WEBVTT
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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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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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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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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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it for sine omega x and cosine
omega x.
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In other words,
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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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That's a pure vibration is
you're imposing pure vibration
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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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together, and make these
decaying oscillations.
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So, we could also have
something like e to the ax times
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sine omega x,
or times cosine omega x.
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Now, the point is,
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all of these together are
really just special cases of one
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general thing,
exponential,
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if you allow the exponent not
to be a real number,
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but to be a complex number.
So, they're all special cases
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of e to the, I'll write it alpha
x,
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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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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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And, if neither is zero,
I have this case.
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But, I don't want to keep
writing (a plus i omega) all the
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time.
So, I'm going to write that
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simply as e to the alpha x.
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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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Okay, so the complex number.
So, the equation we are solving
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is which one?
This pretty purple equation.
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And, we are trying to find a
particular solution of it.
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And, the special functions we
are going to use are these,
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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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Now, it turns out this is
amazingly easy to do because
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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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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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so, there it is.
It's y double prime plus A y
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prime plus B y equals f of x.
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But, I'm going to think of the
left-hand side as the polynomial
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operator, AD plus B.
A and B are constants,
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applied to y equals f of x.
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That's the way I write the
thing.
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And, this part,
I'm going to think of in the
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form.
This is p of D,
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a polynomial in D.
In fact, it's a simple
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quadratic polynomial.
But, most of what I'm going to
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say today would apply equally
well if we were a higher order
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polynomial, a polynomial of
higher degree.
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And, just to reinforce the
idea, I've given you one problem
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in your problem set when p is a
polynomial of higher degree.
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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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time, and because that's the one
you will be most concerned with
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in the problems.
All right, so p of D y equals f
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of x.
And now, there are just a
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couple of basic formulas that
we're going to use all the time.
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The first is that if you apply
p of D to a complex
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exponential, or a real one,
it doesn't matter,
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the answer is you get just what
you started with,
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with D substituted by alpha.
So, it's p of alpha.
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In other words,
put an alpha wherever you saw a
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D in the polynomial.
And, what is this?
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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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But, let's prove it just so you
see, to reinforce things and
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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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times e to the alpha x plus B
times e to the alpha x.
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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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So, it's in fact exactly this.
It's e to the alpha x times
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(alpha squared plus A alpha plus
B).
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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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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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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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taking f of x to be an
exponential.
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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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equation, adequately indicated
this way.
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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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the alpha x, how about that?
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That equation,
y double prime plus a y prime
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plus b y equals e to
the alpha x.
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So, here's the exponential
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input.
The solution is e to the alpha
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x divided by p of alpha.
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Now, that's a very useful
formula.
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In fact, Haynes Miller,
who also teaches this course,
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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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important theorem for these two
weeks, anyway.
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But, you will have others as
well.
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Okay, so that's a theorem.
The theorem is going green.
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You can tell what they are by
their color code.
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Well, in other words,
what I've done is simply write
00:13:12.000 --> 00:13:18.000
down the solution for you,
write down the particular
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solution.
But let's verify it in general.
00:13:19.000 --> 00:13:25.000
So, the proof would be what?
Well, I have to substitute it
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into the equation.
So, the equation is p of D
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applied to y is equal to alpha
x.
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And, I want to know,
when I substitute that
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expression in,
is it the case that when I plug
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it in, that the right-hand side,
I calculate it out,
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apply p of D to it.
Is it the case that I get e to
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the alpha x on the right?
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Well, all you have to do is do
it.
00:13:51.000 --> 00:13:57.000
What is p of D applied to e to
the alpha x divided by p of
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alpha?
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Well, p of D applied to e to
the alpha x is p of alpha times
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e to the alpha x.
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That's the substitution rule.
What about this guy?
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This guy is a constant,
so it just gets dragged along
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because this operator is linear.
If this applied to that is
00:14:17.000 --> 00:14:23.000
this, then if I apply it to one
half that, I get one half the
00:14:21.000 --> 00:14:27.000
answer, and so on.
So, the p of alpha
00:14:24.000 --> 00:14:30.000
is a constant and just gets
dragged along.
00:14:27.000 --> 00:14:33.000
And now, they cancel each
other, and the answer is,
00:14:30.000 --> 00:14:36.000
indeed, e to the alpha x.
00:14:34.000 --> 00:14:40.000
That's not much of a proof.
I hope that to at least half
00:14:39.000 --> 00:14:45.000
this class, you're wondering,
yes, but what if Peter had not
00:14:45.000 --> 00:14:51.000
caught the wolf?
I mean, what if?
00:14:48.000 --> 00:14:54.000
What if?
00:15:01.000 --> 00:15:07.000
I'm looking stern.
Okay, we will take care of it
00:15:04.000 --> 00:15:10.000
in the simplest possible way.
We will assume that p of alpha
00:15:10.000 --> 00:15:16.000
is not zero.
The case p of alpha is zero
00:15:13.000 --> 00:15:19.000
is, in fact,
an extremely important
00:15:17.000 --> 00:15:23.000
case, one that makes the world
go 'round, one that contributes
00:15:22.000 --> 00:15:28.000
to all sorts of catastrophes,
and they occur first here in
00:15:27.000 --> 00:15:33.000
the solution of differential
equations, and that's what
00:15:32.000 --> 00:15:38.000
controls all the catastrophes.
But, there's a good side to it,
00:15:37.000 --> 00:15:43.000
too.
It also makes a lot of good
00:15:39.000 --> 00:15:45.000
things happen.
So, there are no moral
00:15:41.000 --> 00:15:47.000
judgments in mathematics.
For the time being,
00:15:44.000 --> 00:15:50.000
let's assume p of alpha is not
zero.
00:15:46.000 --> 00:15:52.000
And, that proof is okay because
the p of alpha,
00:15:48.000 --> 00:15:54.000
being in the denominator,
it's okay to be in the
00:15:51.000 --> 00:15:57.000
denominator if you're not zero.
Okay, let's work in a simple
00:15:55.000 --> 00:16:01.000
example.
