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The topic for today is how to
change variables.
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So, we're talking about
substitutions and differential
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equations, or changing
variables.
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That might seem like a sort of
fussy thing to talk about in the
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third or fourth lecture,
but the reason is that so far,
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you know how to solve two kinds
of differential equations,
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two kinds of first-order
differential equations,
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one where you can separate
variables, and the linear
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equation that we talked about
last time.
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Now, the sad fact is that in
some sense, those are the only
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two general methods there are,
that those are the only two
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kinds of equations that can
always be solved.
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Well, what about all the
others?
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The answer is that to a great
extent, all the other equations
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that can be solved,
the solution can be done by
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changing the variables in the
equation to reduce it to one of
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the cases that we can already
do.
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Now, I'm going to give you two
examples of that,
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two significant examples of
that today.
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But, ultimately,
as you will see,
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the way the equations are
solved is by changing them into
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a linear equation,
or an equation where the
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variables are separable.
However, that's for a few
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minutes.
The first change of variables
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that I want to talk about is an
almost trivial one.
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But it's the most common kind
there is, and you've already had
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it in physics class.
But I think it's so important
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in the science and engineering
subjects that it's a good idea,
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even in 18.03,
to call attention to it
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explicitly.
So, in that sense,
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the most common change of
variables is the one simple one
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called scaling.
So, again, the kind of equation
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I'm talking about is a general
first-order equation.
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And, scaling simply means to
change the coordinates,
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in effect, or axes,
to change the coordinates on
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the axes to scale the axes,
to either stretch them or
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contract them.
So, what does the change of
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variable actually look like?
Well, it means you introduce
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new variables,
where x1 is equal to x times
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something or times a constant.
I'll write it as divided by a
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constant, since that tends to be
a little bit more the way people
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think of it.
And y, the same.
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So, the new variable y1 is
related to the old one by an
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equation of that form.
So, a, b are constants.
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So, those are the equations.
Why does one do this?
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Well, for a lot of reasons.
But, maybe we can list them.
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You, for example,
could be changing units.
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That's a common reason in
physics.
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Changing the units that he
used, you would have to make a
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change of coordinates of this
form.
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Perhaps the even more important
reason is to,
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sometimes it's used to make the
variables dimensionless.
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In other words,
so that the variables become
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pure numbers,
with no units attached to them.
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Since you are well aware of the
tortures involved in dealing
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with units in physics,
the point of making variables,
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I'm sorry, dimensionless,
I don't have to sell that.
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Dimensionless,
i.e.
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no units, without any units
attached.
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It just represents the number
three, not three seconds,
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or three grahams,
or anything like that.
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And, the third reason is to
reduce or simplify the
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constants: reduce the number or
simplify the constants in the
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equation.
Reduce their number is self
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explanatory.
Simplify means make them less,
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either dimensionless also,
or if you can't do that,
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at least less dependent upon
the critical units than the old
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ones were.
Let me give you a very simple
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example which will illustrate
most of these things.
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It's the equation.
It's a version of the cooling
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law, which applies at very high
temperatures,
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and it runs.
So, it's like Newton's cooling
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laws, except it's the internal
and external temperatures vary,
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what's important is not the
first power as in Newton's Law,
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but the fourth power.
So, it's a constant.
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And, the difference is,
now, it's the external
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temperature, which,
just so there won't be so many
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capital T's in the equation,
I'm going to call M,
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to the forth power minus T to
the forth power.
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So, T is the internal
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temperature, the thing we are
interested in.
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And, M is the external
constant, which I'll assume,
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now, is a constant external
temperature.
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So, this is valid if big
temperature differences,
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Newton's Law,
breaks down and one needs a
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different one.
Now, you are free to solve that
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equation just as it stands,
if you can.
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There are difficulties
connected with it because you're
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dealing with the fourth powers,
of course.
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But, before you do that,
one should scale.
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How should I scale?
Well, I'm going to scale by
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relating T to M.
So, that is very likely to use
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is T1 equals T divided by M.
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This is now dimensionless
because M, of course,
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has the units of temperature,
degrees Celsius,
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degrees absolute,
whatever it is,
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as does T.
And therefore,
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by taking the ratio of the two,
there are no units attached to
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it.
So, this is dimensionless.
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Now, how actually do I change
the variable in the equation?
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Well, watch this.
It's an utterly trivial idea,
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and utterly important.
Don't slog around doing it this
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way, trying to stuff it in,
and divide first.
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Instead, do the inverse.
In other words,
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write it instead as T equals
MT1, the reason being that it's
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T that's facing you in that
equation, and therefore T you
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want to substitute for.
So, let's do it.
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The new equation will be what?
Well, dT-- Since this is a
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constant, the left-hand side
becomes dT1 / dt times M equals
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k times M to the forth minus M
to the forth T1 to the forth,
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so I'm going to factor
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out that M to the forth,
and make it one minus T1 to the
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forth, okay?
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Now, I could divide through by
M and get rid of one of those,
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and so, the new equation,
now, is dT1 / dt,
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d time, is equal to-- Now,
I have k M cubed out
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front here.
I'm going to just give that a
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new name, k1.
Essentially,
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it's the same equation.
It's no harder to solve and no
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easier to solve than the
original one.
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But it's been simplified.
For one, I think it looks
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better.
So, to compare the two,
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I'll put this one up in green,
and this one in green,
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too, just to convince you it's
the same, but indicate that it's
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the same equation.
Notice, so, T1 has been
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rendered, is now dimensionless.
So, I don't have to even ask
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when I solve this equation,
oh, please tell me what the
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units of temperature are.
How you are measuring
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temperature makes no difference
to this equation.
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k1 still has units.
What units does it have?
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It's been simplified because it
now has the units of,
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since this is dimensionless and
this is dimensionless,
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it has the units of inverse
time.
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So, k1, whereas it had units
involving both degrees and
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seconds before,
now it has inverse time as its
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units.
And, moreover,
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there's one less constant.
So, one less constant in the
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equation.
It just looks better.
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This business,
I think you know that k1,
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the process of forming k1 out
of k M cubed is
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called lumping constants.
I think they use standard
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terminology in physics and
engineering courses.
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Try to get all the constants
together like this.
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And then you lump them there.
They are lumped for you,
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and then you just give the lump
a new name.
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So, that's an example of
scaling.
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Watch out for when you can use
that.
