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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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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,

234
00:16:14,000 --> 00:16:20,000
three or four of them.
And, they fought with each

235
00:16:18,000 --> 00:16:24,000
other.
But, they were all smart.

236
00:16:20,000 --> 00:16:26,000
Now, the key trick,
if you like,

237
00:16:23,000 --> 00:16:29,000
method, to solving any
Bernoulli equation,

238
00:16:26,000 --> 00:16:32,000
let me call another thing.
Most important is what's

239
00:16:30,000 --> 00:16:36,000
missing.
It must not have a pure x term

240
00:16:34,000 --> 00:16:40,000
in it.
And that goes for a constant

241
00:16:37,000 --> 00:16:43,000
term.
In other words,

242
00:16:38,000 --> 00:16:44,000
it must look exactly like this.
Everything multiplied by y,

243
00:16:43,000 --> 00:16:49,000
or a power of y,
two terms.

244
00:16:45,000 --> 00:16:51,000
So, for example,
if I add one to this,

245
00:16:48,000 --> 00:16:54,000
the equation becomes
non-doable.

246
00:16:51,000 --> 00:16:57,000
Right, it's very easy to
contaminate it into an equation

247
00:16:55,000 --> 00:17:01,000
that's unsolvable.
It's got to look just like

248
00:16:59,000 --> 00:17:05,000
that.
Now, you've got one on your

249
00:17:03,000 --> 00:17:09,000
homework.
You've got several.

250
00:17:05,000 --> 00:17:11,000
Both part one and part two have
Bernoulli equations on them.

251
00:17:10,000 --> 00:17:16,000
So, this is practical,
in some sense.

252
00:17:13,000 --> 00:17:19,000
What do we got?
The idea is to divide by y to

253
00:17:17,000 --> 00:17:23,000
the n.
Ignore all formulas that you're

254
00:17:20,000 --> 00:17:26,000
given.
Just remember that when you see

255
00:17:23,000 --> 00:17:29,000
something that looks like this,
or something that you can turn

256
00:17:28,000 --> 00:17:34,000
into something that looks like
this, divide through by y to the

257
00:17:34,000 --> 00:17:40,000
nth power, no matter what n is.
All right, so y prime over y to

258
00:17:40,000 --> 00:17:46,000
the n is equal to p of x times
one over y to the n minus one,

259
00:17:44,000 --> 00:17:50,000
right, plus q of x.

260
00:17:49,000 --> 00:17:55,000
Well, that certainly doesn't
look any better than what I

261
00:17:53,000 --> 00:17:59,000
started with.
And, in your terms,

262
00:17:55,000 --> 00:18:01,000
it probably looks somewhat
worse because it's got all those

263
00:17:59,000 --> 00:18:05,000
Y's at the denominator,
and who wants to see them

264
00:18:03,000 --> 00:18:09,000
there?
But, look at it.

265
00:18:06,000 --> 00:18:12,000
In this very slightly
transformed Bernoulli equation

266
00:18:10,000 --> 00:18:16,000
is a linear equation struggling
to be free.

267
00:18:14,000 --> 00:18:20,000
Where is it?
Why is it trying to be a linear

268
00:18:17,000 --> 00:18:23,000
equation?
Make a new variable,

269
00:18:20,000 --> 00:18:26,000
call this hunk of it in new
variable.

270
00:18:23,000 --> 00:18:29,000
Let's call it V.
So, V is equal to one over y to

271
00:18:27,000 --> 00:18:33,000
the n minus one.

272
00:18:30,000 --> 00:18:36,000
Or, if you like,
you can think of that as y to

273
00:18:34,000 --> 00:18:40,000
the one minus n.
What's V prime?

274
00:18:39,000 --> 00:18:45,000
So, this is the direct
substitution I am going to use,

275
00:18:44,000 --> 00:18:50,000
but of course,
the problem is,

276
00:18:46,000 --> 00:18:52,000
what am I going to use on this?
Well, the little miracle

277
00:18:51,000 --> 00:18:57,000
happens.
What's the derivative of this?

278
00:18:54,000 --> 00:19:00,000
It is one minus n times y to
the negative n times y prime

279
00:18:59,000 --> 00:19:05,000
In other words,

280
00:19:04,000 --> 00:19:10,000
up to a constant,
this constant factor,

281
00:19:07,000 --> 00:19:13,000
one minus n,
it's exactly the left-hand side

282
00:19:11,000 --> 00:19:17,000
of the equation.
Well, let's make N not equal

283
00:19:15,000 --> 00:19:21,000
one either.
As I said, you could separate

284
00:19:18,000 --> 00:19:24,000
variables if n equals one.
What's the equation,

285
00:19:22,000 --> 00:19:28,000
then, turned into?
A Bernoulli equation,

286
00:19:28,000 --> 00:19:34,000
divided through in this way,
is then turned into the

287
00:19:36,000 --> 00:19:42,000
equation one minus n,
sorry, V prime divided by one

288
00:19:44,000 --> 00:19:50,000
minus n is equal to p of x times
V plus q of x.

289
00:19:55,000 --> 00:20:01,000
It's linear.

290
00:20:01,000 --> 00:20:07,000
And now, solve it as a linear
equation.

291
00:20:03,000 --> 00:20:09,000
Solve it as a linear equation.
You notice, it's not in

292
00:20:06,000 --> 00:20:12,000
standard form,
not in standard linear form.

293
00:20:09,000 --> 00:20:15,000
To do that, you're going to
have to put the p on the other

294
00:20:13,000 --> 00:20:19,000
side.
That's okay,

295
00:20:14,000 --> 00:20:20,000
that term, on the other side,
solve it, and at the end,

296
00:20:17,000 --> 00:20:23,000
don't forget that you put in
the V.

297
00:20:19,000 --> 00:20:25,000
It wasn't in the original
problem.

