1
00:00:00,000 --> 00:00:06,000
y prime and y double prime.

2
00:00:04,000 --> 00:00:10,000
So, q of x times y
equal zero.

3
00:00:09,000 --> 00:00:15,000
The linearity of the equation,
that is, the form in which it

4
00:00:15,000 --> 00:00:21,000
appears is going to be the key
idea today.

5
00:00:20,000 --> 00:00:26,000
Today is going to be
theoretical, but some of the

6
00:00:25,000 --> 00:00:31,000
ideas in it are the most
important in the course.

7
00:00:32,000 --> 00:00:38,000
So, I don't have to apologize
for the theory.

8
00:00:37,000 --> 00:00:43,000
Remember, the solution method
was to find two independent y

9
00:00:45,000 --> 00:00:51,000
one, y two
independent solutions.

10
00:00:51,000 --> 00:00:57,000
And now, I'll formally write
out what independent means.

11
00:01:00,000 --> 00:01:06,000
There are different ways to say
it.

12
00:01:03,000 --> 00:01:09,000
But for you,
I think the simplest and most

13
00:01:07,000 --> 00:01:13,000
intelligible will be to say that
y2 is not to be a constant

14
00:01:12,000 --> 00:01:18,000
multiple of y1.
And, unfortunately,

15
00:01:15,000 --> 00:01:21,000
it's necessary to add,
nor is y1 to be a constant

16
00:01:20,000 --> 00:01:26,000
multiple.
I have to call it by different

17
00:01:24,000 --> 00:01:30,000
constants.
So, let's call this one c prime

18
00:01:28,000 --> 00:01:34,000
of y2.
Well, I mean,

19
00:01:31,000 --> 00:01:37,000
the most obvious question is,
well, look.

20
00:01:35,000 --> 00:01:41,000
If this is not a constant times
that, this can't be there

21
00:01:41,000 --> 00:01:47,000
because I would just use one
over c if it was.

22
00:01:46,000 --> 00:01:52,000
Unfortunately,
the reason I have to write it

23
00:01:51,000 --> 00:01:57,000
this way is to take account of
the possibility that y1 might be

24
00:01:57,000 --> 00:02:03,000
zero.
If y1 is zero,

25
00:01:59,000 --> 00:02:05,000
so, the bad case that must be
excluded is that y1 equals zero,

26
00:02:06,000 --> 00:02:12,000
y2 nonzero.
I don't want to call those

27
00:02:10,000 --> 00:02:16,000
independent.
But nonetheless,

28
00:02:13,000 --> 00:02:19,000
it is true that y2 is not a
constant multiple of y1.

29
00:02:16,000 --> 00:02:22,000
However, y1 is a constant
multiple of y2,

30
00:02:19,000 --> 00:02:25,000
namely, the multiple zero.
It's just to exclude that case

31
00:02:23,000 --> 00:02:29,000
that I have to say both of those
things.

32
00:02:26,000 --> 00:02:32,000
And, one would not be sufficed.
That's a fine point that I'm

33
00:02:33,000 --> 00:02:39,000
not going to fuss over.
But I just have,

34
00:02:37,000 --> 00:02:43,000
of course.
Now, why do you do that?

35
00:02:41,000 --> 00:02:47,000
That's because,
then, all solutions,

36
00:02:45,000 --> 00:02:51,000
and this is what concerns us
today, are what?

37
00:02:50,000 --> 00:02:56,000
The linear combination with
constant coefficients of these

38
00:02:56,000 --> 00:03:02,000
two, and the fundamental
question we have to answer today

39
00:03:02,000 --> 00:03:08,000
is, why?
Now, there are really two

40
00:03:07,000 --> 00:03:13,000
statements involved in that.
On the one hand,

41
00:03:13,000 --> 00:03:19,000
I'm claiming there is an easier
statement, which is that they

42
00:03:20,000 --> 00:03:26,000
are all solutions.
So, that's question one,

43
00:03:24,000 --> 00:03:30,000
or statement one.
Why are all of these guys

44
00:03:29,000 --> 00:03:35,000
solutions?
That, I could trust you to

45
00:03:33,000 --> 00:03:39,000
answer yourself.
I could not trust you to answer

46
00:03:37,000 --> 00:03:43,000
it elegantly.
And, it's the elegance that's

47
00:03:39,000 --> 00:03:45,000
the most important thing today
because you have to answer it

48
00:03:44,000 --> 00:03:50,000
elegantly.
Otherwise, you can't go on and

49
00:03:46,000 --> 00:03:52,000
do more complicated things.
If you answer it in an ad hoc

50
00:03:50,000 --> 00:03:56,000
basis just by hacking out a
computation, you don't really

51
00:03:54,000 --> 00:04:00,000
see what's going on.
And you can't do more difficult

52
00:03:58,000 --> 00:04:04,000
things later.
So, we have to answer this,

53
00:04:02,000 --> 00:04:08,000
and answer it nicely.
The second question is,

54
00:04:05,000 --> 00:04:11,000
so, if that answers why there
are solutions at all,

55
00:04:09,000 --> 00:04:15,000
why are they all the solutions?
Why all the solutions?

56
00:04:14,000 --> 00:04:20,000
In other words,
to say it as badly as possible,

57
00:04:17,000 --> 00:04:23,000
why are all solutions,
why all the solutions-- Never

58
00:04:21,000 --> 00:04:27,000
mind.
Why are all the solutions.

59
00:04:24,000 --> 00:04:30,000
This is a harder question to
answer, but that should make you

60
00:04:29,000 --> 00:04:35,000
happy because that means it
depends upon a theorem which I'm

61
00:04:34,000 --> 00:04:40,000
not going to prove.
I'll just quote to you.

62
00:04:40,000 --> 00:04:46,000
Let's attack there for problem
one first.

63
00:04:47,000 --> 00:04:53,000
q1 is answered by what's called
the superposition.

64
00:04:56,000 --> 00:05:02,000
The superposition principle
says exactly that.

65
00:05:05,000 --> 00:05:11,000
It says exactly that,
that if y1 and y2 are solutions

66
00:05:09,000 --> 00:05:15,000
to a linear equation,
to a linear homogeneous ODE,

67
00:05:12,000 --> 00:05:18,000
in fact it can be of higher
order, too, although I won't

68
00:05:17,000 --> 00:05:23,000
stress that.
In other words,

69
00:05:19,000 --> 00:05:25,000
you don't have to stop with the
second derivative.

70
00:05:23,000 --> 00:05:29,000
You could add a third
derivative and a fourth

71
00:05:26,000 --> 00:05:32,000
derivative.
As long as the former makes the

72
00:05:30,000 --> 00:05:36,000
same, but that implies
automatically that c1 y1 plus c2

73
00:05:34,000 --> 00:05:40,000
y2 is a solution.
Now, the way to do that nicely

74
00:05:39,000 --> 00:05:45,000
is to take a little detour and
talk a little bit about linear

75
00:05:43,000 --> 00:05:49,000
operators.
And, since we are going to be

76
00:05:46,000 --> 00:05:52,000
using these for the rest of the
term, this is the natural place

77
00:05:51,000 --> 00:05:57,000
for you to learn a little bit
about what they are.

78
00:05:54,000 --> 00:06:00,000
So, I'm going to do it.
Ultimately, I am aimed at a

79
00:05:58,000 --> 00:06:04,000
proof of this statement,
but there are going to be

80
00:06:02,000 --> 00:06:08,000
certain side excursions I have
to make.

81
00:06:06,000 --> 00:06:12,000
The first side side excursion
is to write the differential

82
00:06:10,000 --> 00:06:16,000
equation in a different way.
So, I'm going to just write its

83
00:06:15,000 --> 00:06:21,000
transformations.
The first, I'll simply recopy

84
00:06:19,000 --> 00:06:25,000
what you know it to be,
q y equals zero.

85
00:06:22,000 --> 00:06:28,000
That's the first form.
The second form,

86
00:06:25,000 --> 00:06:31,000
I'm going to replace this by
the differentiation operator.

87
00:06:31,000 --> 00:06:37,000
So, I'm going to write this as
D squared y.

88
00:06:35,000 --> 00:06:41,000
That means differentiate it
twice.

89
00:06:38,000 --> 00:06:44,000
D it, and then D it again.
This one I only have to

90
00:06:42,000 --> 00:06:48,000
differentiate once,
so I'll write that as p D(y),

91
00:06:46,000 --> 00:06:52,000
p times the derivative of Y.
The last one isn't

92
00:06:50,000 --> 00:06:56,000
differentiated at all.
I just recopy it.

