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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.
13
00:00:32 --> 00:00:35
There's a misprint, there's
a minus sign that's wrong.
14
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I claimed last time that
c1 - c2 was + 1/2.
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00:00:39 --> 00:00:40
But, actually, it's - 1/2.
16
00:00:40 --> 00:00:42
If you go through the
calculation that we did with
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00:00:42 --> 00:00:46
the antiderivative of sine x
cosine x, we get these
18
00:00:46 --> 00:00:48
two possible answers.
19
00:00:48 --> 00:00:52
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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00:00:56 --> 00:01:01
So c1 - c2 = 1/2.
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00:01:01 --> 00:01:03
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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00:01:14 --> 00:01:15
So here we are.
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00:01:15 --> 00:01:19
This is our last day of the
second unit, Applications
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00:01:19 --> 00:01:22
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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00:01:32 --> 00:01:35
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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00:01:58 --> 00:02:05
Which is differentials.
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00:02:05 --> 00:02:13
The first and simplest kind of
differential equation dy/dx
37
00:02:13 --> 00:02:16
= some function, f (x).
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00:02:16 --> 00:02:19
Now, that's a perfectly good
differential equation.
39
00:02:19 --> 00:02:25
And we already discussed last
time that the solution; that
40
00:02:25 --> 00:02:27
is, the function y, is going
to be the antiderivative,
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00:02:27 --> 00:02:33
or the integral, of x.
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00:02:33 --> 00:02:36
Now, for the purposes of today,
we're going to consider
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00:02:36 --> 00:02:40
this problem to be solved.
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00:02:40 --> 00:02:41
That is, you can
always do this.
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00:02:41 --> 00:02:44
You can always take
antiderivatives.
46
00:02:44 --> 00:02:52
And for our purposes now, that
is for now, we only have one
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00:02:52 --> 00:03:08
technique to find
antiderivatives.
48
00:03:08 --> 00:03:15
And that's called substitution.
49
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It has a very small
variant, which we called
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advanced guessing.
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00:03:27 --> 00:03:29
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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00:03:32 --> 00:03:35
As a practical matter,
these are the ones
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00:03:35 --> 00:03:36
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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00:03:38 --> 00:03:42
have to make a substitution.
57
00:03:42 --> 00:03:48
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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00:05:35 --> 00:05:38
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.
93
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Namely, we divide by y
and multiply by dx.
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00:06:14 --> 00:06:17
So now we've separated
the equation.
95
00:06:17 --> 00:06:20
We've separated out
the differentials.
96
00:06:20 --> 00:06:23
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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00:06:27 --> 00:06:30
right-hand side is expressed
solely in terms of x.
99
00:06:30 --> 00:06:33
And we'll go through
this in careful detail.
100
00:06:33 --> 00:06:36
So now, the idea is if you've
set up the equation in terms of
101
00:06:36 --> 00:06:40
differentials as opposed to
ratios of differentials, or
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00:06:40 --> 00:06:44
rates of change, now I can
use Leibniz's notation and
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integrate these differentials.
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00:06:47 --> 00:06:55
Take their antiderivatives.
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And we know what
each of these is.
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00:07:02 --> 00:07:17
Namely, the left-hand side is
just - ah. well, that's tough.
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00:07:17 --> 00:07:24
OK.
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00:07:24 --> 00:07:29
I had an au pair who actually
did a lot of Tae Kwan Do.
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00:07:29 --> 00:07:31
She could definitely
defeat any of you in any
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00:07:31 --> 00:07:35
encounter, I promise.
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00:07:35 --> 00:07:35
OK.
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00:07:35 --> 00:07:37
Anyway.
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00:07:37 --> 00:07:38
So, let's go back.
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00:07:38 --> 00:07:41
We want to take the
antiderivative of this.
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So remember, this is
the function whose
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00:07:48 --> 00:07:50
derivative is 1 / y.
117
00:07:50 --> 00:07:52
And now there's a
slight novelty here.
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00:07:52 --> 00:07:55
Here we're differentiating the
variables x, and here we're
119
00:07:55 --> 00:07:58
differentiating the
variable as y.
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00:07:58 --> 00:08:02
So the antiderivative
here is ln y.
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00:08:02 --> 00:08:07
And the antiderivative on the
other side is - x^2 / 2.
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00:08:07 --> 00:08:10
And they differ by a constant.
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00:08:10 --> 00:08:17
So we have this
relationship here.
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00:08:17 --> 00:08:19
Now, that's almost the
end of the story.
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00:08:19 --> 00:08:23
We have to exponentiate to
express y in terms of x.
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00:08:23 --> 00:08:29
So, e ^ ln y = e
^ -x^2 / 2 + c.
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00:08:29 --> 00:08:37
And now I can rewrite that as
y = I'll write as A e ^ -
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00:08:37 --> 00:08:43
x^2 / 2, where a = e ^c.
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00:08:43 --> 00:08:46
And incidentally, we're
just taking the case
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y positive here.
131
00:08:48 --> 00:08:51
We'll talk about what
happens when y is negative
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in a few minutes.
