1
00:00:07,025 --> 00:00:07,650
LINAN CHEN: Hi.

2
00:00:07,650 --> 00:00:09,790
Welcome back to recitation.

3
00:00:09,790 --> 00:00:12,960
In the lecture, you've learned
eigenvalues and eigenvectors

4
00:00:12,960 --> 00:00:14,170
of a matrix.

5
00:00:14,170 --> 00:00:17,250
One of the many important
applications of them

6
00:00:17,250 --> 00:00:20,720
is solving a higher-order
linear differential equation

7
00:00:20,720 --> 00:00:22,760
with constant coefficients.

8
00:00:22,760 --> 00:00:27,340
A typical example is like what
I've written on the board here.

9
00:00:27,340 --> 00:00:32,250
y is a function of t,
and y and its derivatives

10
00:00:32,250 --> 00:00:34,510
satisfy this equation.

11
00:00:34,510 --> 00:00:38,830
As you can see, it involves
y, y prime, and all the way

12
00:00:38,830 --> 00:00:41,370
to its third derivative.

13
00:00:41,370 --> 00:00:45,700
So our first goal is to solve
this differential equation

14
00:00:45,700 --> 00:00:50,440
for its general solution
using the method of matrix.

15
00:00:50,440 --> 00:00:52,830
So the very first
thing that we should do

16
00:00:52,830 --> 00:00:56,640
is to find out which matrix
we should be working with.

17
00:00:56,640 --> 00:01:01,380
So after that, we also
want to say something about

18
00:01:01,380 --> 00:01:05,030
the explanation of
this matrix A*t.

19
00:01:05,030 --> 00:01:08,570
We want to find out the
first column of this matrix

20
00:01:08,570 --> 00:01:10,750
exponential.

21
00:01:10,750 --> 00:01:13,940
Why don't you hit the pause
now, and try to write down

22
00:01:13,940 --> 00:01:16,420
this matrix A by yourself.

23
00:01:16,420 --> 00:01:19,270
But before you continue,
make sure you come back

24
00:01:19,270 --> 00:01:21,370
to this video and
check with me you've

25
00:01:21,370 --> 00:01:23,850
got the correct A. I'll
see you in a while.

26
00:01:33,940 --> 00:01:37,950
OK, let's work together
to transform this problem

27
00:01:37,950 --> 00:01:40,220
into linear algebra.

28
00:01:40,220 --> 00:01:48,770
The idea is to put
y double prime,

29
00:01:48,770 --> 00:01:55,560
y prime, and y
together as a vector.

30
00:01:55,560 --> 00:01:58,180
And let me call this vector u.

31
00:01:58,180 --> 00:02:07,420
So of course, vector u
is also a function in t.

32
00:02:07,420 --> 00:02:09,720
So this is vector u.

33
00:02:09,720 --> 00:02:12,790
If this is u, what's
going to be u prime?

34
00:02:12,790 --> 00:02:18,350
OK, u prime is going to
be-- so I take derivative

35
00:02:18,350 --> 00:02:23,930
of every coordinate here that's
going to be y triple prime,

36
00:02:23,930 --> 00:02:27,170
y double prime, and y prime.

37
00:02:27,170 --> 00:02:31,580
So this is our u prime t.

38
00:02:34,210 --> 00:02:40,670
And my goal is to write
u prime as a matrix,

39
00:02:40,670 --> 00:02:46,710
call it A, times
vector u itself.

40
00:02:46,710 --> 00:02:49,090
So I want to put a matrix here.

41
00:02:52,720 --> 00:02:56,660
And I want to create this
matrix by incorporating

42
00:02:56,660 --> 00:02:59,170
this differential equation.

43
00:02:59,170 --> 00:03:02,590
If you move everything
except y triple prime

44
00:03:02,590 --> 00:03:05,060
to the right-hand
side of the equation,

45
00:03:05,060 --> 00:03:10,290
you can read y triple prime
is equal to negative 2 times

46
00:03:10,290 --> 00:03:14,240
y double prime-- so y
triple prime is negative

47
00:03:14,240 --> 00:03:19,960
2 times y double prime--
plus y prime-- that's

48
00:03:19,960 --> 00:03:24,550
plus 1 times y prime-- plus 2y.

