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PROFESSOR: And this last
little bit is something

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which is not yet on the Web.
 

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But, anyway, when I was walking
out of the room last time, I

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noticed that I'd written down
the wrong formula for c1 - c1.

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There's a misprint, there's
a minus sign that's wrong.

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I claimed last time that
c1 - c2 was + 1/2.

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But, actually, it's - 1/2.
 

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If you go through the
calculation that we did with

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the antiderivative of sine x
cosine x, we get these

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two possible answers.
 

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And if they're to be equal,
then if we just subtract them

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we get c1 - c2 + 1/2 = 0.
 

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So c1 - c2 = 1/2.
 

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So, those are all
of the correction.

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Again, everything here
will be on the Web.

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But just wanted to make
it all clear to you.

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So here we are.
 

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This is our last day of the
second unit, Applications

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of Differentiation.
 

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And I have one of the most fun
topics to introduce to you.

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Which is differential
equations.

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Now, we have a whole course
on differential equations,

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which is called 18.03.
 

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And so we're only going
to do just a little bit.

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But I'm going to teach
you one technique.

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Which fits in precisely with
what we've been doing already.

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Which is differentials.
 

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The first and simplest kind of
differential equation dy/dx

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= some function, f (x).
 

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Now, that's a perfectly good
differential equation.

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And we already discussed last
time that the solution; that

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is, the function y, is going
to be the antiderivative,

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or the integral, of x.
 

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Now, for the purposes of today,
we're going to consider

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this problem to be solved.
 

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That is, you can
always do this.

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You can always take
antiderivatives.

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And for our purposes now, that
is for now, we only have one

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technique to find
antiderivatives.

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And that's called substitution.
 

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It has a very small
variant, which we called

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advanced guessing.
 

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And that works just as well.
 

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And that's basically all that
you'll ever need to do.

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As a practical matter,
these are the ones

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you'll face for now.
 

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Ones that you can actually see
what the answer is, or you'll

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have to make a substitution.
 

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Now, the first tricky example,
or the first maybe interesting

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example of a differential
equation, which I'll call

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Example 2, is going to be the
following. (d / dx + x)y = 0.

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So that's our first
differential equation that

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were going to try to solve.
 

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Apart from this standard
antiderivative approach.

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This operation here has a name.
 

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This actually has a
name, it's called the

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annihilation operator.
 

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And it's called that
in quantum mechanics.

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And there's a corresponding
creation operator where you

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change the sign from
plus to minus.

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And this is one of the simplest
differential equations.

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The reason why it's studied in
quantum mechanics all it that

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it has very simple solutions
that you can just write out.

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So we're going to
solve this equation.

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It's the one that governs
the ground state of the

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harmonic oscillator.
 

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So it has a lot of fancy words
associated with it, but it's a

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fairly simple differential
equation and it works

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perfectly by the method that
we're going to propose.

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So the first step in this
solution is just to rewrite the

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equation by putting one of the
terms on the right-hand side.

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So this is dy / dx = - x y.
 

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Now, here is where you see the
difference between this type of

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equation and the previous type.
 

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In the previous equation, we
just had a function of x

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on the right-hand side.
 

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But here, the rate of change
depends on both x and y.

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So it's not clear at all
that we can solve this

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kind of equation.
 

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But there is a remarkable
trick which works very

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well in this case.
 

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Which is to use multiplication.
 

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To use this idea of
differential that we

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talked about last time.
 

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Namely, we divide by y
and multiply by dx.

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So now we've separated
the equation.

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We've separated out
the differentials.

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And what's going to be
important for us is that the

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left-hand side is expressed
solely in terms of y and the

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right-hand side is expressed
solely in terms of x.

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And we'll go through
this in careful detail.

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So now, the idea is if you've
set up the equation in terms of

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differentials as opposed to
ratios of differentials, or

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rates of change, now I can
use Leibniz's notation and

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integrate these differentials.
 

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Take their antiderivatives.
 

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And we know what
each of these is.

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Namely, the left-hand side is
just - ah. well, that's tough.

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OK.
 

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I had an au pair who actually
did a lot of Tae Kwan Do.

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She could definitely
defeat any of you in any

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encounter, I promise.
 

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OK.
 

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Anyway.
 

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So, let's go back.
 

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We want to take the
antiderivative of this.

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So remember, this is
the function whose

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derivative is 1 / y.
 

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And now there's a
slight novelty here.

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Here we're differentiating the
variables x, and here we're

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differentiating the
variable as y.

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So the antiderivative
here is ln y.

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And the antiderivative on the
other side is - x^2 / 2.

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And they differ by a constant.
 

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So we have this
relationship here.

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Now, that's almost the
end of the story.

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We have to exponentiate to
express y in terms of x.

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So, e ^ ln y = e
^ -x^2 / 2 + c.

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And now I can rewrite that as
y = I'll write as A e ^ -

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x^2 / 2, where a = e ^c.
 

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And incidentally, we're
just taking the case

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y positive here.
 

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We'll talk about what
happens when y is negative

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in a few minutes.
 

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So here's the answer to the
question, almost, except

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for this fact that I
picked out y positive.

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Really, the solution is
y = ae ^ - x^2 / 2.

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Any constant, a; a
positive, negative, or 0.

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Any constant will do.
 

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And we should double-check
that to make sure.

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If you take dy / d x right,
that's going to be a

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d/ dx ( e^ -x ^2 / 2).
 

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And now by the chain rule, you
can see that this is a times

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the factor of - x, that's the
derivative of the exponent,

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with respect to x times
the exponential.

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And now you just
rearrange that.

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That's - x y.
 

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So it does check.
 

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These are solutions
to the question.

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The a didn't matter.
 

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It didn't matter whether it
was positive or negative.

