1
00:00:00 --> 00:00:06
y prime and y double prime.

2
00:00:04 --> 00:00:10
So, q of x times y 
equal zero.

3
00:00:09 --> 00:00:15
The linearity of the equation,
that is, the form in which it

4
00:00:15 --> 00:00:21
appears is going to be the key
idea today.

5
00:00:20 --> 00:00:26
Today is going to be
theoretical, but some of the

6
00:00:25 --> 00:00:31
ideas in it are the most
important in the course.

7
00:00:32 --> 00:00:38
So, I don't have to apologize
for the theory.

8
00:00:37 --> 00:00:43
Remember, the solution method
was to find two independent y

9
00:00:45 --> 00:00:51
one, y two
independent solutions.

10
00:00:51 --> 00:00:57
And now, I'll formally write
out what independent means.

11
00:01:00 --> 00:01:06
There are different ways to say
it.

12
00:01:03 --> 00:01:09
But for you,
I think the simplest and most

13
00:01:07 --> 00:01:13
intelligible will be to say that
y2 is not to be a constant

14
00:01:12 --> 00:01:18
multiple of y1.
And, unfortunately,

15
00:01:15 --> 00:01:21
it's necessary to add,
nor is y1 to be a constant

16
00:01:20 --> 00:01:26
multiple.
I have to call it by different

17
00:01:24 --> 00:01:30
constants.
So, let's call this one c prime

18
00:01:28 --> 00:01:34
of y2.
Well, I mean,

19
00:01:31 --> 00:01:37
the most obvious question is,
well, look.

20
00:01:35 --> 00:01:41
If this is not a constant times
that, this can't be there

21
00:01:41 --> 00:01:47
because I would just use one
over c if it was.

22
00:01:46 --> 00:01:52
Unfortunately,
the reason I have to write it

23
00:01:51 --> 00:01:57
this way is to take account of
the possibility that y1 might be

24
00:01:57 --> 00:02:03
zero.
If y1 is zero,

25
00:01:59 --> 00:02:05
so, the bad case that must be
excluded is that y1 equals zero,

26
00:02:06 --> 00:02:12
y2 nonzero.
I don't want to call those

27
00:02:10 --> 00:02:16
independent.
But nonetheless,

28
00:02:13 --> 00:02:19
it is true that y2 is not a
constant multiple of y1.

29
00:02:16 --> 00:02:22
However, y1 is a constant
multiple of y2,

30
00:02:19 --> 00:02:25
namely, the multiple zero.
It's just to exclude that case

31
00:02:23 --> 00:02:29
that I have to say both of those
things.

32
00:02:26 --> 00:02:32
And, one would not be sufficed.
That's a fine point that I'm

33
00:02:33 --> 00:02:39
not going to fuss over.
But I just have,

34
00:02:37 --> 00:02:43
of course.
Now, why do you do that?

35
00:02:41 --> 00:02:47
That's because,
then, all solutions,

36
00:02:45 --> 00:02:51
and this is what concerns us
today, are what?

37
00:02:50 --> 00:02:56
The linear combination with
constant coefficients of these

38
00:02:56 --> 00:03:02
two, and the fundamental
question we have to answer today

39
00:03:02 --> 00:03:08
is, why?
Now, there are really two

40
00:03:07 --> 00:03:13
statements involved in that.
On the one hand,

41
00:03:13 --> 00:03:19
I'm claiming there is an easier
statement, which is that they

42
00:03:20 --> 00:03:26
are all solutions.
So, that's question one,

43
00:03:24 --> 00:03:30
or statement one.
Why are all of these guys

44
00:03:29 --> 00:03:35
solutions?
That, I could trust you to

45
00:03:33 --> 00:03:39
answer yourself.
I could not trust you to answer

46
00:03:37 --> 00:03:43
it elegantly.
And, it's the elegance that's

47
00:03:39 --> 00:03:45
the most important thing today
because you have to answer it

48
00:03:44 --> 00:03:50
elegantly.
Otherwise, you can't go on and

49
00:03:46 --> 00:03:52
do more complicated things.
If you answer it in an ad hoc

50
00:03:50 --> 00:03:56
basis just by hacking out a
computation, you don't really

51
00:03:54 --> 00:04:00
see what's going on.
And you can't do more difficult

52
00:03:58 --> 00:04:04
things later.
So, we have to answer this,

53
00:04:02 --> 00:04:08
and answer it nicely.
The second question is,

54
00:04:05 --> 00:04:11
so, if that answers why there
are solutions at all,

55
00:04:09 --> 00:04:15
why are they all the solutions?
Why all the solutions?

56
00:04:14 --> 00:04:20
In other words,
to say it as badly as possible,

57
00:04:17 --> 00:04:23
why are all solutions,
why all the solutions-- Never

58
00:04:21 --> 00:04:27
mind.
Why are all the solutions.

59
00:04:24 --> 00:04:30
This is a harder question to
answer, but that should make you

60
00:04:29 --> 00:04:35
happy because that means it
depends upon a theorem which I'm

61
00:04:34 --> 00:04:40
not going to prove.
I'll just quote to you.

62
00:04:40 --> 00:04:46
Let's attack there for problem
one first.

63
00:04:47 --> 00:04:53
q1 is answered by what's called
the superposition.

64
00:04:56 --> 00:05:02
The superposition principle
says exactly that.

65
00:05:05 --> 00:05:11
It says exactly that,
that if y1 and y2 are solutions

66
00:05:09 --> 00:05:15
to a linear equation,
to a linear homogeneous ODE,

67
00:05:12 --> 00:05:18
in fact it can be of higher
order, too, although I won't

68
00:05:17 --> 00:05:23
stress that.
In other words,

69
00:05:19 --> 00:05:25
you don't have to stop with the
second derivative.

70
00:05:23 --> 00:05:29
You could add a third
derivative and a fourth

71
00:05:26 --> 00:05:32
derivative.
As long as the former makes the

72
00:05:30 --> 00:05:36
same, but that implies
automatically that c1 y1 plus c2

73
00:05:34 --> 00:05:40
y2 is a solution.
Now, the way to do that nicely

74
00:05:39 --> 00:05:45
is to take a little detour and
talk a little bit about linear

75
00:05:43 --> 00:05:49
operators.
And, since we are going to be

76
00:05:46 --> 00:05:52
using these for the rest of the
term, this is the natural place

77
00:05:51 --> 00:05:57
for you to learn a little bit
about what they are.

78
00:05:54 --> 00:06:00
So, I'm going to do it.
Ultimately, I am aimed at a

79
00:05:58 --> 00:06:04
proof of this statement,
but there are going to be

80
00:06:02 --> 00:06:08
certain side excursions I have
to make.

81
00:06:06 --> 00:06:12
The first side side excursion
is to write the differential

82
00:06:10 --> 00:06:16
equation in a different way.
So, I'm going to just write its

83
00:06:15 --> 00:06:21
transformations.
The first, I'll simply recopy

84
00:06:19 --> 00:06:25
what you know it to be,
q y equals zero.

85
00:06:22 --> 00:06:28
That's the first form.
The second form,

86
00:06:25 --> 00:06:31
I'm going to replace this by
the differentiation operator.

87
00:06:31 --> 00:06:37
So, I'm going to write this as
D squared y.

88
00:06:35 --> 00:06:41
That means differentiate it
twice.

89
00:06:38 --> 00:06:44
D it, and then D it again.
This one I only have to

90
00:06:42 --> 00:06:48
differentiate once,
so I'll write that as p D(y),

91
00:06:46 --> 00:06:52
p times the derivative of Y.
The last one isn't

92
00:06:50 --> 00:06:56
differentiated at all.
I just recopy it.