93
00:06:53,000 --> 00:06:59,000
Now, I'm going to formally
factor out the y.

94
00:06:57,000 --> 00:07:03,000
So this, I'm going to turn into
D squared plus pD plus q.

95
00:07:02,000 --> 00:07:08,000
Now, everybody reads this as

96
00:07:07,000 --> 00:07:13,000
times y equals zero.
But, what it means is this guy,

97
00:07:13,000 --> 00:07:19,000
it means this is shorthand for
that.

98
00:07:16,000 --> 00:07:22,000
I'm not multiplying.
I'm multiplying q times y.

99
00:07:21,000 --> 00:07:27,000
But, I'm not multiplying D
times y.

100
00:07:25,000 --> 00:07:31,000
I'm applying D to y.
Nonetheless,

101
00:07:28,000 --> 00:07:34,000
the notation suggests this is
very suggestive of that.

102
00:07:32,000 --> 00:07:38,000
And this, in turn,
implies that.

103
00:07:35,000 --> 00:07:41,000
I'm just transforming it.
And now, I'll take the final

104
00:07:39,000 --> 00:07:45,000
step.
I'm going to view this thing in

105
00:07:42,000 --> 00:07:48,000
parentheses as a guy all by
itself, a linear operator.

106
00:07:46,000 --> 00:07:52,000
This is a linear operator,
called a linear operator.

107
00:07:50,000 --> 00:07:56,000
And, I'm going to simply
abbreviate it by the letter L.

108
00:07:54,000 --> 00:08:00,000
And so, the final version of
this equation has been reduced

109
00:07:58,000 --> 00:08:04,000
to nothing but Ly equals zero.

110
00:08:03,000 --> 00:08:09,000
Now, what's L?
You can think of L as,

111
00:08:05,000 --> 00:08:11,000
well, formally,
L you would write as D squared

112
00:08:09,000 --> 00:08:15,000
plus pD plus q.
But, you can think of L,

113
00:08:13,000 --> 00:08:19,000
the way to think of it is as a
black box, a function of what

114
00:08:18,000 --> 00:08:24,000
goes into the black box,
well, if this were a function

115
00:08:22,000 --> 00:08:28,000
box, what would go it would be a
number, and what would come out

116
00:08:27,000 --> 00:08:33,000
with the number.
But it's not that kind of a

117
00:08:31,000 --> 00:08:37,000
black box.
It's an operator box,

118
00:08:34,000 --> 00:08:40,000
and therefore,
what goes in is a function of

119
00:08:38,000 --> 00:08:44,000
x.
And, what comes out is another

120
00:08:41,000 --> 00:08:47,000
function of x,
the result of applying this

121
00:08:44,000 --> 00:08:50,000
operator to that.
So, from this point of view,

122
00:08:48,000 --> 00:08:54,000
differential equations,
trying to solve the

123
00:08:52,000 --> 00:08:58,000
differential equation means,
what should come out you want

124
00:08:57,000 --> 00:09:03,000
to come out zero,
and the question is,

125
00:09:00,000 --> 00:09:06,000
what should you put in?
That's what it means solving

126
00:09:05,000 --> 00:09:11,000
differential equations in an
inverse problem.

127
00:09:08,000 --> 00:09:14,000
The easy thing is to put it a
function, and see what comes

128
00:09:11,000 --> 00:09:17,000
out.
You just calculate.

129
00:09:13,000 --> 00:09:19,000
The hard thing is to ask,
you say, I want such and such a

130
00:09:16,000 --> 00:09:22,000
thing to come out,
for example,

131
00:09:18,000 --> 00:09:24,000
zero; what should I put in?
That's a difficult question,

132
00:09:22,000 --> 00:09:28,000
and that's what we're spending
the term answering.

133
00:09:25,000 --> 00:09:31,000
Now, the key thing about this
is that this is a linear

134
00:09:28,000 --> 00:09:34,000
operator.
And, what that means is that it

135
00:09:32,000 --> 00:09:38,000
behaves in a certain way with
respect to functions.

136
00:09:37,000 --> 00:09:43,000
The easiest way to say it is,
I like to make two laws of it,

137
00:09:42,000 --> 00:09:48,000
that L of u1,
if you have two functions,

138
00:09:45,000 --> 00:09:51,000
I'm not going to put up the
parentheses, x,

139
00:09:49,000 --> 00:09:55,000
because that just makes things
look longer and not any clearer,

140
00:09:55,000 --> 00:10:01,000
actually.
What does L do to the sum of

141
00:09:58,000 --> 00:10:04,000
two functions?
If that's a linear operator,

142
00:10:03,000 --> 00:10:09,000
if you put in the sum of two
functions, what you must get out

143
00:10:08,000 --> 00:10:14,000
is the corresponding L's,
the sum of the corresponding

144
00:10:12,000 --> 00:10:18,000
L's of each.
So, that's a law.

145
00:10:14,000 --> 00:10:20,000
And, the other law,
linearity law,

146
00:10:17,000 --> 00:10:23,000
and this goes for anything in
mathematics and its

147
00:10:21,000 --> 00:10:27,000
applications,
which is called linear,

148
00:10:24,000 --> 00:10:30,000
basically anything is linear if
it does the following thing to

149
00:10:29,000 --> 00:10:35,000
functions or numbers or
whatever.

150
00:10:33,000 --> 00:10:39,000
The other one is of a constant
times any function,

151
00:10:37,000 --> 00:10:43,000
I don't have to give it a
number now because I'm only

152
00:10:41,000 --> 00:10:47,000
using one of them,
should be equal to c times L of

153
00:10:46,000 --> 00:10:52,000
u.
So, here, c is a constant,

154
00:10:48,000 --> 00:10:54,000
and here, of course,
the u is a function,

155
00:10:51,000 --> 00:10:57,000
functions of x.
These are the two laws of

156
00:10:55,000 --> 00:11:01,000
linearity.
An operator is linear if it

157
00:10:58,000 --> 00:11:04,000
satisfies these two laws.
Now, for example,

158
00:11:03,000 --> 00:11:09,000
the differentiation operator is
such an operator.

159
00:11:06,000 --> 00:11:12,000
D is linear,
why?

160
00:11:07,000 --> 00:11:13,000
Well, because of the very first
things you verify after you

161
00:11:11,000 --> 00:11:17,000
learn what the derivative is
because the derivative of,

162
00:11:15,000 --> 00:11:21,000
well, I will write it in the D
form.

163
00:11:17,000 --> 00:11:23,000
I'll write it in the form in
which you know it.

164
00:11:20,000 --> 00:11:26,000
It would be D apply to u1 plus
u2.

165
00:11:23,000 --> 00:11:29,000
How does one write that in
ordinary calculus?

166
00:11:26,000 --> 00:11:32,000
Well, like that.
Or, maybe you write d by dx out

167
00:11:31,000 --> 00:11:37,000
front.
Let's write it this way,

168
00:11:33,000 --> 00:11:39,000
is equal to u1 prime plus u2
prime.

169
00:11:36,000 --> 00:11:42,000
That's a law.
You prove it when you first

170
00:11:39,000 --> 00:11:45,000
study what a derivative is.
It's a property.

171
00:11:42,000 --> 00:11:48,000
From our point of view,
it's a property of the

172
00:11:45,000 --> 00:11:51,000
differentiation operator.
It has this property.

173
00:11:48,000 --> 00:11:54,000
The D of u1 plus u2 is D of u1
plus D of u2.

174
00:11:52,000 --> 00:11:58,000
And, it also has the property

175
00:11:55,000 --> 00:12:01,000
that c u prime,
you can pull out the constant.

176
00:12:00,000 --> 00:12:06,000
That's not affected by the
differentiation.

177
00:12:03,000 --> 00:12:09,000
So, these two familiar laws
from the beginning of calculus

178
00:12:07,000 --> 00:12:13,000
say, in our language,
that D is a linear operator.

179
00:12:11,000 --> 00:12:17,000
What about the multiplication
of law?

180
00:12:14,000 --> 00:12:20,000
That's even more important,
that u1 times u2 prime,

181
00:12:18,000 --> 00:12:24,000
I have nothing whatever to say
about that in here.

182
00:12:22,000 --> 00:12:28,000
In this context,
it's an important law,

183
00:12:25,000 --> 00:12:31,000
but it's not important with
respect to the study of

184
00:12:29,000 --> 00:12:35,000
linearity.
So, there's an example.

185
00:12:33,000 --> 00:12:39,000
Here's a more complicated one
that I'm claiming is the linear

186
00:12:38,000 --> 00:12:44,000
operator.
And, since I don't want to have

187
00:12:41,000 --> 00:12:47,000
to work in this lecture,
the work is left to you.