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This function is known as the
normal distribution, so it fits

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beautifully with a lot of
probability and probabilistic

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interpretation of
quantum mechanics.

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This is sort of the,
where the particle is.

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So next, what I'd like to
do is just go through the

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method in general and
point out when it works.

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And then I'll make a few
comments just to make sure

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that you understand the
technicalities of dealing

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with constants and so forth.
 

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So, first of all, the general
method of separation

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of variables.
 

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And here's when it works.
 

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It works when you're faced
with a differential equation

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of the form f (x) g( y).
 

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That's the situation
that we had.

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And I'll just illustrate that.
 

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Just to remind you here.
 

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Here's our equation.
 

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It's in that form.
 

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And the function f (x) =
- x, and the function

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g( y) is just y.
 

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And now, the way the method
works is, this separation step.

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From here to here,
this is the key step.

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This is the only conceptually
remarkable step, which all has

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to do with the fact that
Leibniz fixed his notations up

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so that this works perfectly.
 

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And so that involves taking the
y, so dividing by g( y), and

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multiplying by dx, it's
comfortable because it feels

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like ordinary arithmetic, even
though these are differentials.

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And then, we just
antidifferentiate.

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So we have a function, h, which
is the integral of dy/ g(

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y), and we have another
function which is F.

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Note they are functions
of completely different

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variables here.
 

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Integral of f ( x) dx.
 

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Now, in our example
we did that.

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We carried out this
antidifferentiation, and this

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function turned out to be ln
y, and this function turned

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out to be - x^2 / 2.
 

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And then we write
the relationship.

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Which is that if these are both
antiderivatives of the same

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thing, then they have to
differ by a constant.

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Or, in other words, H ( y)
has to equal to F(x) + c.

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Where c is constant.
 

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Now, notice that this kind
of equation is what we

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call an implicit equation.
 

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It's not quite a formula
for y, directly.

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It defines y implicitly.
 

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That's that top line up here.
 

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That's the implicit equation.
 

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In order to make it an explicit
equation, which is what is

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underneath, what I have to
do is take the inverse.

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So I write it as y =
= H -1( F ( x) + c).

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Now, in real life the calculus
part is often pretty easy.

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And it can be quite messy to
do the inverse operation.

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So sometimes we just leave it
alone in the implicit form.

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But it's also satisfying,
sometimes, to write it

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in the final form here.
 

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00:14:09 --> 00:14:12
Now I've got to give you
a few little pieces

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of commentary before.
 

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For those of you walked in
a little bit late, this

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will all be on the Web.
 

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So just a few pieces
of commentary.

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So if you like, some remarks.
 

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The first remark is that
I could have written ln

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| y| = - x ^2 / 2 + c.
 

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We learned last time that the
antiderivative works also

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00:15:02 --> 00:15:02
for the negative values.
 

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00:15:02 --> 00:15:08
So this would work for
y not equal to 0.

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Both for positive and
negative values.

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And you can see that that would
have captured most of the

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00:15:14 --> 00:15:15
rest of the solution.
 

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Namely, | y| would be = A e
^ - x^2 / 2, by the same

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00:15:22 --> 00:15:24
reasoning as before.
 

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And then that would mean that y
= plus or minus A e^ - x^2 /

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2, which is really
just what we got.

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Because, in fact, I
didn't bother with this.

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00:15:38 --> 00:15:40
Because actually in most - and
the reason why going through

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this, by the way, carefully
this time, is that you're going

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to be faced with this
very frequently.

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00:15:44 --> 00:15:47
The exponential function
comes up all the time.

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00:15:47 --> 00:15:49
And so, therefore, you want
to be completely comfortable

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00:15:49 --> 00:15:52
dealing with it.
 

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00:15:52 --> 00:15:55
So this time I had the positive
a, while the negative a fits in

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00:15:55 --> 00:15:57
either this way, or
I can throw it in.

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Because I know that that's
going to work that way.

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But of course, I double-checked
to be confident.

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00:16:03 --> 00:16:07
Now, this still leaves
out one value.

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00:16:07 --> 00:16:11
So, this still leaves out.
 

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00:16:11 --> 00:16:13
So, if you like, what I have
here now is a is equal to

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00:16:13 --> 00:16:15
plus or minus capital A.
 

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00:16:15 --> 00:16:18
The capital A1
being positive 1.

242
00:16:18 --> 00:16:20
But this still leaves
out one case.

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00:16:20 --> 00:16:23
Which is y = 0.
 

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00:16:23 --> 00:16:27
Which is an extremely boring
solution, but nevertheless a

245
00:16:27 --> 00:16:28
solution to this problem.
 

246
00:16:28 --> 00:16:32
If you plug in 0 here
for y, you get 0.

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00:16:32 --> 00:16:34
If you plug in 0 here for
y, you get that these

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00:16:34 --> 00:16:36
two sides are equal.
 

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00:16:36 --> 00:16:38
0 = 0.
 

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00:16:38 --> 00:16:40
Not a very interesting
answer to the question.

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00:16:40 --> 00:16:42
But it's still an answer.
 

252
00:16:42 --> 00:16:43
And so y = 0 is left out..
 

253
00:16:43 --> 00:16:52
Well, that's not so surprising
that we missed that solution.

254
00:16:52 --> 00:16:55
Because in the process
of carrying out these

255
00:16:55 --> 00:16:58
operations, I divided by y.
 

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00:16:58 --> 00:17:02
I did that right here.
 

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00:17:02 --> 00:17:03
So, that's what happens.
 

258
00:17:03 --> 00:17:06
If you're going to do various
non-linear operations; in

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00:17:06 --> 00:17:08
particular, if you're going to
divide by something, if it

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happens to be 0 you're going
to miss that solution.