93
00:06:53 --> 00:06:59
Now, I'm going to formally
factor out the y.

94
00:06:57 --> 00:07:03
So this, I'm going to turn into
D squared plus pD plus q.

95
00:07:02 --> 00:07:08
Now, everybody reads this as

96
00:07:07 --> 00:07:13
times y equals zero.
But, what it means is this guy,

97
00:07:13 --> 00:07:19
it means this is shorthand for
that.

98
00:07:16 --> 00:07:22
I'm not multiplying.
I'm multiplying q times y.

99
00:07:21 --> 00:07:27
But, I'm not multiplying D
times y.

100
00:07:25 --> 00:07:31
I'm applying D to y.
Nonetheless,

101
00:07:28 --> 00:07:34
the notation suggests this is
very suggestive of that.

102
00:07:32 --> 00:07:38
And this, in turn,
implies that.

103
00:07:35 --> 00:07:41
I'm just transforming it.
And now, I'll take the final

104
00:07:39 --> 00:07:45
step.
I'm going to view this thing in

105
00:07:42 --> 00:07:48
parentheses as a guy all by
itself, a linear operator.

106
00:07:46 --> 00:07:52
This is a linear operator,
called a linear operator.

107
00:07:50 --> 00:07:56
And, I'm going to simply
abbreviate it by the letter L.

108
00:07:54 --> 00:08:00
And so, the final version of
this equation has been reduced

109
00:07:58 --> 00:08:04
to nothing but Ly equals zero.

110
00:08:03 --> 00:08:09
Now, what's L?
You can think of L as,

111
00:08:05 --> 00:08:11
well, formally,
L you would write as D squared

112
00:08:09 --> 00:08:15
plus pD plus q.
But, you can think of L,

113
00:08:13 --> 00:08:19
the way to think of it is as a
black box, a function of what

114
00:08:18 --> 00:08:24
goes into the black box,
well, if this were a function

115
00:08:22 --> 00:08:28
box, what would go it would be a
number, and what would come out

116
00:08:27 --> 00:08:33
with the number.
But it's not that kind of a

117
00:08:31 --> 00:08:37
black box.
It's an operator box,

118
00:08:34 --> 00:08:40
and therefore,
what goes in is a function of

119
00:08:38 --> 00:08:44
x.
And, what comes out is another

120
00:08:41 --> 00:08:47
function of x,
the result of applying this

121
00:08:44 --> 00:08:50
operator to that.
So, from this point of view,

122
00:08:48 --> 00:08:54
differential equations,
trying to solve the

123
00:08:52 --> 00:08:58
differential equation means,
what should come out you want

124
00:08:57 --> 00:09:03
to come out zero,
and the question is,

125
00:09:00 --> 00:09:06
what should you put in?
That's what it means solving

126
00:09:05 --> 00:09:11
differential equations in an
inverse problem.

127
00:09:08 --> 00:09:14
The easy thing is to put it a
function, and see what comes

128
00:09:11 --> 00:09:17
out.
You just calculate.

129
00:09:13 --> 00:09:19
The hard thing is to ask,
you say, I want such and such a

130
00:09:16 --> 00:09:22
thing to come out,
for example,

131
00:09:18 --> 00:09:24
zero; what should I put in?
That's a difficult question,

132
00:09:22 --> 00:09:28
and that's what we're spending
the term answering.

133
00:09:25 --> 00:09:31
Now, the key thing about this
is that this is a linear

134
00:09:28 --> 00:09:34
operator.
And, what that means is that it

135
00:09:32 --> 00:09:38
behaves in a certain way with
respect to functions.

136
00:09:37 --> 00:09:43
The easiest way to say it is,
I like to make two laws of it,

137
00:09:42 --> 00:09:48
that L of u1,
if you have two functions,

138
00:09:45 --> 00:09:51
I'm not going to put up the
parentheses, x,

139
00:09:49 --> 00:09:55
because that just makes things
look longer and not any clearer,

140
00:09:55 --> 00:10:01
actually.
What does L do to the sum of

141
00:09:58 --> 00:10:04
two functions?
If that's a linear operator,

142
00:10:03 --> 00:10:09
if you put in the sum of two
functions, what you must get out

143
00:10:08 --> 00:10:14
is the corresponding L's,
the sum of the corresponding

144
00:10:12 --> 00:10:18
L's of each.
So, that's a law.

145
00:10:14 --> 00:10:20
And, the other law,
linearity law,

146
00:10:17 --> 00:10:23
and this goes for anything in
mathematics and its

147
00:10:21 --> 00:10:27
applications,
which is called linear,

148
00:10:24 --> 00:10:30
basically anything is linear if
it does the following thing to

149
00:10:29 --> 00:10:35
functions or numbers or
whatever.

150
00:10:33 --> 00:10:39
The other one is of a constant
times any function,

151
00:10:37 --> 00:10:43
I don't have to give it a
number now because I'm only

152
00:10:41 --> 00:10:47
using one of them,
should be equal to c times L of

153
00:10:46 --> 00:10:52
u.
So, here, c is a constant,

154
00:10:48 --> 00:10:54
and here, of course,
the u is a function,

155
00:10:51 --> 00:10:57
functions of x.
These are the two laws of

156
00:10:55 --> 00:11:01
linearity.
An operator is linear if it

157
00:10:58 --> 00:11:04
satisfies these two laws.
Now, for example,

158
00:11:03 --> 00:11:09
the differentiation operator is
such an operator.

159
00:11:06 --> 00:11:12
D is linear,
why?

160
00:11:07 --> 00:11:13
Well, because of the very first
things you verify after you

161
00:11:11 --> 00:11:17
learn what the derivative is
because the derivative of,

162
00:11:15 --> 00:11:21
well, I will write it in the D
form.

163
00:11:17 --> 00:11:23
I'll write it in the form in
which you know it.

164
00:11:20 --> 00:11:26
It would be D apply to u1 plus
u2.

165
00:11:23 --> 00:11:29
How does one write that in
ordinary calculus?

166
00:11:26 --> 00:11:32
Well, like that.
Or, maybe you write d by dx out

167
00:11:31 --> 00:11:37
front.
Let's write it this way,

168
00:11:33 --> 00:11:39
is equal to u1 prime plus u2
prime.

169
00:11:36 --> 00:11:42
That's a law.
You prove it when you first

170
00:11:39 --> 00:11:45
study what a derivative is.
It's a property.

171
00:11:42 --> 00:11:48
From our point of view,
it's a property of the

172
00:11:45 --> 00:11:51
differentiation operator.
It has this property.

173
00:11:48 --> 00:11:54
The D of u1 plus u2 is D of u1
plus D of u2.

174
00:11:52 --> 00:11:58
And, it also has the property

175
00:11:55 --> 00:12:01
that c u prime,
you can pull out the constant.

176
00:12:00 --> 00:12:06
That's not affected by the
differentiation.

177
00:12:03 --> 00:12:09
So, these two familiar laws
from the beginning of calculus

178
00:12:07 --> 00:12:13
say, in our language,
that D is a linear operator.

179
00:12:11 --> 00:12:17
What about the multiplication
of law?

180
00:12:14 --> 00:12:20
That's even more important,
that u1 times u2 prime,

181
00:12:18 --> 00:12:24
I have nothing whatever to say
about that in here.

182
00:12:22 --> 00:12:28
In this context,
it's an important law,

183
00:12:25 --> 00:12:31
but it's not important with
respect to the study of

184
00:12:29 --> 00:12:35
linearity.
So, there's an example.

185
00:12:33 --> 00:12:39
Here's a more complicated one
that I'm claiming is the linear

186
00:12:38 --> 00:12:44
operator.
And, since I don't want to have

187
00:12:41 --> 00:12:47
to work in this lecture,
the work is left to you.