188
00:12:45,000 --> 00:12:51,000
So, the proof,
prove that L is linear,

189
00:12:48,000 --> 00:12:54,000
is this particular operator.
L is linear.

190
00:12:51,000 --> 00:12:57,000
That's in your part one
homework to verify that.

191
00:12:55,000 --> 00:13:01,000
And you will do some simple
exercises in recitation tomorrow

192
00:12:59,000 --> 00:13:05,000
to sort of warm you up for that
if you haven't done it already.

193
00:13:06,000 --> 00:13:12,000
Well, you shouldn't have
because this only goes with this

194
00:13:13,000 --> 00:13:19,000
lecture, actually.
It's forbidden to work ahead in

195
00:13:19,000 --> 00:13:25,000
this class.
All right, where are we?

196
00:13:23,000 --> 00:13:29,000
Well, all that was a prelude to
proving this simple theorem,

197
00:13:31,000 --> 00:13:37,000
superposition principle.
So, finally,

198
00:13:35,000 --> 00:13:41,000
what's the proof?
The proof of the superposition

199
00:13:42,000 --> 00:13:48,000
principle: if you believe that
the operator is linear,

200
00:13:50,000 --> 00:13:56,000
then L of c1,
in other words,

201
00:13:53,000 --> 00:13:59,000
the ODE is L.
L is D squared plus pD plus q.

202
00:13:59,000 --> 00:14:05,000
So, the ODE is Ly equals zero.

203
00:14:04,000 --> 00:14:10,000
And, what am I being asked to

204
00:14:07,000 --> 00:14:13,000
prove?
I'm being asked to prove that

205
00:14:10,000 --> 00:14:16,000
if y1 and y2 are solutions,
then so is that thing.

206
00:14:13,000 --> 00:14:19,000
By the way, that's called a
linear combination.

207
00:14:16,000 --> 00:14:22,000
Put that in your notes.
Maybe I better write it even on

208
00:14:20,000 --> 00:14:26,000
the board because it's something
people say all the time without

209
00:14:24,000 --> 00:14:30,000
realizing they haven't defined
it.

210
00:14:26,000 --> 00:14:32,000
This is called a linear
combination.

211
00:14:30,000 --> 00:14:36,000
This expression is called a
linear combination of y1 and y2.

212
00:14:35,000 --> 00:14:41,000
It means that particular sum
with constant coefficients.

213
00:14:40,000 --> 00:14:46,000
Okay, so, the ODE is Ly equals
zero.

214
00:14:44,000 --> 00:14:50,000
And, I'm trying to prove that
fact about it,

215
00:14:48,000 --> 00:14:54,000
that if y1 and y2 are
solutions, so is a linear

216
00:14:52,000 --> 00:14:58,000
combination of them.
So, the proof,

217
00:14:55,000 --> 00:15:01,000
then, I start with apply L to
c1 y1 plus c2 y2.

218
00:15:00,000 --> 00:15:06,000
Now, because this operator is

219
00:15:05,000 --> 00:15:11,000
linear, it takes the sum of two
functions into the corresponding

220
00:15:10,000 --> 00:15:16,000
sum up what the operator would
be.

221
00:15:13,000 --> 00:15:19,000
So, it would be L of c1 y1 plus
L of c2 y2.

222
00:15:17,000 --> 00:15:23,000
That's because L is a linear

223
00:15:20,000 --> 00:15:26,000
operator.
But, I don't have to stop

224
00:15:23,000 --> 00:15:29,000
there.
Because L is a linear operator,

225
00:15:26,000 --> 00:15:32,000
I can pull the c out front.
So, it's c1 L of y1 plus c2 L

226
00:15:31,000 --> 00:15:37,000
of y2. Now, where am I?

227
00:15:36,000 --> 00:15:42,000
Trying to prove that this is
zero.

228
00:15:38,000 --> 00:15:44,000
Well, what is L of y1?
At this point,

229
00:15:40,000 --> 00:15:46,000
I use the fact that y1 is a
solution.

230
00:15:43,000 --> 00:15:49,000
Because it's a solution,
this is zero.

231
00:15:46,000 --> 00:15:52,000
That's what it means to solve
that differential equation.

232
00:15:50,000 --> 00:15:56,000
It means, when you apply the
linear operator,

233
00:15:53,000 --> 00:15:59,000
L, to the function,
you get zero.

234
00:15:55,000 --> 00:16:01,000
In the same way,
y2 is a solution.

235
00:15:57,000 --> 00:16:03,000
So, that's zero.
And, the sum of c1 times zero

236
00:16:02,000 --> 00:16:08,000
plus c2 times zero is zero.

237
00:16:05,000 --> 00:16:11,000
That's the argument.
Now, you could make the same

238
00:16:08,000 --> 00:16:14,000
argument just by plugging c1 y1,
plugging it into the equation

239
00:16:13,000 --> 00:16:19,000
and calculating,
and calculating,

240
00:16:15,000 --> 00:16:21,000
and calculating,
grouping the terms and so on

241
00:16:18,000 --> 00:16:24,000
and so forth.
But, that's just calculation.

242
00:16:21,000 --> 00:16:27,000
It doesn't show you why it's
so.

243
00:16:24,000 --> 00:16:30,000
Why it's so is because the
operator, this differential

244
00:16:28,000 --> 00:16:34,000
equation is expressed as a
linear operator applied to y is

245
00:16:32,000 --> 00:16:38,000
zero.
And, the only properties that

246
00:16:37,000 --> 00:16:43,000
are really been used as the fact
that this operator is linear.

247
00:16:46,000 --> 00:16:52,000
That's the key point.
L is linear.

248
00:16:50,000 --> 00:16:56,000
It's a linear operator.
Well, that's all there is to

249
00:16:57,000 --> 00:17:03,000
the superposition principle.
As a prelude to answering the

250
00:17:04,000 --> 00:17:10,000
more difficult question,
why are these all the

251
00:17:07,000 --> 00:17:13,000
solutions?
Why are there no other

252
00:17:09,000 --> 00:17:15,000
solutions?
We need a few definitions,

253
00:17:12,000 --> 00:17:18,000
and a few more ideas.
And, they are going to occur in

254
00:17:16,000 --> 00:17:22,000
connection with,
so I'm now addressing,

255
00:17:18,000 --> 00:17:24,000
ultimately, question two.
But, it's not going to be

256
00:17:22,000 --> 00:17:28,000
addressed directly for quite
awhile.

257
00:17:25,000 --> 00:17:31,000
Instead, I'm going to phrase it
in terms of solving the initial

258
00:17:29,000 --> 00:17:35,000
value problem.
So far, we've only talked about

259
00:17:34,000 --> 00:17:40,000
the general solution with those
two arbitrary constants.

260
00:17:39,000 --> 00:17:45,000
But, how do you solve the
initial value problem,

261
00:17:43,000 --> 00:17:49,000
in other words,
fit initial conditions,

262
00:17:46,000 --> 00:17:52,000
find the solution with given
initial values for the function

263
00:17:51,000 --> 00:17:57,000
and its derivatives.
Now, --

264
00:18:04,000 --> 00:18:10,000
-- the theorem is that this
collection of functions with

265
00:18:08,000 --> 00:18:14,000
these arbitrary constants,
these are all the solutions we

266
00:18:12,000 --> 00:18:18,000
have so far.
In fact, they are all the

267
00:18:15,000 --> 00:18:21,000
solutions there are,
but we don't know that yet.

268
00:18:19,000 --> 00:18:25,000
However, if we just focus on
the big class of solutions,

269
00:18:23,000 --> 00:18:29,000
there might be others lurking
out there somewhere lurking out

270
00:18:28,000 --> 00:18:34,000
there.
I don't know.

271
00:18:31,000 --> 00:18:37,000
But let's use what we have,
that just from this family is

272
00:18:39,000 --> 00:18:45,000
enough to satisfy any initial
condition, to satisfy any

273
00:18:47,000 --> 00:18:53,000
initial values.
In other words,

274
00:18:51,000 --> 00:18:57,000
if you give me any initial
values, I will be able to find

275
00:18:59,000 --> 00:19:05,000
the c1 and c2 which work.
Now, why is that?

276
00:19:05,000 --> 00:19:11,000
Well, I'd have to do a song and
dance at this point,

277
00:19:08,000 --> 00:19:14,000
if you hadn't been softened up
by actually calculating for

278
00:19:12,000 --> 00:19:18,000
specific differential equations.
You've had exercises and

279
00:19:15,000 --> 00:19:21,000
actually how to calculate the
values of c1 and c2.