49
00:03:24,550 --> 00:03:28,440
That's 2y, right?

50
00:03:28,440 --> 00:03:30,770
That gives you the first row.

51
00:03:30,770 --> 00:03:33,270
Then look at the
second coordinate,

52
00:03:33,270 --> 00:03:34,970
this y double prime.

53
00:03:34,970 --> 00:03:37,470
y double prime is simply itself.

54
00:03:37,470 --> 00:03:43,260
So you read y double prime is
equal to 1 y double prime, then

55
00:03:43,260 --> 00:03:45,650
0, 0.

56
00:03:45,650 --> 00:03:46,820
That's the second row.

57
00:03:46,820 --> 00:03:49,850
Well, same thing
happens to the last row.

58
00:03:49,850 --> 00:03:52,390
y prime is again itself.

59
00:03:52,390 --> 00:03:58,660
So that's 0, 1, and 0.

60
00:03:58,660 --> 00:04:03,680
That is our matrix A. Did
you get the right answer?

61
00:04:03,680 --> 00:04:07,160
So we have transformed
this equation,

62
00:04:07,160 --> 00:04:11,770
this third-order ordinary
differential equation of y

63
00:04:11,770 --> 00:04:16,540
into a first-order
differential equation of u(t).

64
00:04:16,540 --> 00:04:19,480
Although u(t) is a
vector, but if we

65
00:04:19,480 --> 00:04:22,000
can solve this
equation for u, we

66
00:04:22,000 --> 00:04:25,380
have all the information
we need for y.

67
00:04:25,380 --> 00:04:30,490
So let's plan on
solving this equation.

68
00:04:30,490 --> 00:04:32,855
In order to solve
this equation, we

69
00:04:32,855 --> 00:04:38,240
will need the eigenvalues and
eigenvectors of this matrix A.

70
00:04:38,240 --> 00:04:40,930
Again, this is a good
practice for you.

71
00:04:40,930 --> 00:04:43,220
Why don't you pause
the video again,

72
00:04:43,220 --> 00:04:46,330
and try to complete this
problem on your own.

73
00:04:46,330 --> 00:04:48,970
When you're ready, I'm going
to come back and show you

74
00:04:48,970 --> 00:04:49,820
how I did it.

75
00:05:01,170 --> 00:05:03,520
Let's finish up
everything together.

76
00:05:03,520 --> 00:05:07,540
So as we said, we need the
eigenvalues and eigenvectors

77
00:05:07,540 --> 00:05:13,395
of matrix A, and that involves
computing the determinant

78
00:05:13,395 --> 00:05:15,660
of the following matrix.

79
00:05:15,660 --> 00:05:20,540
So I want to compute the
determinant of A minus lambda

80
00:05:20,540 --> 00:05:23,550
times the identity matrix I.

81
00:05:23,550 --> 00:05:25,180
Let's write it out.

82
00:05:25,180 --> 00:05:36,060
That's the determinant
of -2 minus lambda, 1 2;

83
00:05:36,060 --> 00:05:42,540
1, negative lambda, 0;
and 0, 1, negative lambda.

84
00:05:45,850 --> 00:05:50,290
So we need the determinant of
this three by three matrix.

85
00:05:50,290 --> 00:05:52,220
Do it in your favorite way.

86
00:05:52,220 --> 00:05:55,570
You can either use the
big summation formula,

87
00:05:55,570 --> 00:06:00,710
or you can do by cofactor
along any row or any column.

88
00:06:00,710 --> 00:06:04,140
The correct answer
should be this

89
00:06:04,140 --> 00:06:10,570
is equal to 1 minus
lambda times 1 plus lambda

90
00:06:10,570 --> 00:06:11,900
times 2 plus lambda.

91
00:06:14,830 --> 00:06:21,790
And this polynomial has
three roots: 1, -1, and -2.

