09,000
e to the alpha x.

203
00:14:07,000 --> 00:14:13,000
That's the substitution rule.
What about this guy?

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00:14:10,000 --> 00:14:16,000
This guy is a constant,
so it just gets dragged along

205
00:14:13,000 --> 00:14:19,000
because this operator is linear.
If this applied to that is

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00:14:17,000 --> 00:14:23,000
this, then if I apply it to one
half that, I get one half the

207
00:14:21,000 --> 00:14:27,000
answer, and so on.
So, the p of alpha

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00:14:24,000 --> 00:14:30,000
is a constant and just gets
dragged along.

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00:14:27,000 --> 00:14:33,000
And now, they cancel each
other, and the answer is,

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00:14:30,000 --> 00:14:36,000
indeed, e to the alpha x.

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00:14:34,000 --> 00:14:40,000
That's not much of a proof.
I hope that to at least half

212
00:14:39,000 --> 00:14:45,000
this class, you're wondering,
yes, but what if Peter had not

213
00:14:45,000 --> 00:14:51,000
caught the wolf?
I mean, what if?

214
00:14:48,000 --> 00:14:54,000
What if?

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00:15:01,000 --> 00:15:07,000
I'm looking stern.
Okay, we will take care of it

216
00:15:04,000 --> 00:15:10,000
in the simplest possible way.
We will assume that p of alpha

217
00:15:10,000 --> 00:15:16,000
is not zero.
The case p of alpha is zero

218
00:15:13,000 --> 00:15:19,000
is, in fact,
an extremely important

219
00:15:17,000 --> 00:15:23,000
case, one that makes the world
go 'round, one that contributes

220
00:15:22,000 --> 00:15:28,000
to all sorts of catastrophes,
and they occur first here in

221
00:15:27,000 --> 00:15:33,000
the solution of differential
equations, and that's what

222
00:15:32,000 --> 00:15:38,000
controls all the catastrophes.
But, there's a good side to it,

223
00:15:37,000 --> 00:15:43,000
too.
It also makes a lot of good

224
00:15:39,000 --> 00:15:45,000
things happen.
So, there are no moral

225
00:15:41,000 --> 00:15:47,000
judgments in mathematics.
For the time being,

226
00:15:44,000 --> 00:15:50,000
let's assume p of alpha is not
zero.

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00:15:46,000 --> 00:15:52,000
And, that proof is okay because
the p of alpha,

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00:15:48,000 --> 00:15:54,000
being in the denominator,
it's okay to be in the

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00:15:51,000 --> 00:15:57,000
denominator if you're not zero.
Okay, let's work in a simple

230
00:15:55,000 --> 00:16:01,000
example.
Well, I'm picking the most

231
00:15:56,000 --> 00:16:02,000
complicated example I can think
of.

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00:16:00,000 --> 00:16:06,000
Simple examples,
I'll leave for your practice

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00:16:04,000 --> 00:16:10,000
and for the recitations,
can start off with simple

234
00:16:08,000 --> 00:16:14,000
examples if you are confused by
this.

235
00:16:12,000 --> 00:16:18,000
But, let's solve an equation,
find a particular solution.

236
00:16:17,000 --> 00:16:23,000
So, y double prime plus minus y
prime plus 2y is equal to 10 e

237
00:16:23,000 --> 00:16:29,000
to the minus x sine x.

238
00:16:29,000 --> 00:16:35,000
Gulp.
Okay, so, the input is this

239
00:16:33,000 --> 00:16:39,000
function, 10 e to the minus x,

240
00:16:36,000 --> 00:16:42,000
it's a decaying oscillation.
You're seeing those already on

241
00:16:40,000 --> 00:16:46,000
the computer screen if you
started your homework,

242
00:16:44,000 --> 00:16:50,000
if you've done problem one on
your homework.

243
00:16:47,000 --> 00:16:53,000
It's a decaying exponential,
and I want to find a particular

244
00:16:52,000 --> 00:16:58,000
solution.
Well, let's find a particular

245
00:16:55,000 --> 00:17:01,000
and the general solution.
Find the general solution.

246
00:17:00,000 --> 00:17:06,000
Well, the main part of the work
is finding the particular

247
00:17:05,000 --> 00:17:11,000
solution, but let's quickly,
the general solution,

248
00:17:10,000 --> 00:17:16,000
let's find first the
complementary part of it,

249
00:17:14,000 --> 00:17:20,000
in other words,
the solution to the homogeneous

250
00:17:19,000 --> 00:17:25,000
equation.
That's D squared minus D plus

251
00:17:22,000 --> 00:17:28,000
two. No, let's not.

252
00:17:26,000 --> 00:17:32,000
I don't want to solve messy
quadratics.

253
00:17:31,000 --> 00:17:37,000
Okay, we're going to find a
particular solution.

254
00:17:34,000 --> 00:17:40,000
I thought it was going to come
out easy, and then I realized it

255
00:17:39,000 --> 00:17:45,000
wasn't because I picked the
wrong signs.

256
00:17:43,000 --> 00:17:49,000
Okay, so if you don't like,
just change the problem.

257
00:17:47,000 --> 00:17:53,000
I can do that,
but you cannot.

258
00:17:49,000 --> 00:17:55,000
Don't forget that.
So, we want a particular

259
00:17:53,000 --> 00:17:59,000
solution in our equation.
It is this equals that.

260
00:17:57,000 --> 00:18:03,000
Now, let's complexify it to
make this part of a complex

261
00:18:01,000 --> 00:18:07,000
exponential.
So, the complex exponential

262
00:18:06,000 --> 00:18:12,000
that's relevant is ten times e
to the (minus one plus i),

263
00:18:12,000 --> 00:18:18,000
you see that? x.