280
00:19:19,000 --> 00:19:25,000
So, I'm going to do it now in
general what you have done so

281
00:19:23,000 --> 00:19:29,000
far for particular equations
using particular values of the

282
00:19:26,000 --> 00:19:32,000
initial conditions.
So, I'm relying on that

283
00:19:30,000 --> 00:19:36,000
experience that you've had in
doing the homework to make

284
00:19:35,000 --> 00:19:41,000
intelligible what I'm going to
do now in the abstract using

285
00:19:40,000 --> 00:19:46,000
just letters.
So, why is this so?

286
00:19:43,000 --> 00:19:49,000
Why is that so?
Well, we are going to need it,

287
00:19:47,000 --> 00:19:53,000
by the way, here,
too.

288
00:19:49,000 --> 00:19:55,000
I'll have to,
again, open up parentheses.

289
00:19:52,000 --> 00:19:58,000
But let's go as far as we can.
Well, you just try to do it.

290
00:19:57,000 --> 00:20:03,000
Suppose the initial conditions
are, how will we write them?

291
00:20:04,000 --> 00:20:10,000
So, they're going to be at some
initial point,

292
00:20:06,000 --> 00:20:12,000
x0.
You can take it to be zero if

293
00:20:08,000 --> 00:20:14,000
you want, but I'd like to be,
just for a little while,

294
00:20:12,000 --> 00:20:18,000
a little more general.
So, let's say the initial

295
00:20:15,000 --> 00:20:21,000
conditions, the initial values
are being given at the point x0,

296
00:20:19,000 --> 00:20:25,000
all right, that's going to be
some number.

297
00:20:21,000 --> 00:20:27,000
Let's just call it a.
And, the initial value also has

298
00:20:25,000 --> 00:20:31,000
to specify the velocity or the
value of the derivative there.

299
00:20:30,000 --> 00:20:36,000
Let's say these are the initial
values.

300
00:20:33,000 --> 00:20:39,000
So, the problem is to find the
c which work.

301
00:20:37,000 --> 00:20:43,000
Now, how do you do that?
Well, you know from

302
00:20:41,000 --> 00:20:47,000
calculation.
You write y equals c1 y1 plus

303
00:20:45,000 --> 00:20:51,000
c2 y2.
And you write y prime,

304
00:20:47,000 --> 00:20:53,000
and you take the derivative
underneath that,

305
00:20:51,000 --> 00:20:57,000
which is easy to do.
And now, you plug in x equals

306
00:20:56,000 --> 00:21:02,000
zero.
And, what happens?

307
00:21:00,000 --> 00:21:06,000
Well, these now turn into a set
of equations.

308
00:21:04,000 --> 00:21:10,000
What will they look like?
Well, y of x0 is a,

309
00:21:08,000 --> 00:21:14,000
and this is b.
So, what I get is let me flop

310
00:21:12,000 --> 00:21:18,000
it over onto the other side
because that's the way you're

311
00:21:17,000 --> 00:21:23,000
used to looking at systems of
equations.

312
00:21:21,000 --> 00:21:27,000
So, what we get is c1 times y1
of x0 plus c2 times y2 of x0.

313
00:21:26,000 --> 00:21:32,000
What's that supposed to be

314
00:21:31,000 --> 00:21:37,000
equal to?
Well, that's supposed to be

315
00:21:34,000 --> 00:21:40,000
equal to y of x0.
It's supposed to be equal to

316
00:21:37,000 --> 00:21:43,000
the given number,
a.

317
00:21:39,000 --> 00:21:45,000
And, in the same way,
c1 y1 prime of x0 plus c2 y2

318
00:21:42,000 --> 00:21:48,000
prime of x0, that's supposed to
turn out to

319
00:21:46,000 --> 00:21:52,000
be the number, b.

320
00:21:48,000 --> 00:21:54,000
In the calculations you've done
up to this point,

321
00:21:51,000 --> 00:21:57,000
y1 and y2 were always specific
functions like e to the x

322
00:21:55,000 --> 00:22:01,000
or cosine of 22x,

323
00:21:57,000 --> 00:22:03,000
stuff like that.
Now I'm doing it in the

324
00:22:01,000 --> 00:22:07,000
abstract, just calling them y1
and y2, so as to include all

325
00:22:06,000 --> 00:22:12,000
those possible cases.
Now, what am I supposed to do?

326
00:22:10,000 --> 00:22:16,000
I'm supposed to find c1 and c2.
What kind of things are they?

327
00:22:15,000 --> 00:22:21,000
This is what you studied in
high school, right?

328
00:22:19,000 --> 00:22:25,000
The letters are around us,
but it's a pair of simultaneous

329
00:22:23,000 --> 00:22:29,000
linear equations.
What are the variables?

330
00:22:26,000 --> 00:22:32,000
What are the variables?
What are the variables?

331
00:22:30,000 --> 00:22:36,000
Somebody raise their hand.
If you have a pair of

332
00:22:35,000 --> 00:22:41,000
simultaneous linear equations,
you've got variables and you've

333
00:22:39,000 --> 00:22:45,000
got constants,
right?

334
00:22:41,000 --> 00:22:47,000
And you are trying to find the
answer.

335
00:22:43,000 --> 00:22:49,000
What are the variables?
Yeah?

336
00:22:45,000 --> 00:22:51,000
c1 and c2.
Very good.

337
00:22:47,000 --> 00:22:53,000
I mean, it's extremely
confusing because in the first

338
00:22:50,000 --> 00:22:56,000
place, how can they be the
variables if they occur on the

339
00:22:54,000 --> 00:23:00,000
wrong side?
They're on the wrong side;

340
00:22:57,000 --> 00:23:03,000
they are constants.
How can constants be variables?

341
00:23:02,000 --> 00:23:08,000
Everything about this is wrong.
Nonetheless,

342
00:23:06,000 --> 00:23:12,000
the c1 and the c2 are the
unknowns, if you like the high

343
00:23:10,000 --> 00:23:16,000
school terminology.
c1 and c2 are the unknowns.

344
00:23:14,000 --> 00:23:20,000
These messes are just numbers.
After you've plugged in x0,

345
00:23:19,000 --> 00:23:25,000
this is some number.
You've got four numbers here.

346
00:23:23,000 --> 00:23:29,000
So, c1 and c2 are the
variables.

347
00:23:26,000 --> 00:23:32,000
The two find,
in other words,

348
00:23:28,000 --> 00:23:34,000
to find the values of.
All right, now you know general

349
00:23:34,000 --> 00:23:40,000
theorems from 18.02 about when
can you solve such a system of

350
00:23:39,000 --> 00:23:45,000
equations.
I'm claiming that you can

351
00:23:42,000 --> 00:23:48,000
always find c1 and c2 that work.
But, you know that's not always

352
00:23:48,000 --> 00:23:54,000
the case that a pair of
simultaneous linear equations

353
00:23:52,000 --> 00:23:58,000
can be solved.
There's a condition.

354
00:23:55,000 --> 00:24:01,000
There's a condition which
guarantees their solution,

355
00:23:59,000 --> 00:24:05,000
which is what?
What has to be true about the

356
00:24:04,000 --> 00:24:10,000
coefficients?
These are the coefficients.

357
00:24:08,000 --> 00:24:14,000
What has to be true?
The matrix of coefficients must

358
00:24:13,000 --> 00:24:19,000
be invertible.
The determinant of coefficients

359
00:24:18,000 --> 00:24:24,000
must be nonzero.
So, they are solvable if,

360
00:24:22,000 --> 00:24:28,000
for the c1 and c2,
if this thing,

361
00:24:25,000 --> 00:24:31,000
I'm going to write it.
Since all of these are

362
00:24:29,000 --> 00:24:35,000
evaluated at x0,
I'm going to write it in this

363
00:24:34,000 --> 00:24:40,000
way.
y1, the determinant,

364
00:24:38,000 --> 00:24:44,000
whose entries are y1,
y2, y1 prime,

365
00:24:41,000 --> 00:24:47,000
and y2 prime,

366
00:24:44,000 --> 00:24:50,000
evaluated at zero,
x0, that means that I evaluate

367
00:24:49,000 --> 00:24:55,000
each of the functions in the
determinant at x0.

368
00:24:54,000 --> 00:25:00,000
I'll write it this way.
That should be not zero.

369
00:25:00,000 --> 00:25:06,000
So, in other words,
the key thing which makes this

370
00:25:03,000 --> 00:25:09,000
possible, makes it possible for
us to solve the initial value

371
00:25:08,000 --> 00:25:14,000
problem, is that this funny
determinant should not be zero

372
00:25:13,000 --> 00:25:19,000
at the point at which we are
interested.