298
00:20:22,000 --> 00:20:28,000
So, you have to convert the
problem, the answer,

299
00:20:25,000 --> 00:20:31,000
back in terms of y.
It'll come out in terms of V,

300
00:20:28,000 --> 00:20:34,000
but you must put it back in
terms of y.

301
00:20:32,000 --> 00:20:38,000
Let's do a really simple
example just to illustrate the

302
00:20:38,000 --> 00:20:44,000
method, and to illustrate the
fact that I don't want you to

303
00:20:45,000 --> 00:20:51,000
memorize formulas.
Learn methods,

304
00:20:49,000 --> 00:20:55,000
not final formulas.
So, suppose the equation is,

305
00:20:54,000 --> 00:21:00,000
let's say, y prime equals y
over x minus y squared.

306
00:21:01,000 --> 00:21:07,000
That's a Bernoulli equation.

307
00:21:06,000 --> 00:21:12,000
I could, of course,
have concealed it by writing xy

308
00:21:09,000 --> 00:21:15,000
prime plus xy prime minus xy
equals negative y squared.

309
00:21:13,000 --> 00:21:19,000
Then, it wouldn't look
instantly like a Bernoulli

310
00:21:16,000 --> 00:21:22,000
equation.
You would have to stare at it a

311
00:21:19,000 --> 00:21:25,000
while and say,
hey, that's a Bernoulli

312
00:21:22,000 --> 00:21:28,000
equation.
Okay, but so I'm handing it to

313
00:21:25,000 --> 00:21:31,000
you a silver platter,
as it were.

314
00:21:27,000 --> 00:21:33,000
So, what do we do?
Divide through by y squared.

315
00:21:32,000 --> 00:21:38,000
So, it's y prime over y squared
equals one over x times one over

316
00:21:38,000 --> 00:21:44,000
y minus one.

317
00:21:41,000 --> 00:21:47,000
And now, the substitution,
then, I'm going to make,

318
00:21:46,000 --> 00:21:52,000
is for this thing.
V equals one over y.

319
00:21:51,000 --> 00:21:57,000
It's a direct substitution.

320
00:21:53,000 --> 00:21:59,000
V prime is going to be negative
one over y squared

321
00:21:59,000 --> 00:22:05,000
times y prime.
Don't forget to use the chain

322
00:22:05,000 --> 00:22:11,000
rule when you differentiate with
respect-- because the

323
00:22:08,000 --> 00:22:14,000
differentiation is with respect
to x, of course,

324
00:22:12,000 --> 00:22:18,000
not with respect to y.
Okay, so what's this thing?

325
00:22:15,000 --> 00:22:21,000
That's the left-hand side.
The only thing is it's got a

326
00:22:19,000 --> 00:22:25,000
negative sign.
So, this is minus V prime

327
00:22:22,000 --> 00:22:28,000
equals, one over x stays one
over x, one over y.

328
00:22:26,000 --> 00:22:32,000
So, it's V over x minus one.

329
00:22:30,000 --> 00:22:36,000
So, let's put that in standard
form.

330
00:22:32,000 --> 00:22:38,000
In standard form,
it will look like,

331
00:22:35,000 --> 00:22:41,000
first imagine multiplying it
through by negative one,

332
00:22:39,000 --> 00:22:45,000
and then putting this term on
the other side.

333
00:22:42,000 --> 00:22:48,000
And, it will turn into V prime
plus V over X is equal to one.

334
00:22:47,000 --> 00:22:53,000
So, that's the linear equation

335
00:22:51,000 --> 00:22:57,000
in standard linear form that we
are asked to solve.

336
00:22:54,000 --> 00:23:00,000
And, the solution isn't very
hard.

337
00:22:57,000 --> 00:23:03,000
The integrating factor is,
well, I integrate one over x.

338
00:23:03,000 --> 00:23:09,000
That makes log x.
And, e to the log x,

339
00:23:05,000 --> 00:23:11,000
so, it's e to the log x,
which is, of course,

340
00:23:09,000 --> 00:23:15,000
just x itself.
So, I should multiply this

341
00:23:12,000 --> 00:23:18,000
through by x squared,
be able to integrate it.

342
00:23:15,000 --> 00:23:21,000
Now, some of you,
I would hope,

343
00:23:17,000 --> 00:23:23,000
just can see that right away,
that if you multiply this

344
00:23:21,000 --> 00:23:27,000
through by x,
it's going to look good.

345
00:23:24,000 --> 00:23:30,000
So, after we multiply through
by x, which I get?

346
00:23:27,000 --> 00:23:33,000
(xV) prime for the-- maybe I
shouldn't skip a step.

347
00:23:33,000 --> 00:23:39,000
You are still learning this
stuff, so let's not skip a step.

348
00:23:38,000 --> 00:23:44,000
So, it becomes x V prime plus V
equals x,

349
00:23:44,000 --> 00:23:50,000
okay?
After I multiplied through by

350
00:23:47,000 --> 00:23:53,000
the integrating factor,
this now says this is xV prime,

351
00:23:53,000 --> 00:23:59,000
and I quickly check that that,
in fact, is what it's equal to,

352
00:23:59,000 --> 00:24:05,000
equals x, and therefore xV is
equal to one half x squared plus

353
00:24:05,000 --> 00:24:11,000
a constant. And,

354
00:24:08,000 --> 00:24:14,000
therefore, V is equal to one
half x plus C over x.

355
00:24:14,000 --> 00:24:20,000
You can leave it at that form,

356
00:24:19,000 --> 00:24:25,000
or you can combine terms.
It doesn't matter much.

357
00:24:23,000 --> 00:24:29,000
Am I done?
The answer is,

358
00:24:25,000 --> 00:24:31,000
no I am not done,
because nobody reading this

359
00:24:29,000 --> 00:24:35,000
answer would know what V was.
V wasn't in the original

360
00:24:33,000 --> 00:24:39,000
problem.
It was y that was in the

361
00:24:35,000 --> 00:24:41,000
original problem.
And therefore,

362
00:24:37,000 --> 00:24:43,000
the relation is,
one is the reciprocal of the

363
00:24:40,000 --> 00:24:46,000
other.
And therefore,

364
00:24:41,000 --> 00:24:47,000
I have to turn this expression
upside down.