Well, I'm picking the most
00:15:56.000 --> 00:16:02.000
complicated example I can think
of.
00:16:00.000 --> 00:16:06.000
Simple examples,
I'll leave for your practice
00:16:04.000 --> 00:16:10.000
and for the recitations,
can start off with simple
00:16:08.000 --> 00:16:14.000
examples if you are confused by
this.
00:16:12.000 --> 00:16:18.000
But, let's solve an equation,
find a particular solution.
00:16:17.000 --> 00:16:23.000
So, y double prime plus minus y
prime plus 2y is equal to 10 e
00:16:23.000 --> 00:16:29.000
to the minus x sine x.
00:16:29.000 --> 00:16:35.000
Gulp.
Okay, so, the input is this
00:16:33.000 --> 00:16:39.000
function, 10 e to the minus x,
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it's a decaying oscillation.
You're seeing those already on
00:16:40.000 --> 00:16:46.000
the computer screen if you
started your homework,
00:16:44.000 --> 00:16:50.000
if you've done problem one on
your homework.
00:16:47.000 --> 00:16:53.000
It's a decaying exponential,
and I want to find a particular
00:16:52.000 --> 00:16:58.000
solution.
Well, let's find a particular
00:16:55.000 --> 00:17:01.000
and the general solution.
Find the general solution.
00:17:00.000 --> 00:17:06.000
Well, the main part of the work
is finding the particular
00:17:05.000 --> 00:17:11.000
solution, but let's quickly,
the general solution,
00:17:10.000 --> 00:17:16.000
let's find first the
complementary part of it,
00:17:14.000 --> 00:17:20.000
in other words,
the solution to the homogeneous
00:17:19.000 --> 00:17:25.000
equation.
That's D squared minus D plus
00:17:22.000 --> 00:17:28.000
two. No, let's not.
00:17:26.000 --> 00:17:32.000
I don't want to solve messy
quadratics.
00:17:31.000 --> 00:17:37.000
Okay, we're going to find a
particular solution.
00:17:34.000 --> 00:17:40.000
I thought it was going to come
out easy, and then I realized it
00:17:39.000 --> 00:17:45.000
wasn't because I picked the
wrong signs.
00:17:43.000 --> 00:17:49.000
Okay, so if you don't like,
just change the problem.
00:17:47.000 --> 00:17:53.000
I can do that,
but you cannot.
00:17:49.000 --> 00:17:55.000
Don't forget that.
So, we want a particular
00:17:53.000 --> 00:17:59.000
solution in our equation.
It is this equals that.
00:17:57.000 --> 00:18:03.000
Now, let's complexify it to
make this part of a complex
00:18:01.000 --> 00:18:07.000
exponential.
So, the complex exponential
00:18:06.000 --> 00:18:12.000
that's relevant is ten times e
to the (minus one plus i),
00:18:12.000 --> 00:18:18.000
you see that? x.
00:18:15.000 --> 00:18:21.000
What is this?
This is the imaginary part of
00:18:19.000 --> 00:18:25.000
this complex exponential.
So, this is imaginary part of
00:18:24.000 --> 00:18:30.000
that guy, e to the negative x
times e to the i x,
00:18:29.000 --> 00:18:35.000
and the imaginary part of e to
00:18:33.000 --> 00:18:39.000
the i x.
00:18:39.000 --> 00:18:45.000
The ten, of course,
just comes along for the ride.
00:18:42.000 --> 00:18:48.000
Okay, well, now,
since this is a complex
00:18:44.000 --> 00:18:50.000
equation, I shouldn't call this
y anymore by my notation.
00:18:48.000 --> 00:18:54.000
I like to call it y tilde to
indicate that the solution we
00:18:52.000 --> 00:18:58.000
get to this is not going to be
the original solution to the
00:18:56.000 --> 00:19:02.000
original problem,
but you will have to take the
00:18:59.000 --> 00:19:05.000
imaginary part of it to get it.
So, we are looking,
00:19:03.000 --> 00:19:09.000
now, for the complex solution
to this complexified equation.
00:19:08.000 --> 00:19:14.000
Okay, what is it?
Well, the complex particular
00:19:11.000 --> 00:19:17.000
solution I can write down
immediately.
00:19:14.000 --> 00:19:20.000
It is ten, that,
of course, just gets dragged
00:19:18.000 --> 00:19:24.000
along by linearity,
times e to the (minus one plus
00:19:22.000 --> 00:19:28.000
i) times x.
And, it's over this polynomial
00:19:27.000 --> 00:19:33.000
evaluated at this alpha.
So, just write it down with,
00:19:31.000 --> 00:19:37.000
have faith.
So, what do I get?
00:19:33.000 --> 00:19:39.000
The alpha is minus one plus i.
00:19:36.000 --> 00:19:42.000
I.
square that,
00:19:37.000 --> 00:19:43.000
because I'm substituting this
alpha into that polynomial.
00:19:41.000 --> 00:19:47.000
The reason I'm doing that is
because the formula tells me to
00:19:45.000 --> 00:19:51.000
do it.
That's going to be that
00:19:47.000 --> 00:19:53.000
solution.
Okay, so it's minus one plus i,
00:19:50.000 --> 00:19:56.000
the quantity squared, minus
minus one plus i plus two.
00:19:54.000 --> 00:20:00.000
All I've done is substitute
00:19:58.000 --> 00:20:04.000
minus one plus i for
D in that polynomial,
00:20:01.000 --> 00:20:07.000
the quadratic polynomial.
And now, all I want is the
00:20:07.000 --> 00:20:13.000
imaginary part of this.
The imaginary part of this will
00:20:12.000 --> 00:20:18.000
be the solution to the original
problem because this was the
00:20:18.000 --> 00:20:24.000
right hand side with the
imaginary part of the
00:20:22.000 --> 00:20:28.000
complexified right hand side.
Okay, now, let's make it look a
00:20:28.000 --> 00:20:34.000
little better,
yp tilde.
00:20:32.000 --> 00:20:38.000
Clearly, what we have to do
something nice to the
00:20:34.000 --> 00:20:40.000
denominator.