92
00:06:21,790 --> 00:06:24,380
These are the eigenvalues
we're looking for.

93
00:06:24,380 --> 00:06:26,240
So let me write it here.

94
00:06:26,240 --> 00:06:30,700
Lambda_1 is equal to 1.

95
00:06:30,700 --> 00:06:34,990
Lambda_2 is equal to -1.

96
00:06:34,990 --> 00:06:42,970
And lambda_3 is equal to -2.

97
00:06:42,970 --> 00:06:47,040
So now what we need is the
eigenvector corresponding

98
00:06:47,040 --> 00:06:49,620
to each eigenvalue.

99
00:06:49,620 --> 00:06:52,820
Let's take lambda_1 for example.

100
00:06:52,820 --> 00:06:57,130
The eigenvector of A
corresponding to lambda_1 is

101
00:06:57,130 --> 00:07:02,390
in the null space of the
matrix A minus lambda_1*I,

102
00:07:02,390 --> 00:07:07,230
so in this case it's A minus
I. So it's in the null space

103
00:07:07,230 --> 00:07:08,870
of this matrix.

104
00:07:08,870 --> 00:07:12,360
In other words, we are
looking for a vector,

105
00:07:12,360 --> 00:07:18,600
let's call it [a, b, c], a
column vector a, b, and c, such

106
00:07:18,600 --> 00:07:23,560
that this matrix multiplying
[a,  b, c] gives me 0.

107
00:07:23,560 --> 00:07:25,960
So if you write it
out, that's going

108
00:07:25,960 --> 00:07:39,150
to be A minus I is
[-3, 1, 2; 1,  -1, 0; 0, 1, -1]

109
00:07:39,150 --> 00:07:44,175
times [a; b; c] is equal to 0.

110
00:07:46,960 --> 00:07:47,460
OK.

111
00:07:47,460 --> 00:07:50,340
Could we choose
constants a, b, c

112
00:07:50,340 --> 00:07:53,360
such that this is always true?

113
00:07:53,360 --> 00:07:58,390
Well if you read the last
row, so the last dot product,

114
00:07:58,390 --> 00:08:04,030
it says that b has
to be equal to c.

115
00:08:04,030 --> 00:08:07,090
And if you read the
second row it says

116
00:08:07,090 --> 00:08:09,690
that a has to be equal to b.

117
00:08:12,460 --> 00:08:16,750
Which means a is equal
to b is equal to c.

118
00:08:16,750 --> 00:08:20,140
And if this relation is
true, the first product

119
00:08:20,140 --> 00:08:23,170
is always going to be 0.

120
00:08:23,170 --> 00:08:26,770
So that simply
means we can choose

121
00:08:26,770 --> 00:08:29,780
the first eigenvector,
the eigenvector

122
00:08:29,780 --> 00:08:34,690
corresponding to
lambda_1, to be x_1 is

123
00:08:34,690 --> 00:08:41,100
equal to [1, 1, 1] transpose.

124
00:08:41,100 --> 00:08:43,760
So we choose the
first eigenvector

125
00:08:43,760 --> 00:08:50,290
to be the column vector with
all the coordinates being 1.

126
00:08:50,290 --> 00:08:55,370
And you can do the same thing
to lambda_2 and lambda_3.

127
00:08:55,370 --> 00:08:58,600
But please allow me to
skip the computation here.

128
00:08:58,600 --> 00:09:00,900
I'm going to write out
the answer for you.

129
00:09:00,900 --> 00:09:06,040
So x_2 the eigenvector
corresponding

130
00:09:06,040 --> 00:09:08,340
to the second
eigenvalue, is going

131
00:09:08,340 --> 00:09:13,630
to be equal to 1, -1, and 1.

132
00:09:13,630 --> 00:09:23,190
And x_3 is going
to be 4, -2, and 1.