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You might have problems
with that solution.

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00:17:13 --> 00:17:16
But we have to live with that
because we want to get ahead.

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00:17:16 --> 00:17:20
And we want to get the formulas
for various solutions.

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So that's the first remark
that I wanted to make.

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00:17:22 --> 00:17:30
And now, the second one is
almost related to what I was

266
00:17:30 --> 00:17:33
just discussing right here.
 

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00:17:33 --> 00:17:37
That I'm erasing.
 

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00:17:37 --> 00:17:39
And that's the following.
 

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00:17:39 --> 00:17:52
I could have also written ln
y + c1 = - x^2 / 2 + c2.

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Where c1 and c2 are
different constants.

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00:17:54 --> 00:17:57
When I'm faced with this
antidifferentiation, I just

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00:17:57 --> 00:18:00
taught you last time, that
you want to have an

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00:18:00 --> 00:18:02
arbitrary constant.
 

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00:18:02 --> 00:18:06
Here and there, in both slots.
 

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00:18:06 --> 00:18:09
So I perfectly well could
have written this down.

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00:18:09 --> 00:18:14
But notice that I can
rewrite this as ln y =

277
00:18:14 --> 00:18:20
- x ^2 / 2 + c2 - c1.
 

278
00:18:20 --> 00:18:22
I can subtract.
 

279
00:18:22 --> 00:18:25
And then, if I just combine
these two guys together and

280
00:18:25 --> 00:18:29
name them c, I have a
different constant.

281
00:18:29 --> 00:18:32
In other words, it's
superfluous and redundant to

282
00:18:32 --> 00:18:35
have two arbitrary constants
here, because they can always

283
00:18:35 --> 00:18:38
be combined into one.
 

284
00:18:38 --> 00:18:47
So two constants
are superfluous.

285
00:18:47 --> 00:18:54
Can always be combined.
 

286
00:18:54 --> 00:18:56
So we just never do
it this first way.

287
00:18:56 --> 00:19:05
It's just extra writing,
it's a waste of time.

288
00:19:05 --> 00:19:07
There's one other subtle
remark, which you won't

289
00:19:07 --> 00:19:10
actually appreciate until
you've done several problems

290
00:19:10 --> 00:19:11
in this direction.
 

291
00:19:11 --> 00:19:17
Which is that the constant
appears additive here, in this

292
00:19:17 --> 00:19:18
first solution to the problem.
 

293
00:19:18 --> 00:19:22
But when I do this nonlinear
operation of exponentiation,

294
00:19:22 --> 00:19:26
it now becomes
multiplicative constant.

295
00:19:26 --> 00:19:31
And so, in general, there's
a free constant somewhere

296
00:19:31 --> 00:19:31
in the problem.
 

297
00:19:31 --> 00:19:35
But it's not always an
additive constant.

298
00:19:35 --> 00:19:38
It's only an additive constant
right at the first step when

299
00:19:38 --> 00:19:39
you take the antiderivative.
 

300
00:19:39 --> 00:19:41
And then after that, when you
do all your other nonlinear

301
00:19:41 --> 00:19:45
operations, it can turn
into anything at all.

302
00:19:45 --> 00:19:47
So you should always expect it
to be something slightly more

303
00:19:47 --> 00:19:49
interesting than an
additive constant.

304
00:19:49 --> 00:19:59
Although occasionally it
stays an additive constant.

305
00:19:59 --> 00:20:01
The last little bit of
commentary that I want to

306
00:20:01 --> 00:20:06
make just goes back to the
original problem here.

307
00:20:06 --> 00:20:09
Which is right here.
 

308
00:20:09 --> 00:20:11
The example 1.
 

309
00:20:11 --> 00:20:14
And I want to solve it, even
though this is simpleminded.

310
00:20:14 --> 00:20:21
But example 1 via separation.
 

311
00:20:21 --> 00:20:25
So that you see our variables.
 

312
00:20:25 --> 00:20:28
So that you see what it does.
 

313
00:20:28 --> 00:20:34
The situation is this.
 

314
00:20:34 --> 00:20:36
And the separation just
means you put the dx

315
00:20:36 --> 00:20:38
on the other side.
 

316
00:20:38 --> 00:20:44
So this is dy = = f(x) dx.
 

317
00:20:44 --> 00:20:54
And then we integrate.
 

318
00:20:54 --> 00:20:58
And the antiderivative
of dy is just y.

319
00:20:58 --> 00:21:03
So this is the solution
to the problem.

320
00:21:03 --> 00:21:05
And it's just what we
wrote before; it's

321
00:21:05 --> 00:21:07
just a funny notation.
 

322
00:21:07 --> 00:21:19
And it comes to the same
thing as the antiderivative.

323
00:21:19 --> 00:21:23
OK, so now we're going to go
on to a trickier problem.

324
00:21:23 --> 00:21:24
A trickier example.
 

325
00:21:24 --> 00:21:26
We need one or two more
just to get some practice

326
00:21:26 --> 00:21:29
with this method.
 

327
00:21:29 --> 00:21:31
Everybody happy so far?
 

328
00:21:31 --> 00:21:32
Question.
 

329
00:21:32 --> 00:21:53
STUDENT: [INAUDIBLE]
 

330
00:21:53 --> 00:21:55
PROFESSOR: So, the question
is, how do we deal

331
00:21:55 --> 00:21:58
with this ambiguity.
 

332
00:21:58 --> 00:22:03
I'm summarizing very, very,
briefly what I heard.

333
00:22:03 --> 00:22:05
Well, you know, sometimes
a > 0, sometimes a <

334
00:22:05 --> 00:22:07
0, sometimes it's not.
 

335
00:22:07 --> 00:22:12
So there's a name for this guy.
 

