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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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00:10:00 --> 00:10:06
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
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down the solution for you,
write down the particular
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solution.
But let's verify it in general.
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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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00:13:37 --> 00:13:43
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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00:13:46 --> 00:13:52
the alpha x on the right?
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00:13:49 --> 00:13:55
Well, all you have to do is do
it.
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What is p of D applied to e to
the alpha x divided by p of
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00:13:56 --> 00:14:02
alpha?
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00:14:00 --> 00:14:06
Well, p of D applied to e to
the alpha x is p of alpha times
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00:14:03 --> 00:14:09
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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00:14:13 --> 00:14:19
because this operator is linear.
If this applied to that is
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00:14:17 --> 00:14:23
this, then if I apply it to one
half that, I get one half the
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00:14:21 --> 00:14:27
answer, and so on.
So, the p of alpha
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00:14:24 --> 00:14:30
is a constant and just gets
dragged along.
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00:14:27 --> 00:14:33
And now, they cancel each
other, and the answer is,
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00:14:30 --> 00:14:36
indeed, e to the alpha x.
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00:14:34 --> 00:14:40
That's not much of a proof.
I hope that to at least half
213
00:14:39 --> 00:14:45
this class, you're wondering,
yes, but what if Peter had not
214
00:14:45 --> 00:14:51
caught the wolf?
I mean, what if?
215
00:14:48 --> 00:14:54
What if?
216
00:14:50 --> 00:14:56
217
00:15:01 --> 00:15:07
I'm looking stern.
Okay, we will take care of it
218
00:15:04 --> 00:15:10
in the simplest possible way.
We will assume that p of alpha
219
00:15:10 --> 00:15:16
is not zero.
The case p of alpha is zero
220
00:15:13 --> 00:15:19
is, in fact,
an extremely important
221
00:15:17 --> 00:15:23
case, one that makes the world
go 'round, one that contributes
222
00:15:22 --> 00:15:28
to all sorts of catastrophes,
and they occur first here in
223
00:15:27 --> 00:15:33
the solution of differential
equations, and that's what
224
00:15:32 --> 00:15:38
controls all the catastrophes.
But, there's a good side to it,
225
00:15:37 --> 00:15:43
too.
It also makes a lot of good
226
00:15:39 --> 00:15:45
things happen.
So, there are no moral
227
00:15:41 --> 00:15:47
judgments in mathematics.
For the time being,
228
00:15:44 --> 00:15:50
let's assume p of alpha is not
zero.
229
00:15:46 --> 00:15:52
And, that proof is okay because
the p of alpha,
230
00:15:48 --> 00:15:54
being in the denominator,
it's okay to be in the
231
00:15:51 --> 00:15:57
denominator if you're not zero.
Okay, let's work in a simple
232
00:15:55 --> 00:16:01
example.
Well, I'm picking the most
233
00:15:56 --> 00:16:02
complicated example I can think
of.
234
00:16:00 --> 00:16:06
Simple examples,
I'll leave for your practice
235
00:16:04 --> 00:16:10
and for the recitations,
can start off with simple
236
00:16:08 --> 00:16:14
examples if you are confused by
this.
237
00:16:12 --> 00:16:18
But, let's solve an equation,
find a particular solution.
238
00:16:17 --> 00:16:23
So, y double prime plus minus y
prime plus 2y is equal to 10 e
239
00:16:23 --> 00:16:29
to the minus x sine x.
240
00:16:29 --> 00:16:35
Gulp.
Okay, so, the input is this
241
00:16:33 --> 00:16:39
function, 10 e to the minus x,
242
00:16:36 --> 00:16:42
it's a decaying oscillation.
You're seeing those already on
243
00:16:40 --> 00:16:46
the computer screen if you
started your homework,
244
00:16:44 --> 00:16:50
if you've done problem one on
your homework.
245
00:16:47 --> 00:16:53
It's a decaying exponential,
and I want to find a particular
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00:16:52 --> 00:16:58
solution.
Well, let's find a particular
247
00:16:55 --> 00:17:01
and the general solution.
Find the general solution.
248
00:17:00 --> 00:17:06
Well, the main part of the work
is finding the particular
249
00:17:05 --> 00:17:11
solution, but let's quickly,
the general solution,
250
00:17:10 --> 00:17:16
let's find first the
complementary part of it,
251
00:17:14 --> 00:17:20
in other words,
the solution to the homogeneous
252
00:17:19 --> 00:17:25
equation.
That's D squared minus D plus
253
00:17:22 --> 00:17:28
two. No, let's not.
254
00:17:26 --> 00:17:32
I don't want to solve messy
quadratics.
255
00:17:31 --> 00:17:37
Okay, we're going to find a
particular solution.
256
00:17:34 --> 00:17:40
I thought it was going to come
out easy, and then I realized it
257
00:17:39 --> 00:17:45
wasn't because I picked the
wrong signs.
258
00:17:43 --> 00:17:49
Okay, so if you don't like,
just change the problem.
259
00:17:47 --> 00:17:53
I can do that,
but you cannot.
260
00:17:49 --> 00:17:55
Don't forget that.
So, we want a particular
261
00:17:53 --> 00:17:59
solution in our equation.
It is this equals that.
262
00:17:57 --> 00:18:03
Now, let's complexify it to
make this part of a complex
263
00:18:01 --> 00:18:07
exponential.