264
00:18:15,000 --> 00:18:21,000
What is this?
This is the imaginary part of

265
00:18:19,000 --> 00:18:25,000
this complex exponential.
So, this is imaginary part of

266
00:18:24,000 --> 00:18:30,000
that guy, e to the negative x
times e to the i x,

267
00:18:29,000 --> 00:18:35,000
and the imaginary part of e to

268
00:18:33,000 --> 00:18:39,000
the i x.

269
00:18:39,000 --> 00:18:45,000
The ten, of course,
just comes along for the ride.

270
00:18:42,000 --> 00:18:48,000
Okay, well, now,
since this is a complex

271
00:18:44,000 --> 00:18:50,000
equation, I shouldn't call this
y anymore by my notation.

272
00:18:48,000 --> 00:18:54,000
I like to call it y tilde to
indicate that the solution we

273
00:18:52,000 --> 00:18:58,000
get to this is not going to be
the original solution to the

274
00:18:56,000 --> 00:19:02,000
original problem,
but you will have to take the

275
00:18:59,000 --> 00:19:05,000
imaginary part of it to get it.
So, we are looking,

276
00:19:03,000 --> 00:19:09,000
now, for the complex solution
to this complexified equation.

277
00:19:08,000 --> 00:19:14,000
Okay, what is it?
Well, the complex particular

278
00:19:11,000 --> 00:19:17,000
solution I can write down
immediately.

279
00:19:14,000 --> 00:19:20,000
It is ten, that,
of course, just gets dragged

280
00:19:18,000 --> 00:19:24,000
along by linearity,
times e to the (minus one plus

281
00:19:22,000 --> 00:19:28,000
i) times x.
And, it's over this polynomial

282
00:19:27,000 --> 00:19:33,000
evaluated at this alpha.
So, just write it down with,

283
00:19:31,000 --> 00:19:37,000
have faith.
So, what do I get?

284
00:19:33,000 --> 00:19:39,000
The alpha is minus one plus i.

285
00:19:36,000 --> 00:19:42,000
I.
square that,

286
00:19:37,000 --> 00:19:43,000
because I'm substituting this
alpha into that polynomial.

287
00:19:41,000 --> 00:19:47,000
The reason I'm doing that is
because the formula tells me to

288
00:19:45,000 --> 00:19:51,000
do it.
That's going to be that

289
00:19:47,000 --> 00:19:53,000
solution.
Okay, so it's minus one plus i,

290
00:19:50,000 --> 00:19:56,000
the quantity squared, minus
minus one plus i plus two.

291
00:19:54,000 --> 00:20:00,000
All I've done is substitute

292
00:19:58,000 --> 00:20:04,000
minus one plus i for
D in that polynomial,

293
00:20:01,000 --> 00:20:07,000
the quadratic polynomial.
And now, all I want is the

294
00:20:07,000 --> 00:20:13,000
imaginary part of this.
The imaginary part of this will

295
00:20:12,000 --> 00:20:18,000
be the solution to the original
problem because this was the

296
00:20:18,000 --> 00:20:24,000
right hand side with the
imaginary part of the

297
00:20:22,000 --> 00:20:28,000
complexified right hand side.
Okay, now, let's make it look a

298
00:20:28,000 --> 00:20:34,000
little better,
yp tilde.

299
00:20:32,000 --> 00:20:38,000
Clearly, what we have to do
something nice to the

300
00:20:34,000 --> 00:20:40,000
denominator.
So, I'll copy the numerator.

301
00:20:37,000 --> 00:20:43,000
That's e to the minus (one plus
i) x

302
00:20:40,000 --> 00:20:46,000
and how about the denominator?
Well, again,

303
00:20:43,000 --> 00:20:49,000
don't expand things out because
it's already this long.

304
00:20:46,000 --> 00:20:52,000
And, what's the point of making
it this long?

305
00:20:48,000 --> 00:20:54,000
You want to make it as long,
right?

306
00:20:50,000 --> 00:20:56,000
Okay, then there is room here
for one real number,

307
00:20:53,000 --> 00:20:59,000
and another real number times
i, there's no more room.

308
00:20:57,000 --> 00:21:03,000
Okay, what's the real number?
Okay, we're looking for the

309
00:21:02,000 --> 00:21:08,000
real part of this expression.
So, just put it in and keep it

310
00:21:07,000 --> 00:21:13,000
mentally.
So, minus one squared:

311
00:21:09,000 --> 00:21:15,000
that's one, plus i squared,
that's minus one.

312
00:21:13,000 --> 00:21:19,000
One minus one is zero.
I can forget about that term.

313
00:21:18,000 --> 00:21:24,000
The term gives me plus one for
the real part,

314
00:21:21,000 --> 00:21:27,000
plus two.
The answer is that the real

315
00:21:24,000 --> 00:21:30,000
part is three.
How about the imaginary part?

316
00:21:28,000 --> 00:21:34,000
Well, from here,
there's negative 2i,

317
00:21:31,000 --> 00:21:37,000
negative 2i.
I'm expanding that out by the

318
00:21:38,000 --> 00:21:44,000
binomial theorem,
or whatever you like to call

319
00:21:44,000 --> 00:21:50,000
that, minus 2i minus i makes
minus 3i.

320
00:21:51,000 --> 00:21:57,000
Is that right?
Minus 2i, minus i,

321
00:21:56,000 --> 00:22:02,000
minus 3i.
So, it is ten thirds,

322
00:22:00,000 --> 00:22:06,000
and now in the denominator I
have one minus i.

323
00:22:08,000 --> 00:22:14,000
I'll put that in the numerator,
make it one plus i,

324
00:22:12,000 --> 00:22:18,000
but I have to divide by the
product of one minus i and its

325
00:22:17,000 --> 00:22:23,000
complex conjugate.
In other words,

326
00:22:20,000 --> 00:22:26,000
I'm multiplying both top and
bottom by one plus i.

327
00:22:24,000 --> 00:22:30,000
And so, that makes here one
squared plus one squared is two.

328
00:22:30,000 --> 00:22:36,000
And now, what's left is e to
the negative x times cosine x

329
00:22:35,000 --> 00:22:41,000
plus i sine x.