336
00:22:12 --> 00:22:20
Which is that this is what's
called the general solution.

337
00:22:20 --> 00:22:22
In other words, the whole
family of solutions is the

338
00:22:22 --> 00:22:24
answer to the question.
 

339
00:22:24 --> 00:22:28
Now, it could be that you're
given extra information.

340
00:22:28 --> 00:22:31
If you're given extra
information, that might be, and

341
00:22:31 --> 00:22:34
this is very typical in such
problems, you have the rate of

342
00:22:34 --> 00:22:36
change of the function,
which is what we've given.

343
00:22:36 --> 00:22:39
But you might also have the
place where it starts.

344
00:22:39 --> 00:22:44
Which would be, say,
it starts at 3.

345
00:22:44 --> 00:22:46
Now, if you have that extra
piece of information, then

346
00:22:46 --> 00:22:50
you can nail down exactly
which function it is.

347
00:22:50 --> 00:22:54
If you do that, if you
plug in 3, you see that

348
00:22:54 --> 00:22:57
ae ^ - 0^2 / 2 = 3.
 

349
00:22:57 --> 00:23:00
So a = 3.
 

350
00:23:00 --> 00:23:06
And the answer is y
= 3e ^ -x^2 / 2.

351
00:23:06 --> 00:23:08
And similarly, if it's
negative, if it starts out

352
00:23:08 --> 00:23:10
negative, it'll stay negative.
 

353
00:23:10 --> 00:23:11
For instance.
 

354
00:23:11 --> 00:23:14
If it starts out 0, it'll
stay 0 this particular

355
00:23:14 --> 00:23:16
function here.
 

356
00:23:16 --> 00:23:18
So the answer to your
question is how you deal

357
00:23:18 --> 00:23:19
with the ambiguity.
 

358
00:23:19 --> 00:23:23
The answer is that you simply
say what the solution is.

359
00:23:23 --> 00:23:25
And the solution is not
one function, it's a

360
00:23:25 --> 00:23:26
family of functions.
 

361
00:23:26 --> 00:23:28
It's a list and you have
to have what's known

362
00:23:28 --> 00:23:30
as a parameter.
 

363
00:23:30 --> 00:23:33
And that parameter gets nailed
down if you tell me more

364
00:23:33 --> 00:23:35
information about the function.
 

365
00:23:35 --> 00:23:37
Not the rate of change,
but something about the

366
00:23:37 --> 00:23:38
values of the function.
 

367
00:23:38 --> 00:23:46
 


368
00:23:46 --> 00:23:53
STUDENT: [INAUDIBLE]
 

369
00:23:53 --> 00:23:55
PROFESSOR: The general
solution is this solution.

370
00:23:55 --> 00:23:56
STUDENT: [INAUDIBLE]
 

371
00:23:56 --> 00:23:59
PROFESSOR: And I'm showing you
here that you could get to

372
00:23:59 --> 00:24:00
most of the general solution.
 

373
00:24:00 --> 00:24:04
There's one thing that's left
out, namely the case a = 0.

374
00:24:04 --> 00:24:08
So, in other words, I would
not go through this method.

375
00:24:08 --> 00:24:10
I would only use this,
which is simpler.

376
00:24:10 --> 00:24:13
But then I have to understand
that I haven't gotten all

377
00:24:13 --> 00:24:15
of the solutions this way.
 

378
00:24:15 --> 00:24:19
I'm going to need to throw in
all the rest of the solutions.

379
00:24:19 --> 00:24:21
So in the back of your head,
you always have to have

380
00:24:21 --> 00:24:23
something like this in mind.
 

381
00:24:23 --> 00:24:25
So that you can generate
all the solutions.

382
00:24:25 --> 00:24:28
This is very suggestive, right?
 

383
00:24:28 --> 00:24:31
The restriction, it turns that
the restriction A > 0 is

384
00:24:31 --> 00:24:40
superfluous, is unnecessary.
 

385
00:24:40 --> 00:24:42
But that, we only get
by further thought

386
00:24:42 --> 00:24:46
and by checking.
 

387
00:24:46 --> 00:24:46
Another question?
 

388
00:24:46 --> 00:24:47
Over here.
 

389
00:24:47 --> 00:24:52
STUDENT: [INAUDIBLE]
 

390
00:24:52 --> 00:24:54
PROFESSOR: The aim of
differential equations

391
00:24:54 --> 00:24:55
is to solve them.
 

392
00:24:55 --> 00:24:59
Just as with
algebraic equations.

393
00:24:59 --> 00:25:01
Usually, differential equations
are telling you something

394
00:25:01 --> 00:25:05
about the balance between an
acceleration and a velocity.

395
00:25:05 --> 00:25:09
If you have a falling object,
it might have a resistance.

396
00:25:09 --> 00:25:11
It's telling you something.
 

397
00:25:11 --> 00:25:14
So, actually, sometimes in
applied problems, formulating

398
00:25:14 --> 00:25:16
what differential equation
describe this situation

399
00:25:16 --> 00:25:18
is very important.
 

400
00:25:18 --> 00:25:22
In order to see that that's the
right thing, you have to have

401
00:25:22 --> 00:25:25
solved it to see that it fits
the data that you're getting.

402
00:25:25 --> 00:25:28
STUDENT: [INAUDIBLE]
 

403
00:25:28 --> 00:25:31
PROFESSOR: The question is, can
you solve for x instead of y.

404
00:25:31 --> 00:25:36
The answer is, sure.
 

405
00:25:36 --> 00:25:39
That's the same thing as - so
that would be the inverse

406
00:25:39 --> 00:25:42
function of the function that
we're officially looking for.

407
00:25:42 --> 00:25:43
But yeah, it's legal.
 

