1
00:00:07,160 --> 00:00:09,189
PROFESSOR: Welcome
back to recitation.

2
00:00:09,189 --> 00:00:11,730
Today we're going to work on a
problem involving differential

3
00:00:11,730 --> 00:00:12,430
equations.

4
00:00:12,430 --> 00:00:14,860
I'm going to read it to
you, give you a little hint,

5
00:00:14,860 --> 00:00:16,340
give you some time
to work on it,

6
00:00:16,340 --> 00:00:18,810
and then I'll be back
and work it out for you.

7
00:00:18,810 --> 00:00:22,210
So, the problem is to
find a function y equals

8
00:00:22,210 --> 00:00:26,240
f of x that has the following
two properties: d squared

9
00:00:26,240 --> 00:00:28,119
y dx squared is equal to 6x.

10
00:00:28,119 --> 00:00:30,410
And just to remind you what
this means, this is really,

11
00:00:30,410 --> 00:00:34,940
this is the second derivative
of y with respect to x.

12
00:00:34,940 --> 00:00:38,900
So, the second derivative with
respect to x should be 6x.

13
00:00:38,900 --> 00:00:41,210
And the second
condition's kind of long,

14
00:00:41,210 --> 00:00:46,070
but it says the graph of the
function in the xy-plane passes

15
00:00:46,070 --> 00:00:49,880
through the point (1, 1) with
a horizontal tangent there.

16
00:00:49,880 --> 00:00:51,240
So let me give you one hint.

17
00:00:51,240 --> 00:00:53,920
And that hint is that there are
some initial conditions buried

18
00:00:53,920 --> 00:00:54,420
in here.

19
00:00:54,420 --> 00:00:56,940
That's why we have
this condition.

20
00:00:56,940 --> 00:00:59,390
So I'm going to give
you a little bit of time

21
00:00:59,390 --> 00:01:01,940
to work on it and I'll be back
and I'll work it out with you.

22
00:01:10,530 --> 00:01:11,650
Welcome back.

23
00:01:11,650 --> 00:01:13,370
Hopefully you were
able to start at least

24
00:01:13,370 --> 00:01:16,710
solving the problem
initially, give yourself

25
00:01:16,710 --> 00:01:17,890
a little direction.

26
00:01:17,890 --> 00:01:20,460
So let's see how you did.

27
00:01:20,460 --> 00:01:22,240
OK, so the first thing
I would like to do

28
00:01:22,240 --> 00:01:26,570
is try and figure out maybe
what the first derivative

29
00:01:26,570 --> 00:01:29,490
of the function y
equals f of x is.

30
00:01:29,490 --> 00:01:31,410
I have its second
derivative, so in order

31
00:01:31,410 --> 00:01:33,370
to find the first
derivative I want

32
00:01:33,370 --> 00:01:37,000
to find, ultimately, a function
that when I take its derivative

33
00:01:37,000 --> 00:01:37,500
I get 6x.

34
00:01:37,500 --> 00:01:38,390
Right?

35
00:01:38,390 --> 00:01:40,530
So, we can think about this.

36
00:01:40,530 --> 00:01:42,140
Maybe the easiest
thing for us to do

37
00:01:42,140 --> 00:01:45,970
would actually be to
consider another function

38
00:01:45,970 --> 00:01:47,600
whose derivative is 6x.

39
00:01:47,600 --> 00:01:49,380
And we'll know that's dy dx.

40
00:01:49,380 --> 00:01:55,070
So I'm going to say this: dy--
sorry-- d squared y dx squared,

41
00:01:55,070 --> 00:01:58,780
I'm going to say is the first
derivative of another function,

42
00:01:58,780 --> 00:02:01,402
we'll say dw/dx.

43
00:02:01,402 --> 00:02:02,860
And the reason I'm
going to do that

44
00:02:02,860 --> 00:02:05,100
is so we're not too
nervous about how

45
00:02:05,100 --> 00:02:06,630
we solve this problem.

46
00:02:06,630 --> 00:02:08,800
So let's just assume that.

47
00:02:08,800 --> 00:02:10,890
So I'm introducing
another function w,

48
00:02:10,890 --> 00:02:14,930
which is the first
derivative of y with respect

49
00:02:14,930 --> 00:02:15,980
to x, ultimately.

50
00:02:15,980 --> 00:02:18,600
Because its derivative
is the second derivative

51
00:02:18,600 --> 00:02:20,130
of y with respect to x.

52
00:02:20,130 --> 00:02:25,420
So let's see how to solve the
differential equation dw/dx

53
00:02:25,420 --> 00:02:27,040
equals 6x.

54
00:02:27,040 --> 00:02:28,870
Now you may be able
to do that right away.

55
00:02:28,870 --> 00:02:30,100
You may see what
this is right away.

56
00:02:30,100 --> 00:02:31,516
If you're a little
nervous, we can

57
00:02:31,516 --> 00:02:33,050
do a separation of variables.

58
00:02:33,050 --> 00:02:35,910
So right away, maybe
some of you can

59
00:02:35,910 --> 00:02:40,430
see that this will be
2x-- sorry-- 3 x squared.