188
00:12:45 --> 00:12:51
So, the proof,
prove that L is linear,

189
00:12:48 --> 00:12:54
is this particular operator.
L is linear.

190
00:12:51 --> 00:12:57
That's in your part one
homework to verify that.

191
00:12:55 --> 00:13:01
And you will do some simple
exercises in recitation tomorrow

192
00:12:59 --> 00:13:05
to sort of warm you up for that
if you haven't done it already.

193
00:13:06 --> 00:13:12
Well, you shouldn't have
because this only goes with this

194
00:13:13 --> 00:13:19
lecture, actually.
It's forbidden to work ahead in

195
00:13:19 --> 00:13:25
this class.
All right, where are we?

196
00:13:23 --> 00:13:29
Well, all that was a prelude to
proving this simple theorem,

197
00:13:31 --> 00:13:37
superposition principle.
So, finally,

198
00:13:35 --> 00:13:41
what's the proof?
The proof of the superposition

199
00:13:42 --> 00:13:48
principle: if you believe that
the operator is linear,

200
00:13:50 --> 00:13:56
then L of c1,
in other words,

201
00:13:53 --> 00:13:59
the ODE is L.
L is D squared plus pD plus q.

202
00:13:59 --> 00:14:05
So, the ODE is Ly equals zero.

203
00:14:04 --> 00:14:10
And, what am I being asked to

204
00:14:07 --> 00:14:13
prove?
I'm being asked to prove that

205
00:14:10 --> 00:14:16
if y1 and y2 are solutions,
then so is that thing.

206
00:14:13 --> 00:14:19
By the way, that's called a
linear combination.

207
00:14:16 --> 00:14:22
Put that in your notes.
Maybe I better write it even on

208
00:14:20 --> 00:14:26
the board because it's something
people say all the time without

209
00:14:24 --> 00:14:30
realizing they haven't defined
it.

210
00:14:26 --> 00:14:32
This is called a linear
combination.

211
00:14:30 --> 00:14:36
This expression is called a
linear combination of y1 and y2.

212
00:14:35 --> 00:14:41
It means that particular sum
with constant coefficients.

213
00:14:40 --> 00:14:46
Okay, so, the ODE is Ly equals
zero.

214
00:14:44 --> 00:14:50
And, I'm trying to prove that
fact about it,

215
00:14:48 --> 00:14:54
that if y1 and y2 are
solutions, so is a linear

216
00:14:52 --> 00:14:58
combination of them.
So, the proof,

217
00:14:55 --> 00:15:01
then, I start with apply L to
c1 y1 plus c2 y2.

218
00:15:00 --> 00:15:06
Now, because this operator is

219
00:15:05 --> 00:15:11
linear, it takes the sum of two
functions into the corresponding

220
00:15:10 --> 00:15:16
sum up what the operator would
be.

221
00:15:13 --> 00:15:19
So, it would be L of c1 y1 plus
L of c2 y2.

222
00:15:17 --> 00:15:23
That's because L is a linear

223
00:15:20 --> 00:15:26
operator.
But, I don't have to stop

224
00:15:23 --> 00:15:29
there.
Because L is a linear operator,

225
00:15:26 --> 00:15:32
I can pull the c out front.
So, it's c1 L of y1 plus c2 L

226
00:15:31 --> 00:15:37
of y2. Now, where am I?

227
00:15:36 --> 00:15:42
Trying to prove that this is
zero.

228
00:15:38 --> 00:15:44
Well, what is L of y1?
At this point,

229
00:15:40 --> 00:15:46
I use the fact that y1 is a
solution.

230
00:15:43 --> 00:15:49
Because it's a solution,
this is zero.

231
00:15:46 --> 00:15:52
That's what it means to solve
that differential equation.

232
00:15:50 --> 00:15:56
It means, when you apply the
linear operator,

233
00:15:53 --> 00:15:59
L, to the function,
you get zero.

234
00:15:55 --> 00:16:01
In the same way,
y2 is a solution.

235
00:15:57 --> 00:16:03
So, that's zero.
And, the sum of c1 times zero

236
00:16:02 --> 00:16:08
plus c2 times zero is zero.

237
00:16:05 --> 00:16:11
That's the argument.
Now, you could make the same

238
00:16:08 --> 00:16:14
argument just by plugging c1 y1,
plugging it into the equation

239
00:16:13 --> 00:16:19
and calculating,
and calculating,

240
00:16:15 --> 00:16:21
and calculating,
grouping the terms and so on

241
00:16:18 --> 00:16:24
and so forth.
But, that's just calculation.

242
00:16:21 --> 00:16:27
It doesn't show you why it's
so.

243
00:16:24 --> 00:16:30
Why it's so is because the
operator, this differential

244
00:16:28 --> 00:16:34
equation is expressed as a
linear operator applied to y is

245
00:16:32 --> 00:16:38
zero.
And, the only properties that

246
00:16:37 --> 00:16:43
are really been used as the fact
that this operator is linear.

247
00:16:46 --> 00:16:52
That's the key point.
L is linear.

248
00:16:50 --> 00:16:56
It's a linear operator.
Well, that's all there is to

249
00:16:57 --> 00:17:03
the superposition principle.
As a prelude to answering the

250
00:17:04 --> 00:17:10
more difficult question,
why are these all the

251
00:17:07 --> 00:17:13
solutions?
Why are there no other

252
00:17:09 --> 00:17:15
solutions?
We need a few definitions,

253
00:17:12 --> 00:17:18
and a few more ideas.
And, they are going to occur in

254
00:17:16 --> 00:17:22
connection with,
so I'm now addressing,

255
00:17:18 --> 00:17:24
ultimately, question two.
But, it's not going to be

256
00:17:22 --> 00:17:28
addressed directly for quite
awhile.

257
00:17:25 --> 00:17:31
Instead, I'm going to phrase it
in terms of solving the initial

258
00:17:29 --> 00:17:35
value problem.
So far, we've only talked about

259
00:17:34 --> 00:17:40
the general solution with those
two arbitrary constants.

260
00:17:39 --> 00:17:45
But, how do you solve the
initial value problem,

261
00:17:43 --> 00:17:49
in other words,
fit initial conditions,

262
00:17:46 --> 00:17:52
find the solution with given
initial values for the function

263
00:17:51 --> 00:17:57
and its derivatives.
Now, --

264
00:17:55 --> 00:18:01

265
00:18:04 --> 00:18:10
-- the theorem is that this
collection of functions with

266
00:18:08 --> 00:18:14
these arbitrary constants,
these are all the solutions we

267
00:18:12 --> 00:18:18
have so far.
In fact, they are all the

268
00:18:15 --> 00:18:21
solutions there are,
but we don't know that yet.

269
00:18:19 --> 00:18:25
However, if we just focus on
the big class of solutions,

270
00:18:23 --> 00:18:29
there might be others lurking
out there somewhere lurking out

271
00:18:28 --> 00:18:34
there.
I don't know.

272
00:18:31 --> 00:18:37
But let's use what we have,
that just from this family is

273
00:18:39 --> 00:18:45
enough to satisfy any initial
condition, to satisfy any

274
00:18:47 --> 00:18:53
initial values.
In other words,

275
00:18:51 --> 00:18:57
if you give me any initial
values, I will be able to find

276
00:18:59 --> 00:19:05
the c1 and c2 which work.
Now, why is that?

277
00:19:05 --> 00:19:11
Well, I'd have to do a song and
dance at this point,

278
00:19:08 --> 00:19:14
if you hadn't been softened up
by actually calculating for

279
00:19:12 --> 00:19:18
specific differential equations.
You've had exercises and

280
00:19:15 --> 00:19:21
actually how to calculate the
values of c1 and c2.