So, I'll copy the numerator.
00:20:37.000 --> 00:20:43.000
That's e to the minus (one plus
i) x
00:20:40.000 --> 00:20:46.000
and how about the denominator?
Well, again,
00:20:43.000 --> 00:20:49.000
don't expand things out because
it's already this long.
00:20:46.000 --> 00:20:52.000
And, what's the point of making
it this long?
00:20:48.000 --> 00:20:54.000
You want to make it as long,
right?
00:20:50.000 --> 00:20:56.000
Okay, then there is room here
for one real number,
00:20:53.000 --> 00:20:59.000
and another real number times
i, there's no more room.
00:20:57.000 --> 00:21:03.000
Okay, what's the real number?
Okay, we're looking for the
00:21:02.000 --> 00:21:08.000
real part of this expression.
So, just put it in and keep it
00:21:07.000 --> 00:21:13.000
mentally.
So, minus one squared:
00:21:09.000 --> 00:21:15.000
that's one, plus i squared,
that's minus one.
00:21:13.000 --> 00:21:19.000
One minus one is zero.
I can forget about that term.
00:21:18.000 --> 00:21:24.000
The term gives me plus one for
the real part,
00:21:21.000 --> 00:21:27.000
plus two.
The answer is that the real
00:21:24.000 --> 00:21:30.000
part is three.
How about the imaginary part?
00:21:28.000 --> 00:21:34.000
Well, from here,
there's negative 2i,
00:21:31.000 --> 00:21:37.000
negative 2i.
I'm expanding that out by the
00:21:38.000 --> 00:21:44.000
binomial theorem,
or whatever you like to call
00:21:44.000 --> 00:21:50.000
that, minus 2i minus i makes
minus 3i.
00:21:51.000 --> 00:21:57.000
Is that right?
Minus 2i, minus i,
00:21:56.000 --> 00:22:02.000
minus 3i.
So, it is ten thirds,
00:22:00.000 --> 00:22:06.000
and now in the denominator I
have one minus i.
00:22:08.000 --> 00:22:14.000
I'll put that in the numerator,
make it one plus i,
00:22:12.000 --> 00:22:18.000
but I have to divide by the
product of one minus i and its
00:22:17.000 --> 00:22:23.000
complex conjugate.
In other words,
00:22:20.000 --> 00:22:26.000
I'm multiplying both top and
bottom by one plus i.
00:22:24.000 --> 00:22:30.000
And so, that makes here one
squared plus one squared is two.
00:22:30.000 --> 00:22:36.000
And now, what's left is e to
the negative x times cosine x
00:22:35.000 --> 00:22:41.000
plus i sine x.
00:22:40.000 --> 00:22:46.000
Now, of that,
what we want is just the
00:22:43.000 --> 00:22:49.000
imaginary part.
Well, let's see.
00:22:46.000 --> 00:22:52.000
Two goes into ten makes five,
so that's five thirds.
00:22:52.000 --> 00:22:58.000
So, we're practically at our
solution.
00:22:55.000 --> 00:23:01.000
The solution,
then, finally,
00:22:58.000 --> 00:23:04.000
is going to be yp is the
imaginary part of yp tilde.
00:23:05.000 --> 00:23:11.000
And, what's that?
Well, what's the coefficient
00:23:08.000 --> 00:23:14.000
out front, first of all?
It's five thirds,
00:23:12.000 --> 00:23:18.000
so let's pull out to five
thirds before we forget it.
00:23:16.000 --> 00:23:22.000
And, we'll pull out the e to
the negative x before we forget
00:23:21.000 --> 00:23:27.000
that.
And then, the rest is simply a
00:23:24.000 --> 00:23:30.000
question of seeing what's left.
Well, it's one.
00:23:28.000 --> 00:23:34.000
I want the imaginary part.
So, the imaginary part is going
00:23:33.000 --> 00:23:39.000
to be one times cosine x,
and then the other
00:23:37.000 --> 00:23:43.000
imaginary part comes from these
two pieces, which is one times
00:23:42.000 --> 00:23:48.000
sine x.
And, that should be the
00:23:47.000 --> 00:23:53.000
particular solution.
Notice that most of the work is
00:23:52.000 --> 00:23:58.000
not getting this thing.
It's turning it into something
00:23:56.000 --> 00:24:02.000
human that you can take the real
and imaginary parts of.
00:24:02.000 --> 00:24:08.000
If we don't like this form,
you can put it in the other
00:24:07.000 --> 00:24:13.000
form, which many engineers would
do almost automatically,
00:24:12.000 --> 00:24:18.000
make it five thirds,
e to the negative x,
00:24:16.000 --> 00:24:22.000
and what will that be?
Well, you can use the general
00:24:21.000 --> 00:24:27.000
formula if you want.
Remember, cosine,
00:24:24.000 --> 00:24:30.000
the two coefficients are one
and one, so it's one and one.
00:24:30.000 --> 00:24:36.000
So, this is the square root of
two.
00:24:33.000 --> 00:24:39.000
So, it is times,
this part makes the square root
00:24:37.000 --> 00:24:43.000
of two times cosine of x minus
pi over the angle.
00:24:44.000 --> 00:24:50.000
This is a phi.
So, that's pi over four,
00:24:47.000 --> 00:24:53.000
minus pi over four.
00:24:58.000 --> 00:25:04.000
Okay, all right,
now let's address the case
00:25:02.000 --> 00:25:08.000
which is going to occupy a lot
of the rest of today,
00:25:07.000 --> 00:25:13.000
and in a certain sense,
all of next time.
00:25:11.000 --> 00:25:17.000
What happens when p of alpha is
zero?
00:25:16.000 --> 00:25:22.000
Well, in order to be able to
handle this decently,
00:25:21.000 --> 00:25:27.000
it's necessary to have one more
formula, which is very slightly
00:25:28.000 --> 00:25:34.000
more complicated than the
substitution rule.
00:25:34.000 --> 00:25:40.000
But, it's the same kind of
rule.