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For example,
it would have probably been a
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good thing to use in the first
problem set when you were
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handling this problem of drug
elimination and hormone
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elimination production inside of
the thing.
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You could lump constants,
and as was done to some extent
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on the solutions to get a neater
looking answer,
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one without so many constants
in it.
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Okay, let's now go to serious
stuff, where we are actually
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going to make changes of
variables which we hope will
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render unsolvable equations
suddenly solvable.
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Now, I'm going to do that by
making substitutions.
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But, it's, I think,
quite important to watch up
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there are two kinds of
substitutions.
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There are direct substitutions.
That's where you introduce a
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new variable.
I don't know how to write this
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on the board.
I'll just write it
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schematically.
So, it's one which says that
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the new variable is equal to
some combination of the old
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variables.
The other kind of substitution
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is inverse.
It's just the reverse.
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Here, you say that the old
variables are some combination
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of the new.
Now, often you'll have to stick
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in a few old variables,
too.
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But the basic,
it's what appears on the
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left-hand side.
Is it a new variable that
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appears on the left-hand side by
itself, or is it the old
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variable that appears on the
left-hand side?
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Now, right here,
we have an example.
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If I did it as a direct
substitution,
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I would have written T1 equals
T over M.
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That's the way I define the new
variable, which of course you
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have to do if you're introducing
it.
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But when I actually did the
substitution,
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I did the inverse substitution.
Namely, I used T equals T1,
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M times T1. And,
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the reason for doing that was
because it was the capital T's
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that faced me in the equation
and I had to have something to
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replace them with.
Now, you see this already in
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calculus, this distinction.
But that might have been a year
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and a half ago.
Just let me remind you,
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typically in calculus,
for example,
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when you want to do this kind
of integral, let's say,
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x times the square root of one
minus x squared dx,
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the substitution you'd use for
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that is u equals one minus x
squared,
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right?
And then, you calculate,
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and then you would observe that
this, the x dx,
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more or less makes up du,
apart from a constant factor,
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there.
So, this would be an example of
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direct substitution.
You put it in and convert the
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integral into an integral of u.
What would be an example of
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inverse substitution?
Well, if I take away the x and
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ask you, instead,
to do this integral,
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then you know that the right
thing to do is not to start with
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u, but to start with the x and
write x equals sine or cosine u.
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So, this is a direct
substitution in that integral,
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but this integral calls for an
inverse substitution in order to
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be able to do it.
And notice, they would look
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practically the same.
But, of course,
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as you know from your
experience, they are not.
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They're very different.
Okay, so I'm going to watch for
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that distinction as I do these
examples.
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The first one I want to do is
an example as a direct
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substitution.
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214
215
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So, it applies to the equation
of the form y prime equals,
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there are two kinds of terms on
the right-hand side.
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Let's use p of x,
p of x times y plus q of x
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times any power
whatsoever of y.
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Well, notice,
for example,
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if n were zero,
what kind of equation would
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this be?
y to the n would be
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one, and this would be a linear
equation, which you know how to
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solve.
So, n equals zero we already
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know how to do.
So, let's assume that n is not
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zero, so that we're in new
territory.
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Well, if n were equal to one,
you could separate variables.
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So, that too is not exciting.
But, nonetheless,
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it will be included in what I'm
going to say now.
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If n is two or three,
or n could be one half.
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So anything:
even zero is all right.
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It's just silly.
Any number: it could be
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negative.
n equals minus five.
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That would be fine also.
This kind of equation,
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to give it its name,
is called the Bernoulli
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equation, named after which
Bernoulli, I haven't the
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faintest idea.
There were, I think,
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three or four of them.
And, they fought with each
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other.
But, they were all smart.
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Now, the key trick,
if you like,
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method, to solving any
Bernoulli equation,
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let me call another thing.
Most important is what's
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missing.
It must not have a pure x term
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in it.
And that goes for a constant
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term.
In other words,
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it must look exactly like this.
Everything multiplied by y,
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or a power of y,
two terms.
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So, for example,
if I add one to this,
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the equation becomes
non-doable.
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Right, it's very easy to
contaminate it into an equation
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that's unsolvable.
It's got to look just like
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that.
Now, you've got one on your
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homework.
You've got several.
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Both part one and part two have
Bernoulli equations on them.
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So, this is practical,
in some sense.
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What do we got?
The idea is to divide by y to
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the n.
Ignore all formulas that you're
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given.
Just remember that when you see
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something that looks like this,
or something that you can turn
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into something that looks like
this, divide through by y to the
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nth power, no matter what n is.
All right, so y prime over y to
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the n is equal to p of x times
one over y to the n minus one,
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right, plus q of x.
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Well, that certainly doesn't
look any better than what I
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started with.
And, in your terms,
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it probably looks somewhat
worse because it's got all those
267
00:17:59 --> 00:18:05
Y's at the denominator,
and who wants to see them
268
00:18:03 --> 00:18:09
there?
But, look at it.
269
00:18:06 --> 00:18:12
In this very slightly
transformed Bernoulli equation
270
00:18:10 --> 00:18:16
is a linear equation struggling
to be free.
271
00:18:14 --> 00:18:20
Where is it?
Why is it trying to be a linear
272
00:18:17 --> 00:18:23
equation?
Make a new variable,
273
00:18:20 --> 00:18:26
call this hunk of it in new
variable.
274
00:18:23 --> 00:18:29
Let's call it V.
So, V is equal to one over y to
275
00:18:27 --> 00:18:33
the n minus one.
276
00:18:30 --> 00:18:36
Or, if you like,
you can think of that as y to
277
00:18:34 --> 00:18:40
the one minus n.
What's V prime?
278
00:18:39 --> 00:18:45
So, this is the direct
substitution I am going to use,
279
00:18:44 --> 00:18:50
but of course,
the problem is,
280
00:18:46 --> 00:18:52
what am I going to use on this?
Well, the little miracle
281
00:18:51 --> 00:18:57
happens.
What's the derivative of this?
282
00:18:54 --> 00:19:00
It is one minus n times y to
the negative n times y prime
283
00:18:59 --> 00:19:05
In other words,
284
00:19:04 --> 00:19:10
up to a constant,
this constant factor,
285
00:19:07 --> 00:19:13
one minus n,
it's exactly the left-hand side
286
00:19:11 --> 00:19:17
of the equation.
Well, let's make N not equal
287
00:19:15 --> 00:19:21
one either.