373
00:25:16,000 --> 00:25:22,000
Now, this determinant is
important in 18.03.

374
00:25:19,000 --> 00:25:25,000
It has a name,
and this is when you're going

375
00:25:23,000 --> 00:25:29,000
to learn it, if you don't know
it already.

376
00:25:26,000 --> 00:25:32,000
That determinant is called the
Wronskian.

377
00:25:31,000 --> 00:25:37,000
The Wronskian of what?
If you want to be pompous,

378
00:25:34,000 --> 00:25:40,000
you say this with a V sound
instead of a W.

379
00:25:37,000 --> 00:25:43,000
But, nobody does except people
trying to be pompous.

380
00:25:41,000 --> 00:25:47,000
The Wronskian,
we'll write a W.

381
00:25:43,000 --> 00:25:49,000
Now, notice,
you can only calculate it when

382
00:25:47,000 --> 00:25:53,000
you know what the two functions
are.

383
00:25:49,000 --> 00:25:55,000
So, the Wronskian of the two
functions, y1 and y2,

384
00:25:53,000 --> 00:25:59,000
what's the variable?
It's not a function of two

385
00:25:56,000 --> 00:26:02,000
variables, y1 and y2.
These are just the names of

386
00:26:00,000 --> 00:26:06,000
functions of x.
So, when you do it,

387
00:26:04,000 --> 00:26:10,000
put it in, calculate out that
determinant.

388
00:26:08,000 --> 00:26:14,000
This is a function of x,
a function of the independent

389
00:26:13,000 --> 00:26:19,000
variable after you've done the
calculation.

390
00:26:17,000 --> 00:26:23,000
Anyway, let's write its
definition, y1,

391
00:26:20,000 --> 00:26:26,000
y2, y1 prime,
y2 prime.

392
00:26:23,000 --> 00:26:29,000
Now, in order to do this,

393
00:26:26,000 --> 00:26:32,000
the point is we must know that
that Wronskian is not zero,

394
00:26:32,000 --> 00:26:38,000
that the Wronskian of these two
functions is not zero at the

395
00:26:37,000 --> 00:26:43,000
point x0.
Now, enter a theorem which

396
00:26:42,000 --> 00:26:48,000
you're going to prove for
homework, but this is harder.

397
00:26:47,000 --> 00:26:53,000
So, it's part two homework.
It's not part one homework.

398
00:26:51,000 --> 00:26:57,000
In other words,
I didn't give you the answer.

399
00:26:55,000 --> 00:27:01,000
You've got to find it yourself,
alone or in the company of good

400
00:27:01,000 --> 00:27:07,000
friends.
So, anyway, here's the

401
00:27:05,000 --> 00:27:11,000
Wronskian.
Now, what can we say for sure?

402
00:27:08,000 --> 00:27:14,000
Note, suppose y1 and y,
just to get you a feeling for

403
00:27:13,000 --> 00:27:19,000
it a little bit,
suppose they were not

404
00:27:16,000 --> 00:27:22,000
independent.
The word for not independent is

405
00:27:20,000 --> 00:27:26,000
dependent.
Suppose they were dependent.

406
00:27:24,000 --> 00:27:30,000
In other words,
suppose that y2 were a constant

407
00:27:28,000 --> 00:27:34,000
multiple of y1.
We know that's not the case

408
00:27:33,000 --> 00:27:39,000
because our functions are
supposed to be independent.

409
00:27:37,000 --> 00:27:43,000
But suppose they weren't.
What would the value of the

410
00:27:42,000 --> 00:27:48,000
Wronskian be?
If y2 is a constant times y1,

411
00:27:46,000 --> 00:27:52,000
then y2 prime is that same
constant times y1 prime.

412
00:27:50,000 --> 00:27:56,000
What's the value of the
determinant?

413
00:27:53,000 --> 00:27:59,000
Zero.
For what values of x is it zero

414
00:27:56,000 --> 00:28:02,000
for all values of x?
And now, that's the theorem

415
00:28:01,000 --> 00:28:07,000
that you're going to prove,
that if y1 and y2 are solutions

416
00:28:06,000 --> 00:28:12,000
to the ODE, I won't keep,
say, it's the ODE we've been

417
00:28:10,000 --> 00:28:16,000
talking about,
y double prime plus py.

418
00:28:14,000 --> 00:28:20,000
But the linear homogeneous with

419
00:28:20,000 --> 00:28:26,000
not constant coefficients,
just linear homogeneous second

420
00:28:26,000 --> 00:28:32,000
order.
Our solutions,

421
00:28:29,000 --> 00:28:35,000
as there are only two
possibilities,

422
00:28:33,000 --> 00:28:39,000
either-or.
Either the Wronskian of y,

423
00:28:37,000 --> 00:28:43,000
there are only two
possibilities.

424
00:28:39,000 --> 00:28:45,000
Either the Wronskian of y1 and
y2 is always zero,

425
00:28:44,000 --> 00:28:50,000
identically zero,
zero for all values of x.

426
00:28:47,000 --> 00:28:53,000
This is redundant.
When I say identically,

427
00:28:51,000 --> 00:28:57,000
I mean for all values of x.
But, I am just making assurance

428
00:28:56,000 --> 00:29:02,000
doubly sure.
Or, or the Wronskian is never

429
00:29:01,000 --> 00:29:07,000
zero.
Now, there is no notation as

430
00:29:06,000 --> 00:29:12,000
for that.
I'd better just write it out,

431
00:29:10,000 --> 00:29:16,000
is never zero,
i.e.

432
00:29:13,000 --> 00:29:19,000
for no x is it,
i.e.

433
00:29:15,000 --> 00:29:21,000
for all x.
There's no way to say that.

434
00:29:20,000 --> 00:29:26,000
I mean, for all values of x,
it's not zero.

435
00:29:26,000 --> 00:29:32,000
That means, there is not a
single point for which it's

436
00:29:32,000 --> 00:29:38,000
zero.
In particular,

437
00:29:35,000 --> 00:29:41,000
it's not zero here.
So, this is your homework:

438
00:29:43,000 --> 00:29:49,000
problem five,
part two.

439
00:29:45,000 --> 00:29:51,000
I'll give you a method of
proving it, which was discovered

440
00:29:52,000 --> 00:29:58,000
by the famous Norwegian
mathematician,

441
00:29:57,000 --> 00:30:03,000
Abel, who is,
I guess, the centenary of his

442
00:30:02,000 --> 00:30:08,000
birth, I guess,
was just celebrated last year.

443
00:30:09,000 --> 00:30:15,000
He has one of the truly tragic
stories in mathematics,

444
00:30:13,000 --> 00:30:19,000
which I think you can read.
It must be a Simmons book,

445
00:30:18,000 --> 00:30:24,000
if you have that.
Simmons is very good on

446
00:30:22,000 --> 00:30:28,000
biographies.
Look up Abel.

447
00:30:24,000 --> 00:30:30,000
He'll have a biography of Abel,
and you can weep if you're

448
00:30:29,000 --> 00:30:35,000
feeling sad.
He died at the age of 26 of

449
00:30:34,000 --> 00:30:40,000
tuberculosis,
having done a number of

450
00:30:37,000 --> 00:30:43,000
sensational things,
none of which was recognized in

451
00:30:41,000 --> 00:30:47,000
his lifetime because people
buried his papers under big

452
00:30:46,000 --> 00:30:52,000
piles of papers.
So, he died unknown,

453
00:30:49,000 --> 00:30:55,000
uncelebrated,
and now he's Norway's greatest

454
00:30:53,000 --> 00:30:59,000
culture hero.
In the middle of a park in

455
00:30:56,000 --> 00:31:02,000
Oslo, there's a huge statue.
And, since nobody knows what

456
00:31:03,000 --> 00:31:09,000
Abel looked like,
the statue is way up high so

457
00:31:07,000 --> 00:31:13,000
you can't see very well.
But, the inscription on the

458
00:31:13,000 --> 00:31:19,000
bottom says Niels Henrik Abel,
1801-1826 or something like

459
00:31:19,000 --> 00:31:25,000
that.
Now, --

460
00:31:38,000 --> 00:31:44,000
-- the choice,
I'm still, believe it or not,

461
00:31:41,000 --> 00:31:47,000
aiming at question two,
but I have another big

462
00:31:44,000 --> 00:31:50,000
parentheses to open.
And, when I closed it,

463
00:31:47,000 --> 00:31:53,000
the answer to question two will
be simple.