So, the complex exponential
264
00:18:06 --> 00:18:12
that's relevant is ten times e
to the (minus one plus i),
265
00:18:12 --> 00:18:18
you see that? x.
266
00:18:15 --> 00:18:21
What is this?
This is the imaginary part of
267
00:18:19 --> 00:18:25
this complex exponential.
So, this is imaginary part of
268
00:18:24 --> 00:18:30
that guy, e to the negative x
times e to the i x,
269
00:18:29 --> 00:18:35
and the imaginary part of e to
270
00:18:33 --> 00:18:39
the i x.
271
00:18:39 --> 00:18:45
The ten, of course,
just comes along for the ride.
272
00:18:42 --> 00:18:48
Okay, well, now,
since this is a complex
273
00:18:44 --> 00:18:50
equation, I shouldn't call this
y anymore by my notation.
274
00:18:48 --> 00:18:54
I like to call it y tilde to
indicate that the solution we
275
00:18:52 --> 00:18:58
get to this is not going to be
the original solution to the
276
00:18:56 --> 00:19:02
original problem,
but you will have to take the
277
00:18:59 --> 00:19:05
imaginary part of it to get it.
So, we are looking,
278
00:19:03 --> 00:19:09
now, for the complex solution
to this complexified equation.
279
00:19:08 --> 00:19:14
Okay, what is it?
Well, the complex particular
280
00:19:11 --> 00:19:17
solution I can write down
immediately.
281
00:19:14 --> 00:19:20
It is ten, that,
of course, just gets dragged
282
00:19:18 --> 00:19:24
along by linearity,
times e to the (minus one plus
283
00:19:22 --> 00:19:28
i) times x.
And, it's over this polynomial
284
00:19:27 --> 00:19:33
evaluated at this alpha.
So, just write it down with,
285
00:19:31 --> 00:19:37
have faith.
So, what do I get?
286
00:19:33 --> 00:19:39
The alpha is minus one plus i.
287
00:19:36 --> 00:19:42
I.
square that,
288
00:19:37 --> 00:19:43
because I'm substituting this
alpha into that polynomial.
289
00:19:41 --> 00:19:47
The reason I'm doing that is
because the formula tells me to
290
00:19:45 --> 00:19:51
do it.
That's going to be that
291
00:19:47 --> 00:19:53
solution.
Okay, so it's minus one plus i,
292
00:19:50 --> 00:19:56
the quantity squared, minus
minus one plus i plus two.
293
00:19:54 --> 00:20:00
All I've done is substitute
294
00:19:58 --> 00:20:04
minus one plus i for
D in that polynomial,
295
00:20:01 --> 00:20:07
the quadratic polynomial.
And now, all I want is the
296
00:20:07 --> 00:20:13
imaginary part of this.
The imaginary part of this will
297
00:20:12 --> 00:20:18
be the solution to the original
problem because this was the
298
00:20:18 --> 00:20:24
right hand side with the
imaginary part of the
299
00:20:22 --> 00:20:28
complexified right hand side.
Okay, now, let's make it look a
300
00:20:28 --> 00:20:34
little better,
yp tilde.
301
00:20:32 --> 00:20:38
Clearly, what we have to do
something nice to the
302
00:20:34 --> 00:20:40
denominator.
So, I'll copy the numerator.
303
00:20:37 --> 00:20:43
That's e to the minus (one plus
i) x
304
00:20:40 --> 00:20:46
and how about the denominator?
Well, again,
305
00:20:43 --> 00:20:49
don't expand things out because
it's already this long.
306
00:20:46 --> 00:20:52
And, what's the point of making
it this long?
307
00:20:48 --> 00:20:54
You want to make it as long,
right?
308
00:20:50 --> 00:20:56
Okay, then there is room here
for one real number,
309
00:20:53 --> 00:20:59
and another real number times
i, there's no more room.
310
00:20:57 --> 00:21:03
Okay, what's the real number?
Okay, we're looking for the
311
00:21:02 --> 00:21:08
real part of this expression.
So, just put it in and keep it
312
00:21:07 --> 00:21:13
mentally.
So, minus one squared:
313
00:21:09 --> 00:21:15
that's one, plus i squared,
that's minus one.
314
00:21:13 --> 00:21:19
One minus one is zero.
I can forget about that term.
315
00:21:18 --> 00:21:24
The term gives me plus one for
the real part,
316
00:21:21 --> 00:21:27
plus two.
The answer is that the real
317
00:21:24 --> 00:21:30
part is three.
How about the imaginary part?
318
00:21:28 --> 00:21:34
Well, from here,
there's negative 2i,
319
00:21:31 --> 00:21:37
negative 2i.
I'm expanding that out by the
320
00:21:38 --> 00:21:44
binomial theorem,
or whatever you like to call
321
00:21:44 --> 00:21:50
that, minus 2i minus i makes
minus 3i.
322
00:21:51 --> 00:21:57
Is that right?
Minus 2i, minus i,
323
00:21:56 --> 00:22:02
minus 3i.
So, it is ten thirds,
324
00:22:00 --> 00:22:06
and now in the denominator I
have one minus i.
325
00:22:08 --> 00:22:14
I'll put that in the numerator,
make it one plus i,
326
00:22:12 --> 00:22:18
but I have to divide by the
product of one minus i and its
327
00:22:17 --> 00:22:23
complex conjugate.