As I said, you could separate
288
00:19:18 --> 00:19:24
variables if n equals one.
What's the equation,
289
00:19:22 --> 00:19:28
then, turned into?
A Bernoulli equation,
290
00:19:28 --> 00:19:34
divided through in this way,
is then turned into the
291
00:19:36 --> 00:19:42
equation one minus n,
sorry, V prime divided by one
292
00:19:44 --> 00:19:50
minus n is equal to p of x times
V plus q of x.
293
00:19:55 --> 00:20:01
It's linear.
294
00:20:01 --> 00:20:07
And now, solve it as a linear
equation.
295
00:20:03 --> 00:20:09
Solve it as a linear equation.
You notice, it's not in
296
00:20:06 --> 00:20:12
standard form,
not in standard linear form.
297
00:20:09 --> 00:20:15
To do that, you're going to
have to put the p on the other
298
00:20:13 --> 00:20:19
side.
That's okay,
299
00:20:14 --> 00:20:20
that term, on the other side,
solve it, and at the end,
300
00:20:17 --> 00:20:23
don't forget that you put in
the V.
301
00:20:19 --> 00:20:25
It wasn't in the original
problem.
302
00:20:22 --> 00:20:28
So, you have to convert the
problem, the answer,
303
00:20:25 --> 00:20:31
back in terms of y.
It'll come out in terms of V,
304
00:20:28 --> 00:20:34
but you must put it back in
terms of y.
305
00:20:32 --> 00:20:38
Let's do a really simple
example just to illustrate the
306
00:20:38 --> 00:20:44
method, and to illustrate the
fact that I don't want you to
307
00:20:45 --> 00:20:51
memorize formulas.
Learn methods,
308
00:20:49 --> 00:20:55
not final formulas.
So, suppose the equation is,
309
00:20:54 --> 00:21:00
let's say, y prime equals y
over x minus y squared.
310
00:21:01 --> 00:21:07
That's a Bernoulli equation.
311
00:21:06 --> 00:21:12
I could, of course,
have concealed it by writing xy
312
00:21:09 --> 00:21:15
prime plus xy prime minus xy
equals negative y squared.
313
00:21:13 --> 00:21:19
Then, it wouldn't look
instantly like a Bernoulli
314
00:21:16 --> 00:21:22
equation.
You would have to stare at it a
315
00:21:19 --> 00:21:25
while and say,
hey, that's a Bernoulli
316
00:21:22 --> 00:21:28
equation.
Okay, but so I'm handing it to
317
00:21:25 --> 00:21:31
you a silver platter,
as it were.
318
00:21:27 --> 00:21:33
So, what do we do?
Divide through by y squared.
319
00:21:32 --> 00:21:38
So, it's y prime over y squared
equals one over x times one over
320
00:21:38 --> 00:21:44
y minus one.
321
00:21:41 --> 00:21:47
And now, the substitution,
then, I'm going to make,
322
00:21:46 --> 00:21:52
is for this thing.
V equals one over y.
323
00:21:51 --> 00:21:57
It's a direct substitution.
324
00:21:53 --> 00:21:59
V prime is going to be negative
one over y squared
325
00:21:59 --> 00:22:05
times y prime.
Don't forget to use the chain
326
00:22:05 --> 00:22:11
rule when you differentiate with
respect-- because the
327
00:22:08 --> 00:22:14
differentiation is with respect
to x, of course,
328
00:22:12 --> 00:22:18
not with respect to y.
Okay, so what's this thing?
329
00:22:15 --> 00:22:21
That's the left-hand side.
The only thing is it's got a
330
00:22:19 --> 00:22:25
negative sign.
So, this is minus V prime
331
00:22:22 --> 00:22:28
equals, one over x stays one
over x, one over y.
332
00:22:26 --> 00:22:32
So, it's V over x minus one.
333
00:22:30 --> 00:22:36
So, let's put that in standard
form.
334
00:22:32 --> 00:22:38
In standard form,
it will look like,
335
00:22:35 --> 00:22:41
first imagine multiplying it
through by negative one,
336
00:22:39 --> 00:22:45
and then putting this term on
the other side.
337
00:22:42 --> 00:22:48
And, it will turn into V prime
plus V over X is equal to one.
338
00:22:47 --> 00:22:53
So, that's the linear equation
339
00:22:51 --> 00:22:57
in standard linear form that we
are asked to solve.
340
00:22:54 --> 00:23:00
And, the solution isn't very
hard.
341
00:22:57 --> 00:23:03
The integrating factor is,
well, I integrate one over x.
342
00:23:03 --> 00:23:09
That makes log x.
And, e to the log x,
343
00:23:05 --> 00:23:11
so, it's e to the log x,
which is, of course,
344
00:23:09 --> 00:23:15
just x itself.
So, I should multiply this
345
00:23:12 --> 00:23:18
through by x squared,
be able to integrate it.
346
00:23:15 --> 00:23:21
Now, some of you,
I would hope,
347
00:23:17 --> 00:23:23
just can see that right away,
that if you multiply this
348
00:23:21 --> 00:23:27
through by x,
it's going to look good.
349
00:23:24 --> 00:23:30
So, after we multiply through
by x, which I get?
350
00:23:27 --> 00:23:33
(xV) prime for the-- maybe I
shouldn't skip a step.
351
00:23:33 --> 00:23:39
You are still learning this
stuff, so let's not skip a step.
352
00:23:38 --> 00:23:44
So, it becomes x V prime plus V
equals x,
353
00:23:44 --> 00:23:50
okay?
After I multiplied through by
354
00:23:47 --> 00:23:53
the integrating factor,
this now says this is xV prime,
355
00:23:53 --> 00:23:59
and I quickly check that that,
in fact, is what it's equal to,
356
00:23:59 --> 00:24:05
equals x, and therefore xV is
equal to one half x squared plus
357
00:24:05 --> 00:24:11
a constant. And,
358
00:24:08 --> 00:24:14
therefore, V is equal to one
half x plus C over x.
359
00:24:14 --> 00:24:20
You can leave it at that form,
360
00:24:19 --> 00:24:25
or you can combine terms.
It doesn't matter much.
361
00:24:23 --> 00:24:29
Am I done?
The answer is,
362
00:24:25 --> 00:24:31
no I am not done,
because nobody reading this
363
00:24:29 --> 00:24:35
answer would know what V was.
V wasn't in the original
364
00:24:33 --> 00:24:39
problem.