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00:08:55 --> 00:08:58
So here's the answer to the
question, almost, except
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00:08:58 --> 00:09:02
for this fact that I
picked out y positive.
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00:09:02 --> 00:09:11
Really, the solution is
y = ae ^ - x^2 / 2.
136
00:09:11 --> 00:09:19
Any constant, a; a
positive, negative, or 0.
137
00:09:19 --> 00:09:22
Any constant will do.
138
00:09:22 --> 00:09:24
And we should double-check
that to make sure.
139
00:09:24 --> 00:09:32
If you take dy / d x right,
that's going to be a
140
00:09:32 --> 00:09:36
d/ dx ( e^ -x ^2 / 2).
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00:09:36 --> 00:09:40
And now by the chain rule, you
can see that this is a times
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00:09:40 --> 00:09:44
the factor of - x, that's the
derivative of the exponent,
143
00:09:44 --> 00:09:48
with respect to x times
the exponential.
144
00:09:48 --> 00:09:50
And now you just
rearrange that.
145
00:09:50 --> 00:09:54
That's - x y.
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00:09:54 --> 00:09:55
So it does check.
147
00:09:55 --> 00:09:57
These are solutions
to the question.
148
00:09:57 --> 00:09:58
The a didn't matter.
149
00:09:58 --> 00:10:05
It didn't matter whether it
was positive or negative.
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00:10:05 --> 00:10:09
This function is known as the
normal distribution, so it fits
151
00:10:09 --> 00:10:12
beautifully with a lot of
probability and probabilistic
152
00:10:12 --> 00:10:15
interpretation of
quantum mechanics.
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00:10:15 --> 00:10:22
This is sort of the,
where the particle is.
154
00:10:22 --> 00:10:26
So next, what I'd like to
do is just go through the
155
00:10:26 --> 00:10:30
method in general and
point out when it works.
156
00:10:30 --> 00:10:33
And then I'll make a few
comments just to make sure
157
00:10:33 --> 00:10:37
that you understand the
technicalities of dealing
158
00:10:37 --> 00:10:39
with constants and so forth.
159
00:10:39 --> 00:10:42
So, first of all, the general
method of separation
160
00:10:42 --> 00:10:53
of variables.
161
00:10:53 --> 00:10:55
And here's when it works.
162
00:10:55 --> 00:10:59
It works when you're faced
with a differential equation
163
00:10:59 --> 00:11:03
of the form f (x) g( y).
164
00:11:03 --> 00:11:05
That's the situation
that we had.
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00:11:05 --> 00:11:08
And I'll just illustrate that.
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00:11:08 --> 00:11:09
Just to remind you here.
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00:11:09 --> 00:11:11
Here's our equation.
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00:11:11 --> 00:11:13
It's in that form.
169
00:11:13 --> 00:11:16
And the function f (x) =
- x, and the function
170
00:11:16 --> 00:11:22
g( y) is just y.
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00:11:22 --> 00:11:26
And now, the way the method
works is, this separation step.
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00:11:26 --> 00:11:30
From here to here,
this is the key step.
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00:11:30 --> 00:11:36
This is the only conceptually
remarkable step, which all has
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00:11:36 --> 00:11:39
to do with the fact that
Leibniz fixed his notations up
175
00:11:39 --> 00:11:42
so that this works perfectly.
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00:11:42 --> 00:11:48
And so that involves taking the
y, so dividing by g( y), and
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00:11:48 --> 00:11:53
multiplying by dx, it's
comfortable because it feels
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00:11:53 --> 00:11:59
like ordinary arithmetic, even
though these are differentials.
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00:11:59 --> 00:12:02
And then, we just
antidifferentiate.
180
00:12:02 --> 00:12:09
So we have a function, h, which
is the integral of dy/ g(
181
00:12:09 --> 00:12:13
y), and we have another
function which is F.
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00:12:13 --> 00:12:15
Note they are functions
of completely different
183
00:12:15 --> 00:12:16
variables here.
184
00:12:16 --> 00:12:20
Integral of f ( x) dx.
185
00:12:20 --> 00:12:23
Now, in our example
we did that.
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00:12:23 --> 00:12:26
We carried out this
antidifferentiation, and this
187
00:12:26 --> 00:12:30
function turned out to be ln
y, and this function turned
188
00:12:30 --> 00:12:39
out to be - x^2 / 2.
189
00:12:39 --> 00:12:42
And then we write
the relationship.
190
00:12:42 --> 00:12:46
Which is that if these are both
antiderivatives of the same
191
00:12:46 --> 00:12:48
thing, then they have to
differ by a constant.
192
00:12:48 --> 00:12:55
Or, in other words, H ( y)
has to equal to F(x) + c.
193
00:12:55 --> 00:13:10
Where c is constant.
194
00:13:10 --> 00:13:16
Now, notice that this kind
of equation is what we
195
00:13:16 --> 00:13:20
call an implicit equation.
196
00:13:20 --> 00:13:23
It's not quite a formula
for y, directly.
197
00:13:23 --> 00:13:26
It defines y implicitly.
198
00:13:26 --> 00:13:29
That's that top line up here.