In other words,
328
00:22:20 --> 00:22:26
I'm multiplying both top and
bottom by one plus i.
329
00:22:24 --> 00:22:30
And so, that makes here one
squared plus one squared is two.
330
00:22:30 --> 00:22:36
And now, what's left is e to
the negative x times cosine x
331
00:22:35 --> 00:22:41
plus i sine x.
332
00:22:40 --> 00:22:46
Now, of that,
what we want is just the
333
00:22:43 --> 00:22:49
imaginary part.
Well, let's see.
334
00:22:46 --> 00:22:52
Two goes into ten makes five,
so that's five thirds.
335
00:22:52 --> 00:22:58
So, we're practically at our
solution.
336
00:22:55 --> 00:23:01
The solution,
then, finally,
337
00:22:58 --> 00:23:04
is going to be yp is the
imaginary part of yp tilde.
338
00:23:05 --> 00:23:11
And, what's that?
Well, what's the coefficient
339
00:23:08 --> 00:23:14
out front, first of all?
It's five thirds,
340
00:23:12 --> 00:23:18
so let's pull out to five
thirds before we forget it.
341
00:23:16 --> 00:23:22
And, we'll pull out the e to
the negative x before we forget
342
00:23:21 --> 00:23:27
that.
And then, the rest is simply a
343
00:23:24 --> 00:23:30
question of seeing what's left.
Well, it's one.
344
00:23:28 --> 00:23:34
I want the imaginary part.
So, the imaginary part is going
345
00:23:33 --> 00:23:39
to be one times cosine x,
and then the other
346
00:23:37 --> 00:23:43
imaginary part comes from these
two pieces, which is one times
347
00:23:42 --> 00:23:48
sine x.
And, that should be the
348
00:23:47 --> 00:23:53
particular solution.
Notice that most of the work is
349
00:23:52 --> 00:23:58
not getting this thing.
It's turning it into something
350
00:23:56 --> 00:24:02
human that you can take the real
and imaginary parts of.
351
00:24:02 --> 00:24:08
If we don't like this form,
you can put it in the other
352
00:24:07 --> 00:24:13
form, which many engineers would
do almost automatically,
353
00:24:12 --> 00:24:18
make it five thirds,
e to the negative x,
354
00:24:16 --> 00:24:22
and what will that be?
Well, you can use the general
355
00:24:21 --> 00:24:27
formula if you want.
Remember, cosine,
356
00:24:24 --> 00:24:30
the two coefficients are one
and one, so it's one and one.
357
00:24:30 --> 00:24:36
So, this is the square root of
two.
358
00:24:33 --> 00:24:39
So, it is times,
this part makes the square root
359
00:24:37 --> 00:24:43
of two times cosine of x minus
pi over the angle.
360
00:24:44 --> 00:24:50
This is a phi.
So, that's pi over four,
361
00:24:47 --> 00:24:53
minus pi over four.
362
00:24:49 --> 00:24:55
363
00:24:58 --> 00:25:04
Okay, all right,
now let's address the case
364
00:25:02 --> 00:25:08
which is going to occupy a lot
of the rest of today,
365
00:25:07 --> 00:25:13
and in a certain sense,
all of next time.
366
00:25:11 --> 00:25:17
What happens when p of alpha is
zero?
367
00:25:16 --> 00:25:22
Well, in order to be able to
handle this decently,
368
00:25:21 --> 00:25:27
it's necessary to have one more
formula, which is very slightly
369
00:25:28 --> 00:25:34
more complicated than the
substitution rule.
370
00:25:34 --> 00:25:40
But, it's the same kind of
rule.
371
00:25:36 --> 00:25:42
I'm going to call this,
or it is called the
372
00:25:38 --> 00:25:44
exponential.
So, I'm going to first prove a
373
00:25:41 --> 00:25:47
formula, which is the analog of
that, and then I will prove a
374
00:25:45 --> 00:25:51
green formula,
which is what to do here if p
375
00:25:48 --> 00:25:54
of alpha turns out to be zero.
376
00:25:51 --> 00:25:57
But, in order to be able to
prove that, we're going to be
377
00:25:54 --> 00:26:00
the analog of the orange
formula.
378
00:25:56 --> 00:26:02
And, the analogue of the orange
formula, that tells you how to
379
00:26:00 --> 00:26:06
apply p of D to a
simple exponential.
380
00:26:05 --> 00:26:11
I need a formula which applies
p of D to that simple
381
00:26:11 --> 00:26:17
exponential times another
function.
382
00:26:14 --> 00:26:20
Now, I found I got into trouble
by continuing to call that
383
00:26:20 --> 00:26:26
alpha.
So, I'm now going to change the
384
00:26:24 --> 00:26:30
name of alpha to change alpha's
name to a.
385
00:26:30 --> 00:26:36
But, it's still complex.
I don't mean it's guaranteed to
386
00:26:34 --> 00:26:40
be complex.
I mean it's allowed to be
387
00:26:37 --> 00:26:43
complex.
So, a is now allowed to be a
388
00:26:40 --> 00:26:46
complex number.
I'm thinking of it,
389
00:26:43 --> 00:26:49
in general, as a complex
number, okay?
390
00:26:46 --> 00:26:52
I hope this doesn't upset you
too much, but you know,
391
00:26:51 --> 00:26:57
you change x to t's,
and y's to x's.