It was y that was in the
365
00:24:35 --> 00:24:41
original problem.
And therefore,
366
00:24:37 --> 00:24:43
the relation is,
one is the reciprocal of the
367
00:24:40 --> 00:24:46
other.
And therefore,
368
00:24:41 --> 00:24:47
I have to turn this expression
upside down.
369
00:24:44 --> 00:24:50
Well, if you're going to have
to turn it upside down,
370
00:24:47 --> 00:24:53
you probably should make it
look a little better.
371
00:24:50 --> 00:24:56
Let's rewrite it as x squared
plus 2c,
372
00:24:53 --> 00:24:59
combining fractions,
I think they call it in high
373
00:24:56 --> 00:25:02
school or elementary school,
plus 2c.
374
00:25:00 --> 00:25:06
How's that? x squared
plus 2c divided by 2x.
375
00:25:03 --> 00:25:09
Now, 2c, you don't call it
376
00:25:07 --> 00:25:13
constant 2c because this is just
as arbitrary to call it c1.
377
00:25:12 --> 00:25:18
So, I'll call that,
so, my answer will be y equals
378
00:25:16 --> 00:25:22
2x divided by x squared plus an
arbitrary constant.
379
00:25:20 --> 00:25:26
But, to indicate it's different
from that one,
380
00:25:23 --> 00:25:29
I'll call it C1. C1 is
381
00:25:27 --> 00:25:33
two times the old one,
but that doesn't really matter.
382
00:25:31 --> 00:25:37
So, there's the solution.
It has an arbitrary constant in
383
00:25:37 --> 00:25:43
it, but you note it's not an
additive arbitrary constant.
384
00:25:40 --> 00:25:46
The arbitrary constant is
tucked into the solution.
385
00:25:44 --> 00:25:50
If you had to satisfy an
initial condition,
386
00:25:47 --> 00:25:53
you would take this form,
and starting from this form,
387
00:25:50 --> 00:25:56
figure out what C1 was in order
to satisfy that initial
388
00:25:54 --> 00:26:00
condition.
Thus, Bernoulli equation is
389
00:25:57 --> 00:26:03
solved.
Our first Bernoulli equation:
390
00:25:59 --> 00:26:05
isn't that exciting?
So, here was the equation,
391
00:26:05 --> 00:26:11
and there is its solution.
Now, the one I'm asking you to
392
00:26:11 --> 00:26:17
solve on the problem set in part
two is no harder than this,
393
00:26:18 --> 00:26:24
except I ask you some hard
questions about it,
394
00:26:24 --> 00:26:30
not very hard,
but a little hard about it.
395
00:26:30 --> 00:26:36
I hope you will find them
interesting questions.
396
00:26:33 --> 00:26:39
You already have the
experimental evidence from the
397
00:26:37 --> 00:26:43
first problem set,
and the problem is to explain
398
00:26:40 --> 00:26:46
the experimental evidence by
actually solving the equation in
399
00:26:45 --> 00:26:51
the scene.
I think you'll find it
400
00:26:47 --> 00:26:53
interesting.
But, maybe that's just a pious
401
00:26:51 --> 00:26:57
hope.
Okay, I like,
402
00:26:52 --> 00:26:58
now, to turn to the second
method, where a second class of
403
00:26:56 --> 00:27:02
equations which require inverse
substitution,
404
00:27:00 --> 00:27:06
and those are equations,
which are called homogeneous,
405
00:27:04 --> 00:27:10
a highly overworked word in
differential equations,
406
00:27:08 --> 00:27:14
and in mathematics in general.
But, it's unfortunately just
407
00:27:14 --> 00:27:20
the right word to describe them.
So, these are homogeneous,
408
00:27:19 --> 00:27:25
first-order ODE's.
Now, I already used the word in
409
00:27:23 --> 00:27:29
one context in talking about the
linear equations when zero is
410
00:27:28 --> 00:27:34
the right hand side.
This is different,
411
00:27:32 --> 00:27:38
but nonetheless,
the two uses of the word have
412
00:27:35 --> 00:27:41
the same common source.
The homogeneous differential
413
00:27:39 --> 00:27:45
equation, homogeneous newspeak,
is y prime equals,
414
00:27:43 --> 00:27:49
it's a question of what the
right hand side looks like.
415
00:27:47 --> 00:27:53
And, now, the supposed way to
say it is, you should be able to
416
00:27:52 --> 00:27:58
write the right-hand side as a
function of a combined variable,
417
00:27:57 --> 00:28:03
y divided by x.
In other words,
418
00:28:01 --> 00:28:07
after fooling around with the
right hand side a little bit,
419
00:28:06 --> 00:28:12
you should be able to write it
so that every time a variable
420
00:28:11 --> 00:28:17
appears, it's always in the
combination y over x.
421
00:28:15 --> 00:28:21
Let me give some examples.
For example,
422
00:28:19 --> 00:28:25
suppose y prime were,
let's say, x squared y divided
423
00:28:23 --> 00:28:29
by x cubed plus y cubed.
424
00:28:29 --> 00:28:35
Well, that doesn't look in that
form.
425
00:28:31 --> 00:28:37
Well, yes it is.
Imagine dividing the top and
426
00:28:34 --> 00:28:40
bottom by x cubed.
What would you get?
427
00:28:37 --> 00:28:43
The top would be y over x,
if you divided it by x
428
00:28:40 --> 00:28:46
cubed.
And, if I divide the bottom by
429
00:28:43 --> 00:28:49
x cubed, also,
which, of course,
430
00:28:45 --> 00:28:51
doesn't change the value of the
fraction, as they say in
431
00:28:49 --> 00:28:55
elementary school,
one plus (y over x) cubed.
432
00:28:52 --> 00:28:58
So, this is the way it started
433
00:28:55 --> 00:29:01
out looking, but you just said
ah-ha, that was a homogeneous
434
00:28:59 --> 00:29:05
equation because I could see it
could be written that way.
435
00:29:05 --> 00:29:11
How about another homogeneous
equation?
436
00:29:10 --> 00:29:16
How about x y prime?
Is that a homogeneous equation?
437
00:29:18 --> 00:29:24
Of course it is:
otherwise, why would I be
438
00:29:24 --> 00:29:30
talking about it?
If you divide through by x,
439
00:29:29 --> 00:29:35
you can tuck it inside the
radical, the square root,
440
00:29:33 --> 00:29:39
if you remember to square it
when you do that.