281
00:19:19 --> 00:19:25
So, I'm going to do it now in
general what you have done so

282
00:19:23 --> 00:19:29
far for particular equations
using particular values of the

283
00:19:26 --> 00:19:32
initial conditions.
So, I'm relying on that

284
00:19:30 --> 00:19:36
experience that you've had in
doing the homework to make

285
00:19:35 --> 00:19:41
intelligible what I'm going to
do now in the abstract using

286
00:19:40 --> 00:19:46
just letters.
So, why is this so?

287
00:19:43 --> 00:19:49
Why is that so?
Well, we are going to need it,

288
00:19:47 --> 00:19:53
by the way, here,
too.

289
00:19:49 --> 00:19:55
I'll have to,
again, open up parentheses.

290
00:19:52 --> 00:19:58
But let's go as far as we can.
Well, you just try to do it.

291
00:19:57 --> 00:20:03
Suppose the initial conditions
are, how will we write them?

292
00:20:04 --> 00:20:10
So, they're going to be at some
initial point,

293
00:20:06 --> 00:20:12
x0.
You can take it to be zero if

294
00:20:08 --> 00:20:14
you want, but I'd like to be,
just for a little while,

295
00:20:12 --> 00:20:18
a little more general.
So, let's say the initial

296
00:20:15 --> 00:20:21
conditions, the initial values
are being given at the point x0,

297
00:20:19 --> 00:20:25
all right, that's going to be
some number.

298
00:20:21 --> 00:20:27
Let's just call it a.
And, the initial value also has

299
00:20:25 --> 00:20:31
to specify the velocity or the
value of the derivative there.

300
00:20:30 --> 00:20:36
Let's say these are the initial
values.

301
00:20:33 --> 00:20:39
So, the problem is to find the
c which work.

302
00:20:37 --> 00:20:43
Now, how do you do that?
Well, you know from

303
00:20:41 --> 00:20:47
calculation.
You write y equals c1 y1 plus

304
00:20:45 --> 00:20:51
c2 y2.
And you write y prime,

305
00:20:47 --> 00:20:53
and you take the derivative
underneath that,

306
00:20:51 --> 00:20:57
which is easy to do.
And now, you plug in x equals

307
00:20:56 --> 00:21:02
zero.
And, what happens?

308
00:21:00 --> 00:21:06
Well, these now turn into a set
of equations.

309
00:21:04 --> 00:21:10
What will they look like?
Well, y of x0 is a,

310
00:21:08 --> 00:21:14
and this is b.
So, what I get is let me flop

311
00:21:12 --> 00:21:18
it over onto the other side
because that's the way you're

312
00:21:17 --> 00:21:23
used to looking at systems of
equations.

313
00:21:21 --> 00:21:27
So, what we get is c1 times y1
of x0 plus c2 times y2 of x0.

314
00:21:26 --> 00:21:32
What's that supposed to be

315
00:21:31 --> 00:21:37
equal to?
Well, that's supposed to be

316
00:21:34 --> 00:21:40
equal to y of x0.
It's supposed to be equal to

317
00:21:37 --> 00:21:43
the given number,
a.

318
00:21:39 --> 00:21:45
And, in the same way,
c1 y1 prime of x0 plus c2 y2

319
00:21:42 --> 00:21:48
prime of x0, that's supposed to
turn out to 

320
00:21:46 --> 00:21:52
be the number, b.

321
00:21:48 --> 00:21:54
In the calculations you've done
up to this point,

322
00:21:51 --> 00:21:57
y1 and y2 were always specific
functions like e to the x

323
00:21:55 --> 00:22:01
or cosine of 22x,

324
00:21:57 --> 00:22:03
stuff like that.
Now I'm doing it in the

325
00:22:01 --> 00:22:07
abstract, just calling them y1
and y2, so as to include all

326
00:22:06 --> 00:22:12
those possible cases.
Now, what am I supposed to do?

327
00:22:10 --> 00:22:16
I'm supposed to find c1 and c2.
What kind of things are they?

328
00:22:15 --> 00:22:21
This is what you studied in
high school, right?

329
00:22:19 --> 00:22:25
The letters are around us,
but it's a pair of simultaneous

330
00:22:23 --> 00:22:29
linear equations.
What are the variables?

331
00:22:26 --> 00:22:32
What are the variables?
What are the variables?

332
00:22:30 --> 00:22:36
Somebody raise their hand.
If you have a pair of

333
00:22:35 --> 00:22:41
simultaneous linear equations,
you've got variables and you've

334
00:22:39 --> 00:22:45
got constants,
right?

335
00:22:41 --> 00:22:47
And you are trying to find the
answer.

336
00:22:43 --> 00:22:49
What are the variables?
Yeah?

337
00:22:45 --> 00:22:51
c1 and c2.
Very good.

338
00:22:47 --> 00:22:53
I mean, it's extremely
confusing because in the first

339
00:22:50 --> 00:22:56
place, how can they be the
variables if they occur on the

340
00:22:54 --> 00:23:00
wrong side?
They're on the wrong side;

341
00:22:57 --> 00:23:03
they are constants.
How can constants be variables?

342
00:23:02 --> 00:23:08
Everything about this is wrong.
Nonetheless,

343
00:23:06 --> 00:23:12
the c1 and the c2 are the
unknowns, if you like the high

344
00:23:10 --> 00:23:16
school terminology.
c1 and c2 are the unknowns.

345
00:23:14 --> 00:23:20
These messes are just numbers.
After you've plugged in x0,

346
00:23:19 --> 00:23:25
this is some number.
You've got four numbers here.

347
00:23:23 --> 00:23:29
So, c1 and c2 are the
variables.

348
00:23:26 --> 00:23:32
The two find,
in other words,

349
00:23:28 --> 00:23:34
to find the values of.
All right, now you know general

350
00:23:34 --> 00:23:40
theorems from 18.02 about when
can you solve such a system of

351
00:23:39 --> 00:23:45
equations.
I'm claiming that you can

352
00:23:42 --> 00:23:48
always find c1 and c2 that work.
But, you know that's not always

353
00:23:48 --> 00:23:54
the case that a pair of
simultaneous linear equations

354
00:23:52 --> 00:23:58
can be solved.
There's a condition.

355
00:23:55 --> 00:24:01
There's a condition which
guarantees their solution,

356
00:23:59 --> 00:24:05
which is what?
What has to be true about the

357
00:24:04 --> 00:24:10
coefficients?
These are the coefficients.

358
00:24:08 --> 00:24:14
What has to be true?
The matrix of coefficients must

359
00:24:13 --> 00:24:19
be invertible.
The determinant of coefficients

360
00:24:18 --> 00:24:24
must be nonzero.
So, they are solvable if,

361
00:24:22 --> 00:24:28
for the c1 and c2,
if this thing,

362
00:24:25 --> 00:24:31
I'm going to write it.
Since all of these are

363
00:24:29 --> 00:24:35
evaluated at x0,
I'm going to write it in this

364
00:24:34 --> 00:24:40
way.
y1, the determinant,

365
00:24:38 --> 00:24:44
whose entries are y1,
y2, y1 prime,

366
00:24:41 --> 00:24:47
and y2 prime,

367
00:24:44 --> 00:24:50
evaluated at zero,
x0, that means that I evaluate

368
00:24:49 --> 00:24:55
each of the functions in the
determinant at x0.

369
00:24:54 --> 00:25:00
I'll write it this way.
That should be not zero.

370
00:25:00 --> 00:25:06
So, in other words,
the key thing which makes this

371
00:25:03 --> 00:25:09
possible, makes it possible for
us to solve the initial value

372
00:25:08 --> 00:25:14
problem, is that this funny
determinant should not be zero

373
00:25:13 --> 00:25:19
at the point at which we are
interested.