392
00:26:54 --> 00:27:00
This is no worse.
All right, what we going to do?
393
00:26:58 --> 00:27:04
Well, I'm going to use this
exponential shift rule,
394
00:27:02 --> 00:27:08
I'll call it,
exponential shift rule or
395
00:27:06 --> 00:27:12
formula or law.
That's the substitution rule
396
00:27:11 --> 00:27:17
for me.
So, this is going to be
397
00:27:13 --> 00:27:19
exponential shift law.
And, to apply,
398
00:27:17 --> 00:27:23
it tells you how to apply the
polynomial to not D,
399
00:27:21 --> 00:27:27
not just the exponential,
but the exponential times some
400
00:27:25 --> 00:27:31
function of x.
What's that?
401
00:27:28 --> 00:27:34
And now, the rule is very
simple.
402
00:27:32 --> 00:27:38
See, you can understand the
difficulty.
403
00:27:34 --> 00:27:40
If you try to start
differentiating,
404
00:27:37 --> 00:27:43
you're going to have to
calculate second derivatives of
405
00:27:40 --> 00:27:46
the stuff, and God forbid,
higher order equations.
406
00:27:44 --> 00:27:50
You would have to calculate
fourth derivatives,
407
00:27:47 --> 00:27:53
fifth derivatives.
You barely even want to
408
00:27:50 --> 00:27:56
calculate the first derivative.
That's okay.
409
00:27:53 --> 00:27:59
But, second derivative,
do I have to?
410
00:27:55 --> 00:28:01
No, not if you know the
exponential shift rule,
411
00:27:58 --> 00:28:04
which says you can get rid of
the e to the ax,
412
00:28:01 --> 00:28:07
make it pass to the left of the
operator where it's not in any
413
00:28:06 --> 00:28:12
position to do any harm any
longer, or upset the
414
00:28:09 --> 00:28:15
differentiation.
And, all you have to do is,
415
00:28:14 --> 00:28:20
when it passes over that
operator, it changes D to D plus
416
00:28:18 --> 00:28:24
a. So, the answer is,
417
00:28:21 --> 00:28:27
e to the ax.
There, it's passed over.
418
00:28:24 --> 00:28:30
But, when it did so,
it changed D to D plus a.
419
00:28:28 --> 00:28:34
And, what about the u?
Well, the u just stayed there.
420
00:28:32 --> 00:28:38
Nothing happened to it.
Okay, there's our orange
421
00:28:36 --> 00:28:42
formula.
I guess we better put the thing
422
00:28:40 --> 00:28:46
around the whole business.
Should I prove that,
423
00:28:43 --> 00:28:49
or the proof is quite easy.
So, let's do it just again to
424
00:28:48 --> 00:28:54
give you a chance to try to see,
now, if somebody gives you a
425
00:28:53 --> 00:28:59
formula like that,
you first stare at it.
426
00:28:56 --> 00:29:02
You might try a couple of
special cases,
427
00:28:59 --> 00:29:05
try it on a function and see if
it works, but already,
428
00:29:03 --> 00:29:09
you probably don't want to do
that.
429
00:29:08 --> 00:29:14
I mean, even if you took a
function like x here,
430
00:29:10 --> 00:29:16
you'd have to do a certain
amount of differentiating,
431
00:29:14 --> 00:29:20
and some quadratic thing here.
You'd calculate and calculate
432
00:29:17 --> 00:29:23
away for a little while,
and then if you did it
433
00:29:20 --> 00:29:26
correctly, the two would in fact
turn out to be equal.
434
00:29:23 --> 00:29:29
But, you would not necessarily
feel any the wiser.
435
00:29:26 --> 00:29:32
A better procedure in trying to
understand something like this
436
00:29:30 --> 00:29:36
is say, well,
let's keep the u general.
437
00:29:34 --> 00:29:40
Suppose we make D simple.
For example,
438
00:29:36 --> 00:29:42
well, if D is a constant,
of course there's nothing to
439
00:29:39 --> 00:29:45
happen because if this is just a
constant, both sides of these
440
00:29:43 --> 00:29:49
are the same.
This doesn't make any sense if
441
00:29:46 --> 00:29:52
p doesn't really have a D in it.
Well, what's the simplest
442
00:29:49 --> 00:29:55
polynomial which would have a D
in it?
443
00:29:52 --> 00:29:58
Well, D itself.
So, let's take a special case.
444
00:29:54 --> 00:30:00
p of D equals D,
and check the formula in that
445
00:29:58 --> 00:30:04
case; see if it works.
So, the formula is asking us,
446
00:30:03 --> 00:30:09
what is D, that's the p of D,
of e to the ax times u?
447
00:30:08 --> 00:30:14
I'm not going to put in the
448
00:30:12 --> 00:30:18
variable here because it's just
a waste of chalk.
449
00:30:16 --> 00:30:22
Well, what is that?
I know how to calculate that.
450
00:30:21 --> 00:30:27
I'll use the product rule.
So, it's the derivative.
451
00:30:25 --> 00:30:31
I'll tell you what;
let's do the other order first.
452
00:30:30 --> 00:30:36
So, it's e to the ax times the
derivative of u plus the
453
00:30:35 --> 00:30:41
derivative of e to the ax,
which is a times e to the ax
454
00:30:40 --> 00:30:46
times u.
455
00:30:45 --> 00:30:51
Do you follow that?
This is the product rule.