441
00:29:36 --> 00:29:42
And, it becomes the square root
of x squared over x squared,
442
00:29:41 --> 00:29:47
which is one,
plus y squared over x squared.
443
00:29:44 --> 00:29:50
It's homogeneous.
444
00:29:47 --> 00:29:53
Now, you might say,
hey, this looks like you had to
445
00:29:50 --> 00:29:56
be rather clever to figure out
if an equation is homogeneous.
446
00:29:55 --> 00:30:01
Is there some other way?
Yeah, there is another way,
447
00:29:58 --> 00:30:04
and it's the other way which
explains why it's called
448
00:30:02 --> 00:30:08
homogeneous.
You can think of it this way.
449
00:30:07 --> 00:30:13
It's an equation which is,
in modern speak,
450
00:30:12 --> 00:30:18
invariant, invariant under the
operation zoom.
451
00:30:18 --> 00:30:24
What is zoom?
Zoom is, you increase the scale
452
00:30:23 --> 00:30:29
equally on both axes.
So, the zoom operation is the
453
00:30:30 --> 00:30:36
one which sends x into
a times x,
454
00:30:36 --> 00:30:42
and y into a times y.
455
00:30:42 --> 00:30:48
In other words,
you change the scale on both
456
00:30:46 --> 00:30:52
axes by the same factor,
a.
457
00:30:48 --> 00:30:54
Now, what I say is,
gee, maybe you shouldn't write
458
00:30:53 --> 00:30:59
it like this.
Why don't we say,
459
00:30:56 --> 00:31:02
we introduce,
how about this?
460
00:31:00 --> 00:31:06
So, think of it as a change of
variables.
461
00:31:02 --> 00:31:08
We will write it like that.
So, you can put here an equals
462
00:31:06 --> 00:31:12
sign, if you don't know what
this meaningless arrow means.
463
00:31:10 --> 00:31:16
So, I'm making this change of
variables, and I'm describing it
464
00:31:14 --> 00:31:20
as an inverse substitution.
But of course,
465
00:31:16 --> 00:31:22
it wouldn't make any
difference.
466
00:31:19 --> 00:31:25
It's exactly the same as the
direct substitution I started
467
00:31:22 --> 00:31:28
out with underscaling.
The only difference is,
468
00:31:25 --> 00:31:31
I'm not using different scales
on both axes.
469
00:31:28 --> 00:31:34
I'm expanding them both
equally.
470
00:31:32 --> 00:31:38
That's what I mean by zoom.
Now, what happens to the
471
00:31:36 --> 00:31:42
equation?
Well, what happens to dy over
472
00:31:40 --> 00:31:46
dx?
Well, dx is a dx1.
473
00:31:43 --> 00:31:49
dy is a dy1.
474
00:31:47 --> 00:31:53
And therefore,
the ratio, dy by dx is the same
475
00:31:51 --> 00:31:57
as dy1 over dx1.
476
00:31:54 --> 00:32:00
So, the left-hand side becomes
dy1 over dx1,
477
00:31:58 --> 00:32:04
and the right-hand side becomes
F of, well, y over x is the same
478
00:32:04 --> 00:32:10
as y over, since I've scaled
them equally,
479
00:32:08 --> 00:32:14
this is the same as y1 over x1.
480
00:32:14 --> 00:32:20
So, it's y1 over x1,
and the net effect is I simply,
481
00:32:18 --> 00:32:24
everywhere I have an x,
I change it to x1,
482
00:32:22 --> 00:32:28
and wherever I have a y,
I change it to y1,
483
00:32:25 --> 00:32:31
which, what's in a name?
It's the identical equation.
484
00:32:31 --> 00:32:37
So, I haven't changed the
equation at all via zoom
485
00:32:35 --> 00:32:41
transformation.
And, that's what makes it
486
00:32:38 --> 00:32:44
homogeneous.
That's not an uncommon use of
487
00:32:42 --> 00:32:48
the word homogeneous.
When you say space is
488
00:32:45 --> 00:32:51
homogeneous, every direction,
well, that means,
489
00:32:49 --> 00:32:55
I don't know.
It means, okay,
490
00:32:51 --> 00:32:57
I'm getting into trouble there.
I'll let it go since I can't
491
00:32:56 --> 00:33:02
prepare any better,
I haven't prepared any better
492
00:33:00 --> 00:33:06
explanation, but this is a
pretty good one.
493
00:33:06 --> 00:33:12
Okay, so, suppose we've got a
homogeneous equation.
494
00:33:13 --> 00:33:19
How do we solve it?
So, here's our equation,
495
00:33:20 --> 00:33:26
F of y over x.
Well, what substitution would
496
00:33:29 --> 00:33:35
you like to make?
Obviously, we should make a
497
00:33:34 --> 00:33:40
direct substitution,
z equals y over x.
498
00:33:38 --> 00:33:44
So, why did he say that this
was going to be an example of
499
00:33:42 --> 00:33:48
inverse substitution?
Because I wanted to confuse
500
00:33:45 --> 00:33:51
you.
But look, that's fine.
501
00:33:47 --> 00:33:53
If you write it in that form,
you'll know exactly what to do
502
00:33:51 --> 00:33:57
with the right-hand side.
And, this is why everybody
503
00:33:55 --> 00:34:01
loves to do that.
But for Charlie,
504
00:33:57 --> 00:34:03
you have to substitute into the
left-hand side as well.
505
00:34:03 --> 00:34:09
And, I can testify,
for many years of looking with
506
00:34:06 --> 00:34:12
sinking heart at examination
papers, what happens if you try
507
00:34:10 --> 00:34:16
to make a direct substitution
like this?
508
00:34:13 --> 00:34:19
You say, oh,
I need z prime.
509
00:34:15 --> 00:34:21
z prime equals,
well, I better use the quotient
510
00:34:18 --> 00:34:24
rule for differentiating that.
And, it comes out this long,
511
00:34:22 --> 00:34:28
and then either a long pause,
what do I do now?
512
00:34:26 --> 00:34:32
Because it's not at all obvious
what to do at that point.
513
00:34:30 --> 00:34:36
Or, much worse,
two pages of frantic
514
00:34:32 --> 00:34:38
calculations,
and giving up in total despair.
515
00:34:37 --> 00:34:43
Now, the reason for that is
because you tried to do it
516
00:34:40 --> 00:34:46
making a direct substitution.