00:25:36.000 --> 00:25:42.000
I'm going to call this,
or it is called the
00:25:38.000 --> 00:25:44.000
exponential.
So, I'm going to first prove a
00:25:41.000 --> 00:25:47.000
formula, which is the analog of
that, and then I will prove a
00:25:45.000 --> 00:25:51.000
green formula,
which is what to do here if p
00:25:48.000 --> 00:25:54.000
of alpha turns out to be zero.
00:25:51.000 --> 00:25:57.000
But, in order to be able to
prove that, we're going to be
00:25:54.000 --> 00:26:00.000
the analog of the orange
formula.
00:25:56.000 --> 00:26:02.000
And, the analogue of the orange
formula, that tells you how to
00:26:00.000 --> 00:26:06.000
apply p of D to a
simple exponential.
00:26:05.000 --> 00:26:11.000
I need a formula which applies
p of D to that simple
00:26:11.000 --> 00:26:17.000
exponential times another
function.
00:26:14.000 --> 00:26:20.000
Now, I found I got into trouble
by continuing to call that
00:26:20.000 --> 00:26:26.000
alpha.
So, I'm now going to change the
00:26:24.000 --> 00:26:30.000
name of alpha to change alpha's
name to a.
00:26:30.000 --> 00:26:36.000
But, it's still complex.
I don't mean it's guaranteed to
00:26:34.000 --> 00:26:40.000
be complex.
I mean it's allowed to be
00:26:37.000 --> 00:26:43.000
complex.
So, a is now allowed to be a
00:26:40.000 --> 00:26:46.000
complex number.
I'm thinking of it,
00:26:43.000 --> 00:26:49.000
in general, as a complex
number, okay?
00:26:46.000 --> 00:26:52.000
I hope this doesn't upset you
too much, but you know,
00:26:51.000 --> 00:26:57.000
you change x to t's,
and y's to x's.
00:26:54.000 --> 00:27:00.000
This is no worse.
All right, what we going to do?
00:26:58.000 --> 00:27:04.000
Well, I'm going to use this
exponential shift rule,
00:27:02.000 --> 00:27:08.000
I'll call it,
exponential shift rule or
00:27:06.000 --> 00:27:12.000
formula or law.
That's the substitution rule
00:27:11.000 --> 00:27:17.000
for me.
So, this is going to be
00:27:13.000 --> 00:27:19.000
exponential shift law.
And, to apply,
00:27:17.000 --> 00:27:23.000
it tells you how to apply the
polynomial to not D,
00:27:21.000 --> 00:27:27.000
not just the exponential,
but the exponential times some
00:27:25.000 --> 00:27:31.000
function of x.
What's that?
00:27:28.000 --> 00:27:34.000
And now, the rule is very
simple.
00:27:32.000 --> 00:27:38.000
See, you can understand the
difficulty.
00:27:34.000 --> 00:27:40.000
If you try to start
differentiating,
00:27:37.000 --> 00:27:43.000
you're going to have to
calculate second derivatives of
00:27:40.000 --> 00:27:46.000
the stuff, and God forbid,
higher order equations.
00:27:44.000 --> 00:27:50.000
You would have to calculate
fourth derivatives,
00:27:47.000 --> 00:27:53.000
fifth derivatives.
You barely even want to
00:27:50.000 --> 00:27:56.000
calculate the first derivative.
That's okay.
00:27:53.000 --> 00:27:59.000
But, second derivative,
do I have to?
00:27:55.000 --> 00:28:01.000
No, not if you know the
exponential shift rule,
00:27:58.000 --> 00:28:04.000
which says you can get rid of
the e to the ax,
00:28:01.000 --> 00:28:07.000
make it pass to the left of the
operator where it's not in any
00:28:06.000 --> 00:28:12.000
position to do any harm any
longer, or upset the
00:28:09.000 --> 00:28:15.000
differentiation.
And, all you have to do is,
00:28:14.000 --> 00:28:20.000
when it passes over that
operator, it changes D to D plus
00:28:18.000 --> 00:28:24.000
a. So, the answer is,
00:28:21.000 --> 00:28:27.000
e to the ax.
There, it's passed over.
00:28:24.000 --> 00:28:30.000
But, when it did so,
it changed D to D plus a.
00:28:28.000 --> 00:28:34.000
And, what about the u?
Well, the u just stayed there.
00:28:32.000 --> 00:28:38.000
Nothing happened to it.
Okay, there's our orange
00:28:36.000 --> 00:28:42.000
formula.
I guess we better put the thing
00:28:40.000 --> 00:28:46.000
around the whole business.
Should I prove that,
00:28:43.000 --> 00:28:49.000
or the proof is quite easy.
So, let's do it just again to
00:28:48.000 --> 00:28:54.000
give you a chance to try to see,
now, if somebody gives you a
00:28:53.000 --> 00:28:59.000
formula like that,
you first stare at it.
00:28:56.000 --> 00:29:02.000
You might try a couple of
special cases,
00:28:59.000 --> 00:29:05.000
try it on a function and see if
it works, but already,
00:29:03.000 --> 00:29:09.000
you probably don't want to do
that.
00:29:08.000 --> 00:29:14.000
I mean, even if you took a
function like x here,
00:29:10.000 --> 00:29:16.000
you'd have to do a certain
amount of differentiating,
00:29:14.000 --> 00:29:20.000
and some quadratic thing here.
You'd calculate and calculate
00:29:17.000 --> 00:29:23.000
away for a little while,
and then if you did it
00:29:20.000 --> 00:29:26.000
correctly, the two would in fact
turn out to be equal.
00:29:23.000 --> 00:29:29.000
But, you would not necessarily
feel any the wiser.
00:29:26.000 --> 00:29:32.000
A better procedure in trying to
understand something like this
00:29:30.000 --> 00:29:36.000
is say, well,
let's keep the u general.
00:29:34.000 --> 00:29:40.000
Suppose we make D simple.