133
00:09:23,190 --> 00:09:26,030
Now we've got everything
we need in order

134
00:09:26,030 --> 00:09:30,130
to create the general
solutions for u(t)

135
00:09:30,130 --> 00:09:34,720
So we have eigenvalues, we have
the corresponding eigenvectors.

136
00:09:34,720 --> 00:09:36,660
What should be u(t)?

137
00:09:36,660 --> 00:09:45,550
The general solution for u(t)
is equal to some constant C_1

138
00:09:45,550 --> 00:09:49,280
times e to the power
lambda_1*t-- so in this case,

139
00:09:49,280 --> 00:09:50,890
e to the power t.

140
00:09:50,890 --> 00:09:56,250
Then times the first
eigenvector, x_1.

141
00:09:56,250 --> 00:10:01,160
Plus some other constant
C_2 times e to the power

142
00:10:01,160 --> 00:10:02,140
lambda_2*t--

143
00:10:02,140 --> 00:10:07,750
so e to the power
negative t-- times x_2.

144
00:10:07,750 --> 00:10:09,840
That's the second eigenvector.

145
00:10:09,840 --> 00:10:16,870
Then plus some other constant,
C_3 times e to the power

146
00:10:16,870 --> 00:10:22,190
lambda_3t-- so negative
2t-- times x_3.

147
00:10:22,190 --> 00:10:25,740
That gives you the
general solution for u.

148
00:10:25,740 --> 00:10:29,080
As we just said, if
you know what u is,

149
00:10:29,080 --> 00:10:33,030
you have all the
information you need for y.

150
00:10:33,030 --> 00:10:36,340
Just in case you're
curious about what y is,

151
00:10:36,340 --> 00:10:38,850
you can just read
the last coordinate

152
00:10:38,850 --> 00:10:43,160
of x_1, x_2, and x_3.

153
00:10:43,160 --> 00:10:46,920
And you can see that
all of them are 1.

154
00:10:46,920 --> 00:10:57,986
So y(t) is simply equal to C_1
e to the power lambda t plus C_2

155
00:10:57,986 --> 00:11:06,290
e to the power negative t plus
C_3 e to the power negative 2t.

156
00:11:06,290 --> 00:11:12,300
And the choice of C_1, C_2, and
C_3 is completely arbitrary.

157
00:11:12,300 --> 00:11:16,030
So that completes the first
part of this question.

158
00:11:16,030 --> 00:11:19,490
In the second part, we
want to say something about

159
00:11:19,490 --> 00:11:21,840
the exponential of A*t.

160
00:11:21,840 --> 00:11:25,180
So let me first give you
the recipe to cook up

161
00:11:25,180 --> 00:11:28,680
the exponential of A*t.

162
00:11:28,680 --> 00:11:36,320
The exponential of A*t is
equal to the product of three

163
00:11:36,320 --> 00:11:38,050
matrices.

164
00:11:38,050 --> 00:11:42,990
So you usually we
denote them by S times e

165
00:11:42,990 --> 00:11:49,140
to the power capital
lambda t times S inverse.

166
00:11:49,140 --> 00:11:53,320
And you may ask what S is,
and what this matrix is.

167
00:11:53,320 --> 00:12:00,370
So S is the matrix that has x_1,
x_2, and x_3 being its column

168
00:12:00,370 --> 00:12:01,560
vectors.

169
00:12:01,560 --> 00:12:09,270
So S is x_1, x_2, and x_3.

170
00:12:09,270 --> 00:12:11,600
Let me copy it down here.

171
00:12:11,600 --> 00:12:19,870
So that's 1, 1, 1;
1 -1, 1; 4, -2, 1.

172
00:12:22,460 --> 00:12:26,780
And the matrix in the middle,
e to the power lambda*t is

173
00:12:26,780 --> 00:12:28,610
a diagonal matrix.

174
00:12:28,610 --> 00:12:34,060
So e to the power lambda*t,
it's a diagonal matrix,

175
00:12:34,060 --> 00:12:39,300
and its diagonal entries
are given by e to the power

176
00:12:39,300 --> 00:12:42,532
lambda_1*t-- so that's
e to the power t--

177
00:12:42,532 --> 00:12:45,770
then e to the power
lambda_2*t-- negative t--

178
00:12:45,770 --> 00:12:48,330
and e to the power
lambda_3*t-- negative 2t.