All you have to do instead is
517
00:34:45 --> 00:34:51
use it, treat it as an inverse
substitution,
518
00:34:48 --> 00:34:54
write y equals zx.
What's the motivation for doing
519
00:34:51 --> 00:34:57
that?
It's clear from the equation.
520
00:34:54 --> 00:35:00
This goes through all of
mathematics.
521
00:34:57 --> 00:35:03
Whenever you have to change a
variable, excuse me,
522
00:35:00 --> 00:35:06
whenever you have to change a
variable, look at what you have
523
00:35:05 --> 00:35:11
to substitute for,
and focus your attention on
524
00:35:08 --> 00:35:14
that.
I need to know what y prime is.
525
00:35:12 --> 00:35:18
Okay, well, then I better know
what y is.
526
00:35:15 --> 00:35:21
If I know what y is,
do I know what y prime is?
527
00:35:19 --> 00:35:25
Oh, of course.
y prime is z prime x plus z
528
00:35:22 --> 00:35:28
times the derivative of this
factor, which is one.
529
00:35:26 --> 00:35:32
And now, I turned with that one
530
00:35:31 --> 00:35:37
stroke, the equation has now
become z prime x plus z is equal
531
00:35:36 --> 00:35:42
to F of z.
Well, I don't know.
532
00:35:40 --> 00:35:46
Can I solve that?
Sure.
533
00:35:42 --> 00:35:48
That can be solved because this
is x times dz / dx.
534
00:35:48 --> 00:35:54
Just put the z on the other
side, it's F of z minus z.
535
00:35:53 --> 00:35:59
And now, this side is just a
536
00:35:55 --> 00:36:01
function of z.
Separate variables.
537
00:36:00 --> 00:36:06
And, the only thing to watch
out for is, at the end,
538
00:36:03 --> 00:36:09
the z was your business.
You've got to put the answer
539
00:36:07 --> 00:36:13
back in terms of z and y.
Okay, let's work an example of
540
00:36:11 --> 00:36:17
this.
Since I haven't done any
541
00:36:13 --> 00:36:19
modeling yet this period,
let's make a little model,
542
00:36:17 --> 00:36:23
differential equations model.
It's a physical situation,
543
00:36:21 --> 00:36:27
which will be solved by an
equation.
544
00:36:24 --> 00:36:30
And, guess what?
The equation will turn out to
545
00:36:27 --> 00:36:33
be homogeneous.
Okay, so the situation is as
546
00:36:32 --> 00:36:38
follows.
We are in the Caribbean
547
00:36:34 --> 00:36:40
somewhere, a little isolated
island somewhere with a little
548
00:36:39 --> 00:36:45
lighthouse on it at the origin,
and a beam of light shines from
549
00:36:44 --> 00:36:50
the lighthouse.
The beam of light can rotate
550
00:36:48 --> 00:36:54
the way the lighthouse beams.
But, this particular beam is
551
00:36:53 --> 00:36:59
being controlled by a guy in the
lighthouse who can aim it
552
00:36:57 --> 00:37:03
wherever he wants.
And, the reason he's interested
553
00:37:01 --> 00:37:07
in aiming it wherever he wants
is there's a drug boat here,
554
00:37:06 --> 00:37:12
[LAUGHTER] which has just been
caught in the beam of light.
555
00:37:13 --> 00:37:19
So, the drug boat,
which has just been caught in a
556
00:37:16 --> 00:37:22
beam of light,
and feels it'd a better escape.
557
00:37:20 --> 00:37:26
Now, the lighthouse keeper
wants to keep the drug boat;
558
00:37:24 --> 00:37:30
the light is shining on it so
that the U.S.
559
00:37:27 --> 00:37:33
Coast Guard helicopters can
zoom over it and do whatever
560
00:37:31 --> 00:37:37
they do to drug boats,
--
561
00:37:34 --> 00:37:40
-- I don't know.
So, the drug boat immediately
562
00:37:37 --> 00:37:43
has to follow an escape
strategy.
563
00:37:39 --> 00:37:45
And, the only one that occurs
to him is, well,
564
00:37:42 --> 00:37:48
he wants to go further away,
of course, from the lighthouse.
565
00:37:47 --> 00:37:53
On the other hand,
it doesn't seem sensible to do
566
00:37:50 --> 00:37:56
it in a straight line because
the beam will keep shining on
567
00:37:54 --> 00:38:00
him.
So, he fixes the boat at some
568
00:37:57 --> 00:38:03
angle, let's say,
and goes off so that the angle
569
00:38:00 --> 00:38:06
stays 45 degrees.
So, it goes so that the angle
570
00:38:05 --> 00:38:11
between the beam and maybe,
draw the beam a little less
571
00:38:11 --> 00:38:17
like a 45 degree angle.
So, the angle between the beam
572
00:38:16 --> 00:38:22
and the boat,
the boat's path is always 45
573
574
575
00:38:20 --> 00:38:26
degrees, goes at a constant 45
degree angle to the beam,
576
00:38:26 --> 00:38:32
hoping thereby to escape.
On the other hand,
577
00:38:30 --> 00:38:36
of course, the lighthouse guy
keeps the beam always on the
578
00:38:36 --> 00:38:42
boat.
So, it's not clear it's a good
579
00:38:40 --> 00:38:46
strategy, but this is a
differential equations class.
580
00:38:44 --> 00:38:50
The question is,
what's the path of the boat?
581
00:38:48 --> 00:38:54
What's the boat's path?
Now, an obvious question is,
582
00:38:52 --> 00:38:58
why is this a problem in
differential equations at all?
583
00:38:57 --> 00:39:03
In other words,
looking at this,
584
00:38:59 --> 00:39:05
you might scratch your head and
try to think of different ways
585
00:39:04 --> 00:39:10
to solve it.
But, what suggests that it's
586
00:39:09 --> 00:39:15
going to be a problem in
differential equations?
587
00:39:13 --> 00:39:19
The answer is,
you're looking for a path.
588
00:39:17 --> 00:39:23
The answer is going to be a
curve.
589
00:39:20 --> 00:39:26
A curve means a function.
We are looking for an unknown
590
00:39:24 --> 00:39:30
function, in other words.
And, what type of information
591
00:39:29 --> 00:39:35
do we have about the function?