365
00:24:44,000 --> 00:24:50,000
Well, if you're going to have
to turn it upside down,

366
00:24:47,000 --> 00:24:53,000
you probably should make it
look a little better.

367
00:24:50,000 --> 00:24:56,000
Let's rewrite it as x squared
plus 2c,

368
00:24:53,000 --> 00:24:59,000
combining fractions,
I think they call it in high

369
00:24:56,000 --> 00:25:02,000
school or elementary school,
plus 2c.

370
00:25:00,000 --> 00:25:06,000
How's that? x squared
plus 2c divided by 2x.

371
00:25:03,000 --> 00:25:09,000
Now, 2c, you don't call it

372
00:25:07,000 --> 00:25:13,000
constant 2c because this is just
as arbitrary to call it c1.

373
00:25:12,000 --> 00:25:18,000
So, I'll call that,
so, my answer will be y equals

374
00:25:16,000 --> 00:25:22,000
2x divided by x squared plus an
arbitrary constant.

375
00:25:20,000 --> 00:25:26,000
But, to indicate it's different
from that one,

376
00:25:23,000 --> 00:25:29,000
I'll call it C1. C1 is

377
00:25:27,000 --> 00:25:33,000
two times the old one,
but that doesn't really matter.

378
00:25:31,000 --> 00:25:37,000
So, there's the solution.
It has an arbitrary constant in

379
00:25:37,000 --> 00:25:43,000
it, but you note it's not an
additive arbitrary constant.

380
00:25:40,000 --> 00:25:46,000
The arbitrary constant is
tucked into the solution.

381
00:25:44,000 --> 00:25:50,000
If you had to satisfy an
initial condition,

382
00:25:47,000 --> 00:25:53,000
you would take this form,
and starting from this form,

383
00:25:50,000 --> 00:25:56,000
figure out what C1 was in order
to satisfy that initial

384
00:25:54,000 --> 00:26:00,000
condition.
Thus, Bernoulli equation is

385
00:25:57,000 --> 00:26:03,000
solved.
Our first Bernoulli equation:

386
00:25:59,000 --> 00:26:05,000
isn't that exciting?
So, here was the equation,

387
00:26:05,000 --> 00:26:11,000
and there is its solution.
Now, the one I'm asking you to

388
00:26:11,000 --> 00:26:17,000
solve on the problem set in part
two is no harder than this,

389
00:26:18,000 --> 00:26:24,000
except I ask you some hard
questions about it,

390
00:26:24,000 --> 00:26:30,000
not very hard,
but a little hard about it.

391
00:26:30,000 --> 00:26:36,000
I hope you will find them
interesting questions.

392
00:26:33,000 --> 00:26:39,000
You already have the
experimental evidence from the

393
00:26:37,000 --> 00:26:43,000
first problem set,
and the problem is to explain

394
00:26:40,000 --> 00:26:46,000
the experimental evidence by
actually solving the equation in

395
00:26:45,000 --> 00:26:51,000
the scene.
I think you'll find it

396
00:26:47,000 --> 00:26:53,000
interesting.
But, maybe that's just a pious

397
00:26:51,000 --> 00:26:57,000
hope.
Okay, I like,

398
00:26:52,000 --> 00:26:58,000
now, to turn to the second
method, where a second class of

399
00:26:56,000 --> 00:27:02,000
equations which require inverse
substitution,

400
00:27:00,000 --> 00:27:06,000
and those are equations,
which are called homogeneous,

401
00:27:04,000 --> 00:27:10,000
a highly overworked word in
differential equations,

402
00:27:08,000 --> 00:27:14,000
and in mathematics in general.
But, it's unfortunately just

403
00:27:14,000 --> 00:27:20,000
the right word to describe them.
So, these are homogeneous,

404
00:27:19,000 --> 00:27:25,000
first-order ODE's.
Now, I already used the word in

405
00:27:23,000 --> 00:27:29,000
one context in talking about the
linear equations when zero is

406
00:27:28,000 --> 00:27:34,000
the right hand side.
This is different,

407
00:27:32,000 --> 00:27:38,000
but nonetheless,
the two uses of the word have

408
00:27:35,000 --> 00:27:41,000
the same common source.
The homogeneous differential

409
00:27:39,000 --> 00:27:45,000
equation, homogeneous newspeak,
is y prime equals,

410
00:27:43,000 --> 00:27:49,000
it's a question of what the
right hand side looks like.

411
00:27:47,000 --> 00:27:53,000
And, now, the supposed way to
say it is, you should be able to

412
00:27:52,000 --> 00:27:58,000
write the right-hand side as a
function of a combined variable,

413
00:27:57,000 --> 00:28:03,000
y divided by x.
In other words,

414
00:28:01,000 --> 00:28:07,000
after fooling around with the
right hand side a little bit,

415
00:28:06,000 --> 00:28:12,000
you should be able to write it
so that every time a variable

416
00:28:11,000 --> 00:28:17,000
appears, it's always in the
combination y over x.

417
00:28:15,000 --> 00:28:21,000
Let me give some examples.
For example,

418
00:28:19,000 --> 00:28:25,000
suppose y prime were,
let's say, x squared y divided

419
00:28:23,000 --> 00:28:29,000
by x cubed plus y cubed.

420
00:28:29,000 --> 00:28:35,000
Well, that doesn't look in that
form.

421
00:28:31,000 --> 00:28:37,000
Well, yes it is.
Imagine dividing the top and

422
00:28:34,000 --> 00:28:40,000
bottom by x cubed.
What would you get?