179
00:12:51,490 --> 00:12:53,190
0 everywhere else.

180
00:12:53,190 --> 00:12:55,850
So that's e to the
power lambda*t.

181
00:12:55,850 --> 00:12:58,860
Then the exponential
of this At is

182
00:12:58,860 --> 00:13:02,490
given by the product of
these three matrices.

183
00:13:02,490 --> 00:13:05,310
It looks a bit complicated
because it involves

184
00:13:05,310 --> 00:13:09,010
the inverse of S.
But luckily, we only

185
00:13:09,010 --> 00:13:12,400
want the first
column of the result.

186
00:13:12,400 --> 00:13:22,710
So if we consider this
product, we can see:

187
00:13:22,710 --> 00:13:25,370
the product of the
first two matrices

188
00:13:25,370 --> 00:13:30,220
is relatively easy, because
this is a diagonal matrix,

189
00:13:30,220 --> 00:13:33,730
and we know that S is
given by these columns.

190
00:13:33,730 --> 00:13:36,510
So the result of the
product of these two

191
00:13:36,510 --> 00:13:39,130
is simply multiplying
the columns

192
00:13:39,130 --> 00:13:44,220
of S by these
coefficients respectively.

193
00:13:44,220 --> 00:13:47,610
So you expect to get e
to the power lambda_t

194
00:13:47,610 --> 00:13:52,230
x_1 times e to
the power-- sorry.

195
00:13:52,230 --> 00:13:56,220
The second column should be e
to the power negative t, x_2.

196
00:13:56,220 --> 00:13:59,556
The third column should be e
to the power negative 2t, x_3.

197
00:14:02,870 --> 00:14:07,570
And here, what we
should put is S inverse.

198
00:14:07,570 --> 00:14:10,680
But we don't need
everything from S inverse,

199
00:14:10,680 --> 00:14:13,770
because as we just
said, we only need

200
00:14:13,770 --> 00:14:17,660
the first column of this
result. And the first column

201
00:14:17,660 --> 00:14:20,050
of this product is
going to be given

202
00:14:20,050 --> 00:14:23,940
by linear combinations
of these columns,

203
00:14:23,940 --> 00:14:27,790
and the coefficients are going
to be given by the first column

204
00:14:27,790 --> 00:14:29,430
S inverse.

205
00:14:29,430 --> 00:14:32,720
So our goal should
be just to get

206
00:14:32,720 --> 00:14:37,320
the first column of S inverse.

207
00:14:37,320 --> 00:14:40,290
Then what is the first
column of S inverse?

208
00:14:40,290 --> 00:14:44,040
Well, the formula
for S inverse is

209
00:14:44,040 --> 00:14:47,510
S inverse is going to be the
reciprocal of the determinant

210
00:14:47,510 --> 00:14:50,770
of S, so 1 over
determinant of S,

211
00:14:50,770 --> 00:14:57,130
times the transpose of a
matrix C. This matrix C,

212
00:14:57,130 --> 00:15:02,680
the entries of this matrix C are
given by cofactors of matrix S.

213
00:15:02,680 --> 00:15:05,660
And then you take transpose,
you divide everything

214
00:15:05,660 --> 00:15:11,270
by the determinant of S. The
result will be S inverse.

215
00:15:11,270 --> 00:15:15,700
And we only need the first
column of this matrix.

216
00:15:15,700 --> 00:15:18,220
Let's try to write
the first column out.

217
00:15:18,220 --> 00:15:20,930
Well again, do it
in your favorite way

218
00:15:20,930 --> 00:15:25,830
to compute the determinant of S.
The result should be 1 over 6.

219
00:15:25,830 --> 00:15:28,650
So the determinant of S is 6.

220
00:15:28,650 --> 00:15:32,080
Then what is the first
column of C transpose?