The only information we have
592
00:39:34 --> 00:39:40
about the function is something
about its slope,
593
00:39:38 --> 00:39:44
that its slope makes a constant
45° angle with the lighthouse
594
00:39:44 --> 00:39:50
beam.
Its slope makes a constant
595
00:39:53 --> 00:39:59
known angle to a known angle.
Well, if you are trying to find
596
00:40:04 --> 00:40:10
a function, and all you know is
something about its slope,
597
00:40:09 --> 00:40:15
that is a problem in
differential equations.
598
00:40:13 --> 00:40:19
Well, let's try to solve it.
Well, let's see.
599
00:40:16 --> 00:40:22
Well, let me draw just a little
bit.
600
00:40:19 --> 00:40:25
So, here's the horizontal.
Let's introduce the
601
00:40:23 --> 00:40:29
coordinates.
In other words,
602
00:40:25 --> 00:40:31
there's the horizontal and
here's the boat to indicate
603
00:40:30 --> 00:40:36
where I am with respect to the
picture.
604
00:40:35 --> 00:40:41
So, here's the boat.
Here's the beam,
605
00:40:38 --> 00:40:44
and the path of the boat is
going to make a 45° angle with
606
00:40:44 --> 00:40:50
it.
So, this is the path that we
607
00:40:47 --> 00:40:53
are talking about.
And now, let's label what I
608
00:40:51 --> 00:40:57
know.
Well, this angle is 45°.
609
00:40:54 --> 00:41:00
This angle, I don't know,
but of course I can calculate
610
00:41:00 --> 00:41:06
it easily enough because it has
to do with, if I know the
611
00:41:05 --> 00:41:11
coordinates of this point,
(x, y), then of course that
612
00:41:11 --> 00:41:17
horizontal angle,
I know the slope of this line,
613
00:41:15 --> 00:41:21
and that angle will be related
to the slope.
614
00:41:22 --> 00:41:28
So, let's call this alpha.
And now, what I want to know is
615
00:41:29 --> 00:41:35
what the slope of the whole path
is.
616
00:41:35 --> 00:41:41
So, y prime-- let's call y
equals y of x,
617
00:41:42 --> 00:41:48
the unknown function whose
path, whose graph is going to be
618
00:41:50 --> 00:41:56
the boat's path,
unknown graph.
619
00:41:54 --> 00:42:00
What's its slope?
Well, its slope is the tangent
620
00:42:00 --> 00:42:06
of the sum of these two angles,
alpha plus 45°.
621
00:42:08 --> 00:42:14
Now, what do I know?
Well, I know that the tangent
622
00:42:11 --> 00:42:17
of alpha is how much?
That's y over x.
623
00:42:14 --> 00:42:20
In other words,
624
00:42:16 --> 00:42:22
if this was the point,
x over y, this is the angle it
625
00:42:19 --> 00:42:25
makes with a horizontal,
if you think of it over here.
626
00:42:23 --> 00:42:29
So, this angle is the same as
that one, and it's y over,
627
00:42:26 --> 00:42:32
its slope of that line is y
over x.
628
00:42:30 --> 00:42:36
So, the tangent of the angle is
y over x.
629
00:42:32 --> 00:42:38
How about the tangent of 45°?
That's one, and there's a
630
00:42:36 --> 00:42:42
formula.
This is the hard part.
631
00:42:38 --> 00:42:44
All you have to know is that
the formula exists,
632
00:42:41 --> 00:42:47
and then you will look it up if
you have forgotten it,
633
00:42:44 --> 00:42:50
relating the tangent or giving
you the tangent of the sum of
634
00:42:48 --> 00:42:54
two angles, and you can,
if you like,
635
00:42:50 --> 00:42:56
clever, derive it from the
formula for the sign and cosine
636
00:42:54 --> 00:43:00
of the sum of two angles.
But, one peak is worth a
637
00:42:57 --> 00:43:03
thousand finesses.
So, it is the tangent of alpha
638
00:43:02 --> 00:43:08
plus the tangent of 45°.
Let me read it out in all its
639
00:43:06 --> 00:43:12
gory details,
divided by one,
640
00:43:08 --> 00:43:14
so you'll at least learn the
formula, one minus tangent alpha
641
00:43:12 --> 00:43:18
times tangent 45°.
642
00:43:15 --> 00:43:21
This would work for the tangent
of the sum of any two angles.
643
00:43:20 --> 00:43:26
That's the formula.
So, what do I get then?
644
00:43:23 --> 00:43:29
y prime is equal to the tangent
of alpha, which is y over x,
645
00:43:27 --> 00:43:33
oh, I like that combination,
plus one, divided by (one minus
646
00:43:32 --> 00:43:38
y over x times one).
647
00:43:37 --> 00:43:43
Now, there is no reason for
doing anything to it,
648
00:43:40 --> 00:43:46
but let's make it look a little
prettier, and thereby,
649
00:43:44 --> 00:43:50
make it less obvious that it's
a homogeneous equation.
650
00:43:48 --> 00:43:54
If I multiply top and bottom by
x, it looks prettier.
651
00:43:52 --> 00:43:58
x plus y over x minus y equals
y prime.
652
00:43:55 --> 00:44:01
That's our
differential equation.
653
00:44:00 --> 00:44:06
But, notice,
that let step to make it look
654
00:44:02 --> 00:44:08
pretty has undone the good work.
It's fine if you immediately
655
00:44:06 --> 00:44:12
recognize this as being a
homogeneous equation because you
656
00:44:10 --> 00:44:16
can divide the top and bottom by
x.
657
00:44:12 --> 00:44:18
But here, it's a lot clearer
that it's a homogeneous equation
658
00:44:16 --> 00:44:22
because it's already been
written in the right form.
659
00:44:20 --> 00:44:26
Okay, let's solve it now,
since we know what to do.
660
00:44:23 --> 00:44:29
We're going to use as the new
variable, z equals y over x.
661
00:44:27 --> 00:44:33
And, as I wrote up there for y
662
00:44:33 --> 00:44:39
prime, we'll substitute z prime
x plus z.
663
00:44:40 --> 00:44:46
And, with that,
let's solve.
664
00:44:44 --> 00:44:50
Let's solve it.
The equation becomes z prime x
665
00:44:50 --> 00:44:56
plus z is equal to z plus one
over one minus z.
666
00:44:57 --> 00:45:03
We want to separate variables,
667
00:45:04 --> 00:45:10
so you have to put all the z's
on one side.