330
00:22:40,000 --> 00:22:46,000
Now, of that,
what we want is just the

331
00:22:43,000 --> 00:22:49,000
imaginary part.
Well, let's see.

332
00:22:46,000 --> 00:22:52,000
Two goes into ten makes five,
so that's five thirds.

333
00:22:52,000 --> 00:22:58,000
So, we're practically at our
solution.

334
00:22:55,000 --> 00:23:01,000
The solution,
then, finally,

335
00:22:58,000 --> 00:23:04,000
is going to be yp is the
imaginary part of yp tilde.

336
00:23:05,000 --> 00:23:11,000
And, what's that?
Well, what's the coefficient

337
00:23:08,000 --> 00:23:14,000
out front, first of all?
It's five thirds,

338
00:23:12,000 --> 00:23:18,000
so let's pull out to five
thirds before we forget it.

339
00:23:16,000 --> 00:23:22,000
And, we'll pull out the e to
the negative x before we forget

340
00:23:21,000 --> 00:23:27,000
that.
And then, the rest is simply a

341
00:23:24,000 --> 00:23:30,000
question of seeing what's left.
Well, it's one.

342
00:23:28,000 --> 00:23:34,000
I want the imaginary part.
So, the imaginary part is going

343
00:23:33,000 --> 00:23:39,000
to be one times cosine x,
and then the other

344
00:23:37,000 --> 00:23:43,000
imaginary part comes from these
two pieces, which is one times

345
00:23:42,000 --> 00:23:48,000
sine x.
And, that should be the

346
00:23:47,000 --> 00:23:53,000
particular solution.
Notice that most of the work is

347
00:23:52,000 --> 00:23:58,000
not getting this thing.
It's turning it into something

348
00:23:56,000 --> 00:24:02,000
human that you can take the real
and imaginary parts of.

349
00:24:02,000 --> 00:24:08,000
If we don't like this form,
you can put it in the other

350
00:24:07,000 --> 00:24:13,000
form, which many engineers would
do almost automatically,

351
00:24:12,000 --> 00:24:18,000
make it five thirds,
e to the negative x,

352
00:24:16,000 --> 00:24:22,000
and what will that be?
Well, you can use the general

353
00:24:21,000 --> 00:24:27,000
formula if you want.
Remember, cosine,

354
00:24:24,000 --> 00:24:30,000
the two coefficients are one
and one, so it's one and one.

355
00:24:30,000 --> 00:24:36,000
So, this is the square root of
two.

356
00:24:33,000 --> 00:24:39,000
So, it is times,
this part makes the square root

357
00:24:37,000 --> 00:24:43,000
of two times cosine of x minus
pi over the angle.

358
00:24:44,000 --> 00:24:50,000
This is a phi.
So, that's pi over four,

359
00:24:47,000 --> 00:24:53,000
minus pi over four.

360
00:24:58,000 --> 00:25:04,000
Okay, all right,
now let's address the case

361
00:25:02,000 --> 00:25:08,000
which is going to occupy a lot
of the rest of today,

362
00:25:07,000 --> 00:25:13,000
and in a certain sense,
all of next time.

363
00:25:11,000 --> 00:25:17,000
What happens when p of alpha is
zero?

364
00:25:16,000 --> 00:25:22,000
Well, in order to be able to
handle this decently,

365
00:25:21,000 --> 00:25:27,000
it's necessary to have one more
formula, which is very slightly

366
00:25:28,000 --> 00:25:34,000
more complicated than the
substitution rule.

367
00:25:34,000 --> 00:25:40,000
But, it's the same kind of
rule.

368
00:25:36,000 --> 00:25:42,000
I'm going to call this,
or it is called the

369
00:25:38,000 --> 00:25:44,000
exponential.
So, I'm going to first prove a

370
00:25:41,000 --> 00:25:47,000
formula, which is the analog of
that, and then I will prove a

371
00:25:45,000 --> 00:25:51,000
green formula,
which is what to do here if p

372
00:25:48,000 --> 00:25:54,000
of alpha turns out to be zero.

373
00:25:51,000 --> 00:25:57,000
But, in order to be able to
prove that, we're going to be

374
00:25:54,000 --> 00:26:00,000
the analog of the orange
formula.

375
00:25:56,000 --> 00:26:02,000
And, the analogue of the orange
formula, that tells you how to

376
00:26:00,000 --> 00:26:06,000
apply p of D to a
simple exponential.

377
00:26:05,000 --> 00:26:11,000
I need a formula which applies
p of D to that simple

378
00:26:11,000 --> 00:26:17,000
exponential times another
function.

379
00:26:14,000 --> 00:26:20,000
Now, I found I got into trouble
by continuing to call that

380
00:26:20,000 --> 00:26:26,000
alpha.
So, I'm now going to change the

381
00:26:24,000 --> 00:26:30,000
name of alpha to change alpha's
name to a.

382
00:26:30,000 --> 00:26:36,000
But, it's still complex.
I don't mean it's guaranteed to

383
00:26:34,000 --> 00:26:40,000
be complex.
I mean it's allowed to be

384
00:26:37,000 --> 00:26:43,000
complex.
So, a is now allowed to be a

385
00:26:40,000 --> 00:26:46,000
complex number.
I'm thinking of it,

386
00:26:43,000 --> 00:26:49,000
in general, as a complex
number, okay?

387
00:26:46,000 --> 00:26:52,000
I hope this doesn't upset you
too much, but you know,

388
00:26:51,000 --> 00:26:57,000
you change x to t's,
and y's to x's.

389
00:26:54,000 --> 00:27:00,000
This is no worse.
All right, what we going to do?