456
00:30:48 --> 00:30:54
It's e to the ax times the
derivative of u plus the
457
00:30:51 --> 00:30:57
derivative of e to the ax,
which is this thing times u,
458
00:30:55 --> 00:31:01
the other factor.
459
00:30:58 --> 00:31:04
Now, is that right?
I want to make it look like
460
00:31:01 --> 00:31:07
that.
Well, to make it look like that
461
00:31:04 --> 00:31:10
I should first factor e to the
ax out.
462
00:31:07 --> 00:31:13
And now, what's left?
Well, if I factor e to the ax
463
00:31:11 --> 00:31:17
out, what's left is Du plus au,
which is exactly (D plus a)
464
00:31:15 --> 00:31:21
operating on u,
D u plus a u.
465
00:31:19 --> 00:31:25
Well, hey, that's just what the
formula said it should be.
466
00:31:23 --> 00:31:29
If you make e to the x pass
over D, it changes D to
467
00:31:27 --> 00:31:33
D plus a.
Okay, now here's the main thing
468
00:31:32 --> 00:31:38
I want to show you.
All right, now,
469
00:31:34 --> 00:31:40
well let's try,
if this is true,
470
00:31:37 --> 00:31:43
also works out for D squared,
then the formula is
471
00:31:41 --> 00:31:47
clearly true by linearity
because an arbitrary p of D
472
00:31:45 --> 00:31:51
is just a linear
combination with constant
473
00:31:49 --> 00:31:55
coefficients of D,
D squared, and that constant
474
00:31:53 --> 00:31:59
thing, which we agreed there was
nothing to prove about.
475
00:31:57 --> 00:32:03
Now, hack, you're a hack if you
take D squared and start
476
00:32:01 --> 00:32:07
calculating the second
derivative of this.
477
00:32:06 --> 00:32:12
Okay, it's question about
hacks.
478
00:32:08 --> 00:32:14
I mean, it's just,
you haven't learned the right
479
00:32:12 --> 00:32:18
thing to do.
Okay, that will work,
480
00:32:14 --> 00:32:20
but it's not what you want to
do.
481
00:32:17 --> 00:32:23
Instead, you bootstrap your way
up.
482
00:32:20 --> 00:32:26
I have already a formula
telling me how to handle this.
483
00:32:24 --> 00:32:30
And, you can be anything.
Look at this not as D squared
484
00:32:28 --> 00:32:34
all by itself.
Calculate, instead,
485
00:32:32 --> 00:32:38
D squared e to the ax times u.
486
00:32:36 --> 00:32:42
Think of that as D,
the derivative of the
487
00:32:39 --> 00:32:45
derivative of e to the a x u.
488
00:32:42 --> 00:32:48
In other words,
we will do it one step at a
489
00:32:45 --> 00:32:51
time.
But you see now immediately the
490
00:32:48 --> 00:32:54
advantage of this.
What's D of e to the a x u?
491
00:32:51 --> 00:32:57
Well, I just calculated that.
492
00:32:55 --> 00:33:01
Now, don't go back to the
beginning.
493
00:32:57 --> 00:33:03
Don't go back to here.
Use the formula.
494
00:33:02 --> 00:33:08
After all, you worked to
calculate it,
495
00:33:04 --> 00:33:10
or I did.
So, it's D of,
496
00:33:06 --> 00:33:12
and what's this inside?
It's e to the ax times (D plus
497
00:33:10 --> 00:33:16
a) times u.
Well, that looks like a mess,
498
00:33:15 --> 00:33:21
but it isn't because I'm taking
D of e to the ax times
499
00:33:19 --> 00:33:25
something.
And I already know how to take
500
00:33:23 --> 00:33:29
D of e to the ax times
something.
501
00:33:25 --> 00:33:31
It doesn't matter what that
something is.
502
00:33:30 --> 00:33:36
Here, the something was u.
Here, the something is D plus
503
00:33:35 --> 00:33:41
(a times u) operating on u.
But, the principle is the same,
504
00:33:40 --> 00:33:46
and the answer is what?
Well, to take D of e to the ax
505
00:33:46 --> 00:33:52
times something,
you pass the e to the ax over
506
00:33:50 --> 00:33:56
the D.
That changes D to D plus a.
507
00:33:53 --> 00:33:59
And, you apply that to the
508
00:33:57 --> 00:34:03
other guy, which is (D plus a)
applied to u.
509
00:34:03 --> 00:34:09
What's the answer?
e to the a x times (D plus a)
510
00:34:06 --> 00:34:12
squared u.
511
00:34:09 --> 00:34:15
It's just what you would have
gotten if you had taken e to the
512
00:34:14 --> 00:34:20
e to the x, pass it over,
and then changed D to D plus a.
513
00:34:18 --> 00:34:24
Now, another advantage to doing
514
00:34:22 --> 00:34:28
it this way is you can see that
this argument is going to
515
00:34:26 --> 00:34:32
generalize to D cubed.
In other words,
516
00:34:31 --> 00:34:37
you would continue on in the
same way by the process of
517
00:34:36 --> 00:34:42
mathematical,
one word, mathematical,
518
00:34:39 --> 00:34:45
begins with an I,
induction.
519
00:34:41 --> 00:34:47
By induction,
you would prove the same
520
00:34:44 --> 00:34:50
formula for the nth derivative.