221
00:15:32,080 --> 00:15:37,360
Well we can read it from here.

222
00:15:37,360 --> 00:15:40,600
This spot, the (1,
1) spot, should be

223
00:15:40,600 --> 00:15:44,560
the cofactor of this spot here.

224
00:15:44,560 --> 00:15:49,800
That negative 1 minus
negative 2, which is 1,

225
00:15:49,800 --> 00:15:51,390
so we put 1 here.

226
00:15:55,230 --> 00:16:03,660
Now this spot will be the
cofactor of this entry here.

227
00:16:03,660 --> 00:16:08,390
so that's 1 minus
negative 2, that's 3.

228
00:16:08,390 --> 00:16:11,940
But this is (1, 2) entry, so
you should put a negative sign

229
00:16:11,940 --> 00:16:12,700
in the front.

230
00:16:16,850 --> 00:16:22,590
Then the last spot should be
the cofactor of this entry here,

231
00:16:22,590 --> 00:16:26,520
which is 1 minus
negative 1, that's 2.

232
00:16:29,220 --> 00:16:30,360
Something else here.

233
00:16:34,240 --> 00:16:35,710
Two warnings.

234
00:16:35,710 --> 00:16:39,700
First, don't forget
this transpose sign.

235
00:16:39,700 --> 00:16:44,430
Second, don't forget
this negative sign.

236
00:16:44,430 --> 00:16:47,360
We've got the first
column of S inverse,

237
00:16:47,360 --> 00:16:49,320
and that's all we need.

238
00:16:49,320 --> 00:16:50,720
So we put it here.

239
00:16:50,720 --> 00:16:57,780
That's 1 over 6, -1/2, and 1/3.

240
00:17:02,750 --> 00:17:04,210
That's good enough for me.

241
00:17:04,210 --> 00:17:08,700
Now I can read out the first
column of exponential of A*t.

242
00:17:08,700 --> 00:17:13,050
So the first column of
the exponential of A*t,

243
00:17:13,050 --> 00:17:16,060
I'm going to write it here.

244
00:17:16,060 --> 00:17:20,690
That's going to be equal
to the linear combination

245
00:17:20,690 --> 00:17:21,990
of these columns.

246
00:17:21,990 --> 00:17:25,154
So that's 1/6 of
the first column,

247
00:17:25,154 --> 00:17:30,580
that's e to the power
t over 6 times x_1.

248
00:17:30,580 --> 00:17:37,700
Plus this times this, so that's
going to be minus 1/2 of e

249
00:17:37,700 --> 00:17:42,040
to the power
negative t times x_2.

250
00:17:42,040 --> 00:17:50,660
Then plus 1/3 of e to the
power negative 2t times x_3.

251
00:17:50,660 --> 00:17:53,940
That's the first column
of the exponential A*t.

252
00:17:53,940 --> 00:17:55,980
And then with the
other two columns.

253
00:18:00,750 --> 00:18:02,520
That's the answer.

254
00:18:02,520 --> 00:18:07,740
If you want more practice, you
can certainly complete this S

255
00:18:07,740 --> 00:18:11,720
inverse, and then you can
also complete the exponential

256
00:18:11,720 --> 00:18:13,280
of A*t.

257
00:18:13,280 --> 00:18:16,360
But I will leave
the rest to you.

258
00:18:16,360 --> 00:18:19,120
OK, I hope this
example shows you

259
00:18:19,120 --> 00:18:23,160
that linear algebra can be
a powerful tool in solving

260
00:18:23,160 --> 00:18:25,760
higher-order ordinary
differential equations

261
00:18:25,760 --> 00:18:27,890
with constant coefficients.

262
00:18:27,890 --> 00:18:32,110
And we have demonstrated the
standard procedure to do it,

263
00:18:32,110 --> 00:18:34,720
and we also practiced
how to calculate

264
00:18:34,720 --> 00:18:37,140
the exponential of a matrix.

265
00:18:37,140 --> 00:18:40,140
Thanks for watching,
and see you next time.