390
00:26:58,000 --> 00:27:04,000
Well, I'm going to use this
exponential shift rule,

391
00:27:02,000 --> 00:27:08,000
I'll call it,
exponential shift rule or

392
00:27:06,000 --> 00:27:12,000
formula or law.
That's the substitution rule

393
00:27:11,000 --> 00:27:17,000
for me.
So, this is going to be

394
00:27:13,000 --> 00:27:19,000
exponential shift law.
And, to apply,

395
00:27:17,000 --> 00:27:23,000
it tells you how to apply the
polynomial to not D,

396
00:27:21,000 --> 00:27:27,000
not just the exponential,
but the exponential times some

397
00:27:25,000 --> 00:27:31,000
function of x.
What's that?

398
00:27:28,000 --> 00:27:34,000
And now, the rule is very
simple.

399
00:27:32,000 --> 00:27:38,000
See, you can understand the
difficulty.

400
00:27:34,000 --> 00:27:40,000
If you try to start
differentiating,

401
00:27:37,000 --> 00:27:43,000
you're going to have to
calculate second derivatives of

402
00:27:40,000 --> 00:27:46,000
the stuff, and God forbid,
higher order equations.

403
00:27:44,000 --> 00:27:50,000
You would have to calculate
fourth derivatives,

404
00:27:47,000 --> 00:27:53,000
fifth derivatives.
You barely even want to

405
00:27:50,000 --> 00:27:56,000
calculate the first derivative.
That's okay.

406
00:27:53,000 --> 00:27:59,000
But, second derivative,
do I have to?

407
00:27:55,000 --> 00:28:01,000
No, not if you know the
exponential shift rule,

408
00:27:58,000 --> 00:28:04,000
which says you can get rid of
the e to the ax,

409
00:28:01,000 --> 00:28:07,000
make it pass to the left of the
operator where it's not in any

410
00:28:06,000 --> 00:28:12,000
position to do any harm any
longer, or upset the

411
00:28:09,000 --> 00:28:15,000
differentiation.
And, all you have to do is,

412
00:28:14,000 --> 00:28:20,000
when it passes over that
operator, it changes D to D plus

413
00:28:18,000 --> 00:28:24,000
a. So, the answer is,

414
00:28:21,000 --> 00:28:27,000
e to the ax.
There, it's passed over.

415
00:28:24,000 --> 00:28:30,000
But, when it did so,
it changed D to D plus a.

416
00:28:28,000 --> 00:28:34,000
And, what about the u?
Well, the u just stayed there.

417
00:28:32,000 --> 00:28:38,000
Nothing happened to it.
Okay, there's our orange

418
00:28:36,000 --> 00:28:42,000
formula.
I guess we better put the thing

419
00:28:40,000 --> 00:28:46,000
around the whole business.
Should I prove that,

420
00:28:43,000 --> 00:28:49,000
or the proof is quite easy.
So, let's do it just again to

421
00:28:48,000 --> 00:28:54,000
give you a chance to try to see,
now, if somebody gives you a

422
00:28:53,000 --> 00:28:59,000
formula like that,
you first stare at it.

423
00:28:56,000 --> 00:29:02,000
You might try a couple of
special cases,

424
00:28:59,000 --> 00:29:05,000
try it on a function and see if
it works, but already,

425
00:29:03,000 --> 00:29:09,000
you probably don't want to do
that.

426
00:29:08,000 --> 00:29:14,000
I mean, even if you took a
function like x here,

427
00:29:10,000 --> 00:29:16,000
you'd have to do a certain
amount of differentiating,

428
00:29:14,000 --> 00:29:20,000
and some quadratic thing here.
You'd calculate and calculate

429
00:29:17,000 --> 00:29:23,000
away for a little while,
and then if you did it

430
00:29:20,000 --> 00:29:26,000
correctly, the two would in fact
turn out to be equal.

431
00:29:23,000 --> 00:29:29,000
But, you would not necessarily
feel any the wiser.

432
00:29:26,000 --> 00:29:32,000
A better procedure in trying to
understand something like this

433
00:29:30,000 --> 00:29:36,000
is say, well,
let's keep the u general.

434
00:29:34,000 --> 00:29:40,000
Suppose we make D simple.
For example,

435
00:29:36,000 --> 00:29:42,000
well, if D is a constant,
of course there's nothing to

436
00:29:39,000 --> 00:29:45,000
happen because if this is just a
constant, both sides of these

437
00:29:43,000 --> 00:29:49,000
are the same.
This doesn't make any sense if

438
00:29:46,000 --> 00:29:52,000
p doesn't really have a D in it.
Well, what's the simplest

439
00:29:49,000 --> 00:29:55,000
polynomial which would have a D
in it?

440
00:29:52,000 --> 00:29:58,000
Well, D itself.
So, let's take a special case.

441
00:29:54,000 --> 00:30:00,000
p of D equals D,
and check the formula in that

442
00:29:58,000 --> 00:30:04,000
case; see if it works.
So, the formula is asking us,

443
00:30:03,000 --> 00:30:09,000
what is D, that's the p of D,
of e to the ax times u?

444
00:30:08,000 --> 00:30:14,000
I'm not going to put in the

445
00:30:12,000 --> 00:30:18,000
variable here because it's just
a waste of chalk.

446
00:30:16,000 --> 00:30:22,000
Well, what is that?
I know how to calculate that.

447
00:30:21,000 --> 00:30:27,000
I'll use the product rule.
So, it's the derivative.

448
00:30:25,000 --> 00:30:31,000
I'll tell you what;
let's do the other order first.

449
00:30:30,000 --> 00:30:36,000
So, it's e to the ax times the
derivative of u plus the

450
00:30:35,000 --> 00:30:41,000
derivative of e to the ax,
which is a times e to the ax

451
00:30:40,000 --> 00:30:46,000
times u.

452
00:30:45,000 --> 00:30:51,000
Do you follow that?
This is the product rule.

453
00:30:48,000 --> 00:30:54,000
It's e to the ax times the
derivative of u plus the

454
00:30:51,000 --> 00:30:57,000
derivative of e to the ax,
which is this thing times u,

455
00:30:55,000 --> 00:31:01,000
the other factor.