If you don't know what
521
00:34:49 --> 00:34:55
mathematical induction is,
shame on you.
522
00:34:53 --> 00:34:59
But it's okay.
A lot of you will be able to go
523
00:34:57 --> 00:35:03
through life without ever having
to learn what it is.
524
00:35:03 --> 00:35:09
And, the rest of you will be
computer scientists.
525
00:35:08 --> 00:35:14
Okay, so that's the idea of
this rule.
526
00:35:13 --> 00:35:19
Now, we can use it to calculate
something.
527
00:35:18 --> 00:35:24
Let's see, I'm going to need
green for this,
528
00:35:23 --> 00:35:29
I guess, for our better
formula.
529
00:35:27 --> 00:35:33
The formula,
now, that tells you what to do
530
00:35:32 --> 00:35:38
if p of alpha is zero.
531
00:35:39 --> 00:35:45
So, we're trying to solve the
equation, D squared plus A D,
532
00:35:45 --> 00:35:51
we are trying to find a
533
00:35:49 --> 00:35:55
particular solution,
e to the ax,
534
00:35:54 --> 00:36:00
let's say.
Remember, a is complex.
535
00:35:58 --> 00:36:04
a could be complex.
It doesn't have to be real.
536
00:36:05 --> 00:36:11
But, the problem is that p of
alpha is zero.
537
00:36:09 --> 00:36:15
How do I get a particular
solution?
538
00:36:12 --> 00:36:18
Well, I will write it down for
you.
539
00:36:14 --> 00:36:20
So, this is part of that
exponential input theorem.
540
00:36:18 --> 00:36:24
I think that's the way it is in
the notes.
541
00:36:21 --> 00:36:27
I gave all the cases together,
but I thought pedagogically
542
00:36:25 --> 00:36:31
it's probably a little better to
do the simplest case first,
543
00:36:30 --> 00:36:36
and then build up on the
complexity.
544
00:36:34 --> 00:36:40
So, what's yp?
The answer is yp is e to the a,
545
00:36:38 --> 00:36:44
except now you have to multiply
546
00:36:43 --> 00:36:49
it by x out front.
Where have you done something
547
00:36:48 --> 00:36:54
like that before?
Yes, don't tell me.
548
00:36:51 --> 00:36:57
I know you know.
But, what should go in the
549
00:36:56 --> 00:37:02
denominator?
Clearly not p of alpha.
550
00:36:59 --> 00:37:05
What goes in the denominator is
551
00:37:04 --> 00:37:10
the derivative.
Okay, but what if p prime of
552
00:37:09 --> 00:37:15
alpha is zero?
Couldn't that happen?
553
00:37:14 --> 00:37:20
Yes, it could happen.
So, we better make cases.
554
00:37:18 --> 00:37:24
This case is,
the case where this is okay
555
00:37:22 --> 00:37:28
corresponds to the case where
we're going to assume that alpha
556
00:37:27 --> 00:37:33
is a simple root,
is a simple root of the
557
00:37:31 --> 00:37:37
polynomial, p.
I don't know what to call the
558
00:37:35 --> 00:37:41
variable, p of D is okay.
A simple zero,
559
00:37:40 --> 00:37:46
in other words,
it's not double.
560
00:37:43 --> 00:37:49
Well, suppose is double.
One of the consequences you
561
00:37:48 --> 00:37:54
will see just in a second,
if it's a simple zero,
562
00:37:52 --> 00:37:58
that means this derivative is
not going to be zero.
563
00:37:57 --> 00:38:03
That's automatic.
Yeah, well, suppose it's not as
564
00:38:01 --> 00:38:07
simple.
Well, suppose is a double root.
565
00:38:05 --> 00:38:11
How did a-- How did that get
changed to, argh!
566
00:38:11 --> 00:38:17
[LAUGHTER] That's not an alpha.
Oh, well, yes it is,
567
00:38:16 --> 00:38:22
obviously.
Change!
568
00:38:18 --> 00:38:24
All of you, I want you to
change.
569
00:38:21 --> 00:38:27
They should have something like
in a search key where
570
00:38:27 --> 00:38:33
occurrences of alpha have been
changed to a with a stroke of a,
571
00:38:34 --> 00:38:40
just your thumb.
They don't have that for the
572
00:38:39 --> 00:38:45
blackboard, unfortunately.
Well, too bad,
573
00:38:42 --> 00:38:48
for the future.
Correctable blackboards.
574
00:38:46 --> 00:38:52
Well, what if a is double root?
It can't be more than a double
575
00:38:51 --> 00:38:57
root because you've only got a
quadratic polynomial.
576
00:38:55 --> 00:39:01
Quadratic polynomials only have
two roots.
577
00:39:00 --> 00:39:06
So, the worst that can happen
is that both of them are a.
578
00:39:04 --> 00:39:10
All right, in that case the
formula should be yp is equal
579
00:39:09 --> 00:39:15
to, you are now going to need x
squared up there times e
580
00:39:14 --> 00:39:20
to the ax,
and in the denominator what you
581
00:39:18 --> 00:39:24
are going to need is the second
derivative of,
582
00:39:22 --> 00:39:28
evaluated at a.
Now, you can guess the way this
583
00:39:25 --> 00:39:31
going to go on.