408
00:25:43 --> 00:25:46
In other words, oftentimes
we're stuck with just the

409
00:25:46 --> 00:25:49
implicit, some implicit formula
and sometimes we're stuck with

410
00:25:49 --> 00:25:54
a formula x is a function of y
versus y as a function of x.

411
00:25:54 --> 00:25:58
The way in which the function
is specified is something

412
00:25:58 --> 00:26:00
that can be complicated.
 

413
00:26:00 --> 00:26:04
As you'll see in the next
example, it's not necessarily

414
00:26:04 --> 00:26:07
the best thing to think about a
function, y as a function of x.

415
00:26:07 --> 00:26:12
Well, in the fourth example.
 

416
00:26:12 --> 00:26:27
Alright, we're going to go on
and do our next example here.

417
00:26:27 --> 00:26:32
So the third example is
going to be taken as a

418
00:26:32 --> 00:26:36
kind of geometry problem.
 

419
00:26:36 --> 00:26:38
I'll draw a picture of it.
 

420
00:26:38 --> 00:26:44
Suppose you have a curve with
the following property.

421
00:26:44 --> 00:26:50
If you take a point on the
curve, and you take the ray,

422
00:26:50 --> 00:26:55
you take the ray from the
origin to the curve, well,

423
00:26:55 --> 00:26:57
that's not going to
be one that I want.

424
00:26:57 --> 00:27:00
I think I'm going to want
something which is steeper.

425
00:27:00 --> 00:27:02
Because what I'm going to
insist is that the tangent

426
00:27:02 --> 00:27:10
line be twice as steep as
the ray from the origin.

427
00:27:10 --> 00:27:22
So, in other words, slope of
tangent line equals twice

428
00:27:22 --> 00:27:31
slope of ray from origin.
 

429
00:27:31 --> 00:27:34
So the slope of this orange
line is twice the slope

430
00:27:34 --> 00:27:39
of the pink line.
 

431
00:27:39 --> 00:27:41
Now, these kinds of geometric
problems can be written

432
00:27:41 --> 00:27:48
very succinctly with
differential equations.

433
00:27:48 --> 00:27:52
Namely, it's just the
following. dy / dx, that's the

434
00:27:52 --> 00:27:56
slope of the tangent line, is
equal to, well remember what

435
00:27:56 --> 00:28:00
the slope of this ray is, if
this point, I need a notation.

436
00:28:00 --> 00:28:04
At this point is (x, y) which
is a point on the curve.

437
00:28:04 --> 00:28:07
So the slope of this
pink line is what?

438
00:28:07 --> 00:28:09
STUDENT: [INAUDIBLE]
 

439
00:28:09 --> 00:28:12
PROFESSOR: y / x.
 

440
00:28:12 --> 00:28:20
So if it's twice it,
there's the equation.

441
00:28:20 --> 00:28:25
OK, now, we only have
one method for solving

442
00:28:25 --> 00:28:28
these equations.
 

443
00:28:28 --> 00:28:29
So let's use it.
 

444
00:28:29 --> 00:28:31
It says to separate variables.
 

445
00:28:31 --> 00:28:41
So I write dy / y
here = 2 dx / x.

446
00:28:41 --> 00:28:42
That's the basic separation.
 

447
00:28:42 --> 00:28:47
That's the procedure that
we're always going to use.

448
00:28:47 --> 00:28:58
And now if I integrate that, I
find that on the right-hand

449
00:28:58 --> 00:29:03
side I have the logarithm of y.
 

450
00:29:03 --> 00:29:06
And - sorry, on the left-hand
side I have the logarithm of y.

451
00:29:06 --> 00:29:10
On the right-hand side, I
have twice the logarithm

452
00:29:10 --> 00:29:20
of x, plus a constant.
 

453
00:29:20 --> 00:29:27
So let's see what happens
to this example.

454
00:29:27 --> 00:29:30
This is an implicit equation,
and of course we have the

455
00:29:30 --> 00:29:32
problems of the plus or minus
signs, which I'm not going

456
00:29:32 --> 00:29:38
to worry about until later.
 

457
00:29:38 --> 00:29:40
So let's exponentiate
and see what happens.

458
00:29:40 --> 00:29:47
We get e ^ ln y =
e ^ 2 ln x + c.

459
00:29:47 --> 00:29:51
So, again, this is y on
the left-hand side.

460
00:29:51 --> 00:29:54
And on the right-hand side,
if you think about it for a

461
00:29:54 --> 00:29:58
second, it's (e ^ ln x)^2.
 

462
00:29:59 --> 00:30:00
Which is x^2.
 

463
00:30:00 --> 00:30:02
So this is x ^2, and
then there's an e^ c.

464
00:30:02 --> 00:30:06
So that's another one of
these A factors here.

465
00:30:06 --> 00:30:13
A = e^ c.
 

466
00:30:13 --> 00:30:20
So the answer is, well,
I'll draw the picture.

467
00:30:20 --> 00:30:22
And I'm going to cheat
as I did before.

468
00:30:22 --> 00:30:24
We skipped the case y negative.
 

469
00:30:24 --> 00:30:30
We really only did the
case y positive, so far.

470
00:30:30 --> 00:30:32
But if you think about it for a
second, and we'll check it in

471
00:30:32 --> 00:30:36
a second, you're going to get
all of these parabolas here.

472
00:30:36 --> 00:30:40
So the solution is this
family of functions.

473
00:30:40 --> 00:30:44
And they can be bending down.
 

474
00:30:44 --> 00:30:45
As well as up.
 

475
00:30:45 --> 00:30:48
So these are the solutions
to this equation.

476
00:30:48 --> 00:30:50
Every single one of these
curves has the property that if

477
00:30:50 --> 00:30:53
you pick a point on it, the
tangent line has twice the

478
00:30:53 --> 00:30:58
slope of the ray to the origin.
 