456
00:30:58,000 --> 00:31:04,000
Now, is that right?
I want to make it look like

457
00:31:01,000 --> 00:31:07,000
that.
Well, to make it look like that

458
00:31:04,000 --> 00:31:10,000
I should first factor e to the
ax out.

459
00:31:07,000 --> 00:31:13,000
And now, what's left?
Well, if I factor e to the ax

460
00:31:11,000 --> 00:31:17,000
out, what's left is Du plus au,
which is exactly (D plus a)

461
00:31:15,000 --> 00:31:21,000
operating on u,
D u plus a u.

462
00:31:19,000 --> 00:31:25,000
Well, hey, that's just what the
formula said it should be.

463
00:31:23,000 --> 00:31:29,000
If you make e to the x pass
over D, it changes D to

464
00:31:27,000 --> 00:31:33,000
D plus a.
Okay, now here's the main thing

465
00:31:32,000 --> 00:31:38,000
I want to show you.
All right, now,

466
00:31:34,000 --> 00:31:40,000
well let's try,
if this is true,

467
00:31:37,000 --> 00:31:43,000
also works out for D squared,
then the formula is

468
00:31:41,000 --> 00:31:47,000
clearly true by linearity
because an arbitrary p of D

469
00:31:45,000 --> 00:31:51,000
is just a linear
combination with constant

470
00:31:49,000 --> 00:31:55,000
coefficients of D,
D squared, and that constant

471
00:31:53,000 --> 00:31:59,000
thing, which we agreed there was
nothing to prove about.

472
00:31:57,000 --> 00:32:03,000
Now, hack, you're a hack if you
take D squared and start

473
00:32:01,000 --> 00:32:07,000
calculating the second
derivative of this.

474
00:32:06,000 --> 00:32:12,000
Okay, it's question about
hacks.

475
00:32:08,000 --> 00:32:14,000
I mean, it's just,
you haven't learned the right

476
00:32:12,000 --> 00:32:18,000
thing to do.
Okay, that will work,

477
00:32:14,000 --> 00:32:20,000
but it's not what you want to
do.

478
00:32:17,000 --> 00:32:23,000
Instead, you bootstrap your way
up.

479
00:32:20,000 --> 00:32:26,000
I have already a formula
telling me how to handle this.

480
00:32:24,000 --> 00:32:30,000
And, you can be anything.
Look at this not as D squared

481
00:32:28,000 --> 00:32:34,000
all by itself.
Calculate, instead,

482
00:32:32,000 --> 00:32:38,000
D squared e to the ax times u.

483
00:32:36,000 --> 00:32:42,000
Think of that as D,
the derivative of the

484
00:32:39,000 --> 00:32:45,000
derivative of e to the a x u.

485
00:32:42,000 --> 00:32:48,000
In other words,
we will do it one step at a

486
00:32:45,000 --> 00:32:51,000
time.
But you see now immediately the

487
00:32:48,000 --> 00:32:54,000
advantage of this.
What's D of e to the a x u?

488
00:32:51,000 --> 00:32:57,000
Well, I just calculated that.

489
00:32:55,000 --> 00:33:01,000
Now, don't go back to the
beginning.

490
00:32:57,000 --> 00:33:03,000
Don't go back to here.
Use the formula.

491
00:33:02,000 --> 00:33:08,000
After all, you worked to
calculate it,

492
00:33:04,000 --> 00:33:10,000
or I did.
So, it's D of,

493
00:33:06,000 --> 00:33:12,000
and what's this inside?
It's e to the ax times (D plus

494
00:33:10,000 --> 00:33:16,000
a) times u.
Well, that looks like a mess,

495
00:33:15,000 --> 00:33:21,000
but it isn't because I'm taking
D of e to the ax times

496
00:33:19,000 --> 00:33:25,000
something.
And I already know how to take

497
00:33:23,000 --> 00:33:29,000
D of e to the ax times
something.

498
00:33:25,000 --> 00:33:31,000
It doesn't matter what that
something is.

499
00:33:30,000 --> 00:33:36,000
Here, the something was u.
Here, the something is D plus

500
00:33:35,000 --> 00:33:41,000
(a times u) operating on u.
But, the principle is the same,

501
00:33:40,000 --> 00:33:46,000
and the answer is what?
Well, to take D of e to the ax

502
00:33:46,000 --> 00:33:52,000
times something,
you pass the e to the ax over

503
00:33:50,000 --> 00:33:56,000
the D.
That changes D to D plus a.

504
00:33:53,000 --> 00:33:59,000
And, you apply that to the

505
00:33:57,000 --> 00:34:03,000
other guy, which is (D plus a)
applied to u.

506
00:34:03,000 --> 00:34:09,000
What's the answer?
e to the a x times (D plus a)

507
00:34:06,000 --> 00:34:12,000
squared u.

508
00:34:09,000 --> 00:34:15,000
It's just what you would have
gotten if you had taken e to the

509
00:34:14,000 --> 00:34:20,000
e to the x, pass it over,
and then changed D to D plus a.

510
00:34:18,000 --> 00:34:24,000
Now, another advantage to doing

511
00:34:22,000 --> 00:34:28,000
it this way is you can see that
this argument is going to

512
00:34:26,000 --> 00:34:32,000
generalize to D cubed.
In other words,

513
00:34:31,000 --> 00:34:37,000
you would continue on in the
same way by the process of

514
00:34:36,000 --> 00:34:42,000
mathematical,
one word, mathematical,

515
00:34:39,000 --> 00:34:45,000
begins with an I,
induction.

516
00:34:41,000 --> 00:34:47,000
By induction,
you would prove the same

517
00:34:44,000 --> 00:34:50,000
formula for the nth derivative.
If you don't know what

518
00:34:49,000 --> 00:34:55,000
mathematical induction is,
shame on you.