199
00:13:29 --> 00:13:33
That's the implicit equation.
200
00:13:33 --> 00:13:35
In order to make it an explicit
equation, which is what is
201
00:13:35 --> 00:13:38
underneath, what I have to
do is take the inverse.
202
00:13:38 --> 00:13:45
So I write it as y =
= H -1( F ( x) + c).
203
00:13:45 --> 00:13:48
Now, in real life the calculus
part is often pretty easy.
204
00:13:48 --> 00:13:52
And it can be quite messy to
do the inverse operation.
205
00:13:52 --> 00:13:55
So sometimes we just leave it
alone in the implicit form.
206
00:13:55 --> 00:13:58
But it's also satisfying,
sometimes, to write it
207
00:13:58 --> 00:14:09
in the final form here.
208
00:14:09 --> 00:14:12
Now I've got to give you
a few little pieces
209
00:14:12 --> 00:14:14
of commentary before.
210
00:14:14 --> 00:14:16
For those of you walked in
a little bit late, this
211
00:14:16 --> 00:14:25
will all be on the Web.
212
00:14:25 --> 00:14:31
So just a few pieces
of commentary.
213
00:14:31 --> 00:14:36
So if you like, some remarks.
214
00:14:36 --> 00:14:51
The first remark is that
I could have written ln
215
00:14:51 --> 00:14:58
| y| = - x ^2 / 2 + c.
216
00:14:58 --> 00:15:02
We learned last time that the
antiderivative works also
217
00:15:02 --> 00:15:02
for the negative values.
218
00:15:02 --> 00:15:08
So this would work for
y not equal to 0.
219
00:15:08 --> 00:15:10
Both for positive and
negative values.
220
00:15:10 --> 00:15:14
And you can see that that would
have captured most of the
221
00:15:14 --> 00:15:15
rest of the solution.
222
00:15:15 --> 00:15:22
Namely, | y| would be = A e
^ - x^2 / 2, by the same
223
00:15:22 --> 00:15:24
reasoning as before.
224
00:15:24 --> 00:15:30
And then that would mean that y
= plus or minus A e^ - x^2 /
225
00:15:30 --> 00:15:34
2, which is really
just what we got.
226
00:15:34 --> 00:15:38
Because, in fact, I
didn't bother with this.
227
00:15:38 --> 00:15:40
Because actually in most - and
the reason why going through
228
00:15:40 --> 00:15:43
this, by the way, carefully
this time, is that you're going
229
00:15:43 --> 00:15:44
to be faced with this
very frequently.
230
00:15:44 --> 00:15:47
The exponential function
comes up all the time.
231
00:15:47 --> 00:15:49
And so, therefore, you want
to be completely comfortable
232
00:15:49 --> 00:15:52
dealing with it.
233
00:15:52 --> 00:15:55
So this time I had the positive
a, while the negative a fits in
234
00:15:55 --> 00:15:57
either this way, or
I can throw it in.
235
00:15:57 --> 00:15:59
Because I know that that's
going to work that way.
236
00:15:59 --> 00:16:03
But of course, I double-checked
to be confident.
237
00:16:03 --> 00:16:07
Now, this still leaves
out one value.
238
00:16:07 --> 00:16:11
So, this still leaves out.
239
00:16:11 --> 00:16:13
So, if you like, what I have
here now is a is equal to
240
00:16:13 --> 00:16:15
plus or minus capital A.
241
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.
243
00:16:20 --> 00:16:23
Which is y = 0.
244
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.
247
00:16:32 --> 00:16:34
If you plug in 0 here for
y, you get that these
248
00:16:34 --> 00:16:36
two sides are equal.
249
00:16:36 --> 00:16:38
0 = 0.
250
00:16:38 --> 00:16:40
Not a very interesting
answer to the question.
251
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.
256
00:16:58 --> 00:17:02
I did that right here.
257
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
259
00:17:06 --> 00:17:08
particular, if you're going to
divide by something, if it
260
00:17:08 --> 00:17:10
happens to be 0 you're going
to miss that solution.
261
00:17:10 --> 00:17:13
You might have problems
with that solution.
262
00:17:13 --> 00:17:16
But we have to live with that
because we want to get ahead.
263
00:17:16 --> 00:17:20
And we want to get the formulas
for various solutions.
264
00:17:20 --> 00:17:22
So that's the first remark
that I wanted to make.
265
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.
267
00:17:33 --> 00:17:37
That I'm erasing.
268
00:17:37 --> 00:17:39
And that's the following.
269
00:17:39 --> 00:17:52
I could have also written ln
y + c1 = - x^2 / 2 + c2.
270
00:17:52 --> 00:17:54
Where c1 and c2 are
different constants.
271
00:17:54 --> 00:17:57
When I'm faced with this
antidifferentiation, I just
272
00:17:57 --> 00:18:00
taught you last time, that
you want to have an
273
00:18:00 --> 00:18:02
arbitrary constant.
274
00:18:02 --> 00:18:06
Here and there, in both slots.
275
00:18:06 --> 00:18:09
So I perfectly well could
have written this down.
276
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