For higher degree things,
584
00:39:29 --> 00:39:35
if you've got a triple root,
you will need here x cubed,
585
00:39:33 --> 00:39:39
586
00:39:37 --> 00:39:43
except you're going to need a
factorial there,
587
00:39:41 --> 00:39:47
too.
So, don't worry about it.
588
00:39:45 --> 00:39:51
It's in the notes,
but I'm not going to give you
589
00:39:49 --> 00:39:55
that for higher roots.
I don't even know if I will
590
00:39:53 --> 00:39:59
give it to you for double root.
Yes, I already did,
591
00:39:57 --> 00:40:03
so it's too late.
It's too late.
592
00:40:01 --> 00:40:07
Okay, so we will make this two
formulas according to whether a
593
00:40:06 --> 00:40:12
is a single or a double root.
Okay, let's prove one of these,
594
00:40:11 --> 00:40:17
and all of that will be good
enough for my conscience.
595
00:40:16 --> 00:40:22
Let's prove the first one.
Mostly, it's an exercise in
596
00:40:21 --> 00:40:27
using the first exponential
shift rule.
597
00:40:24 --> 00:40:30
Okay, this will be a first
example actually seeing a work
598
00:40:29 --> 00:40:35
in practice as opposed to
proving it.
599
00:40:34 --> 00:40:40
Okay, so what does that thing
look like?
600
00:40:37 --> 00:40:43
So, what does the polynomial
look like, which has a as a
601
00:40:42 --> 00:40:48
simple root?
So, we're going to try to prove
602
00:40:46 --> 00:40:52
the simple root case.
So, I'm just going to calculate
603
00:40:50 --> 00:40:56
what those guys actually look
like.
604
00:40:53 --> 00:40:59
What does p of D look
like if a is a simple root?
605
00:41:00 --> 00:41:06
Well, if it's a simple root,
that means it has a factor.
606
00:41:04 --> 00:41:10
When it factors,
it factors into the product of
607
00:41:08 --> 00:41:14
(D minus a) times something
which isn't, D minus some other
608
00:41:13 --> 00:41:19
root.
And, the point is that b is not
609
00:41:16 --> 00:41:22
equal to A.
The roots are really distinct.
610
00:41:20 --> 00:41:26
Okay, what's,
then, p prime,
611
00:41:23 --> 00:41:29
I'm going to have to calculate
p prime of a.
612
00:41:27 --> 00:41:33
What is that?
Well, let's calculate p prime
613
00:41:32 --> 00:41:38
of D first.
It is, well,
614
00:41:34 --> 00:41:40
by the ordinary product rule,
it's the derivative of this
615
00:41:38 --> 00:41:44
times, which is one times (D
minus a) plus,
616
00:41:41 --> 00:41:47
that's one thing plus the same
thing on the other side,
617
00:41:45 --> 00:41:51
the derivative of this,
which is one times (D minus b).
618
00:41:49 --> 00:41:55
So, that's p prime.
And therefore,
619
00:41:51 --> 00:41:57
what's p prime of a?
It's nothing but,
620
00:41:55 --> 00:42:01
this part is zero,
and that's a minus b.
621
00:41:59 --> 00:42:05
Of course, this is not zero
because it's a simple root.
622
00:42:02 --> 00:42:08
And, that's the proof for you
if you want, that if the root is
623
00:42:06 --> 00:42:12
simple, that p prime of a
is guaranteed not to
624
00:42:10 --> 00:42:16
be zero.
And, you can see,
625
00:42:12 --> 00:42:18
it's going to be zero exactly
when b equals a,
626
00:42:15 --> 00:42:21
and that root occurs twice.
But, I'm assuming that didn't
627
00:42:19 --> 00:42:25
happen.
Okay, then all the rest we have
628
00:42:22 --> 00:42:28
to do is calculate,
do the calculation.
629
00:42:24 --> 00:42:30
So, what I want to prove now is
that with this p of D,
630
00:42:28 --> 00:42:34
what I'm trying to calculate
that p of D times that guy,
631
00:42:32 --> 00:42:38
x e to the a x,
except I'm going to
632
00:42:36 --> 00:42:42
write it as e to the a x times
x, guess why,
633
00:42:39 --> 00:42:45
divided by p prime of a.
634
00:42:44 --> 00:42:50
This is my proposed particular
solution.
635
00:42:47 --> 00:42:53
So, what I have to do is
calculate it,
636
00:42:50 --> 00:42:56
and see that it turns out to
be, what do I hope it turns out
637
00:42:54 --> 00:43:00
to be?
What the right hand side of the
638
00:42:57 --> 00:43:03
equation, the input?
The input is e to the ax.
639
00:43:01 --> 00:43:07
If this is true,
640
00:43:03 --> 00:43:09
then yp, a particular solution,
indeed, nothing will be a
641
00:43:07 --> 00:43:13
particular solution.
Of course, there could be
642
00:43:11 --> 00:43:17
others, but in this game,
I only have to find one
643
00:43:14 --> 00:43:20
particular solution,
and that certainly by far is
644
00:43:18 --> 00:43:24
the simple as one you could
possibly find.
645
00:43:21 --> 00:43:27
So, I have to calculate this.
And now, you see why I did the
646
00:43:25 --> 00:43:31
exponential shift rule because
this is begging to be
647
00:43:29 --> 00:43:35
differentiated by something
simpler than hack.