519
00:34:53,000 --> 00:34:59,000
But it's okay.
A lot of you will be able to go

520
00:34:57,000 --> 00:35:03,000
through life without ever having
to learn what it is.

521
00:35:03,000 --> 00:35:09,000
And, the rest of you will be
computer scientists.

522
00:35:08,000 --> 00:35:14,000
Okay, so that's the idea of
this rule.

523
00:35:13,000 --> 00:35:19,000
Now, we can use it to calculate
something.

524
00:35:18,000 --> 00:35:24,000
Let's see, I'm going to need
green for this,

525
00:35:23,000 --> 00:35:29,000
I guess, for our better
formula.

526
00:35:27,000 --> 00:35:33,000
The formula,
now, that tells you what to do

527
00:35:32,000 --> 00:35:38,000
if p of alpha is zero.

528
00:35:39,000 --> 00:35:45,000
So, we're trying to solve the
equation, D squared plus A D,

529
00:35:45,000 --> 00:35:51,000
we are trying to find a

530
00:35:49,000 --> 00:35:55,000
particular solution,
e to the ax,

531
00:35:54,000 --> 00:36:00,000
let's say.
Remember, a is complex.

532
00:35:58,000 --> 00:36:04,000
a could be complex.
It doesn't have to be real.

533
00:36:05,000 --> 00:36:11,000
But, the problem is that p of
alpha is zero.

534
00:36:09,000 --> 00:36:15,000
How do I get a particular
solution?

535
00:36:12,000 --> 00:36:18,000
Well, I will write it down for
you.

536
00:36:14,000 --> 00:36:20,000
So, this is part of that
exponential input theorem.

537
00:36:18,000 --> 00:36:24,000
I think that's the way it is in
the notes.

538
00:36:21,000 --> 00:36:27,000
I gave all the cases together,
but I thought pedagogically

539
00:36:25,000 --> 00:36:31,000
it's probably a little better to
do the simplest case first,

540
00:36:30,000 --> 00:36:36,000
and then build up on the
complexity.

541
00:36:34,000 --> 00:36:40,000
So, what's yp?
The answer is yp is e to the a,

542
00:36:38,000 --> 00:36:44,000
except now you have to multiply

543
00:36:43,000 --> 00:36:49,000
it by x out front.
Where have you done something

544
00:36:48,000 --> 00:36:54,000
like that before?
Yes, don't tell me.

545
00:36:51,000 --> 00:36:57,000
I know you know.
But, what should go in the

546
00:36:56,000 --> 00:37:02,000
denominator?
Clearly not p of alpha.

547
00:36:59,000 --> 00:37:05,000
What goes in the denominator is

548
00:37:04,000 --> 00:37:10,000
the derivative.
Okay, but what if p prime of

549
00:37:09,000 --> 00:37:15,000
alpha is zero?
Couldn't that happen?

550
00:37:14,000 --> 00:37:20,000
Yes, it could happen.
So, we better make cases.

551
00:37:18,000 --> 00:37:24,000
This case is,
the case where this is okay

552
00:37:22,000 --> 00:37:28,000
corresponds to the case where
we're going to assume that alpha

553
00:37:27,000 --> 00:37:33,000
is a simple root,
is a simple root of the

554
00:37:31,000 --> 00:37:37,000
polynomial, p.
I don't know what to call the

555
00:37:35,000 --> 00:37:41,000
variable, p of D is okay.
A simple zero,

556
00:37:40,000 --> 00:37:46,000
in other words,
it's not double.

557
00:37:43,000 --> 00:37:49,000
Well, suppose is double.
One of the consequences you

558
00:37:48,000 --> 00:37:54,000
will see just in a second,
if it's a simple zero,

559
00:37:52,000 --> 00:37:58,000
that means this derivative is
not going to be zero.

560
00:37:57,000 --> 00:38:03,000
That's automatic.
Yeah, well, suppose it's not as

561
00:38:01,000 --> 00:38:07,000
simple.
Well, suppose is a double root.

562
00:38:05,000 --> 00:38:11,000
How did a-- How did that get
changed to, argh!

563
00:38:11,000 --> 00:38:17,000
[LAUGHTER] That's not an alpha.
Oh, well, yes it is,

564
00:38:16,000 --> 00:38:22,000
obviously.
Change!

565
00:38:18,000 --> 00:38:24,000
All of you, I want you to
change.

566
00:38:21,000 --> 00:38:27,000
They should have something like
in a search key where

567
00:38:27,000 --> 00:38:33,000
occurrences of alpha have been
changed to a with a stroke of a,

568
00:38:34,000 --> 00:38:40,000
just your thumb.
They don't have that for the

569
00:38:39,000 --> 00:38:45,000
blackboard, unfortunately.
Well, too bad,

570
00:38:42,000 --> 00:38:48,000
for the future.
Correctable blackboards.

571
00:38:46,000 --> 00:38:52,000
Well, what if a is double root?
It can't be more than a double

572
00:38:51,000 --> 00:38:57,000
root because you've only got a
quadratic polynomial.

573
00:38:55,000 --> 00:39:01,000
Quadratic polynomials only have
two roots.

574
00:39:00,000 --> 00:39:06,000
So, the worst that can happen
is that both of them are a.

575
00:39:04,000 --> 00:39:10,000
All right, in that case the
formula should be yp is equal

576
00:39:09,000 --> 00:39:15,000
to, you are now going to need x
squared up there times e

577
00:39:14,000 --> 00:39:20,000
to the ax,
and in the denominator what you

578
00:39:18,000 --> 00:39:24,000
are going to need is the second
derivative of,

579
00:39:22,000 --> 00:39:28,000
evaluated at a.
Now, you can guess the way this

580
00:39:25,000 --> 00:39:31,000
going to go on.
For higher degree things,

581
00:39:29,000 --> 00:39:35,000
if you've got a triple root,
you will need here x cubed,

582
00:39:37,000 --> 00:39:43,000
except you're going to need a
factorial there,

583
00:39:41,000 --> 00:39:47,000
too.
So, don't worry about it.