423
00:28:37,000 --> 00:28:43,000
The top would be y over x,
if you divided it by x

424
00:28:40,000 --> 00:28:46,000
cubed.
And, if I divide the bottom by

425
00:28:43,000 --> 00:28:49,000
x cubed, also,
which, of course,

426
00:28:45,000 --> 00:28:51,000
doesn't change the value of the
fraction, as they say in

427
00:28:49,000 --> 00:28:55,000
elementary school,
one plus (y over x) cubed.

428
00:28:52,000 --> 00:28:58,000
So, this is the way it started

429
00:28:55,000 --> 00:29:01,000
out looking, but you just said
ah-ha, that was a homogeneous

430
00:28:59,000 --> 00:29:05,000
equation because I could see it
could be written that way.

431
00:29:05,000 --> 00:29:11,000
How about another homogeneous
equation?

432
00:29:10,000 --> 00:29:16,000
How about x y prime?
Is that a homogeneous equation?

433
00:29:18,000 --> 00:29:24,000
Of course it is:
otherwise, why would I be

434
00:29:24,000 --> 00:29:30,000
talking about it?
If you divide through by x,

435
00:29:29,000 --> 00:29:35,000
you can tuck it inside the
radical, the square root,

436
00:29:33,000 --> 00:29:39,000
if you remember to square it
when you do that.

437
00:29:36,000 --> 00:29:42,000
And, it becomes the square root
of x squared over x squared,

438
00:29:41,000 --> 00:29:47,000
which is one,
plus y squared over x squared.

439
00:29:44,000 --> 00:29:50,000
It's homogeneous.

440
00:29:47,000 --> 00:29:53,000
Now, you might say,
hey, this looks like you had to

441
00:29:50,000 --> 00:29:56,000
be rather clever to figure out
if an equation is homogeneous.

442
00:29:55,000 --> 00:30:01,000
Is there some other way?
Yeah, there is another way,

443
00:29:58,000 --> 00:30:04,000
and it's the other way which
explains why it's called

444
00:30:02,000 --> 00:30:08,000
homogeneous.
You can think of it this way.

445
00:30:07,000 --> 00:30:13,000
It's an equation which is,
in modern speak,

446
00:30:12,000 --> 00:30:18,000
invariant, invariant under the
operation zoom.

447
00:30:18,000 --> 00:30:24,000
What is zoom?
Zoom is, you increase the scale

448
00:30:23,000 --> 00:30:29,000
equally on both axes.
So, the zoom operation is the

449
00:30:30,000 --> 00:30:36,000
one which sends x into
a times x,

450
00:30:36,000 --> 00:30:42,000
and y into a times y.

451
00:30:42,000 --> 00:30:48,000
In other words,
you change the scale on both

452
00:30:46,000 --> 00:30:52,000
axes by the same factor,
a.

453
00:30:48,000 --> 00:30:54,000
Now, what I say is,
gee, maybe you shouldn't write

454
00:30:53,000 --> 00:30:59,000
it like this.
Why don't we say,

455
00:30:56,000 --> 00:31:02,000
we introduce,
how about this?

456
00:31:00,000 --> 00:31:06,000
So, think of it as a change of
variables.

457
00:31:02,000 --> 00:31:08,000
We will write it like that.
So, you can put here an equals

458
00:31:06,000 --> 00:31:12,000
sign, if you don't know what
this meaningless arrow means.

459
00:31:10,000 --> 00:31:16,000
So, I'm making this change of
variables, and I'm describing it

460
00:31:14,000 --> 00:31:20,000
as an inverse substitution.
But of course,

461
00:31:16,000 --> 00:31:22,000
it wouldn't make any
difference.

462
00:31:19,000 --> 00:31:25,000
It's exactly the same as the
direct substitution I started

463
00:31:22,000 --> 00:31:28,000
out with underscaling.
The only difference is,

464
00:31:25,000 --> 00:31:31,000
I'm not using different scales
on both axes.

465
00:31:28,000 --> 00:31:34,000
I'm expanding them both
equally.

466
00:31:32,000 --> 00:31:38,000
That's what I mean by zoom.
Now, what happens to the

467
00:31:36,000 --> 00:31:42,000
equation?
Well, what happens to dy over

468
00:31:40,000 --> 00:31:46,000
dx?
Well, dx is a dx1.

469
00:31:43,000 --> 00:31:49,000
dy is a dy1.

470
00:31:47,000 --> 00:31:53,000
And therefore,
the ratio, dy by dx is the same

471
00:31:51,000 --> 00:31:57,000
as dy1 over dx1.

472
00:31:54,000 --> 00:32:00,000
So, the left-hand side becomes
dy1 over dx1,

473
00:31:58,000 --> 00:32:04,000
and the right-hand side becomes
F of, well, y over x is the same

474
00:32:04,000 --> 00:32:10,000
as y over, since I've scaled
them equally,

475
00:32:08,000 --> 00:32:14,000
this is the same as y1 over x1.

476
00:32:14,000 --> 00:32:20,000
So, it's y1 over x1,
and the net effect is I simply,

477
00:32:18,000 --> 00:32:24,000
everywhere I have an x,
I change it to x1,

478
00:32:22,000 --> 00:32:28,000
and wherever I have a y,
I change it to y1,

479
00:32:25,000 --> 00:32:31,000
which, what's in a name?
It's the identical equation.

480
00:32:31,000 --> 00:32:37,000
So, I haven't changed the
equation at all via zoom

481
00:32:35,000 --> 00:32:41,000
transformation.
And, that's what makes it

482
00:32:38,000 --> 00:32:44,000
homogeneous.
That's not an uncommon use of

483
00:32:42,000 --> 00:32:48,000
the word homogeneous.
When you say space is

484
00:32:45,000 --> 00:32:51,000
homogeneous, every direction,
well, that means,

485
00:32:49,000 --> 00:32:55,000
I don't know.
It means, okay,

486
00:32:51,000 --> 00:32:57,000
I'm getting into trouble there.
I'll let it go since I can't

487
00:32:56,000 --> 00:33:02,000
prepare any better,
I haven't prepared any better

488
00:33:00,000 --> 00:33:06,000
explanation, but this is a
pretty good one.