For example,
00:29:36.000 --> 00:29:42.000
well, if D is a constant,
of course there's nothing to
00:29:39.000 --> 00:29:45.000
happen because if this is just a
constant, both sides of these
00:29:43.000 --> 00:29:49.000
are the same.
This doesn't make any sense if
00:29:46.000 --> 00:29:52.000
p doesn't really have a D in it.
Well, what's the simplest
00:29:49.000 --> 00:29:55.000
polynomial which would have a D
in it?
00:29:52.000 --> 00:29:58.000
Well, D itself.
So, let's take a special case.
00:29:54.000 --> 00:30:00.000
p of D equals D,
and check the formula in that
00:29:58.000 --> 00:30:04.000
case; see if it works.
So, the formula is asking us,
00:30:03.000 --> 00:30:09.000
what is D, that's the p of D,
of e to the ax times u?
00:30:08.000 --> 00:30:14.000
I'm not going to put in the
00:30:12.000 --> 00:30:18.000
variable here because it's just
a waste of chalk.
00:30:16.000 --> 00:30:22.000
Well, what is that?
I know how to calculate that.
00:30:21.000 --> 00:30:27.000
I'll use the product rule.
So, it's the derivative.
00:30:25.000 --> 00:30:31.000
I'll tell you what;
let's do the other order first.
00:30:30.000 --> 00:30:36.000
So, it's e to the ax times the
derivative of u plus the
00:30:35.000 --> 00:30:41.000
derivative of e to the ax,
which is a times e to the ax
00:30:40.000 --> 00:30:46.000
times u.
00:30:45.000 --> 00:30:51.000
Do you follow that?
This is the product rule.
00:30:48.000 --> 00:30:54.000
It's e to the ax times the
derivative of u plus the
00:30:51.000 --> 00:30:57.000
derivative of e to the ax,
which is this thing times u,
00:30:55.000 --> 00:31:01.000
the other factor.
00:30:58.000 --> 00:31:04.000
Now, is that right?
I want to make it look like
00:31:01.000 --> 00:31:07.000
that.
Well, to make it look like that
00:31:04.000 --> 00:31:10.000
I should first factor e to the
ax out.
00:31:07.000 --> 00:31:13.000
And now, what's left?
Well, if I factor e to the ax
00:31:11.000 --> 00:31:17.000
out, what's left is Du plus au,
which is exactly (D plus a)
00:31:15.000 --> 00:31:21.000
operating on u,
D u plus a u.
00:31:19.000 --> 00:31:25.000
Well, hey, that's just what the
formula said it should be.
00:31:23.000 --> 00:31:29.000
If you make e to the x pass
over D, it changes D to
00:31:27.000 --> 00:31:33.000
D plus a.
Okay, now here's the main thing
00:31:32.000 --> 00:31:38.000
I want to show you.
All right, now,
00:31:34.000 --> 00:31:40.000
well let's try,
if this is true,
00:31:37.000 --> 00:31:43.000
also works out for D squared,
then the formula is
00:31:41.000 --> 00:31:47.000
clearly true by linearity
because an arbitrary p of D
00:31:45.000 --> 00:31:51.000
is just a linear
combination with constant
00:31:49.000 --> 00:31:55.000
coefficients of D,
D squared, and that constant
00:31:53.000 --> 00:31:59.000
thing, which we agreed there was
nothing to prove about.
00:31:57.000 --> 00:32:03.000
Now, hack, you're a hack if you
take D squared and start
00:32:01.000 --> 00:32:07.000
calculating the second
derivative of this.
00:32:06.000 --> 00:32:12.000
Okay, it's question about
hacks.
00:32:08.000 --> 00:32:14.000
I mean, it's just,
you haven't learned the right
00:32:12.000 --> 00:32:18.000
thing to do.
Okay, that will work,
00:32:14.000 --> 00:32:20.000
but it's not what you want to
do.
00:32:17.000 --> 00:32:23.000
Instead, you bootstrap your way
up.
00:32:20.000 --> 00:32:26.000
I have already a formula
telling me how to handle this.
00:32:24.000 --> 00:32:30.000
And, you can be anything.
Look at this not as D squared
00:32:28.000 --> 00:32:34.000
all by itself.
Calculate, instead,
00:32:32.000 --> 00:32:38.000
D squared e to the ax times u.
00:32:36.000 --> 00:32:42.000
Think of that as D,
the derivative of the
00:32:39.000 --> 00:32:45.000
derivative of e to the a x u.
00:32:42.000 --> 00:32:48.000
In other words,
we will do it one step at a
00:32:45.000 --> 00:32:51.000
time.
But you see now immediately the
00:32:48.000 --> 00:32:54.000
advantage of this.
What's D of e to the a x u?
00:32:51.000 --> 00:32:57.000
Well, I just calculated that.
00:32:55.000 --> 00:33:01.000
Now, don't go back to the
beginning.
00:32:57.000 --> 00:33:03.000
Don't go back to here.
Use the formula.
00:33:02.000 --> 00:33:08.000
After all, you worked to
calculate it,
00:33:04.000 --> 00:33:10.000
or I did.
So, it's D of,
00:33:06.000 --> 00:33:12.000
and what's this inside?
It's e to the ax times (D plus
00:33:10.000 --> 00:33:16.000
a) times u.
Well, that looks like a mess,
00:33:15.000 --> 00:33:21.000
but it isn't because I'm taking
D of e to the ax times
00:33:19.000 --> 00:33:25.000
something.
And I already know how to take
00:33:23.000 --> 00:33:29.000
D of e to the ax times
something.
00:33:25.000 --> 00:33:31.000
It doesn't matter what that
something is.
00:33:30.000 --> 00:33:36.000
Here, the something was u.
Here, the something is D plus
00:33:35.000 --> 00:33:41.000
(a times u) operating on u.
But, the principle is the same,
00:33:40.000 --> 00:33:46.000
and the answer is what?
Well, to take D of e to the ax
00:33:46.000 --> 00:33:52.000
times something,
you pass the e to the ax over
00:33:50.000 --> 00:33:56.000
the D.
That changes D to D plus a.