584
00:39:45,000 --> 00:39:51,000
It's in the notes,
but I'm not going to give you

585
00:39:49,000 --> 00:39:55,000
that for higher roots.
I don't even know if I will

586
00:39:53,000 --> 00:39:59,000
give it to you for double root.
Yes, I already did,

587
00:39:57,000 --> 00:40:03,000
so it's too late.
It's too late.

588
00:40:01,000 --> 00:40:07,000
Okay, so we will make this two
formulas according to whether a

589
00:40:06,000 --> 00:40:12,000
is a single or a double root.
Okay, let's prove one of these,

590
00:40:11,000 --> 00:40:17,000
and all of that will be good
enough for my conscience.

591
00:40:16,000 --> 00:40:22,000
Let's prove the first one.
Mostly, it's an exercise in

592
00:40:21,000 --> 00:40:27,000
using the first exponential
shift rule.

593
00:40:24,000 --> 00:40:30,000
Okay, this will be a first
example actually seeing a work

594
00:40:29,000 --> 00:40:35,000
in practice as opposed to
proving it.

595
00:40:34,000 --> 00:40:40,000
Okay, so what does that thing
look like?

596
00:40:37,000 --> 00:40:43,000
So, what does the polynomial
look like, which has a as a

597
00:40:42,000 --> 00:40:48,000
simple root?
So, we're going to try to prove

598
00:40:46,000 --> 00:40:52,000
the simple root case.
So, I'm just going to calculate

599
00:40:50,000 --> 00:40:56,000
what those guys actually look
like.

600
00:40:53,000 --> 00:40:59,000
What does p of D look
like if a is a simple root?

601
00:41:00,000 --> 00:41:06,000
Well, if it's a simple root,
that means it has a factor.

602
00:41:04,000 --> 00:41:10,000
When it factors,
it factors into the product of

603
00:41:08,000 --> 00:41:14,000
(D minus a) times something
which isn't, D minus some other

604
00:41:13,000 --> 00:41:19,000
root.
And, the point is that b is not

605
00:41:16,000 --> 00:41:22,000
equal to A.
The roots are really distinct.

606
00:41:20,000 --> 00:41:26,000
Okay, what's,
then, p prime,

607
00:41:23,000 --> 00:41:29,000
I'm going to have to calculate
p prime of a.

608
00:41:27,000 --> 00:41:33,000
What is that?
Well, let's calculate p prime

609
00:41:32,000 --> 00:41:38,000
of D first.
It is, well,

610
00:41:34,000 --> 00:41:40,000
by the ordinary product rule,
it's the derivative of this

611
00:41:38,000 --> 00:41:44,000
times, which is one times (D
minus a) plus,

612
00:41:41,000 --> 00:41:47,000
that's one thing plus the same
thing on the other side,

613
00:41:45,000 --> 00:41:51,000
the derivative of this,
which is one times (D minus b).

614
00:41:49,000 --> 00:41:55,000
So, that's p prime.
And therefore,

615
00:41:51,000 --> 00:41:57,000
what's p prime of a?
It's nothing but,

616
00:41:55,000 --> 00:42:01,000
this part is zero,
and that's a minus b.

617
00:41:59,000 --> 00:42:05,000
Of course, this is not zero
because it's a simple root.

618
00:42:02,000 --> 00:42:08,000
And, that's the proof for you
if you want, that if the root is

619
00:42:06,000 --> 00:42:12,000
simple, that p prime of a
is guaranteed not to

620
00:42:10,000 --> 00:42:16,000
be zero.
And, you can see,

621
00:42:12,000 --> 00:42:18,000
it's going to be zero exactly
when b equals a,

622
00:42:15,000 --> 00:42:21,000
and that root occurs twice.
But, I'm assuming that didn't

623
00:42:19,000 --> 00:42:25,000
happen.
Okay, then all the rest we have

624
00:42:22,000 --> 00:42:28,000
to do is calculate,
do the calculation.

625
00:42:24,000 --> 00:42:30,000
So, what I want to prove now is
that with this p of D,

626
00:42:28,000 --> 00:42:34,000
what I'm trying to calculate
that p of D times that guy,

627
00:42:32,000 --> 00:42:38,000
x e to the a x,
except I'm going to

628
00:42:36,000 --> 00:42:42,000
write it as e to the a x times
x, guess why,

629
00:42:39,000 --> 00:42:45,000
divided by p prime of a.

630
00:42:44,000 --> 00:42:50,000
This is my proposed particular
solution.

631
00:42:47,000 --> 00:42:53,000
So, what I have to do is
calculate it,

632
00:42:50,000 --> 00:42:56,000
and see that it turns out to
be, what do I hope it turns out

633
00:42:54,000 --> 00:43:00,000
to be?
What the right hand side of the

634
00:42:57,000 --> 00:43:03,000
equation, the input?
The input is e to the ax.

635
00:43:01,000 --> 00:43:07,000
If this is true,

636
00:43:03,000 --> 00:43:09,000
then yp, a particular solution,
indeed, nothing will be a

637
00:43:07,000 --> 00:43:13,000
particular solution.
Of course, there could be

638
00:43:11,000 --> 00:43:17,000
others, but in this game,
I only have to find one

639
00:43:14,000 --> 00:43:20,000
particular solution,
and that certainly by far is

640
00:43:18,000 --> 00:43:24,000
the simple as one you could
possibly find.

641
00:43:21,000 --> 00:43:27,000
So, I have to calculate this.
And now, you see why I did the

642
00:43:25,000 --> 00:43:31,000
exponential shift rule because
this is begging to be

643
00:43:29,000 --> 00:43:35,000
differentiated by something
simpler than hack.