489
00:33:06,000 --> 00:33:12,000
Okay, so, suppose we've got a
homogeneous equation.

490
00:33:13,000 --> 00:33:19,000
How do we solve it?
So, here's our equation,

491
00:33:20,000 --> 00:33:26,000
F of y over x.
Well, what substitution would

492
00:33:29,000 --> 00:33:35,000
you like to make?
Obviously, we should make a

493
00:33:34,000 --> 00:33:40,000
direct substitution,
z equals y over x.

494
00:33:38,000 --> 00:33:44,000
So, why did he say that this
was going to be an example of

495
00:33:42,000 --> 00:33:48,000
inverse substitution?
Because I wanted to confuse

496
00:33:45,000 --> 00:33:51,000
you.
But look, that's fine.

497
00:33:47,000 --> 00:33:53,000
If you write it in that form,
you'll know exactly what to do

498
00:33:51,000 --> 00:33:57,000
with the right-hand side.
And, this is why everybody

499
00:33:55,000 --> 00:34:01,000
loves to do that.
But for Charlie,

500
00:33:57,000 --> 00:34:03,000
you have to substitute into the
left-hand side as well.

501
00:34:03,000 --> 00:34:09,000
And, I can testify,
for many years of looking with

502
00:34:06,000 --> 00:34:12,000
sinking heart at examination
papers, what happens if you try

503
00:34:10,000 --> 00:34:16,000
to make a direct substitution
like this?

504
00:34:13,000 --> 00:34:19,000
You say, oh,
I need z prime.

505
00:34:15,000 --> 00:34:21,000
z prime equals,
well, I better use the quotient

506
00:34:18,000 --> 00:34:24,000
rule for differentiating that.
And, it comes out this long,

507
00:34:22,000 --> 00:34:28,000
and then either a long pause,
what do I do now?

508
00:34:26,000 --> 00:34:32,000
Because it's not at all obvious
what to do at that point.

509
00:34:30,000 --> 00:34:36,000
Or, much worse,
two pages of frantic

510
00:34:32,000 --> 00:34:38,000
calculations,
and giving up in total despair.

511
00:34:37,000 --> 00:34:43,000
Now, the reason for that is
because you tried to do it

512
00:34:40,000 --> 00:34:46,000
making a direct substitution.
All you have to do instead is

513
00:34:45,000 --> 00:34:51,000
use it, treat it as an inverse
substitution,

514
00:34:48,000 --> 00:34:54,000
write y equals zx.
What's the motivation for doing

515
00:34:51,000 --> 00:34:57,000
that?
It's clear from the equation.

516
00:34:54,000 --> 00:35:00,000
This goes through all of
mathematics.

517
00:34:57,000 --> 00:35:03,000
Whenever you have to change a
variable, excuse me,

518
00:35:00,000 --> 00:35:06,000
whenever you have to change a
variable, look at what you have

519
00:35:05,000 --> 00:35:11,000
to substitute for,
and focus your attention on

520
00:35:08,000 --> 00:35:14,000
that.
I need to know what y prime is.

521
00:35:12,000 --> 00:35:18,000
Okay, well, then I better know
what y is.

522
00:35:15,000 --> 00:35:21,000
If I know what y is,
do I know what y prime is?

523
00:35:19,000 --> 00:35:25,000
Oh, of course.
y prime is z prime x plus z

524
00:35:22,000 --> 00:35:28,000
times the derivative of this
factor, which is one.

525
00:35:26,000 --> 00:35:32,000
And now, I turned with that one

526
00:35:31,000 --> 00:35:37,000
stroke, the equation has now
become z prime x plus z is equal

527
00:35:36,000 --> 00:35:42,000
to F of z.
Well, I don't know.

528
00:35:40,000 --> 00:35:46,000
Can I solve that?
Sure.

529
00:35:42,000 --> 00:35:48,000
That can be solved because this
is x times dz / dx.

530
00:35:48,000 --> 00:35:54,000
Just put the z on the other
side, it's F of z minus z.

531
00:35:53,000 --> 00:35:59,000
And now, this side is just a

532
00:35:55,000 --> 00:36:01,000
function of z.
Separate variables.

533
00:36:00,000 --> 00:36:06,000
And, the only thing to watch
out for is, at the end,

534
00:36:03,000 --> 00:36:09,000
the z was your business.
You've got to put the answer

535
00:36:07,000 --> 00:36:13,000
back in terms of z and y.
Okay, let's work an example of

536
00:36:11,000 --> 00:36:17,000
this.
Since I haven't done any

537
00:36:13,000 --> 00:36:19,000
modeling yet this period,
let's make a little model,

538
00:36:17,000 --> 00:36:23,000
differential equations model.
It's a physical situation,

539
00:36:21,000 --> 00:36:27,000
which will be solved by an
equation.

540
00:36:24,000 --> 00:36:30,000
And, guess what?
The equation will turn out to

541
00:36:27,000 --> 00:36:33,000
be homogeneous.
Okay, so the situation is as

542
00:36:32,000 --> 00:36:38,000
follows.
We are in the Caribbean

543
00:36:34,000 --> 00:36:40,000
somewhere, a little isolated
island somewhere with a little

544
00:36:39,000 --> 00:36:45,000
lighthouse on it at the origin,
and a beam of light shines from

545
00:36:44,000 --> 00:36:50,000
the lighthouse.
The beam of light can rotate

546
00:36:48,000 --> 00:36:54,000
the way the lighthouse beams.
But, this particular beam is

547
00:36:53,000 --> 00:36:59,000
being controlled by a guy in the
lighthouse who can aim it

548
00:36:57,000 --> 00:37:03,000
wherever he wants.
And, the reason he's interested

549
00:37:01,000 --> 00:37:07,000
in aiming it wherever he wants
is there's a drug boat here,

550
00:37:06,000 --> 00:37:12,000
[LAUGHTER] which has just been
caught in the beam of light.