479
00:30:58 --> 00:31:01
And the formula, if you like,
of the general solution is y

480
00:31:01 --> 00:31:08
= ax^2, a is any constant.
 

481
00:31:08 --> 00:31:09
Question?
 

482
00:31:09 --> 00:31:21
STUDENT: [INAUDIBLE]
 

483
00:31:21 --> 00:31:22
PROFESSOR: Yeah.
 

484
00:31:22 --> 00:31:29
So again - so first of
all, so there are two

485
00:31:29 --> 00:31:30
approaches to this.
 

486
00:31:30 --> 00:31:32
One is to check it, and
make sure that it's right.

487
00:31:32 --> 00:31:35
When a formula works for some
family of values, sometimes

488
00:31:35 --> 00:31:36
it works for others.
 

489
00:31:36 --> 00:31:39
But another one is to realize
that these things will

490
00:31:39 --> 00:31:40
usually work out this way.
 

491
00:31:40 --> 00:31:45
Because in this argument here,
I allow the absolute value.

492
00:31:45 --> 00:31:47
And that would have
been a perfectly legal

493
00:31:47 --> 00:31:47
thing for me to do.
 

494
00:31:47 --> 00:31:51
I could have put in
absolute values here.

495
00:31:51 --> 00:31:55
In which case, I would've
gotten that the absolute value

496
00:31:55 --> 00:31:56
of this was equal to that.
 

497
00:31:56 --> 00:32:02
And now you see I've covered
the plus and minus cases.

498
00:32:02 --> 00:32:03
So it's that same idea.
 

499
00:32:03 --> 00:32:11
This implies that y = either
+ A x^2 or - Ax^, depending

500
00:32:11 --> 00:32:14
on which sign you pick.
 

501
00:32:14 --> 00:32:21
So that allows me for the
curves above and curves below.

502
00:32:21 --> 00:32:24
Because it's really true
that the antiderivative

503
00:32:24 --> 00:32:26
here is the function.
 

504
00:32:26 --> 00:32:28
It's defined for y negative.
 

505
00:32:28 --> 00:32:33
So let's just double-check.
 

506
00:32:33 --> 00:32:39
In this case, what's happening,
we have y = ax^2 and we want to

507
00:32:39 --> 00:32:44
compute dy / dx to make sure
that it satisfies the equation

508
00:32:44 --> 00:32:46
that I started out with.
 

509
00:32:46 --> 00:32:50
And what I see here
is that this is 2ax.

510
00:32:50 --> 00:32:53
And now I'm going to write
this in a suggestive way.

511
00:32:53 --> 00:33:00
I'm going to write
it as 2ax^2 / x.

512
00:33:00 --> 00:33:06
And, sure enough,
this is 2y / x.

513
00:33:06 --> 00:33:14
It does not matter whether
a, it works for a positive,

514
00:33:14 --> 00:33:17
a negative, a = 0.
 

515
00:33:17 --> 00:33:24
It's OK.
 

516
00:33:24 --> 00:33:29
Again, we didn't pick up by
this method the a = 0 case.

517
00:33:29 --> 00:33:35
And that's not surprising
because we divided by y.

518
00:33:35 --> 00:33:39
There's another thing to watch
out about, about this example.

519
00:33:39 --> 00:33:41
So there's another warning.
 

520
00:33:41 --> 00:33:44
Which I have to give you.
 

521
00:33:44 --> 00:33:48
And this is a subtlety which
you definitely won't get to in

522
00:33:48 --> 00:33:52
any detail until you get to a
higher level ordinary

523
00:33:52 --> 00:33:54
differential equations course,
but I do want to warn

524
00:33:54 --> 00:33:56
you about it right now.
 

525
00:33:56 --> 00:34:06
Which is that if you look at
the equation, you need to

526
00:34:06 --> 00:34:14
watch out that it's
undefined at x = 0.

527
00:34:14 --> 00:34:15
It's undefined at x = 0.
 

528
00:34:15 --> 00:34:20
We also divided by x, and
x is also a problem.

529
00:34:20 --> 00:34:24
Now, that actually has an
important consequence.

530
00:34:24 --> 00:34:27
Which is that, strangely,
knowing the value here and

531
00:34:27 --> 00:34:31
knowing the rate of change
doesn't specify this function.

532
00:34:31 --> 00:34:33
This is bad.
 

533
00:34:33 --> 00:34:36
And it violates one of
our pieces of intuition.

534
00:34:36 --> 00:34:38
And what's going wrong is
that the rate of change

535
00:34:38 --> 00:34:40
was not specified.
 

536
00:34:40 --> 00:34:43
It's undefined at x = 0.
 

537
00:34:43 --> 00:34:46
So there's a problem here, and
in fact if you think carefully

538
00:34:46 --> 00:34:49
about what this function is
doing, it could come in on

539
00:34:49 --> 00:34:56
one branch and leave on a
completely different branch.

540
00:34:56 --> 00:35:01
It doesn't really have to
obey any rule across x = 0.

541
00:35:01 --> 00:35:03
So you should really be
thinking of these things as

542
00:35:03 --> 00:35:05
rays emanating from the origin.
 

543
00:35:05 --> 00:35:10
The origin was a special point
in the whole geometric problem.

544
00:35:10 --> 00:35:15
Rather than as being
complete parabolas.

545
00:35:15 --> 00:35:16
But that's a very subtle point.
 

546
00:35:16 --> 00:35:23
I don't expect you to be able
to say anything about it.

547
00:35:23 --> 00:35:26
But I just want to warn you
that it really is true that

548
00:35:26 --> 00:35:33
when x = 0 there's a problem
for this differential equation.