644
00:43:34,000 --> 00:43:40,000
Okay, you can also see why I
violated the natural order of

645
00:43:38,000 --> 00:43:44,000
things and put the e to the ax
on the left in order

646
00:43:43,000 --> 00:43:49,000
that it pass over more easily.
So, the answer on the left-hand

647
00:43:48,000 --> 00:43:54,000
side is e to the ax times p of
(D plus a).

648
00:43:51,000 --> 00:43:57,000
Now, what is (p of D) plus a?
Write it in this form.

649
00:43:56,000 --> 00:44:02,000
It's going to be a minus b.

650
00:44:00,000 --> 00:44:06,000
So, p of (D plus a)
is, change D to D plus a.

651
00:44:04,000 --> 00:44:10,000
So, the first factor is going

652
00:44:07,000 --> 00:44:13,000
to be D plus a minus b.

653
00:44:10,000 --> 00:44:16,000
And, what's the second factor?
Change D to D plus a.

654
00:44:15,000 --> 00:44:21,000
It turns into D.

655
00:44:16,000 --> 00:44:22,000
All this is the result of
taking that p of D,

656
00:44:21,000 --> 00:44:27,000
and changing D to D plus a.
And now, this is to be applied

657
00:44:25,000 --> 00:44:31,000
to what?
Well, e to the a x

658
00:44:28,000 --> 00:44:34,000
is already passed over.
So, what's left is x.

659
00:44:33,000 --> 00:44:39,000
And, that's to be divided by
the constant,

660
00:44:36,000 --> 00:44:42,000
p prime of a.
Now, what does this all come

661
00:44:41,000 --> 00:44:47,000
out to be?
e to the ax,

662
00:44:44,000 --> 00:44:50,000
what's D applied to x?
One, right?

663
00:44:48,000 --> 00:44:54,000
And now, what's this thing
applied to the constant one?

664
00:44:53,000 --> 00:44:59,000
Well, the D kills it,
so it has no effect.

665
00:44:57,000 --> 00:45:03,000
It makes it zero.
The rest just multiplies it by

666
00:45:02,000 --> 00:45:08,000
a minus b.
So, the answer to the top is (a

667
00:45:07,000 --> 00:45:13,000
minus b) times one.
And, the answer to the bottom

668
00:45:12,000 --> 00:45:18,000
is p prime of a,
which I showed you by just

669
00:45:17,000 --> 00:45:23,000
explicit calculation is a minus
b.

670
00:45:21,000 --> 00:45:27,000
And so the answer is,
e to the a x comes

671
00:45:25,000 --> 00:45:31,000
out right.
Now, the other one,

672
00:45:29,000 --> 00:45:35,000
the other formula comes out the
same way.

673
00:45:31,000 --> 00:45:37,000
I'll leave that as an exercise.
Also, I don't dare do it

674
00:45:34,000 --> 00:45:40,000
because it's much too close to
the problem I asked you to do

675
00:45:38,000 --> 00:45:44,000
for homework.
So, let's by way of conclusion,

676
00:45:41,000 --> 00:45:47,000
I'll do one more simple
example, okay?

677
00:45:43,000 --> 00:45:49,000
And then, you can feel you
understand something.

678
00:45:46,000 --> 00:45:52,000
I'm sort of bothered that I
haven't done any examples of

679
00:45:49,000 --> 00:45:55,000
this more complicated case.
So, I'll pick an easy version

680
00:45:53,000 --> 00:45:59,000
instead of the one that you have
in your notes,

681
00:45:56,000 --> 00:46:02,000
which is the one you have for
homework, which is even easier.

682
00:46:01,000 --> 00:46:07,000
So, this one's epsilon less
easy.

683
00:46:04,000 --> 00:46:10,000
Y double prime minus 3 y prime
plus 2 y equals e to the x.

684
00:46:10,000 --> 00:46:16,000
Okay, notice that one is a

685
00:46:14,000 --> 00:46:20,000
simple root.
The one I'm talking about is

686
00:46:18,000 --> 00:46:24,000
the a here, which is one.
One is a simple root of the

687
00:46:23,000 --> 00:46:29,000
polynomial D squared minus 3D
plus two,

688
00:46:28,000 --> 00:46:34,000
isn't it?
It's a zero.

689
00:46:32,000 --> 00:46:38,000
Put D equal one and you get
one minus three plus

690
00:46:35,000 --> 00:46:41,000
two equals zero.
It's a simple root because

691
00:46:38,000 --> 00:46:44,000
anybody can see that one is not
a double root because you know

692
00:46:41,000 --> 00:46:47,000
from critical damping,
if one were a double root,

693
00:46:44,000 --> 00:46:50,000
you know just what the
polynomial would look like,

694
00:46:47,000 --> 00:46:53,000
and it wouldn't look like that
at all.

695
00:46:49,000 --> 00:46:55,000
It would not look like D
squared minus 3D plus two.

696
00:46:52,000 --> 00:46:58,000
It would look differently.

697
00:46:54,000 --> 00:47:00,000
Therefore, that proves that one
is a simple root.

698
00:46:57,000 --> 00:47:03,000
Okay, what's the particular
solution, therefore?

699
00:47:01,000 --> 00:47:07,000
The particular solution is x
times e to the x divided by the

700
00:47:06,000 --> 00:47:12,000
derivative, the derivative
evaluated at the point,

701
00:47:11,000 --> 00:47:17,000
so, what's p prime of D?
It is 2D minus three.

702
00:47:15,000 --> 00:47:21,000
If I evaluate it at the point,
one, it is negative one.

703
00:47:20,000 --> 00:47:26,000
So, if this is to be divided by
negative one,

704
00:47:25,000 --> 00:47:31,000
in other words,
it's minus x e to the X.

705
00:47:30,000 --> 00:47:36,000
And, if you don't believe it,
you could plug it in and check

706
00:47:35,000 --> 00:47:41,000
it out.
Okay, I'm letting you out one

707
00:47:39,000 --> 00:47:45,000
minute early.
Remember that.

708
00:47:41,000 --> 00:47:47,000
I'm trying to pay off the
accumulated debt.