551
00:37:13,000 --> 00:37:19,000
So, the drug boat,
which has just been caught in a

552
00:37:16,000 --> 00:37:22,000
beam of light,
and feels it'd a better escape.

553
00:37:20,000 --> 00:37:26,000
Now, the lighthouse keeper
wants to keep the drug boat;

554
00:37:24,000 --> 00:37:30,000
the light is shining on it so
that the U.S.

555
00:37:27,000 --> 00:37:33,000
Coast Guard helicopters can
zoom over it and do whatever

556
00:37:31,000 --> 00:37:37,000
they do to drug boats,
--

557
00:37:34,000 --> 00:37:40,000
-- I don't know.
So, the drug boat immediately

558
00:37:37,000 --> 00:37:43,000
has to follow an escape
strategy.

559
00:37:39,000 --> 00:37:45,000
And, the only one that occurs
to him is, well,

560
00:37:42,000 --> 00:37:48,000
he wants to go further away,
of course, from the lighthouse.

561
00:37:47,000 --> 00:37:53,000
On the other hand,
it doesn't seem sensible to do

562
00:37:50,000 --> 00:37:56,000
it in a straight line because
the beam will keep shining on

563
00:37:54,000 --> 00:38:00,000
him.
So, he fixes the boat at some

564
00:37:57,000 --> 00:38:03,000
angle, let's say,
and goes off so that the angle

565
00:38:00,000 --> 00:38:06,000
stays 45 degrees.
So, it goes so that the angle

566
00:38:05,000 --> 00:38:11,000
between the beam and maybe,
draw the beam a little less

567
00:38:11,000 --> 00:38:17,000
like a 45 degree angle.
So, the angle between the beam

568
00:38:16,000 --> 00:38:22,000
and the boat,
the boat's path is always 45

569
00:38:20,000 --> 00:38:26,000
degrees, goes at a constant 45
degree angle to the beam,

570
00:38:26,000 --> 00:38:32,000
hoping thereby to escape.
On the other hand,

571
00:38:30,000 --> 00:38:36,000
of course, the lighthouse guy
keeps the beam always on the

572
00:38:36,000 --> 00:38:42,000
boat.
So, it's not clear it's a good

573
00:38:40,000 --> 00:38:46,000
strategy, but this is a
differential equations class.

574
00:38:44,000 --> 00:38:50,000
The question is,
what's the path of the boat?

575
00:38:48,000 --> 00:38:54,000
What's the boat's path?
Now, an obvious question is,

576
00:38:52,000 --> 00:38:58,000
why is this a problem in
differential equations at all?

577
00:38:57,000 --> 00:39:03,000
In other words,
looking at this,

578
00:38:59,000 --> 00:39:05,000
you might scratch your head and
try to think of different ways

579
00:39:04,000 --> 00:39:10,000
to solve it.
But, what suggests that it's

580
00:39:09,000 --> 00:39:15,000
going to be a problem in
differential equations?

581
00:39:13,000 --> 00:39:19,000
The answer is,
you're looking for a path.

582
00:39:17,000 --> 00:39:23,000
The answer is going to be a
curve.

583
00:39:20,000 --> 00:39:26,000
A curve means a function.
We are looking for an unknown

584
00:39:24,000 --> 00:39:30,000
function, in other words.
And, what type of information

585
00:39:29,000 --> 00:39:35,000
do we have about the function?
The only information we have

586
00:39:34,000 --> 00:39:40,000
about the function is something
about its slope,

587
00:39:38,000 --> 00:39:44,000
that its slope makes a constant
45° angle with the lighthouse

588
00:39:44,000 --> 00:39:50,000
beam.
Its slope makes a constant

589
00:39:53,000 --> 00:39:59,000
known angle to a known angle.
Well, if you are trying to find

590
00:40:04,000 --> 00:40:10,000
a function, and all you know is
something about its slope,

591
00:40:09,000 --> 00:40:15,000
that is a problem in
differential equations.

592
00:40:13,000 --> 00:40:19,000
Well, let's try to solve it.
Well, let's see.

593
00:40:16,000 --> 00:40:22,000
Well, let me draw just a little
bit.

594
00:40:19,000 --> 00:40:25,000
So, here's the horizontal.
Let's introduce the

595
00:40:23,000 --> 00:40:29,000
coordinates.
In other words,

596
00:40:25,000 --> 00:40:31,000
there's the horizontal and
here's the boat to indicate

597
00:40:30,000 --> 00:40:36,000
where I am with respect to the
picture.

598
00:40:35,000 --> 00:40:41,000
So, here's the boat.
Here's the beam,

599
00:40:38,000 --> 00:40:44,000
and the path of the boat is
going to make a 45° angle with

600
00:40:44,000 --> 00:40:50,000
it.
So, this is the path that we

601
00:40:47,000 --> 00:40:53,000
are talking about.
And now, let's label what I

602
00:40:51,000 --> 00:40:57,000
know.
Well, this angle is 45°.

603
00:40:54,000 --> 00:41:00,000
This angle, I don't know,
but of course I can calculate

604
00:41:00,000 --> 00:41:06,000
it easily enough because it has
to do with, if I know the

605
00:41:05,000 --> 00:41:11,000
coordinates of this point,
(x, y), then of course that

606
00:41:11,000 --> 00:41:17,000
horizontal angle,
I know the slope of this line,

607
00:41:15,000 --> 00:41:21,000
and that angle will be related
to the slope.

608
00:41:22,000 --> 00:41:28,000
So, let's call this alpha.
And now, what I want to know is

609
00:41:29,000 --> 00:41:35,000
what the slope of the whole path
is.

610
00:41:35,000 --> 00:41:41,000
So, y prime-- let's call y
equals y of x,

611
00:41:42,000 --> 00:41:48,000
the unknown function whose
path, whose graph is going to be

612
00:41:50,000 --> 00:41:56,000
the boat's path,
unknown graph.