464
00:31:50,000 --> 00:31:56,000
But, I think it's very
desirable that you get this

465
00:31:53,000 --> 00:31:59,000
second big parentheses.
It will help you to understand

466
00:31:57,000 --> 00:32:03,000
something important.
It will help you on your

467
00:32:02,000 --> 00:32:08,000
problem set tomorrow night.
I don't have to apologize.

468
00:32:08,000 --> 00:32:14,000
I'm just going to do it.
So, the question is,

469
00:32:12,000 --> 00:32:18,000
the thing you have to
understand is that when I write

470
00:32:18,000 --> 00:32:24,000
this combination,
I'm claiming that these are all

471
00:32:23,000 --> 00:32:29,000
the solutions.
I haven't proved that yet.

472
00:32:27,000 --> 00:32:33,000
But, they are going to be all
the solutions

473
00:32:33,000 --> 00:32:39,000
The point is,
there's nothing sacrosanct

474
00:32:36,000 --> 00:32:42,000
about the y1 and y2.
This is exactly the same

475
00:32:41,000 --> 00:32:47,000
collection as a collection which
I would write using other

476
00:32:46,000 --> 00:32:52,000
constants.
Let's call them u1 and u2.

477
00:32:49,000 --> 00:32:55,000
They are exactly the same,
where u1 and u2 are any other

478
00:32:55,000 --> 00:33:01,000
pair of linearly independent
solutions.

479
00:33:00,000 --> 00:33:06,000
Any other pair of independent
solutions, they must be

480
00:33:05,000 --> 00:33:11,000
independent, either a constant
multiple of each other.

481
00:33:10,000 --> 00:33:16,000
In other words,
u1 is some combination,

482
00:33:13,000 --> 00:33:19,000
now I'm really stuck because I
don't know how to,

483
00:33:18,000 --> 00:33:24,000
c1 bar, let's say,
that means a special value of

484
00:33:22,000 --> 00:33:28,000
c1, and a special value of c2,
and u2 is some other special

485
00:33:28,000 --> 00:33:34,000
value, oh my God,
c1 double bar,

486
00:33:31,000 --> 00:33:37,000
how's that?
The notation is getting worse

487
00:33:36,000 --> 00:33:42,000
and worse.
I apologize for it.

488
00:33:38,000 --> 00:33:44,000
In other words,
I could pick y1 and y2 and make

489
00:33:42,000 --> 00:33:48,000
up all of these.
And, I'd get a bunch of

490
00:33:45,000 --> 00:33:51,000
solutions.
But, I could also pick some

491
00:33:48,000 --> 00:33:54,000
other family,
some other two guys in this

492
00:33:51,000 --> 00:33:57,000
family, and just as well express
the solutions in terms of u1 and

493
00:33:56,000 --> 00:34:02,000
u2.
Now, well, why is he telling us

494
00:33:59,000 --> 00:34:05,000
that?
Well, the point is that the y1

495
00:34:03,000 --> 00:34:09,000
and the y2 are typically the
ones you get easily from solving

496
00:34:08,000 --> 00:34:14,000
the equations,
like e to the x and e

497
00:34:11,000 --> 00:34:17,000
to the 2x.
That's what you've gotten,

498
00:34:15,000 --> 00:34:21,000
or cosine x and sine x,

499
00:34:18,000 --> 00:34:24,000
something like that.
But, for certain ways,

500
00:34:22,000 --> 00:34:28,000
they might not be the best way
of writing the solutions.

501
00:34:26,000 --> 00:34:32,000
There is another way of writing
those that you should learn,

502
00:34:31,000 --> 00:34:37,000
and that's called finding
normalized, the normalized.

503
00:34:35,000 --> 00:34:41,000
They are okay,
but they are not normalized.

504
00:34:40,000 --> 00:34:46,000
For some things,
the normalized solutions are

505
00:34:42,000 --> 00:34:48,000
the best.
I'll explain to you what they

506
00:34:44,000 --> 00:34:50,000
are, and I'll explain to you
what they're good for.

507
00:34:48,000 --> 00:34:54,000
You'll see immediately what
they're good for.

508
00:34:50,000 --> 00:34:56,000
Normalized solutions,
now, you have to specify the

509
00:34:53,000 --> 00:34:59,000
point at which you're
normalizing.

510
00:34:55,000 --> 00:35:01,000
In general, it would be x
nought,

511
00:34:58,000 --> 00:35:04,000
but let's, at this point,
since I don't have an infinity

512
00:35:01,000 --> 00:35:07,000
of time, to simplify things,
let's say zero.

513
00:35:05,000 --> 00:35:11,000
It could be x nought,
any point would do just as

514
00:35:08,000 --> 00:35:14,000
well.
But, zero is the most common

515
00:35:11,000 --> 00:35:17,000
choice.
What are the normalized

516
00:35:13,000 --> 00:35:19,000
solutions?
Well, first of all,

517
00:35:15,000 --> 00:35:21,000
I have to give them names.
I want to still call them y.

518
00:35:19,000 --> 00:35:25,000
So, I'll call them capital Y1
and Y2.

519
00:35:21,000 --> 00:35:27,000
And, what they are,
are the solutions which satisfy

520
00:35:25,000 --> 00:35:31,000
certain, special,
very special,

521
00:35:27,000 --> 00:35:33,000
initial conditions.
And, what are those?

522
00:35:31,000 --> 00:35:37,000
So, they're the ones which
satisfy, the initial conditions

523
00:35:37,000 --> 00:35:43,000
for Y1 are, of course there are
going to be guys that look like

524
00:35:43,000 --> 00:35:49,000
this.
The only thing that's going to

525
00:35:46,000 --> 00:35:52,000
make them distinctive is the
initial conditions they satisfy.

526
00:35:51,000 --> 00:35:57,000
Y1 has to satisfy at zero.
Its value should be one,

527
00:35:56,000 --> 00:36:02,000
and the value of its derivative
should be zero.

528
00:36:02,000 --> 00:36:08,000
For Y2, it's just the opposite.
Here, the value of the function

529
00:36:07,000 --> 00:36:13,000
should be zero at zero.
But, the value of its

530
00:36:11,000 --> 00:36:17,000
derivative, now,
I want to be one.

531
00:36:14,000 --> 00:36:20,000
Let me give you a trivial
example of this,

532
00:36:17,000 --> 00:36:23,000
and then one,
which is a little less trivial,

533
00:36:21,000 --> 00:36:27,000
so you'll have some feeling for
what I'm asking for.

534
00:36:25,000 --> 00:36:31,000
Suppose the equation,
for example,

535
00:36:28,000 --> 00:36:34,000
is y double prime plus,
well, let's really make it

536
00:36:33,000 --> 00:36:39,000
simple.
Okay, you know the standard

537
00:36:36,000 --> 00:36:42,000
solutions are y1 is cosine x,

538
00:36:39,000 --> 00:36:45,000
and y2 is sine x.

539
00:36:41,000 --> 00:36:47,000
These are functions,
which, when you take the second

540
00:36:44,000 --> 00:36:50,000
derivative, they turn into their
negative.

541
00:36:46,000 --> 00:36:52,000
You know, you could go the
complex roots are i and minus i,

542
00:36:50,000 --> 00:36:56,000
and blah, blah,
blah, blah, blah,

543
00:36:52,000 --> 00:36:58,000
blah.
If you do it that way,

544
00:36:53,000 --> 00:36:59,000
fine.
But at some point in the course

545
00:36:56,000 --> 00:37:02,000
you have to be able to write
down and, right away,

546
00:36:59,000 --> 00:37:05,000
oh, yeah, cosine x,
sine x.

547
00:37:01,000 --> 00:37:07,000
Okay, what are the normalized

548
00:37:05,000 --> 00:37:11,000
things?
Well, what's the value of this

549
00:37:09,000 --> 00:37:15,000
at zero?
It is one.

550
00:37:11,000 --> 00:37:17,000
What's the value of its
derivative at zero?

551
00:37:15,000 --> 00:37:21,000
Zero.
This is Y1.

552
00:37:16,000 --> 00:37:22,000
This is the only case in which
you locked on immediately to the

553
00:37:22,000 --> 00:37:28,000
normalized solutions.
In the same way,

554
00:37:26,000 --> 00:37:32,000
this guy is Y2 because its
value at zero is zero.

555
00:37:32,000 --> 00:37:38,000
It's value of its derivative at
zero is one.

556
00:37:35,000 --> 00:37:41,000
So, this is Y2.
Okay, now let's look at a case

557
00:37:39,000 --> 00:37:45,000
where you don't immediately lock
on to the normalized solutions.