648
00:43:34 --> 00:43:40
Okay, you can also see why I
violated the natural order of
649
00:43:38 --> 00:43:44
things and put the e to the ax
on the left in order
650
00:43:43 --> 00:43:49
that it pass over more easily.
So, the answer on the left-hand
651
00:43:48 --> 00:43:54
side is e to the ax times p of
(D plus a).
652
00:43:51 --> 00:43:57
Now, what is (p of D) plus a?
Write it in this form.
653
00:43:56 --> 00:44:02
It's going to be a minus b.
654
00:44:00 --> 00:44:06
So, p of (D plus a)
is, change D to D plus a.
655
00:44:04 --> 00:44:10
So, the first factor is going
656
00:44:07 --> 00:44:13
to be D plus a minus b.
657
00:44:10 --> 00:44:16
And, what's the second factor?
Change D to D plus a.
658
00:44:15 --> 00:44:21
It turns into D.
659
00:44:16 --> 00:44:22
All this is the result of
taking that p of D,
660
00:44:21 --> 00:44:27
and changing D to D plus a.
And now, this is to be applied
661
00:44:25 --> 00:44:31
to what?
Well, e to the a x
662
00:44:28 --> 00:44:34
is already passed over.
So, what's left is x.
663
00:44:33 --> 00:44:39
And, that's to be divided by
the constant,
664
00:44:36 --> 00:44:42
p prime of a.
Now, what does this all come
665
00:44:41 --> 00:44:47
out to be?
e to the ax,
666
00:44:44 --> 00:44:50
what's D applied to x?
One, right?
667
00:44:48 --> 00:44:54
And now, what's this thing
applied to the constant one?
668
00:44:53 --> 00:44:59
Well, the D kills it,
so it has no effect.
669
00:44:57 --> 00:45:03
It makes it zero.
The rest just multiplies it by
670
00:45:02 --> 00:45:08
a minus b.
So, the answer to the top is (a
671
00:45:07 --> 00:45:13
minus b) times one.
And, the answer to the bottom
672
00:45:12 --> 00:45:18
is p prime of a,
which I showed you by just
673
00:45:17 --> 00:45:23
explicit calculation is a minus
b.
674
00:45:21 --> 00:45:27
And so the answer is,
e to the a x comes
675
00:45:25 --> 00:45:31
out right.
Now, the other one,
676
00:45:29 --> 00:45:35
the other formula comes out the
same way.
677
00:45:31 --> 00:45:37
I'll leave that as an exercise.
Also, I don't dare do it
678
00:45:34 --> 00:45:40
because it's much too close to
the problem I asked you to do
679
00:45:38 --> 00:45:44
for homework.
So, let's by way of conclusion,
680
00:45:41 --> 00:45:47
I'll do one more simple
example, okay?
681
00:45:43 --> 00:45:49
And then, you can feel you
understand something.
682
00:45:46 --> 00:45:52
I'm sort of bothered that I
haven't done any examples of
683
00:45:49 --> 00:45:55
this more complicated case.
So, I'll pick an easy version
684
00:45:53 --> 00:45:59
instead of the one that you have
in your notes,
685
00:45:56 --> 00:46:02
which is the one you have for
homework, which is even easier.
686
00:46:01 --> 00:46:07
So, this one's epsilon less
easy.
687
00:46:04 --> 00:46:10
Y double prime minus 3 y prime
plus 2 y equals e to the x.
688
00:46:10 --> 00:46:16
Okay, notice that one is a
689
00:46:14 --> 00:46:20
simple root.
The one I'm talking about is
690
00:46:18 --> 00:46:24
the a here, which is one.
One is a simple root of the
691
00:46:23 --> 00:46:29
polynomial D squared minus 3D
plus two,
692
00:46:28 --> 00:46:34
isn't it?
It's a zero.
693
00:46:32 --> 00:46:38
Put D equal one and you get
one minus three plus
694
00:46:35 --> 00:46:41
two equals zero.
It's a simple root because
695
00:46:38 --> 00:46:44
anybody can see that one is not
a double root because you know
696
00:46:41 --> 00:46:47
from critical damping,
if one were a double root,
697
00:46:44 --> 00:46:50
you know just what the
polynomial would look like,
698
00:46:47 --> 00:46:53
and it wouldn't look like that
at all.
699
00:46:49 --> 00:46:55
It would not look like D
squared minus 3D plus two.
700
00:46:52 --> 00:46:58
It would look differently.
701
00:46:54 --> 00:47:00
Therefore, that proves that one
is a simple root.
702
00:46:57 --> 00:47:03
Okay, what's the particular
solution, therefore?
703
00:47:01 --> 00:47:07
The particular solution is x
times e to the x divided by the
704
00:47:06 --> 00:47:12
derivative, the derivative
evaluated at the point,
705
00:47:11 --> 00:47:17
so, what's p prime of D?
It is 2D minus three.
706
00:47:15 --> 00:47:21
If I evaluate it at the point,
one, it is negative one.
707
00:47:20 --> 00:47:26
So, if this is to be divided by
negative one,
708
00:47:25 --> 00:47:31
in other words,
it's minus x e to the X.
709
00:47:30 --> 00:47:36
And, if you don't believe it,
you could plug it in and check
710
00:47:35 --> 00:47:41
it out.
Okay, I'm letting you out one
711
00:47:39 --> 00:47:45
minute early.
Remember that.
712
00:47:41 --> 00:47:47
I'm trying to pay off the
accumulated debt.