668
00:45:07 --> 00:45:13
So, this is going to be x,
dz / dx equals this thing minus
669
00:45:11 --> 00:45:17
z, which is (z plus one) over
(one minus z) minus z.
670
00:45:14 --> 00:45:20
And now, as you realize,
671
00:45:18 --> 00:45:24
putting it on the other side,
I'm going to have to turn it
672
00:45:22 --> 00:45:28
upside down.
Just as before,
673
00:45:24 --> 00:45:30
if you have to turn something
upside down, it's better to
674
00:45:28 --> 00:45:34
combine the terms,
and make it one tiny little
675
00:45:31 --> 00:45:37
fraction.
Otherwise, you are in for quite
676
00:45:35 --> 00:45:41
a lot of mess if you don't do
this nicely.
677
00:45:39 --> 00:45:45
So, z plus one minus z,
that gets rid of the z's.
678
00:45:43 --> 00:45:49
The numerator is one minus z
squared over one minus z,
679
00:45:48 --> 00:45:54
I hope, one,
is that right,
680
00:45:50 --> 00:45:56
(one plus z squared) over (one
minus z).
681
00:45:56 --> 00:46:02
And so, the question is dz,
682
00:45:58 --> 00:46:04
and put this on the other side
and turn it upside down.
683
00:46:05 --> 00:46:11
So, that will be (one minus z)
over (one plus z squared) on the
684
00:46:10 --> 00:46:16
left-hand side and on the
right-hand side,
685
00:46:14 --> 00:46:20
dx over x.
Well, that's ready to be
686
00:46:17 --> 00:46:23
integrated just as it stands.
The right-hand side integrates
687
00:46:23 --> 00:46:29
to be log x.
The left-hand side is the sum
688
00:46:26 --> 00:46:32
of two terms.
The integral of one over one
689
00:46:30 --> 00:46:36
plus z squared is the arc
tangent of z,
690
00:46:34 --> 00:46:40
maybe?
The derivative of this is one
691
00:46:37 --> 00:46:43
over one plus z squared.
692
00:46:40 --> 00:46:46
How about the term z over one
plus z squared?
693
00:46:44 --> 00:46:50
Well, that integrates to be a
694
00:46:46 --> 00:46:52
logarithm.
It is more or less the
695
00:46:48 --> 00:46:54
logarithm of one plus
z squared.
696
00:46:51 --> 00:46:57
If I differentiate this,
I get one over one plus z^2
697
00:46:54 --> 00:47:00
times 2z,
but I wish I had negative z
698
00:46:58 --> 00:47:04
there instead.
Therefore, I should put a minus
699
00:47:01 --> 00:47:07
sign, and I should multiply that
by half to make it come out
700
00:47:05 --> 00:47:11
right.
And, this is log x on the right
701
00:47:09 --> 00:47:15
hand side plus,
put in that arbitrary constant.
702
00:47:13 --> 00:47:19
And now what?
Well, let's now fool around
703
00:47:16 --> 00:47:22
with it a little bit.
The arc tangent,
704
00:47:19 --> 00:47:25
I'm going to simultaneously,
no, two steps.
705
00:47:22 --> 00:47:28
I have to remember your
innocence, although probably a
706
00:47:26 --> 00:47:32
lot of you are better
calculators than I am.
707
00:47:31 --> 00:47:37
I'm going to change this,
use as many laws of logarithms
708
00:47:35 --> 00:47:41
as possible.
I'm going to put this in the
709
00:47:38 --> 00:47:44
exponent, and put this on the
other side.
710
00:47:41 --> 00:47:47
That's going to turn it into
the log of the square root of
711
00:47:45 --> 00:47:51
one plus z squared.
712
00:47:48 --> 00:47:54
And, this is going to be plus
the log of x plus c.
713
00:47:53 --> 00:47:59
And, now I'm going to make,
714
00:47:55 --> 00:48:01
go back and remember that z
equals y over x.
715
00:48:00 --> 00:48:06
So, this becomes the arc
tangent of y over x equals.
716
00:48:04 --> 00:48:10
Now, I combine the logarithms.
717
00:48:09 --> 00:48:15
This is the log of x times this
square root, right,
718
00:48:12 --> 00:48:18
make one logarithm out of it,
and then put z equals y over z.
719
00:48:16 --> 00:48:22
And, you see that if you do
720
00:48:19 --> 00:48:25
that, it'll be the log of x
times the square root of one
721
00:48:23 --> 00:48:29
plus (y over x) squared,
722
00:48:27 --> 00:48:33
and what is that?
Well, if I put this over x
723
00:48:30 --> 00:48:36
squared and take it out,
it cancels that.
724
00:48:34 --> 00:48:40
And, what you are left with is
the log of the square root of x
725
00:48:38 --> 00:48:44
squared plus y squared plus a
constant.
726
00:48:42 --> 00:48:48
Now, technically,
727
00:48:43 --> 00:48:49
you have solved the equation,
but not morally because,
728
00:48:47 --> 00:48:53
I mean, my God,
what a mess!
729
00:48:49 --> 00:48:55
Incredible path.
It tells me absolutely nothing.
730
00:48:52 --> 00:48:58
Wow, what is the screaming?
Change me to polar coordinates.
731
00:48:56 --> 00:49:02
What's the arc tangent of y
over x?
732
00:49:00 --> 00:49:06
Theta.
In polar coordinates it's
733
00:49:02 --> 00:49:08
theta.
This is r.
734
00:49:04 --> 00:49:10
So, the curve is theta equals
the log of r plus a constant.
735
00:49:09 --> 00:49:15
And, I can make even that
little better if I exponentiate
736
00:49:14 --> 00:49:20
everything, exponentiate both
sides, combine this in the usual
737
00:49:19 --> 00:49:25
way, the and what you get is
that r is equal to some other
738
00:49:24 --> 00:49:30
constant times e to the theta.
739
00:49:30 --> 00:49:36
That's the curve.
It's called an exponential
740
00:49:33 --> 00:49:39
spiral, and that's what our
little boat goes in.
741
00:49:37 --> 00:49:43
And notice, probably if I had
set up the problem in polar
742
00:49:42 --> 00:49:48
coordinates from the beginning,
nobody would have been able to
743
00:49:48 --> 00:49:54
solve it.
But, anyone who did would have
744
00:49:51 --> 00:49:57
gotten that answer immediately.
Thanks.