558
00:37:44,000 --> 00:37:50,000
Very simple:
all I have to do is change the

559
00:37:47,000 --> 00:37:53,000
sign.
Here, you know,

560
00:37:49,000 --> 00:37:55,000
think through r squared minus
one equals zero.

561
00:37:54,000 --> 00:38:00,000
The characteristic roots are
plus one and minus one,

562
00:37:58,000 --> 00:38:04,000
right?
And therefore,

563
00:38:00,000 --> 00:38:06,000
the solution is e to the x,
and e to the negative x.

564
00:38:04,000 --> 00:38:10,000
So, the solutions you find by

565
00:38:07,000 --> 00:38:13,000
the usual way of solving it is
y1 equals e to the x,

566
00:38:11,000 --> 00:38:17,000
and y2 equals e to the
negative x.

567
00:38:15,000 --> 00:38:21,000
Those are the standard
solution.

568
00:38:17,000 --> 00:38:23,000
So, the general solution is of
the form.

569
00:38:19,000 --> 00:38:25,000
So, the general solution is of
the form c1 e to the x plus c2 e

570
00:38:23,000 --> 00:38:29,000
to the negative x.

571
00:38:26,000 --> 00:38:32,000
Now, what I want to find out is
what is Y1 and Y2?

572
00:38:31,000 --> 00:38:37,000
How do I find out what Y1 is?
Well, I have to satisfy initial

573
00:38:36,000 --> 00:38:42,000
conditions.
So, if this is y,

574
00:38:38,000 --> 00:38:44,000
let's write down here,
if you can still see that,

575
00:38:43,000 --> 00:38:49,000
y prime is c1 e to the x minus
c2 e to the negative x .

576
00:38:48,000 --> 00:38:54,000
So, if I plug in,

577
00:38:51,000 --> 00:38:57,000
I want y of zero to be one,
I want this guy at the point

578
00:38:56,000 --> 00:39:02,000
zero to be one.
What equation does that give

579
00:39:00,000 --> 00:39:06,000
me?
That gives me c1 plus c2,

580
00:39:04,000 --> 00:39:10,000
c1 plus c2,
plugging in x equals zero,

581
00:39:08,000 --> 00:39:14,000
equals the value of this thing
at zero.

582
00:39:11,000 --> 00:39:17,000
So, that's supposed to be one.

583
00:39:14,000 --> 00:39:20,000
How about the other guy?
The value of its derivative is

584
00:39:19,000 --> 00:39:25,000
supposed to come out to be zero.
And, what is its derivative?

585
00:39:24,000 --> 00:39:30,000
Well, plug into this
expression.

586
00:39:26,000 --> 00:39:32,000
It's c1 minus c2.
Okay, what's the solution to

587
00:39:32,000 --> 00:39:38,000
those pair of equations?
c2 has to be equal to c1.

588
00:39:36,000 --> 00:39:42,000
The sum of the two of them has
to be one.

589
00:39:39,000 --> 00:39:45,000
Each one, therefore,
is equal to one half.

590
00:39:42,000 --> 00:39:48,000
And so, what's the value of Y1?
Y1, therefore,

591
00:39:46,000 --> 00:39:52,000
is the function where c1 and c2
are one half.

592
00:39:50,000 --> 00:39:56,000
It's the function e to the x
plus e to the negative x divided

593
00:39:55,000 --> 00:40:01,000
by two.

594
00:39:59,000 --> 00:40:05,000
In the same way,
I won't repeat the calculation.

595
00:40:03,000 --> 00:40:09,000
You can do yourself.
Same calculation shows that Y2,

596
00:40:07,000 --> 00:40:13,000
so, put in the initial
conditions.

597
00:40:10,000 --> 00:40:16,000
The answer will be that Y2 is
equal to e to the x minus e to

598
00:40:15,000 --> 00:40:21,000
the minus x divided by two.

599
00:40:19,000 --> 00:40:25,000
These are the special
functions.

600
00:40:22,000 --> 00:40:28,000
For this equation,
these are the normalized

601
00:40:25,000 --> 00:40:31,000
solutions.
They are better than the

602
00:40:28,000 --> 00:40:34,000
original solutions because their
initial values are nicer.

603
00:40:35,000 --> 00:40:41,000
Just check it.
The initial value,

604
00:40:37,000 --> 00:40:43,000
when x is equal to zero,
the initial value,

605
00:40:40,000 --> 00:40:46,000
this has is zero.
Here, when x is equal to zero,

606
00:40:44,000 --> 00:40:50,000
the value of the function is
zero.

607
00:40:47,000 --> 00:40:53,000
But, the value of its
derivative, these cancel,

608
00:40:51,000 --> 00:40:57,000
is one.
So, these are the good guys.

609
00:40:54,000 --> 00:41:00,000
Okay, there's no colored chalk
this period.

610
00:40:57,000 --> 00:41:03,000
Okay, there was colored chalk.
There's one.

611
00:41:01,000 --> 00:41:07,000
So, for this equation,
these are the good guys.

612
00:41:04,000 --> 00:41:10,000
These are our best solutions.
e to the x and e to the

613
00:41:07,000 --> 00:41:13,000
minus x are good solutions.

614
00:41:10,000 --> 00:41:16,000
But, these are our better
solutions.

615
00:41:12,000 --> 00:41:18,000
And, this one,
of course, is the function

616
00:41:14,000 --> 00:41:20,000
which is called hyperbolic sine
of x, and this is the one which

617
00:41:18,000 --> 00:41:24,000
is called hyperbolic cosine of
x.

618
00:41:20,000 --> 00:41:26,000
This is one of the most
important ways in which they

619
00:41:23,000 --> 00:41:29,000
enter into mathematics.
And, this is why the engineers

620
00:41:26,000 --> 00:41:32,000
want them.
Now, why do the engineers want

621
00:41:28,000 --> 00:41:34,000
normalized solutions?
Well, I didn't explain that.

622
00:41:34,000 --> 00:41:40,000
So, what's so good about
normalized solutions?

623
00:41:41,000 --> 00:41:47,000
Very simple:
if Y1 and Y2 are normalized at

624
00:41:47,000 --> 00:41:53,000
zero, let's say,
then the solution to the IVP,

625
00:41:53,000 --> 00:41:59,000
in other words,
the ODE plus the initial

626
00:41:59,000 --> 00:42:05,000
values, y of zero equals,
let's say, a and y

627
00:42:07,000 --> 00:42:13,000
prime of zero equals b.

628
00:42:14,000 --> 00:42:20,000
So, the ODE I'm not repeating.
It's the one we've been talking

629
00:42:17,000 --> 00:42:23,000
about all term since the
beginning of the period.

630
00:42:20,000 --> 00:42:26,000
It's the one with the p of x
and q of x.

631
00:42:23,000 --> 00:42:29,000
And, here are the initial
values.

632
00:42:25,000 --> 00:42:31,000
I'm going to call them a and b.
You can also call them,

633
00:42:28,000 --> 00:42:34,000
if you like,
maybe that's better to call

634
00:42:30,000 --> 00:42:36,000
them y0, as they are individual
in the homework.

635
00:42:34,000 --> 00:42:40,000
They are called,
I'm using the,

636
00:42:36,000 --> 00:42:42,000
let's use those.
What is the solution?

637
00:42:39,000 --> 00:42:45,000
I say the solution is,
if you use y1 and y2,

638
00:42:43,000 --> 00:42:49,000
the solution is y0,
in other words,

639
00:42:46,000 --> 00:42:52,000
the a times Y1,
plus y0 prime,

640
00:42:49,000 --> 00:42:55,000
in other words,
b times Y2.

641
00:42:53,000 --> 00:42:59,000
In other words,
you can write down instantly

642
00:42:57,000 --> 00:43:03,000
the solution to the initial
value problem,

643
00:43:00,000 --> 00:43:06,000
if instead of using the
functions, you started out with

644
00:43:05,000 --> 00:43:11,000
the little Y1 and Y2,
you use these better functions.

645
00:43:12,000 --> 00:43:18,000
The thing that's better about
them is that they instantly

646
00:43:16,000 --> 00:43:22,000
solve for you the initial value
problem.

647
00:43:18,000 --> 00:43:24,000
All you do is use this number,
initial condition as the

648
00:43:22,000 --> 00:43:28,000
coefficient of Y1,
and use this number as the

649
00:43:26,000 --> 00:43:32,000
coefficient of Y2.
Now, just check that by looking

650
00:43:29,000 --> 00:43:35,000
at it.
Why is that so?