549
00:35:33 --> 00:35:46
So now, let me say
our next problem.

550
00:35:46 --> 00:35:47
Next example.
 

551
00:35:47 --> 00:35:52
Just another geometry question.
 

552
00:35:52 --> 00:36:01
So here's Example 4.
 

553
00:36:01 --> 00:36:04
I'm just going to use the
example that we've already got.

554
00:36:04 --> 00:36:09
Because there's only so
much time left here.

555
00:36:09 --> 00:36:23
The fourth example is to take
the curves perpendicular

556
00:36:23 --> 00:36:31
to the parabolas.
 

557
00:36:31 --> 00:36:33
This is another
geometry problem.

558
00:36:33 --> 00:36:35
And by specifying that the the
curves are perpendicular to

559
00:36:35 --> 00:36:44
these parabolas, I'm telling
you what their slope is.

560
00:36:44 --> 00:36:47
So let's think about that.
 

561
00:36:47 --> 00:36:48
What's the new equation?
 

562
00:36:48 --> 00:36:56
The new diff. eq.
is the following.

563
00:36:56 --> 00:37:01
It's that the slope is equal to
the negative reciprocal of the

564
00:37:01 --> 00:37:05
slope of the tangent line.
 

565
00:37:05 --> 00:37:14
Of tangent to the parabola.
 

566
00:37:14 --> 00:37:16
So that's the equation.
 

567
00:37:16 --> 00:37:19
That's actually fairly easy
to write down, because

568
00:37:19 --> 00:37:26
it's - 1 / 2(y / x).
 

569
00:37:26 --> 00:37:32
That's the slope
of the parabola.

570
00:37:32 --> 00:37:36
2y / x.
 

571
00:37:36 --> 00:37:38
So let's rewrite that.
 

572
00:37:38 --> 00:37:52
Now, this is the x goes in the
numerator, so it's - x / 2y.

573
00:37:52 --> 00:37:57
And now I want to
solve this one.

574
00:37:57 --> 00:38:01
Well, again, there's
only one technique.

575
00:38:01 --> 00:38:10
Which is we're going to
separate variables.

576
00:38:10 --> 00:38:12
And we separate the
differentials here, so

577
00:38:12 --> 00:38:18
we get 2y dy = - x dx.
 

578
00:38:18 --> 00:38:20
That's just looking at the
equation that I have, which is

579
00:38:20 --> 00:38:26
right over here. dy / dx = - x
/ 2y, and cross-multiplying

580
00:38:26 --> 00:38:30
to get this.
 

581
00:38:30 --> 00:38:33
And now I can take
the antiderivative.

582
00:38:33 --> 00:38:33
This is y^2.
 

583
00:38:35 --> 00:38:40
And the antiderivative over
here is - x^2 / 2 + c.

584
00:38:40 --> 00:38:44
 


585
00:38:44 --> 00:38:57
And so, the solutions are
x^2 / 2 + y^2 = some c.

586
00:38:57 --> 00:39:02
Some constant c.
 

587
00:39:02 --> 00:39:06
Now, this time, things
don't work the same.

588
00:39:06 --> 00:39:09
And you can't expect them
always to work the same.

589
00:39:09 --> 00:39:11
I claimed that this
must be true.

590
00:39:11 --> 00:39:16
But unfortunately I cannot
insist that every c will work.

591
00:39:16 --> 00:39:19
As you can see here, only
the positive numbers

592
00:39:19 --> 00:39:24
c can work here.
 

593
00:39:24 --> 00:39:28
So the picture is that
something slightly

594
00:39:28 --> 00:39:29
different happened here.
 

595
00:39:29 --> 00:39:31
And you have to live with this.
 

596
00:39:31 --> 00:39:33
Is that sometimes not all
the constants will work.

597
00:39:33 --> 00:39:36
Because there's more to
the problem than just

598
00:39:36 --> 00:39:37
the antidifferentiation.
 

599
00:39:37 --> 00:39:40
And here there are
fewer answers rather

600
00:39:40 --> 00:39:40
than more answers.
 

601
00:39:40 --> 00:39:43
In the other case we had to
add in some answers, here

602
00:39:43 --> 00:39:45
we have to take them away.
 

603
00:39:45 --> 00:39:46
Some of them don't
make any sense.

604
00:39:46 --> 00:39:49
And the only ones we can get
are the ones which are of this

605
00:39:49 --> 00:39:54
form, where this is, say,
some radius squared.

606
00:39:54 --> 00:39:55
Well maybe I shouldn't
call it a radius.

607
00:39:55 --> 00:39:56
I'll just call it
a parameter, a^2.

608
00:39:58 --> 00:40:05
And these are of
course ellipses.

609
00:40:05 --> 00:40:11
And you can see that the
ellipses, the length here is

610
00:40:11 --> 00:40:16
the square root of 2a and the
semi-axis, vertical

611
00:40:16 --> 00:40:18
semi-axis, is a.
 

612
00:40:18 --> 00:40:20
So this is the kind of
ellipse that we've got.

613
00:40:20 --> 00:40:24
And I draw it on the previous
diagram, I think it's

614
00:40:24 --> 00:40:28
somewhat suggestive here.
 

615
00:40:28 --> 00:40:30
There, ellipses
are kind of eggs.

616
00:40:30 --> 00:40:32
They're a little bit longer
than they are high.

617
00:40:32 --> 00:40:37
And they go like this.
 

618
00:40:37 --> 00:40:41
And if I drew them pretty
much right, they should

619
00:40:41 --> 00:40:43
be making right angles.
 

620
00:40:43 --> 00:40:49
At all of these places.
 

621
00:40:49 --> 00:40:53
OK, last little bit here.
 