00:33:53.000 --> 00:33:59.000
And, you apply that to the
00:33:57.000 --> 00:34:03.000
other guy, which is (D plus a)
applied to u.
00:34:03.000 --> 00:34:09.000
What's the answer?
e to the a x times (D plus a)
00:34:06.000 --> 00:34:12.000
squared u.
00:34:09.000 --> 00:34:15.000
It's just what you would have
gotten if you had taken e to the
00:34:14.000 --> 00:34:20.000
e to the x, pass it over,
and then changed D to D plus a.
00:34:18.000 --> 00:34:24.000
Now, another advantage to doing
00:34:22.000 --> 00:34:28.000
it this way is you can see that
this argument is going to
00:34:26.000 --> 00:34:32.000
generalize to D cubed.
In other words,
00:34:31.000 --> 00:34:37.000
you would continue on in the
same way by the process of
00:34:36.000 --> 00:34:42.000
mathematical,
one word, mathematical,
00:34:39.000 --> 00:34:45.000
begins with an I,
induction.
00:34:41.000 --> 00:34:47.000
By induction,
you would prove the same
00:34:44.000 --> 00:34:50.000
formula for the nth derivative.
If you don't know what
00:34:49.000 --> 00:34:55.000
mathematical induction is,
shame on you.
00:34:53.000 --> 00:34:59.000
But it's okay.
A lot of you will be able to go
00:34:57.000 --> 00:35:03.000
through life without ever having
to learn what it is.
00:35:03.000 --> 00:35:09.000
And, the rest of you will be
computer scientists.
00:35:08.000 --> 00:35:14.000
Okay, so that's the idea of
this rule.
00:35:13.000 --> 00:35:19.000
Now, we can use it to calculate
something.
00:35:18.000 --> 00:35:24.000
Let's see, I'm going to need
green for this,
00:35:23.000 --> 00:35:29.000
I guess, for our better
formula.
00:35:27.000 --> 00:35:33.000
The formula,
now, that tells you what to do
00:35:32.000 --> 00:35:38.000
if p of alpha is zero.
00:35:39.000 --> 00:35:45.000
So, we're trying to solve the
equation, D squared plus A D,
00:35:45.000 --> 00:35:51.000
we are trying to find a
00:35:49.000 --> 00:35:55.000
particular solution,
e to the ax,
00:35:54.000 --> 00:36:00.000
let's say.
Remember, a is complex.
00:35:58.000 --> 00:36:04.000
a could be complex.
It doesn't have to be real.
00:36:05.000 --> 00:36:11.000
But, the problem is that p of
alpha is zero.
00:36:09.000 --> 00:36:15.000
How do I get a particular
solution?
00:36:12.000 --> 00:36:18.000
Well, I will write it down for
you.
00:36:14.000 --> 00:36:20.000
So, this is part of that
exponential input theorem.
00:36:18.000 --> 00:36:24.000
I think that's the way it is in
the notes.
00:36:21.000 --> 00:36:27.000
I gave all the cases together,
but I thought pedagogically
00:36:25.000 --> 00:36:31.000
it's probably a little better to
do the simplest case first,
00:36:30.000 --> 00:36:36.000
and then build up on the
complexity.
00:36:34.000 --> 00:36:40.000
So, what's yp?
The answer is yp is e to the a,
00:36:38.000 --> 00:36:44.000
except now you have to multiply
00:36:43.000 --> 00:36:49.000
it by x out front.
Where have you done something
00:36:48.000 --> 00:36:54.000
like that before?
Yes, don't tell me.
00:36:51.000 --> 00:36:57.000
I know you know.
But, what should go in the
00:36:56.000 --> 00:37:02.000
denominator?
Clearly not p of alpha.
00:36:59.000 --> 00:37:05.000
What goes in the denominator is
00:37:04.000 --> 00:37:10.000
the derivative.
Okay, but what if p prime of
00:37:09.000 --> 00:37:15.000
alpha is zero?
Couldn't that happen?
00:37:14.000 --> 00:37:20.000
Yes, it could happen.
So, we better make cases.
00:37:18.000 --> 00:37:24.000
This case is,
the case where this is okay
00:37:22.000 --> 00:37:28.000
corresponds to the case where
we're going to assume that alpha
00:37:27.000 --> 00:37:33.000
is a simple root,
is a simple root of the
00:37:31.000 --> 00:37:37.000
polynomial, p.
I don't know what to call the
00:37:35.000 --> 00:37:41.000
variable, p of D is okay.
A simple zero,
00:37:40.000 --> 00:37:46.000
in other words,
it's not double.
00:37:43.000 --> 00:37:49.000
Well, suppose is double.
One of the consequences you
00:37:48.000 --> 00:37:54.000
will see just in a second,
if it's a simple zero,
00:37:52.000 --> 00:37:58.000
that means this derivative is
not going to be zero.
00:37:57.000 --> 00:38:03.000
That's automatic.
Yeah, well, suppose it's not as
00:38:01.000 --> 00:38:07.000
simple.
Well, suppose is a double root.
00:38:05.000 --> 00:38:11.000
How did a-- How did that get
changed to, argh!
00:38:11.000 --> 00:38:17.000
[LAUGHTER] That's not an alpha.
Oh, well, yes it is,
00:38:16.000 --> 00:38:22.000
obviously.
Change!
00:38:18.000 --> 00:38:24.000
All of you, I want you to
change.
00:38:21.000 --> 00:38:27.000
They should have something like
in a search key where
00:38:27.000 --> 00:38:33.000
occurrences of alpha have been
changed to a with a stroke of a,
00:38:34.000 --> 00:38:40.000
just your thumb.
They don't have that for the
00:38:39.000 --> 00:38:45.000
blackboard, unfortunately.
Well, too bad,
00:38:42.000 --> 00:38:48.000
for the future.
Correctable blackboards.
00:38:46.000 --> 00:38:52.000
Well, what if a is double root?
It can't be more than a double
00:38:51.000 --> 00:38:57.000
root because you've only got a
quadratic polynomial.