613
00:41:54,000 --> 00:42:00,000
What's its slope?
Well, its slope is the tangent

614
00:42:00,000 --> 00:42:06,000
of the sum of these two angles,
alpha plus 45°.

615
00:42:08,000 --> 00:42:14,000
Now, what do I know?
Well, I know that the tangent

616
00:42:11,000 --> 00:42:17,000
of alpha is how much?
That's y over x.

617
00:42:14,000 --> 00:42:20,000
In other words,

618
00:42:16,000 --> 00:42:22,000
if this was the point,
x over y, this is the angle it

619
00:42:19,000 --> 00:42:25,000
makes with a horizontal,
if you think of it over here.

620
00:42:23,000 --> 00:42:29,000
So, this angle is the same as
that one, and it's y over,

621
00:42:26,000 --> 00:42:32,000
its slope of that line is y
over x.

622
00:42:30,000 --> 00:42:36,000
So, the tangent of the angle is
y over x.

623
00:42:32,000 --> 00:42:38,000
How about the tangent of 45°?
That's one, and there's a

624
00:42:36,000 --> 00:42:42,000
formula.
This is the hard part.

625
00:42:38,000 --> 00:42:44,000
All you have to know is that
the formula exists,

626
00:42:41,000 --> 00:42:47,000
and then you will look it up if
you have forgotten it,

627
00:42:44,000 --> 00:42:50,000
relating the tangent or giving
you the tangent of the sum of

628
00:42:48,000 --> 00:42:54,000
two angles, and you can,
if you like,

629
00:42:50,000 --> 00:42:56,000
clever, derive it from the
formula for the sign and cosine

630
00:42:54,000 --> 00:43:00,000
of the sum of two angles.
But, one peak is worth a

631
00:42:57,000 --> 00:43:03,000
thousand finesses.
So, it is the tangent of alpha

632
00:43:02,000 --> 00:43:08,000
plus the tangent of 45°.
Let me read it out in all its

633
00:43:06,000 --> 00:43:12,000
gory details,
divided by one,

634
00:43:08,000 --> 00:43:14,000
so you'll at least learn the
formula, one minus tangent alpha

635
00:43:12,000 --> 00:43:18,000
times tangent 45°.

636
00:43:15,000 --> 00:43:21,000
This would work for the tangent
of the sum of any two angles.

637
00:43:20,000 --> 00:43:26,000
That's the formula.
So, what do I get then?

638
00:43:23,000 --> 00:43:29,000
y prime is equal to the tangent
of alpha, which is y over x,

639
00:43:27,000 --> 00:43:33,000
oh, I like that combination,
plus one, divided by (one minus

640
00:43:32,000 --> 00:43:38,000
y over x times one).

641
00:43:37,000 --> 00:43:43,000
Now, there is no reason for
doing anything to it,

642
00:43:40,000 --> 00:43:46,000
but let's make it look a little
prettier, and thereby,

643
00:43:44,000 --> 00:43:50,000
make it less obvious that it's
a homogeneous equation.

644
00:43:48,000 --> 00:43:54,000
If I multiply top and bottom by
x, it looks prettier.

645
00:43:52,000 --> 00:43:58,000
x plus y over x minus y equals
y prime.

646
00:43:55,000 --> 00:44:01,000
That's our
differential equation.

647
00:44:00,000 --> 00:44:06,000
But, notice,
that let step to make it look

648
00:44:02,000 --> 00:44:08,000
pretty has undone the good work.
It's fine if you immediately

649
00:44:06,000 --> 00:44:12,000
recognize this as being a
homogeneous equation because you

650
00:44:10,000 --> 00:44:16,000
can divide the top and bottom by
x.

651
00:44:12,000 --> 00:44:18,000
But here, it's a lot clearer
that it's a homogeneous equation

652
00:44:16,000 --> 00:44:22,000
because it's already been
written in the right form.

653
00:44:20,000 --> 00:44:26,000
Okay, let's solve it now,
since we know what to do.

654
00:44:23,000 --> 00:44:29,000
We're going to use as the new
variable, z equals y over x.

655
00:44:27,000 --> 00:44:33,000
And, as I wrote up there for y

656
00:44:33,000 --> 00:44:39,000
prime, we'll substitute z prime
x plus z.

657
00:44:40,000 --> 00:44:46,000
And, with that,
let's solve.

658
00:44:44,000 --> 00:44:50,000
Let's solve it.
The equation becomes z prime x

659
00:44:50,000 --> 00:44:56,000
plus z is equal to z plus one
over one minus z.

660
00:44:57,000 --> 00:45:03,000
We want to separate variables,

661
00:45:04,000 --> 00:45:10,000
so you have to put all the z's
on one side.

662
00:45:07,000 --> 00:45:13,000
So, this is going to be x,
dz / dx equals this thing minus

663
00:45:11,000 --> 00:45:17,000
z, which is (z plus one) over
(one minus z) minus z.

664
00:45:14,000 --> 00:45:20,000
And now, as you realize,

665
00:45:18,000 --> 00:45:24,000
putting it on the other side,
I'm going to have to turn it

666
00:45:22,000 --> 00:45:28,000
upside down.
Just as before,

667
00:45:24,000 --> 00:45:30,000
if you have to turn something
upside down, it's better to

668
00:45:28,000 --> 00:45:34,000
combine the terms,
and make it one tiny little

669
00:45:31,000 --> 00:45:37,000
fraction.
Otherwise, you are in for quite

670
00:45:35,000 --> 00:45:41,000
a lot of mess if you don't do
this nicely.

671
00:45:39,000 --> 00:45:45,000
So, z plus one minus z,
that gets rid of the z's.

672
00:45:43,000 --> 00:45:49,000
The numerator is one minus z
squared over one minus z,

673
00:45:48,000 --> 00:45:54,000
I hope, one,
is that right,

674
00:45:50,000 --> 00:45:56,000
(one plus z squared) over (one
minus z).