374
00:25:16 --> 00:25:22
Now, this determinant is
important in 18.03.

375
00:25:19 --> 00:25:25
It has a name,
and this is when you're going

376
00:25:23 --> 00:25:29
to learn it, if you don't know
it already.

377
00:25:26 --> 00:25:32
That determinant is called the
Wronskian.

378
00:25:31 --> 00:25:37
The Wronskian of what?
If you want to be pompous,

379
00:25:34 --> 00:25:40
you say this with a V sound
instead of a W.

380
00:25:37 --> 00:25:43
But, nobody does except people
trying to be pompous.

381
00:25:41 --> 00:25:47
The Wronskian,
we'll write a W.

382
00:25:43 --> 00:25:49
Now, notice,
you can only calculate it when

383
00:25:47 --> 00:25:53
you know what the two functions
are.

384
00:25:49 --> 00:25:55
So, the Wronskian of the two
functions, y1 and y2,

385
00:25:53 --> 00:25:59
what's the variable?
It's not a function of two

386
00:25:56 --> 00:26:02
variables, y1 and y2.
These are just the names of

387
00:26:00 --> 00:26:06
functions of x.
So, when you do it,

388
00:26:04 --> 00:26:10
put it in, calculate out that
determinant.

389
00:26:08 --> 00:26:14
This is a function of x,
a function of the independent

390
00:26:13 --> 00:26:19
variable after you've done the
calculation.

391
00:26:17 --> 00:26:23
Anyway, let's write its
definition, y1,

392
00:26:20 --> 00:26:26
y2, y1 prime,
y2 prime.

393
00:26:23 --> 00:26:29
Now, in order to do this,

394
00:26:26 --> 00:26:32
the point is we must know that
that Wronskian is not zero,

395
00:26:32 --> 00:26:38
that the Wronskian of these two
functions is not zero at the

396
00:26:37 --> 00:26:43
point x0.
Now, enter a theorem which

397
00:26:42 --> 00:26:48
you're going to prove for
homework, but this is harder.

398
00:26:47 --> 00:26:53
So, it's part two homework.
It's not part one homework.

399
00:26:51 --> 00:26:57
In other words,
I didn't give you the answer.

400
00:26:55 --> 00:27:01
You've got to find it yourself,
alone or in the company of good

401
00:27:01 --> 00:27:07
friends.
So, anyway, here's the

402
00:27:05 --> 00:27:11
Wronskian.
Now, what can we say for sure?

403
00:27:08 --> 00:27:14
Note, suppose y1 and y,
just to get you a feeling for

404
00:27:13 --> 00:27:19
it a little bit,
suppose they were not

405
00:27:16 --> 00:27:22
independent.
The word for not independent is

406
00:27:20 --> 00:27:26
dependent.
Suppose they were dependent.

407
00:27:24 --> 00:27:30
In other words,
suppose that y2 were a constant

408
00:27:28 --> 00:27:34
multiple of y1.
We know that's not the case

409
00:27:33 --> 00:27:39
because our functions are
supposed to be independent.

410
00:27:37 --> 00:27:43
But suppose they weren't.
What would the value of the

411
00:27:42 --> 00:27:48
Wronskian be?
If y2 is a constant times y1,

412
00:27:46 --> 00:27:52
then y2 prime is that same
constant times y1 prime.

413
00:27:50 --> 00:27:56
What's the value of the
determinant?

414
00:27:53 --> 00:27:59
Zero.
For what values of x is it zero

415
00:27:56 --> 00:28:02
for all values of x?
And now, that's the theorem

416
00:28:01 --> 00:28:07
that you're going to prove,
that if y1 and y2 are solutions

417
00:28:06 --> 00:28:12
to the ODE, I won't keep,
say, it's the ODE we've been

418
00:28:10 --> 00:28:16
talking about,
y double prime plus py.

419
00:28:14 --> 00:28:20
But the linear homogeneous with

420
00:28:20 --> 00:28:26
not constant coefficients,
just linear homogeneous second

421
00:28:26 --> 00:28:32
order.
Our solutions,

422
00:28:29 --> 00:28:35
as there are only two
possibilities,

423
00:28:33 --> 00:28:39
either-or.
Either the Wronskian of y,

424
00:28:37 --> 00:28:43
there are only two
possibilities.

425
00:28:39 --> 00:28:45
Either the Wronskian of y1 and
y2 is always zero,

426
00:28:44 --> 00:28:50
identically zero,
zero for all values of x.

427
00:28:47 --> 00:28:53
This is redundant.
When I say identically,

428
00:28:51 --> 00:28:57
I mean for all values of x.
But, I am just making assurance

429
00:28:56 --> 00:29:02
doubly sure.
Or, or the Wronskian is never

430
00:29:01 --> 00:29:07
zero.
Now, there is no notation as

431
00:29:06 --> 00:29:12
for that.
I'd better just write it out,

432
00:29:10 --> 00:29:16
is never zero,
i.e.

433
00:29:13 --> 00:29:19
for no x is it,
i.e.

434
00:29:15 --> 00:29:21
for all x.
There's no way to say that.

435
00:29:20 --> 00:29:26
I mean, for all values of x,
it's not zero.

436
00:29:26 --> 00:29:32
That means, there is not a
single point for which it's

437
00:29:32 --> 00:29:38
zero.
In particular,

438
00:29:35 --> 00:29:41
it's not zero here.
So, this is your homework:

439
00:29:43 --> 00:29:49
problem five,
part two.

440
00:29:45 --> 00:29:51
I'll give you a method of
proving it, which was discovered

441
00:29:52 --> 00:29:58
by the famous Norwegian
mathematician,

442
00:29:57 --> 00:30:03
Abel, who is,
I guess, the centenary of his

443
00:30:02 --> 00:30:08
birth, I guess,
was just celebrated last year.

444
00:30:09 --> 00:30:15
He has one of the truly tragic
stories in mathematics,

445
00:30:13 --> 00:30:19
which I think you can read.
It must be a Simmons book,

446
00:30:18 --> 00:30:24
if you have that.
Simmons is very good on

447
00:30:22 --> 00:30:28
biographies.
Look up Abel.

448
00:30:24 --> 00:30:30
He'll have a biography of Abel,
and you can weep if you're

449
00:30:29 --> 00:30:35
feeling sad.
He died at the age of 26 of

450
00:30:34 --> 00:30:40
tuberculosis,
having done a number of

451
00:30:37 --> 00:30:43
sensational things,
none of which was recognized in

452
00:30:41 --> 00:30:47
his lifetime because people
buried his papers under big

453
00:30:46 --> 00:30:52
piles of papers.
So, he died unknown,

454
00:30:49 --> 00:30:55
uncelebrated,
and now he's Norway's greatest

455
00:30:53 --> 00:30:59
culture hero.
In the middle of a park in

456
00:30:56 --> 00:31:02
Oslo, there's a huge statue.
And, since nobody knows what

457
00:31:03 --> 00:31:09
Abel looked like,
the statue is way up high so

458
00:31:07 --> 00:31:13
you can't see very well.
But, the inscription on the

459
00:31:13 --> 00:31:19
bottom says Niels Henrik Abel,
1801-1826 or something like

460
00:31:19 --> 00:31:25
that.
Now, --

461
00:31:22 --> 00:31:28

462
00:31:38 --> 00:31:44
-- the choice,
I'm still, believe it or not,

463
00:31:41 --> 00:31:47
aiming at question two,
but I have another big

464
00:31:44 --> 00:31:50
parentheses to open.
And, when I closed it,

465
00:31:47 --> 00:31:53
the answer to question two will
be simple.