622
00:40:53 --> 00:40:57
Again, you've got to be very
careful with these solutions.

623
00:40:57 --> 00:41:06
And so there's a
warning here too.

624
00:41:06 --> 00:41:10
So let's take a look at, this
is the implicit solution

625
00:41:10 --> 00:41:10
to the equation.
 

626
00:41:10 --> 00:41:13
And this is the one that
tells us what the shape is.

627
00:41:13 --> 00:41:16
But we can also have
the explicit solution.

628
00:41:16 --> 00:41:20
And if I solve for the explicit
solution, it's either y = +

629
00:41:20 --> 00:41:27
square root of a^2 - x^2 / 2,
or y = - square root

630
00:41:27 --> 00:41:32
of a^2 - x^2 / 2.
 

631
00:41:32 --> 00:41:39
These are the
explicit solutions.

632
00:41:39 --> 00:41:41
And now, we notice something
that we should have

633
00:41:41 --> 00:41:43
noticed before.
 

634
00:41:43 --> 00:41:50
Which is that an ellipse
is not a function.

635
00:41:50 --> 00:41:55
It's only the top half, if you
like, that's giving you a

636
00:41:55 --> 00:41:56
solution to this equation.
 

637
00:41:56 --> 00:41:58
Or maybe the bottom half
that's giving it the

638
00:41:58 --> 00:42:00
solution to the equation.
 

639
00:42:00 --> 00:42:07
So the one over here, this
one is the top halves.

640
00:42:07 --> 00:42:15
And this one over here
is the bottom halves.

641
00:42:15 --> 00:42:18
And there's something
else that's interesting.

642
00:42:18 --> 00:42:31
Which is that we have a problem
at y = 0. y = 0 is where

643
00:42:31 --> 00:42:32
x = square root of 2a.
 

644
00:42:33 --> 00:42:35
That's when we get
to this end here.

645
00:42:35 --> 00:42:37
And what happens is
the solution comes

646
00:42:37 --> 00:42:38
around and it stops.
 

647
00:42:38 --> 00:42:44
It has a vertical slope.
 

648
00:42:44 --> 00:42:48
Vertical slope.
 

649
00:42:48 --> 00:42:56
And the solution stops.
 

650
00:42:56 --> 00:43:00
But really, that's
not so unreasonable.

651
00:43:00 --> 00:43:01
After all, look at the formula.
 

652
00:43:01 --> 00:43:03
There was a y in the
denominator here.

653
00:43:03 --> 00:43:08
When y = 0, the slope
should be infinite.

654
00:43:08 --> 00:43:12
And so this equation is
just giving us what it

655
00:43:12 --> 00:43:15
geometrically and intuitively
should be giving us.

656
00:43:15 --> 00:43:22
At that stage.
 

657
00:43:22 --> 00:43:25
So that is the
introduction to ordinary

658
00:43:25 --> 00:43:26
differential equations.
 

659
00:43:26 --> 00:43:30
Again, there's only one
technique which is - we're not

660
00:43:30 --> 00:43:33
done yet, we have a whole four
minutes left and we're

661
00:43:33 --> 00:43:34
going to review.
 

662
00:43:34 --> 00:43:39
Now, so fortunately, this
review is very short.

663
00:43:39 --> 00:43:43
Fortunately for you, I handed
out to you exactly what you're

664
00:43:43 --> 00:43:44
going to be covering
on the test.

665
00:43:44 --> 00:43:48
And it's what's printed here
but there's a whole two

666
00:43:48 --> 00:43:51
pages of discussion.
 

667
00:43:51 --> 00:43:59
And I want to give you
very, very clear-cut

668
00:43:59 --> 00:44:00
instructions here.
 

669
00:44:00 --> 00:44:04
This is usually the hardest
test of this course.

670
00:44:04 --> 00:44:07
People usually do
terribly on it.

671
00:44:07 --> 00:44:13
And I'm going to try to
stop that by making it

672
00:44:13 --> 00:44:14
a little bit easier.
 

673
00:44:14 --> 00:44:17
And now here's what
we're going to do.

674
00:44:17 --> 00:44:21
I'm telling you exactly
what type of problems are

675
00:44:21 --> 00:44:22
going to be on the test.
 

676
00:44:22 --> 00:44:23
These are these six.
 

677
00:44:23 --> 00:44:25
It's also written on your
sheet, your handout.

678
00:44:25 --> 00:44:29
It's also just what was
asked on last year's test.

679
00:44:29 --> 00:44:31
You should go and you should
look at last year's test

680
00:44:31 --> 00:44:33
and see what types of
problems they are.

681
00:44:33 --> 00:44:36
I really, really, am going to
ask the same questions, or

682
00:44:36 --> 00:44:38
the same type of questions.
 

683
00:44:38 --> 00:44:41
Not the same questions.
 

684
00:44:41 --> 00:44:44
So that's what's going
to happen on the test.

685
00:44:44 --> 00:44:49
And let me just tell you,
say one thing, which is the

686
00:44:49 --> 00:44:51
main theme of the class.
 

687
00:44:51 --> 00:44:52
And I will open up.
 

688
00:44:52 --> 00:44:54
We'll have time for one
question after that.

689
00:44:54 --> 00:44:58
The main theme of this unit
is simply the following.

690
00:44:58 --> 00:45:06
That information about
derivative and sometimes maybe

691
00:45:06 --> 00:45:17
the second derivative, tells us
information about f itself.

692
00:45:17 --> 00:45:19
And that's just what were
doing here with ordinary

693
00:45:19 --> 00:45:20
differential equations.
 

694
00:45:20 --> 00:45:22
And that was what we were doing
way at the beginning when

695
00:45:22 --> 00:45:24
we did approximations.
 

696
00:45:24 --> 00:45:24