00:38:55.000 --> 00:39:01.000
Quadratic polynomials only have
two roots.
00:39:00.000 --> 00:39:06.000
So, the worst that can happen
is that both of them are a.
00:39:04.000 --> 00:39:10.000
All right, in that case the
formula should be yp is equal
00:39:09.000 --> 00:39:15.000
to, you are now going to need x
squared up there times e
00:39:14.000 --> 00:39:20.000
to the ax,
and in the denominator what you
00:39:18.000 --> 00:39:24.000
are going to need is the second
derivative of,
00:39:22.000 --> 00:39:28.000
evaluated at a.
Now, you can guess the way this
00:39:25.000 --> 00:39:31.000
going to go on.
For higher degree things,
00:39:29.000 --> 00:39:35.000
if you've got a triple root,
you will need here x cubed,
00:39:37.000 --> 00:39:43.000
except you're going to need a
factorial there,
00:39:41.000 --> 00:39:47.000
too.
So, don't worry about it.
00:39:45.000 --> 00:39:51.000
It's in the notes,
but I'm not going to give you
00:39:49.000 --> 00:39:55.000
that for higher roots.
I don't even know if I will
00:39:53.000 --> 00:39:59.000
give it to you for double root.
Yes, I already did,
00:39:57.000 --> 00:40:03.000
so it's too late.
It's too late.
00:40:01.000 --> 00:40:07.000
Okay, so we will make this two
formulas according to whether a
00:40:06.000 --> 00:40:12.000
is a single or a double root.
Okay, let's prove one of these,
00:40:11.000 --> 00:40:17.000
and all of that will be good
enough for my conscience.
00:40:16.000 --> 00:40:22.000
Let's prove the first one.
Mostly, it's an exercise in
00:40:21.000 --> 00:40:27.000
using the first exponential
shift rule.
00:40:24.000 --> 00:40:30.000
Okay, this will be a first
example actually seeing a work
00:40:29.000 --> 00:40:35.000
in practice as opposed to
proving it.
00:40:34.000 --> 00:40:40.000
Okay, so what does that thing
look like?
00:40:37.000 --> 00:40:43.000
So, what does the polynomial
look like, which has a as a
00:40:42.000 --> 00:40:48.000
simple root?
So, we're going to try to prove
00:40:46.000 --> 00:40:52.000
the simple root case.
So, I'm just going to calculate
00:40:50.000 --> 00:40:56.000
what those guys actually look
like.
00:40:53.000 --> 00:40:59.000
What does p of D look
like if a is a simple root?
00:41:00.000 --> 00:41:06.000
Well, if it's a simple root,
that means it has a factor.
00:41:04.000 --> 00:41:10.000
When it factors,
it factors into the product of
00:41:08.000 --> 00:41:14.000
(D minus a) times something
which isn't, D minus some other
00:41:13.000 --> 00:41:19.000
root.
And, the point is that b is not
00:41:16.000 --> 00:41:22.000
equal to A.
The roots are really distinct.
00:41:20.000 --> 00:41:26.000
Okay, what's,
then, p prime,
00:41:23.000 --> 00:41:29.000
I'm going to have to calculate
p prime of a.
00:41:27.000 --> 00:41:33.000
What is that?
Well, let's calculate p prime
00:41:32.000 --> 00:41:38.000
of D first.
It is, well,
00:41:34.000 --> 00:41:40.000
by the ordinary product rule,
it's the derivative of this
00:41:38.000 --> 00:41:44.000
times, which is one times (D
minus a) plus,
00:41:41.000 --> 00:41:47.000
that's one thing plus the same
thing on the other side,
00:41:45.000 --> 00:41:51.000
the derivative of this,
which is one times (D minus b).
00:41:49.000 --> 00:41:55.000
So, that's p prime.
And therefore,
00:41:51.000 --> 00:41:57.000
what's p prime of a?
It's nothing but,
00:41:55.000 --> 00:42:01.000
this part is zero,
and that's a minus b.
00:41:59.000 --> 00:42:05.000
Of course, this is not zero
because it's a simple root.
00:42:02.000 --> 00:42:08.000
And, that's the proof for you
if you want, that if the root is
00:42:06.000 --> 00:42:12.000
simple, that p prime of a
is guaranteed not to
00:42:10.000 --> 00:42:16.000
be zero.
And, you can see,
00:42:12.000 --> 00:42:18.000
it's going to be zero exactly
when b equals a,
00:42:15.000 --> 00:42:21.000
and that root occurs twice.
But, I'm assuming that didn't
00:42:19.000 --> 00:42:25.000
happen.
Okay, then all the rest we have
00:42:22.000 --> 00:42:28.000
to do is calculate,
do the calculation.
00:42:24.000 --> 00:42:30.000
So, what I want to prove now is
that with this p of D,
00:42:28.000 --> 00:42:34.000
what I'm trying to calculate
that p of D times that guy,
00:42:32.000 --> 00:42:38.000
x e to the a x,
except I'm going to
00:42:36.000 --> 00:42:42.000
write it as e to the a x times
x, guess why,
00:42:39.000 --> 00:42:45.000
divided by p prime of a.
00:42:44.000 --> 00:42:50.000
This is my proposed particular
solution.
00:42:47.000 --> 00:42:53.000
So, what I have to do is
calculate it,
00:42:50.000 --> 00:42:56.000
and see that it turns out to
be, what do I hope it turns out
00:42:54.000 --> 00:43:00.000
to be?
What the right hand side of the
00:42:57.000 --> 00:43:03.000
equation, the input?
The input is e to the ax.
00:43:01.000 --> 00:43:07.000
If this is true,
00:43:03.000 --> 00:43:09.000
then yp, a particular solution,
indeed, nothing will be a
00:43:07.000 --> 00:43:13.000
particular solution.
Of course, there could be
00:43:11.000 --> 00:43:17.000
others, but in this game,
I only have to find one
00:43:14.000 --> 00:43:20.000
particular solution,
and that certainly by far is
00:43:18.000 --> 00:43:24.000
the simple as one you could
possibly find.