466
00:31:50 --> 00:31:56
But, I think it's very
desirable that you get this

467
00:31:53 --> 00:31:59
second big parentheses.
It will help you to understand

468
00:31:57 --> 00:32:03
something important.
It will help you on your

469
00:32:02 --> 00:32:08
problem set tomorrow night.
I don't have to apologize.

470
00:32:08 --> 00:32:14
I'm just going to do it.
So, the question is,

471
00:32:12 --> 00:32:18
the thing you have to
understand is that when I write

472
00:32:18 --> 00:32:24
this combination,
I'm claiming that these are all

473
00:32:23 --> 00:32:29
the solutions.
I haven't proved that yet.

474
00:32:27 --> 00:32:33
But, they are going to be all
the solutions

475
00:32:33 --> 00:32:39
The point is,
there's nothing sacrosanct

476
00:32:36 --> 00:32:42
about the y1 and y2.
This is exactly the same

477
00:32:41 --> 00:32:47
collection as a collection which
I would write using other

478
00:32:46 --> 00:32:52
constants.
Let's call them u1 and u2.

479
00:32:49 --> 00:32:55
They are exactly the same,
where u1 and u2 are any other

480
00:32:55 --> 00:33:01
pair of linearly independent
solutions.

481
00:33:00 --> 00:33:06
Any other pair of independent
solutions, they must be

482
00:33:05 --> 00:33:11
independent, either a constant
multiple of each other.

483
00:33:10 --> 00:33:16
In other words,
u1 is some combination,

484
00:33:13 --> 00:33:19
now I'm really stuck because I
don't know how to,

485
00:33:18 --> 00:33:24
c1 bar, let's say,
that means a special value of

486
00:33:22 --> 00:33:28
c1, and a special value of c2,
and u2 is some other special

487
00:33:28 --> 00:33:34
value, oh my God,
c1 double bar,

488
00:33:31 --> 00:33:37
how's that?
The notation is getting worse

489
00:33:36 --> 00:33:42
and worse.
I apologize for it.

490
00:33:38 --> 00:33:44
In other words,
I could pick y1 and y2 and make

491
00:33:42 --> 00:33:48
up all of these.
And, I'd get a bunch of

492
00:33:45 --> 00:33:51
solutions.
But, I could also pick some

493
00:33:48 --> 00:33:54
other family,
some other two guys in this

494
00:33:51 --> 00:33:57
family, and just as well express
the solutions in terms of u1 and

495
00:33:56 --> 00:34:02
u2.
Now, well, why is he telling us

496
00:33:59 --> 00:34:05
that?
Well, the point is that the y1

497
00:34:03 --> 00:34:09
and the y2 are typically the
ones you get easily from solving

498
00:34:08 --> 00:34:14
the equations,
like e to the x and e

499
00:34:11 --> 00:34:17
to the 2x.
That's what you've gotten,

500
00:34:15 --> 00:34:21
or cosine x and sine x,

501
00:34:18 --> 00:34:24
something like that.
But, for certain ways,

502
00:34:22 --> 00:34:28
they might not be the best way
of writing the solutions.

503
00:34:26 --> 00:34:32
There is another way of writing
those that you should learn,

504
00:34:31 --> 00:34:37
and that's called finding
normalized, the normalized.

505
00:34:35 --> 00:34:41
They are okay,
but they are not normalized.

506
00:34:40 --> 00:34:46
For some things,
the normalized solutions are

507
00:34:42 --> 00:34:48
the best.
I'll explain to you what they

508
00:34:44 --> 00:34:50
are, and I'll explain to you
what they're good for.

509
00:34:48 --> 00:34:54
You'll see immediately what
they're good for.

510
00:34:50 --> 00:34:56
Normalized solutions,
now, you have to specify the

511
00:34:53 --> 00:34:59
point at which you're
normalizing.

512
00:34:55 --> 00:35:01
In general, it would be x
nought,

513
00:34:58 --> 00:35:04
but let's, at this point,
since I don't have an infinity

514
00:35:01 --> 00:35:07
of time, to simplify things,
let's say zero.

515
00:35:05 --> 00:35:11
It could be x nought,
any point would do just as

516
00:35:08 --> 00:35:14
well.
But, zero is the most common

517
00:35:11 --> 00:35:17
choice.
What are the normalized

518
00:35:13 --> 00:35:19
solutions?
Well, first of all,

519
00:35:15 --> 00:35:21
I have to give them names.
I want to still call them y.

520
00:35:19 --> 00:35:25
So, I'll call them capital Y1
and Y2.

521
00:35:21 --> 00:35:27
And, what they are,
are the solutions which satisfy

522
00:35:25 --> 00:35:31
certain, special,
very special,

523
00:35:27 --> 00:35:33
initial conditions.
And, what are those?

524
00:35:31 --> 00:35:37
So, they're the ones which
satisfy, the initial conditions

525
00:35:37 --> 00:35:43
for Y1 are, of course there are
going to be guys that look like

526
00:35:43 --> 00:35:49
this.
The only thing that's going to

527
00:35:46 --> 00:35:52
make them distinctive is the
initial conditions they satisfy.

528
00:35:51 --> 00:35:57
Y1 has to satisfy at zero.
Its value should be one,

529
00:35:56 --> 00:36:02
and the value of its derivative
should be zero.

530
00:36:02 --> 00:36:08
For Y2, it's just the opposite.
Here, the value of the function

531
00:36:07 --> 00:36:13
should be zero at zero.
But, the value of its

532
00:36:11 --> 00:36:17
derivative, now,
I want to be one.

533
00:36:14 --> 00:36:20
Let me give you a trivial
example of this,

534
00:36:17 --> 00:36:23
and then one,
which is a little less trivial,

535
00:36:21 --> 00:36:27
so you'll have some feeling for
what I'm asking for.

536
00:36:25 --> 00:36:31
Suppose the equation,
for example,

537
00:36:28 --> 00:36:34
is y double prime plus,
well, let's really make it

538
00:36:33 --> 00:36:39
simple.
Okay, you know the standard

539
00:36:36 --> 00:36:42
solutions are y1 is cosine x,

540
00:36:39 --> 00:36:45
and y2 is sine x.

541
00:36:41 --> 00:36:47
These are functions,
which, when you take the second

542
00:36:44 --> 00:36:50
derivative, they turn into their
negative.

543
00:36:46 --> 00:36:52
You know, you could go the
complex roots are i and minus i,

544
00:36:50 --> 00:36:56
and blah, blah,
blah, blah, blah,

545
00:36:52 --> 00:36:58
blah.
If you do it that way,

546
00:36:53 --> 00:36:59
fine.
But at some point in the course

547
00:36:56 --> 00:37:02
you have to be able to write
down and, right away,

548
00:36:59 --> 00:37:05
oh, yeah, cosine x,
sine x.

549
00:37:01 --> 00:37:07
Okay, what are the normalized

550
00:37:05 --> 00:37:11
things?
Well, what's the value of this

551
00:37:09 --> 00:37:15
at zero?
It is one.

552
00:37:11 --> 00:37:17
What's the value of its
derivative at zero?

553
00:37:15 --> 00:37:21
Zero.
This is Y1.

554
00:37:16 --> 00:37:22
This is the only case in which
you locked on immediately to the

555
00:37:22 --> 00:37:28
normalized solutions.
In the same way,

556
00:37:26 --> 00:37:32
this guy is Y2 because its
value at zero is zero.

557
00:37:32 --> 00:37:38
It's value of its derivative at
zero is one.

558
00:37:35 --> 00:37:41
So, this is Y2.
Okay, now let's look at a case

559
00:37:39 --> 00:37:45
where you don't immediately lock
on to the normalized solutions.