651
00:43:32,000 --> 00:43:38,000
Well, for example,
let's check.

652
00:43:34,000 --> 00:43:40,000
What is its value of this
function at zero?

653
00:43:37,000 --> 00:43:43,000
Well, the value of this guy at
zero is one.

654
00:43:40,000 --> 00:43:46,000
So, the answer is y0 times one,
and the value of this guy at

655
00:43:44,000 --> 00:43:50,000
zero is zero.
So, this term disappears.

656
00:43:47,000 --> 00:43:53,000
And, it's exactly the same with
the derivative.

657
00:43:50,000 --> 00:43:56,000
What's the value of the
derivative at zero?

658
00:43:54,000 --> 00:44:00,000
The value of the derivative of
this thing is zero.

659
00:43:57,000 --> 00:44:03,000
So, this term disappears.
The value of this derivative at

660
00:44:01,000 --> 00:44:07,000
zero is one. And so,
the answer is y0 prime.

661
00:44:06,000 --> 00:44:12,000
So, check, check,

662
00:44:08,000 --> 00:44:14,000
this works.
So, these better solutions have

663
00:44:11,000 --> 00:44:17,000
the property,
what's good about them,

664
00:44:14,000 --> 00:44:20,000
and why scientists and
engineers like them,

665
00:44:18,000 --> 00:44:24,000
is that they enable you
immediately to write down the

666
00:44:22,000 --> 00:44:28,000
answer to the initial value
problem without having to go

667
00:44:26,000 --> 00:44:32,000
through this business,
which I buried down here,

668
00:44:30,000 --> 00:44:36,000
of solving simultaneous linear
equations.

669
00:44:35,000 --> 00:44:41,000
Okay, now, believe it or not,
that's all the work.

670
00:44:40,000 --> 00:44:46,000
We are ready to answer question
number two: why are these all

671
00:44:46,000 --> 00:44:52,000
the solutions?
Of course, I have to invoke a

672
00:44:50,000 --> 00:44:56,000
big theorem.
A big theorem:

673
00:44:53,000 --> 00:44:59,000
where shall I invoke a big
theorem?

674
00:44:56,000 --> 00:45:02,000
Let's see if we can do it here.
The big theorem says,

675
00:45:02,000 --> 00:45:08,000
it's called the existence and
uniqueness theorem.

676
00:45:05,000 --> 00:45:11,000
It's the last thing that's
proved at the end of an analysis

677
00:45:08,000 --> 00:45:14,000
course, at which real analysis
courses, over which students

678
00:45:11,000 --> 00:45:17,000
sweat for one whole semester,
and their reward at the end is,

679
00:45:15,000 --> 00:45:21,000
if they are very lucky,
and if they have been very good

680
00:45:18,000 --> 00:45:24,000
students, they get to see the
proof of the existence and

681
00:45:21,000 --> 00:45:27,000
uniqueness theorem for
differential equations.

682
00:45:24,000 --> 00:45:30,000
But, I can at least say what it
is for the linear equation

683
00:45:27,000 --> 00:45:33,000
because it's so simple.
It says, so,

684
00:45:31,000 --> 00:45:37,000
the equation we are talking
about is the usual one,

685
00:45:35,000 --> 00:45:41,000
homogeneous equation,
and I'm going to assume,

686
00:45:39,000 --> 00:45:45,000
you have to have assumptions
that p and q are continuous for

687
00:45:44,000 --> 00:45:50,000
all x.
So, they're good-looking

688
00:45:46,000 --> 00:45:52,000
functions.
Coefficients aren't allowed to

689
00:45:50,000 --> 00:45:56,000
blow up anywhere.
They've got to look nice.

690
00:45:55,000 --> 00:46:01,000
Then, the theorem says there is
one and only one solution,

691
00:46:03,000 --> 00:46:09,000
one and only solution
satisfying, given initial values

692
00:46:11,000 --> 00:46:17,000
such that y of zero,
let's say y of zero is equal to

693
00:46:19,000 --> 00:46:25,000
some given number, A,

694
00:46:24,000 --> 00:46:30,000
and y, let's make y0,
and y prime of zero equals B.

695
00:46:31,000 --> 00:46:37,000
The initial value problem has

696
00:46:37,000 --> 00:46:43,000
one and only one solution.
The existence is,

697
00:46:42,000 --> 00:46:48,000
it has a solution.
The uniqueness is,

698
00:46:45,000 --> 00:46:51,000
it only has one solution.
If you specify the initial

699
00:46:51,000 --> 00:46:57,000
conditions, there's only one
function which satisfies them

700
00:46:56,000 --> 00:47:02,000
and at the same time satisfies
that differential equation.

701
00:47:02,000 --> 00:47:08,000
Now, this answers our question.
This answers our question,

702
00:47:08,000 --> 00:47:14,000
because, look,
what I want is all solutions.

703
00:47:14,000 --> 00:47:20,000
What we want are all solutions
to the ODE.

704
00:47:21,000 --> 00:47:27,000
And now, here's what I say:
a claim that this collection of

705
00:47:33,000 --> 00:47:39,000
functions, c1 Y1 plus c2 Y2
are all

706
00:47:42,000 --> 00:47:48,000
solutions.
Of course, I began a period by

707
00:47:48,000 --> 00:47:54,000
saying I'd show you that c1
little y1 c little y2 are all

708
00:47:52,000 --> 00:47:58,000
the solutions.
But, it's the case that these

709
00:47:55,000 --> 00:48:01,000
two families are the same.
So, the family that I started

710
00:48:00,000 --> 00:48:06,000
with would be exactly the same
as the family c1 prime Y1

711
00:48:04,000 --> 00:48:10,000
because,
after all, these are two

712
00:48:07,000 --> 00:48:13,000
special guys from that
collection.

713
00:48:11,000 --> 00:48:17,000
So, it doesn't matter whether I
talk about the original ones,

714
00:48:15,000 --> 00:48:21,000
or these.
The theorem is still the same.

715
00:48:19,000 --> 00:48:25,000
The final step,
therefore, if you give me one

716
00:48:22,000 --> 00:48:28,000
more minute, I think that will
be quite enough.

717
00:48:26,000 --> 00:48:32,000
Why are these all the
solutions?

718
00:48:30,000 --> 00:48:36,000
Well, I have to take an
arbitrary solution and show you

719
00:48:35,000 --> 00:48:41,000
that it's one of these.
So, the proof is,

720
00:48:39,000 --> 00:48:45,000
given a solution,
u(x), what are its values?

721
00:48:43,000 --> 00:48:49,000
Well, u of x zero is u zero,

722
00:48:48,000 --> 00:48:54,000
and u prime of x zero, zero,

723
00:48:51,000 --> 00:48:57,000
let's say, is equal to u zero,
is equal to some other

724
00:48:58,000 --> 00:49:04,000
number.
Now, what's the solution?

725
00:49:02,000 --> 00:49:08,000
Write down what's the solution
of these using the Y1's?

726
00:49:07,000 --> 00:49:13,000
Then, I know I've just shown
you that u zero times Y1 plus u

727
00:49:12,000 --> 00:49:18,000
zero prime Y2
satisfies the same initial

728
00:49:17,000 --> 00:49:23,000
conditions, satisfies these
initial conditions,

729
00:49:21,000 --> 00:49:27,000
initial values.
In other words,

730
00:49:24,000 --> 00:49:30,000
I started with my little
solution.

731
00:49:27,000 --> 00:49:33,000
u of x walks up to and
says, hi there.

732
00:49:31,000 --> 00:49:37,000
Hi there, and the differential
equation looks at it and says,

733
00:49:37,000 --> 00:49:43,000
who are you?
You say, oh,

734
00:49:41,000 --> 00:49:47,000
I satisfy you and my initial,
and then it says what are your

735
00:49:46,000 --> 00:49:52,000
initial values?
It says, my initial values are

736
00:49:50,000 --> 00:49:56,000
u0 and u0 prime.
And, it said,

737
00:49:54,000 --> 00:50:00,000
sorry, but we've got one of
ours who satisfies the same

738
00:49:59,000 --> 00:50:05,000
initial conditions.
We don't need you because the

739
00:50:03,000 --> 00:50:09,000
existence and uniqueness theorem
says that there can only be one

740
00:50:09,000 --> 00:50:15,000
function which does that.
And therefore,

741
00:50:13,000 --> 00:50:19,000
you must be equal to this guy
by the uniqueness theorem.

742
00:50:20,000 --> 00:50:26,000
Okay, we'll talk more about
stuff next time,

743
00:50:23,000 --> 00:50:29,000
linear equation next time.