60
00:02:40,430 --> 00:02:42,100
But let's just double-check.

61
00:02:42,100 --> 00:02:44,450
So, separation of
variables, on this side

62
00:02:44,450 --> 00:02:45,810
we get the integral of dw.

63
00:02:45,810 --> 00:02:48,000
On this side we get
the integral of 6x dx.

64
00:02:50,590 --> 00:02:53,480
Here we get a w.

65
00:02:53,480 --> 00:02:56,290
And here, again we
get, we should really

66
00:02:56,290 --> 00:03:01,740
do 6 x squared over
2 plus a constant.

67
00:03:01,740 --> 00:03:07,060
So that's 3 x squared
plus a constant.

68
00:03:07,060 --> 00:03:08,940
I'm going to write
c_1 here because we're

69
00:03:08,940 --> 00:03:12,990
went to need a little
bit of information.

70
00:03:12,990 --> 00:03:15,370
We're going to need
another constant later.

71
00:03:15,370 --> 00:03:19,860
So we actually now know the
derivative of y with respect

72
00:03:19,860 --> 00:03:20,360
to x.

73
00:03:20,360 --> 00:03:27,342
What we found here, I'll
just write that in, is dy/dx.

74
00:03:27,342 --> 00:03:28,550
Again, let me remind you why.

75
00:03:28,550 --> 00:03:32,540
We had-- we were saying
the second derivative of y

76
00:03:32,540 --> 00:03:36,170
with respect to x we're
going to call dw/dx.

77
00:03:36,170 --> 00:03:42,200
So when I took an integral
of dw/dx, I got w.

78
00:03:42,200 --> 00:03:44,410
So that means I've taken
one antiderivative here

79
00:03:44,410 --> 00:03:46,120
and so now I have dy/dx.

80
00:03:46,120 --> 00:03:48,241
So this is dy/dx.

81
00:03:48,241 --> 00:03:50,830
I'm going to draw a line
and now we want to find y.

82
00:03:50,830 --> 00:03:53,250
And again we can use
separation of variables.

83
00:03:53,250 --> 00:03:55,350
Or this time I'm
just going to do

84
00:03:55,350 --> 00:03:57,610
the problem without
separating variables,

85
00:03:57,610 --> 00:04:01,456
because I actually know what an
antiderivative is of this, what

86
00:04:01,456 --> 00:04:03,410
an antiderivative
is of this, and then

87
00:04:03,410 --> 00:04:05,150
I need to add one more constant.

88
00:04:05,150 --> 00:04:10,340
So I can say that y is
definitely equal to x cubed

89
00:04:10,340 --> 00:04:14,600
plus c_1*x plus c_2.

90
00:04:14,600 --> 00:04:17,315
And let me just again,
let's see why that is.

91
00:04:17,315 --> 00:04:18,330
Right?

92
00:04:18,330 --> 00:04:22,840
This antiderivative
of this is x cubed.

93
00:04:22,840 --> 00:04:25,530
Antiderivative of
this is-- it's a

94
00:04:25,530 --> 00:04:29,130
constant so its
antiderivative is c_1 times x.

95
00:04:29,130 --> 00:04:30,930
And then I have to add
on another constant

96
00:04:30,930 --> 00:04:34,510
because I have a whole
family of possible solutions.

97
00:04:34,510 --> 00:04:35,990
So here I have y.

98
00:04:35,990 --> 00:04:39,850
And now I need to figure
out how to use number two.

99
00:04:39,850 --> 00:04:41,860
Maybe before you even go
on, you want to check,

100
00:04:41,860 --> 00:04:44,700
does this really
satisfy number one?

101
00:04:44,700 --> 00:04:46,420
So if we wanted to
check that, we just

102
00:04:46,420 --> 00:04:48,100
take two derivatives
of this expression

103
00:04:48,100 --> 00:04:49,460
on the right-hand side.

104
00:04:49,460 --> 00:04:51,570
Two derivatives of this is 0.

105
00:04:51,570 --> 00:04:53,370
Two derivative of this is 0.

106
00:04:53,370 --> 00:04:55,240
And two derivatives
of this is 6x.

107
00:04:55,240 --> 00:04:57,600
So I do indeed get what I want.

108
00:04:57,600 --> 00:04:59,480
So now we definitely
can go on to number 2.

109
00:04:59,480 --> 00:05:01,137
OK?

110
00:05:01,137 --> 00:05:01,970
So, what do we have?

111
00:05:01,970 --> 00:05:04,510
Number two, it says we have
some initial conditions here.

112
00:05:04,510 --> 00:05:07,510
The graph in the xy-plane
passes through the point (1, 1)

113
00:05:07,510 --> 00:05:10,550
and it has a horizontal
tangent there.

114
00:05:10,550 --> 00:05:11,960
Now what does that
actually mean?

115
00:05:11,960 --> 00:05:12,918
Let's think about that.

116
00:05:12,918 --> 00:05:21,215
That actually means that
two says f of 1 equals 1.

117
00:05:21,215 --> 00:05:22,970
Right?

118
00:05:22,970 --> 00:05:24,410
y is equal to f of x.