560
00:37:44 --> 00:37:50
Very simple:
all I have to do is change the

561
00:37:47 --> 00:37:53
sign.
Here, you know,

562
00:37:49 --> 00:37:55
think through r squared minus
one equals zero.

563
00:37:54 --> 00:38:00
The characteristic roots are
plus one and minus one,

564
00:37:58 --> 00:38:04
right?
And therefore,

565
00:38:00 --> 00:38:06
the solution is e to the x,
and e to the negative x.

566
00:38:04 --> 00:38:10
So, the solutions you find by

567
00:38:07 --> 00:38:13
the usual way of solving it is
y1 equals e to the x,

568
00:38:11 --> 00:38:17
and y2 equals e to the
negative x.

569
00:38:15 --> 00:38:21
Those are the standard
solution.

570
00:38:17 --> 00:38:23
So, the general solution is of
the form.

571
00:38:19 --> 00:38:25
So, the general solution is of
the form c1 e to the x plus c2 e

572
00:38:23 --> 00:38:29
to the negative x.

573
00:38:26 --> 00:38:32
Now, what I want to find out is
what is Y1 and Y2?

574
00:38:31 --> 00:38:37
How do I find out what Y1 is?
Well, I have to satisfy initial

575
00:38:36 --> 00:38:42
conditions.
So, if this is y,

576
00:38:38 --> 00:38:44
let's write down here,
if you can still see that,

577
00:38:43 --> 00:38:49
y prime is c1 e to the x minus
c2 e to the negative x .

578
00:38:48 --> 00:38:54
So, if I plug in,

579
00:38:51 --> 00:38:57
I want y of zero to be one,
I want this guy at the point

580
00:38:56 --> 00:39:02
zero to be one.
What equation does that give

581
00:39:00 --> 00:39:06
me?
That gives me c1 plus c2,

582
00:39:04 --> 00:39:10
c1 plus c2,
plugging in x equals zero,

583
00:39:08 --> 00:39:14
equals the value of this thing
at zero.

584
00:39:11 --> 00:39:17
So, that's supposed to be one.

585
00:39:14 --> 00:39:20
How about the other guy?
The value of its derivative is

586
00:39:19 --> 00:39:25
supposed to come out to be zero.
And, what is its derivative?

587
00:39:24 --> 00:39:30
Well, plug into this
expression.

588
00:39:26 --> 00:39:32
It's c1 minus c2.
Okay, what's the solution to

589
00:39:32 --> 00:39:38
those pair of equations?
c2 has to be equal to c1.

590
00:39:36 --> 00:39:42
The sum of the two of them has
to be one.

591
00:39:39 --> 00:39:45
Each one, therefore,
is equal to one half.

592
00:39:42 --> 00:39:48
And so, what's the value of Y1?
Y1, therefore,

593
00:39:46 --> 00:39:52
is the function where c1 and c2
are one half.

594
00:39:50 --> 00:39:56
It's the function e to the x
plus e to the negative x divided

595
00:39:55 --> 00:40:01
by two.

596
00:39:59 --> 00:40:05
In the same way,
I won't repeat the calculation.

597
00:40:03 --> 00:40:09
You can do yourself.
Same calculation shows that Y2,

598
00:40:07 --> 00:40:13
so, put in the initial
conditions.

599
00:40:10 --> 00:40:16
The answer will be that Y2 is
equal to e to the x minus e to

600
00:40:15 --> 00:40:21
the minus x divided by two.

601
00:40:19 --> 00:40:25
These are the special
functions.

602
00:40:22 --> 00:40:28
For this equation,
these are the normalized

603
00:40:25 --> 00:40:31
solutions.
They are better than the

604
00:40:28 --> 00:40:34
original solutions because their
initial values are nicer.

605
00:40:35 --> 00:40:41
Just check it.
The initial value,

606
00:40:37 --> 00:40:43
when x is equal to zero,
the initial value,

607
00:40:40 --> 00:40:46
this has is zero.
Here, when x is equal to zero,

608
00:40:44 --> 00:40:50
the value of the function is
zero.

609
00:40:47 --> 00:40:53
But, the value of its
derivative, these cancel,

610
00:40:51 --> 00:40:57
is one.
So, these are the good guys.

611
00:40:54 --> 00:41:00
Okay, there's no colored chalk
this period.

612
00:40:57 --> 00:41:03
Okay, there was colored chalk.
There's one.

613
00:41:01 --> 00:41:07
So, for this equation,
these are the good guys.

614
00:41:04 --> 00:41:10
These are our best solutions.
e to the x and e to the

615
00:41:07 --> 00:41:13
minus x are good solutions.

616
00:41:10 --> 00:41:16
But, these are our better
solutions.

617
00:41:12 --> 00:41:18
And, this one,
of course, is the function

618
00:41:14 --> 00:41:20
which is called hyperbolic sine
of x, and this is the one which

619
00:41:18 --> 00:41:24
is called hyperbolic cosine of
x.

620
00:41:20 --> 00:41:26
This is one of the most
important ways in which they

621
00:41:23 --> 00:41:29
enter into mathematics.
And, this is why the engineers

622
00:41:26 --> 00:41:32
want them.
Now, why do the engineers want

623
00:41:28 --> 00:41:34
normalized solutions?
Well, I didn't explain that.

624
00:41:34 --> 00:41:40
So, what's so good about
normalized solutions?

625
00:41:41 --> 00:41:47
Very simple:
if Y1 and Y2 are normalized at

626
00:41:47 --> 00:41:53
zero, let's say,
then the solution to the IVP,

627
00:41:53 --> 00:41:59
in other words,
the ODE plus the initial

628
00:41:59 --> 00:42:05
values, y of zero equals,
let's say, a and y

629
00:42:07 --> 00:42:13
prime of zero equals b.

630
00:42:14 --> 00:42:20
So, the ODE I'm not repeating.
It's the one we've been talking

631
00:42:17 --> 00:42:23
about all term since the
beginning of the period.

632
00:42:20 --> 00:42:26
It's the one with the p of x
and q of x.

633
00:42:23 --> 00:42:29
And, here are the initial
values.

634
00:42:25 --> 00:42:31
I'm going to call them a and b.
You can also call them,

635
00:42:28 --> 00:42:34
if you like,
maybe that's better to call

636
00:42:30 --> 00:42:36
them y0, as they are individual
in the homework.

637
00:42:34 --> 00:42:40
They are called,
I'm using the,

638
00:42:36 --> 00:42:42
let's use those.
What is the solution?

639
00:42:39 --> 00:42:45
I say the solution is,
if you use y1 and y2,

640
00:42:43 --> 00:42:49
the solution is y0,
in other words,

641
00:42:46 --> 00:42:52
the a times Y1,
plus y0 prime,

642
00:42:49 --> 00:42:55
in other words,
b times Y2.

643
00:42:53 --> 00:42:59
In other words,
you can write down instantly

644
00:42:57 --> 00:43:03
the solution to the initial
value problem,

645
00:43:00 --> 00:43:06
if instead of using the
functions, you started out with

646
00:43:05 --> 00:43:11
the little Y1 and Y2,
you use these better functions.

647
00:43:12 --> 00:43:18
The thing that's better about
them is that they instantly

648
00:43:16 --> 00:43:22
solve for you the initial value
problem.

649
00:43:18 --> 00:43:24
All you do is use this number,
initial condition as the

650
00:43:22 --> 00:43:28
coefficient of Y1,
and use this number as the

651
00:43:26 --> 00:43:32
coefficient of Y2.
Now, just check that by looking

652
00:43:29 --> 00:43:35
at it.
Why is that so?