00:43:21.000 --> 00:43:27.000
So, I have to calculate this.
And now, you see why I did the
00:43:25.000 --> 00:43:31.000
exponential shift rule because
this is begging to be
00:43:29.000 --> 00:43:35.000
differentiated by something
simpler than hack.
00:43:34.000 --> 00:43:40.000
Okay, you can also see why I
violated the natural order of
00:43:38.000 --> 00:43:44.000
things and put the e to the ax
on the left in order
00:43:43.000 --> 00:43:49.000
that it pass over more easily.
So, the answer on the left-hand
00:43:48.000 --> 00:43:54.000
side is e to the ax times p of
(D plus a).
00:43:51.000 --> 00:43:57.000
Now, what is (p of D) plus a?
Write it in this form.
00:43:56.000 --> 00:44:02.000
It's going to be a minus b.
00:44:00.000 --> 00:44:06.000
So, p of (D plus a)
is, change D to D plus a.
00:44:04.000 --> 00:44:10.000
So, the first factor is going
00:44:07.000 --> 00:44:13.000
to be D plus a minus b.
00:44:10.000 --> 00:44:16.000
And, what's the second factor?
Change D to D plus a.
00:44:15.000 --> 00:44:21.000
It turns into D.
00:44:16.000 --> 00:44:22.000
All this is the result of
taking that p of D,
00:44:21.000 --> 00:44:27.000
and changing D to D plus a.
And now, this is to be applied
00:44:25.000 --> 00:44:31.000
to what?
Well, e to the a x
00:44:28.000 --> 00:44:34.000
is already passed over.
So, what's left is x.
00:44:33.000 --> 00:44:39.000
And, that's to be divided by
the constant,
00:44:36.000 --> 00:44:42.000
p prime of a.
Now, what does this all come
00:44:41.000 --> 00:44:47.000
out to be?
e to the ax,
00:44:44.000 --> 00:44:50.000
what's D applied to x?
One, right?
00:44:48.000 --> 00:44:54.000
And now, what's this thing
applied to the constant one?
00:44:53.000 --> 00:44:59.000
Well, the D kills it,
so it has no effect.
00:44:57.000 --> 00:45:03.000
It makes it zero.
The rest just multiplies it by
00:45:02.000 --> 00:45:08.000
a minus b.
So, the answer to the top is (a
00:45:07.000 --> 00:45:13.000
minus b) times one.
And, the answer to the bottom
00:45:12.000 --> 00:45:18.000
is p prime of a,
which I showed you by just
00:45:17.000 --> 00:45:23.000
explicit calculation is a minus
b.
00:45:21.000 --> 00:45:27.000
And so the answer is,
e to the a x comes
00:45:25.000 --> 00:45:31.000
out right.
Now, the other one,
00:45:29.000 --> 00:45:35.000
the other formula comes out the
same way.
00:45:31.000 --> 00:45:37.000
I'll leave that as an exercise.
Also, I don't dare do it
00:45:34.000 --> 00:45:40.000
because it's much too close to
the problem I asked you to do
00:45:38.000 --> 00:45:44.000
for homework.
So, let's by way of conclusion,
00:45:41.000 --> 00:45:47.000
I'll do one more simple
example, okay?
00:45:43.000 --> 00:45:49.000
And then, you can feel you
understand something.
00:45:46.000 --> 00:45:52.000
I'm sort of bothered that I
haven't done any examples of
00:45:49.000 --> 00:45:55.000
this more complicated case.
So, I'll pick an easy version
00:45:53.000 --> 00:45:59.000
instead of the one that you have
in your notes,
00:45:56.000 --> 00:46:02.000
which is the one you have for
homework, which is even easier.
00:46:01.000 --> 00:46:07.000
So, this one's epsilon less
easy.
00:46:04.000 --> 00:46:10.000
Y double prime minus 3 y prime
plus 2 y equals e to the x.
00:46:10.000 --> 00:46:16.000
Okay, notice that one is a
00:46:14.000 --> 00:46:20.000
simple root.
The one I'm talking about is
00:46:18.000 --> 00:46:24.000
the a here, which is one.
One is a simple root of the
00:46:23.000 --> 00:46:29.000
polynomial D squared minus 3D
plus two,
00:46:28.000 --> 00:46:34.000
isn't it?
It's a zero.
00:46:32.000 --> 00:46:38.000
Put D equal one and you get
one minus three plus
00:46:35.000 --> 00:46:41.000
two equals zero.
It's a simple root because
00:46:38.000 --> 00:46:44.000
anybody can see that one is not
a double root because you know
00:46:41.000 --> 00:46:47.000
from critical damping,
if one were a double root,
00:46:44.000 --> 00:46:50.000
you know just what the
polynomial would look like,
00:46:47.000 --> 00:46:53.000
and it wouldn't look like that
at all.
00:46:49.000 --> 00:46:55.000
It would not look like D
squared minus 3D plus two.
00:46:52.000 --> 00:46:58.000
It would look differently.
00:46:54.000 --> 00:47:00.000
Therefore, that proves that one
is a simple root.
00:46:57.000 --> 00:47:03.000
Okay, what's the particular
solution, therefore?
00:47:01.000 --> 00:47:07.000
The particular solution is x
times e to the x divided by the
00:47:06.000 --> 00:47:12.000
derivative, the derivative
evaluated at the point,
00:47:11.000 --> 00:47:17.000
so, what's p prime of D?
It is 2D minus three.
00:47:15.000 --> 00:47:21.000
If I evaluate it at the point,
one, it is negative one.
00:47:20.000 --> 00:47:26.000
So, if this is to be divided by
negative one,
00:47:25.000 --> 00:47:31.000
in other words,
it's minus x e to the X.
00:47:30.000 --> 00:47:36.000
And, if you don't believe it,
you could plug it in and check
00:47:35.000 --> 00:47:41.000
it out.
Okay, I'm letting you out one
00:47:39.000 --> 00:47:45.000
minute early.
Remember that.
00:47:41.000 --> 00:47:47.000
I'm trying to pay off the
accumulated debt.