675
00:45:56,000 --> 00:46:02,000
And so, the question is dz,

676
00:45:58,000 --> 00:46:04,000
and put this on the other side
and turn it upside down.

677
00:46:05,000 --> 00:46:11,000
So, that will be (one minus z)
over (one plus z squared) on the

678
00:46:10,000 --> 00:46:16,000
left-hand side and on the
right-hand side,

679
00:46:14,000 --> 00:46:20,000
dx over x.
Well, that's ready to be

680
00:46:17,000 --> 00:46:23,000
integrated just as it stands.
The right-hand side integrates

681
00:46:23,000 --> 00:46:29,000
to be log x.
The left-hand side is the sum

682
00:46:26,000 --> 00:46:32,000
of two terms.
The integral of one over one

683
00:46:30,000 --> 00:46:36,000
plus z squared is the arc
tangent of z,

684
00:46:34,000 --> 00:46:40,000
maybe?
The derivative of this is one

685
00:46:37,000 --> 00:46:43,000
over one plus z squared.

686
00:46:40,000 --> 00:46:46,000
How about the term z over one
plus z squared?

687
00:46:44,000 --> 00:46:50,000
Well, that integrates to be a

688
00:46:46,000 --> 00:46:52,000
logarithm.
It is more or less the

689
00:46:48,000 --> 00:46:54,000
logarithm of one plus
z squared.

690
00:46:51,000 --> 00:46:57,000
If I differentiate this,
I get one over one plus z^2

691
00:46:54,000 --> 00:47:00,000
times 2z,
but I wish I had negative z

692
00:46:58,000 --> 00:47:04,000
there instead.
Therefore, I should put a minus

693
00:47:01,000 --> 00:47:07,000
sign, and I should multiply that
by half to make it come out

694
00:47:05,000 --> 00:47:11,000
right.
And, this is log x on the right

695
00:47:09,000 --> 00:47:15,000
hand side plus,
put in that arbitrary constant.

696
00:47:13,000 --> 00:47:19,000
And now what?
Well, let's now fool around

697
00:47:16,000 --> 00:47:22,000
with it a little bit.
The arc tangent,

698
00:47:19,000 --> 00:47:25,000
I'm going to simultaneously,
no, two steps.

699
00:47:22,000 --> 00:47:28,000
I have to remember your
innocence, although probably a

700
00:47:26,000 --> 00:47:32,000
lot of you are better
calculators than I am.

701
00:47:31,000 --> 00:47:37,000
I'm going to change this,
use as many laws of logarithms

702
00:47:35,000 --> 00:47:41,000
as possible.
I'm going to put this in the

703
00:47:38,000 --> 00:47:44,000
exponent, and put this on the
other side.

704
00:47:41,000 --> 00:47:47,000
That's going to turn it into
the log of the square root of

705
00:47:45,000 --> 00:47:51,000
one plus z squared.

706
00:47:48,000 --> 00:47:54,000
And, this is going to be plus
the log of x plus c.

707
00:47:53,000 --> 00:47:59,000
And, now I'm going to make,

708
00:47:55,000 --> 00:48:01,000
go back and remember that z
equals y over x.

709
00:48:00,000 --> 00:48:06,000
So, this becomes the arc
tangent of y over x equals.

710
00:48:04,000 --> 00:48:10,000
Now, I combine the logarithms.

711
00:48:09,000 --> 00:48:15,000
This is the log of x times this
square root, right,

712
00:48:12,000 --> 00:48:18,000
make one logarithm out of it,
and then put z equals y over z.

713
00:48:16,000 --> 00:48:22,000
And, you see that if you do

714
00:48:19,000 --> 00:48:25,000
that, it'll be the log of x
times the square root of one

715
00:48:23,000 --> 00:48:29,000
plus (y over x) squared,

716
00:48:27,000 --> 00:48:33,000
and what is that?
Well, if I put this over x

717
00:48:30,000 --> 00:48:36,000
squared and take it out,
it cancels that.

718
00:48:34,000 --> 00:48:40,000
And, what you are left with is
the log of the square root of x

719
00:48:38,000 --> 00:48:44,000
squared plus y squared plus a
constant.

720
00:48:42,000 --> 00:48:48,000
Now, technically,

721
00:48:43,000 --> 00:48:49,000
you have solved the equation,
but not morally because,

722
00:48:47,000 --> 00:48:53,000
I mean, my God,
what a mess!

723
00:48:49,000 --> 00:48:55,000
Incredible path.
It tells me absolutely nothing.

724
00:48:52,000 --> 00:48:58,000
Wow, what is the screaming?
Change me to polar coordinates.

725
00:48:56,000 --> 00:49:02,000
What's the arc tangent of y
over x?

726
00:49:00,000 --> 00:49:06,000
Theta.
In polar coordinates it's

727
00:49:02,000 --> 00:49:08,000
theta.
This is r.

728
00:49:04,000 --> 00:49:10,000
So, the curve is theta equals
the log of r plus a constant.

729
00:49:09,000 --> 00:49:15,000
And, I can make even that
little better if I exponentiate

730
00:49:14,000 --> 00:49:20,000
everything, exponentiate both
sides, combine this in the usual

731
00:49:19,000 --> 00:49:25,000
way, the and what you get is
that r is equal to some other

732
00:49:24,000 --> 00:49:30,000
constant times e to the theta.

733
00:49:30,000 --> 00:49:36,000
That's the curve.
It's called an exponential

734
00:49:33,000 --> 00:49:39,000
spiral, and that's what our
little boat goes in.

735
00:49:37,000 --> 00:49:43,000
And notice, probably if I had
set up the problem in polar

736
00:49:42,000 --> 00:49:48,000
coordinates from the beginning,
nobody would have been able to

737
00:49:48,000 --> 00:49:54,000
solve it.
But, anyone who did would have

738
00:49:51,000 --> 00:49:57,000
gotten that answer immediately.
Thanks.