653
00:43:32 --> 00:43:38
Well, for example,
let's check.

654
00:43:34 --> 00:43:40
What is its value of this
function at zero?

655
00:43:37 --> 00:43:43
Well, the value of this guy at
zero is one.

656
00:43:40 --> 00:43:46
So, the answer is y0 times one,
and the value of this guy at

657
00:43:44 --> 00:43:50
zero is zero.
So, this term disappears.

658
00:43:47 --> 00:43:53
And, it's exactly the same with
the derivative.

659
00:43:50 --> 00:43:56
What's the value of the
derivative at zero?

660
00:43:54 --> 00:44:00
The value of the derivative of
this thing is zero.

661
00:43:57 --> 00:44:03
So, this term disappears.
The value of this derivative at

662
00:44:01 --> 00:44:07
zero is one. And so, 
the answer is y0 prime.

663
00:44:06 --> 00:44:12
So, check, check,

664
00:44:08 --> 00:44:14
this works.
So, these better solutions have

665
00:44:11 --> 00:44:17
the property,
what's good about them,

666
00:44:14 --> 00:44:20
and why scientists and
engineers like them,

667
00:44:18 --> 00:44:24
is that they enable you
immediately to write down the

668
00:44:22 --> 00:44:28
answer to the initial value
problem without having to go

669
00:44:26 --> 00:44:32
through this business,
which I buried down here,

670
00:44:30 --> 00:44:36
of solving simultaneous linear
equations.

671
00:44:35 --> 00:44:41
Okay, now, believe it or not,
that's all the work.

672
00:44:40 --> 00:44:46
We are ready to answer question
number two: why are these all

673
00:44:46 --> 00:44:52
the solutions?
Of course, I have to invoke a

674
00:44:50 --> 00:44:56
big theorem.
A big theorem:

675
00:44:53 --> 00:44:59
where shall I invoke a big
theorem?

676
00:44:56 --> 00:45:02
Let's see if we can do it here.
The big theorem says,

677
00:45:02 --> 00:45:08
it's called the existence and
uniqueness theorem.

678
00:45:05 --> 00:45:11
It's the last thing that's
proved at the end of an analysis

679
00:45:08 --> 00:45:14
course, at which real analysis
courses, over which students

680
00:45:11 --> 00:45:17
sweat for one whole semester,
and their reward at the end is,

681
00:45:15 --> 00:45:21
if they are very lucky,
and if they have been very good

682
00:45:18 --> 00:45:24
students, they get to see the
proof of the existence and

683
00:45:21 --> 00:45:27
uniqueness theorem for
differential equations.

684
00:45:24 --> 00:45:30
But, I can at least say what it
is for the linear equation

685
00:45:27 --> 00:45:33
because it's so simple.
It says, so,

686
00:45:31 --> 00:45:37
the equation we are talking
about is the usual one,

687
00:45:35 --> 00:45:41
homogeneous equation,
and I'm going to assume,

688
00:45:39 --> 00:45:45
you have to have assumptions
that p and q are continuous for

689
00:45:44 --> 00:45:50
all x.
So, they're good-looking

690
00:45:46 --> 00:45:52
functions.
Coefficients aren't allowed to

691
00:45:50 --> 00:45:56
blow up anywhere.
They've got to look nice.

692
00:45:55 --> 00:46:01
Then, the theorem says there is
one and only one solution,

693
00:46:03 --> 00:46:09
one and only solution
satisfying, given initial values

694
00:46:11 --> 00:46:17
such that y of zero,
let's say y of zero is equal to

695
00:46:19 --> 00:46:25
some given number, A,

696
00:46:24 --> 00:46:30
and y, let's make y0,
and y prime of zero equals B.

697
00:46:31 --> 00:46:37
The initial value problem has

698
00:46:37 --> 00:46:43
one and only one solution.
The existence is,

699
00:46:42 --> 00:46:48
it has a solution.
The uniqueness is,

700
00:46:45 --> 00:46:51
it only has one solution.
If you specify the initial

701
00:46:51 --> 00:46:57
conditions, there's only one
function which satisfies them

702
00:46:56 --> 00:47:02
and at the same time satisfies
that differential equation.

703
00:47:02 --> 00:47:08
Now, this answers our question.
This answers our question,

704
00:47:08 --> 00:47:14
because, look,
what I want is all solutions.

705
00:47:14 --> 00:47:20
What we want are all solutions
to the ODE.

706
00:47:21 --> 00:47:27
And now, here's what I say:
a claim that this collection of

707
00:47:33 --> 00:47:39
functions, c1 Y1 plus c2 Y2
are all

708
00:47:42 --> 00:47:48
solutions.
Of course, I began a period by

709
00:47:48 --> 00:47:54
saying I'd show you that c1
little y1 c little y2 are all

710
00:47:52 --> 00:47:58
the solutions.
But, it's the case that these

711
00:47:55 --> 00:48:01
two families are the same.
So, the family that I started

712
00:48:00 --> 00:48:06
with would be exactly the same
as the family c1 prime Y1

713
00:48:04 --> 00:48:10
because,
after all, these are two

714
00:48:07 --> 00:48:13
special guys from that
collection.

715
00:48:11 --> 00:48:17
So, it doesn't matter whether I
talk about the original ones,

716
00:48:15 --> 00:48:21
or these.
The theorem is still the same.

717
00:48:19 --> 00:48:25
The final step,
therefore, if you give me one

718
00:48:22 --> 00:48:28
more minute, I think that will
be quite enough.

719
00:48:26 --> 00:48:32
Why are these all the
solutions?

720
00:48:30 --> 00:48:36
Well, I have to take an
arbitrary solution and show you

721
00:48:35 --> 00:48:41
that it's one of these.
So, the proof is,

722
00:48:39 --> 00:48:45
given a solution,
u(x), what are its values?

723
00:48:43 --> 00:48:49
Well, u of x zero is u zero,

724
00:48:48 --> 00:48:54
and u prime of x zero, zero,

725
00:48:51 --> 00:48:57
let's say, is equal to u zero,
is equal to some other

726
00:48:58 --> 00:49:04
number.
Now, what's the solution?

727
00:49:02 --> 00:49:08
Write down what's the solution
of these using the Y1's?

728
00:49:07 --> 00:49:13
Then, I know I've just shown
you that u zero times Y1 plus u

729
00:49:12 --> 00:49:18
zero prime Y2
satisfies the same initial

730
00:49:17 --> 00:49:23
conditions, satisfies these
initial conditions,

731
00:49:21 --> 00:49:27
initial values.
In other words,

732
00:49:24 --> 00:49:30
I started with my little
solution.

733
00:49:27 --> 00:49:33
u of x walks up to and
says, hi there.

734
00:49:31 --> 00:49:37
Hi there, and the differential
equation looks at it and says,

735
00:49:37 --> 00:49:43
who are you?
You say, oh,

736
00:49:41 --> 00:49:47
I satisfy you and my initial,
and then it says what are your

737
00:49:46 --> 00:49:52
initial values?
It says, my initial values are

738
00:49:50 --> 00:49:56
u0 and u0 prime.
And, it said,

739
00:49:54 --> 00:50:00
sorry, but we've got one of
ours who satisfies the same

740
00:49:59 --> 00:50:05
initial conditions.
We don't need you because the

741
00:50:03 --> 00:50:09
existence and uniqueness theorem
says that there can only be one

742
00:50:09 --> 00:50:15
function which does that.
And therefore,

743
00:50:13 --> 00:50:19
you must be equal to this guy
by the uniqueness theorem.

744
00:50:20 --> 00:50:26
Okay, we'll talk more about
stuff next time,

745
00:50:23 --> 00:50:29
linear equation next time.