119
00:05:24,410 --> 00:05:27,060
So we can write
this also as f of x.

120
00:05:27,060 --> 00:05:30,410
So two, the first part
says that f of 1 is 1.

121
00:05:30,410 --> 00:05:31,910
And what does this
second condition?

122
00:05:31,910 --> 00:05:33,326
Let's check this
second condition.

123
00:05:33,326 --> 00:05:35,670
It says it has a
horizontal tangent there.

124
00:05:35,670 --> 00:05:39,760
Horizontal tangent means that
its derivative at that x-value

125
00:05:39,760 --> 00:05:41,040
is 0.

126
00:05:41,040 --> 00:05:43,790
So let me write down
that in a nice form.

127
00:05:43,790 --> 00:05:47,904
The derivative at that
x-value is equal to 0.

128
00:05:47,904 --> 00:05:49,320
This is a little
different, maybe,

129
00:05:49,320 --> 00:05:50,653
from what we've seen previously.

130
00:05:50,653 --> 00:05:54,100
In the lecture you saw examples,
at least-- certainly, where

131
00:05:54,100 --> 00:05:55,760
you had one initial condition.

132
00:05:55,760 --> 00:05:57,460
Here we need two
initial conditions.

133
00:05:57,460 --> 00:06:00,450
And you see why, is because we
actually have two constants.

134
00:06:00,450 --> 00:06:01,587
Where did that come from?

135
00:06:01,587 --> 00:06:03,920
It's because we started with
a second derivative instead

136
00:06:03,920 --> 00:06:05,370
of just the first derivative.

137
00:06:05,370 --> 00:06:06,810
So that's kind
of, that's why you

138
00:06:06,810 --> 00:06:11,330
see more initial conditions than
maybe you've seen previously.

139
00:06:11,330 --> 00:06:13,570
So, let's plug these in
and let's see what we get.

140
00:06:13,570 --> 00:06:17,400
If f of 1 equals 1, then
let's evaluate that.

141
00:06:17,400 --> 00:06:26,250
f of 1 is equal to 1 cubed
plus 1 times c_1 plus 1 times--

142
00:06:26,250 --> 00:06:28,910
or just 1, or c_2 there, sorry.

143
00:06:28,910 --> 00:06:30,780
There's no x there, so just c_2.

144
00:06:30,780 --> 00:06:33,454
And that all has to equal 1.

145
00:06:33,454 --> 00:06:35,370
And then let's look at
what the derivative is.

146
00:06:35,370 --> 00:06:38,520
The derivative is
still over here.

147
00:06:38,520 --> 00:06:44,600
So f prime at 1 is equal to 3
times 1 squared, so 3 times 1,

148
00:06:44,600 --> 00:06:46,880
plus c_1.

149
00:06:46,880 --> 00:06:51,410
And the condition says
that equals 0, equals 0.

150
00:06:51,410 --> 00:06:52,490
So we can read off.

151
00:06:52,490 --> 00:06:55,045
The nice thing is this is system
of equations but one of them

152
00:06:55,045 --> 00:06:58,460
is very easy to solve initially,
then we can substitute.

153
00:06:58,460 --> 00:06:59,420
So what does this say?

154
00:06:59,420 --> 00:07:01,990
Well 3 plus c_1 equals 0.

155
00:07:01,990 --> 00:07:06,660
So c_1 is equal to
minus 3, negative 3.

156
00:07:06,660 --> 00:07:10,374
If I plug in negative 3 for
c_1, in this expression,

157
00:07:10,374 --> 00:07:15,010
in this equation up here, I
get negative 3 plus 1 plus c_2

158
00:07:15,010 --> 00:07:16,910
has to equal 1.

159
00:07:16,910 --> 00:07:18,640
If I subtract the
1's from both sides

160
00:07:18,640 --> 00:07:21,580
I get negative 3 plus
c_2 has to equal 0.

161
00:07:21,580 --> 00:07:25,610
So c_2 has to actually equal 3.

162
00:07:25,610 --> 00:07:30,076
So the final, final
answer is evaluating,

163
00:07:30,076 --> 00:07:35,150
or plugging in the c1 and the
c2 in for the constants there.

164
00:07:35,150 --> 00:07:42,820
The final, final answer is
x cubed minus 3x plus 3.

165
00:07:46,270 --> 00:07:49,290
So we started with a
differential equation and two

166
00:07:49,290 --> 00:07:50,720
sets of initial conditions.

167
00:07:50,720 --> 00:07:54,690
And we came up with one
solution that satisfies that.

168
00:07:54,690 --> 00:07:56,730
Now you can look
at this, you could

169
00:07:56,730 --> 00:07:59,020
graph this on a
calculator or computer

170
00:07:59,020 --> 00:08:03,020
and look and see if it satisfies
that it actually passes

171
00:08:03,020 --> 00:08:06,250
through the point (1, 1) and
has a horizontal tangent there.

172
00:08:06,250 --> 00:08:08,840
But we know, based on our
work, that that actually

173
00:08:08,840 --> 00:08:10,230
should happen.

174
00:08:10,230 --> 00:08:12,993
So I think that's
where I'll stop.