1
00:00:04,790 --> 00:00:08,029
PROFESSOR: Welcome to this
session on separable equations.

2
00:00:08,029 --> 00:00:11,000
So in this problem, you're
asked in the first question

3
00:00:11,000 --> 00:00:13,920
to solve the initial
value problem

4
00:00:13,920 --> 00:00:18,150
dy/dx equals y square with the
initial condition y of zero

5
00:00:18,150 --> 00:00:19,024
equals 1.

6
00:00:19,024 --> 00:00:20,440
In the second part
of the problem,

7
00:00:20,440 --> 00:00:22,290
you're asked to find
the general solution

8
00:00:22,290 --> 00:00:24,620
where no initial
condition is imposed.

9
00:00:24,620 --> 00:00:26,890
So here you need to remember
your method of separation

10
00:00:26,890 --> 00:00:29,980
of variables to tackle
the first question.

11
00:00:29,980 --> 00:00:32,369
And then in the second
part of the problem,

12
00:00:32,369 --> 00:00:35,300
remember all the types of
solutions and conditions

13
00:00:35,300 --> 00:00:40,230
that you applied the first part
to recover the lost solutions.

14
00:00:40,230 --> 00:00:42,720
So why don't you take a
minute, pause the video,

15
00:00:42,720 --> 00:00:44,570
work through questions a and b.

16
00:00:44,570 --> 00:00:46,990
And then we'll continue
together when I come back.

17
00:00:59,200 --> 00:01:00,470
Welcome back.

18
00:01:00,470 --> 00:01:04,250
So in the first
part of the problem,

19
00:01:04,250 --> 00:01:07,350
we'll be solving the equation
dy/dx equals y squared.

20
00:01:07,350 --> 00:01:10,780
So here the method of
separation of variables

21
00:01:10,780 --> 00:01:13,630
tells us that we should regroup
the variables of the same kind,

22
00:01:13,630 --> 00:01:18,480
so all the y variables on
one side and dx variable

23
00:01:18,480 --> 00:01:20,300
on the other side
of the equation

24
00:01:20,300 --> 00:01:22,500
and then integrate
from this point.

25
00:01:22,500 --> 00:01:27,030
So here for this step,
notice that I divided by y

26
00:01:27,030 --> 00:01:28,940
squared, which
means that we need

27
00:01:28,940 --> 00:01:32,080
to impose the condition y not
equal to zero from now on.

28
00:01:35,680 --> 00:01:39,910
So from this step, we just
use indefinite integrals

29
00:01:39,910 --> 00:01:44,030
to integrate both
sides of the equation.

30
00:01:44,030 --> 00:01:48,310
So the left-hand side is the
integral dy over y squared.

31
00:01:48,310 --> 00:01:53,380
So integral of this
gives us minus 1 over y.

32
00:01:53,380 --> 00:01:57,820
And the right-hand side,
integral dx, is just x.

33
00:01:57,820 --> 00:02:00,465
Both sides would give us
constant of integrations,

34
00:02:00,465 --> 00:02:02,610
but we only need
one because this

35
00:02:02,610 --> 00:02:04,640
is a first-order
differential equation.

36
00:02:04,640 --> 00:02:06,850
And so we group them together
on the right-hand side

37
00:02:06,850 --> 00:02:07,670
with constant c.

38
00:02:18,850 --> 00:02:22,800
So from this point, given that
we're interested in variable y,

39
00:02:22,800 --> 00:02:25,045
we need just to
invert the expression.

40
00:02:28,230 --> 00:02:30,880
And that gives us
partial answer,

41
00:02:30,880 --> 00:02:34,190
which is y of x equals
minus 1 over x plus c.

42
00:02:34,190 --> 00:02:37,110
So now we need to use our
initial condition, y of zero

43
00:02:37,110 --> 00:02:38,860
equals to 1, to
determine the value

44
00:02:38,860 --> 00:02:41,100
of c for this particular
initial value problem.

45
00:02:49,400 --> 00:02:55,700
So our initial condition
was y of zero equals 1.

46
00:02:55,700 --> 00:02:57,630
So if we substitute
this in the expression

47
00:02:57,630 --> 00:03:03,780
that we just
obtained, we just have

48
00:03:03,780 --> 00:03:11,480
zero plus c, which then
only gives us c equal to 1.

49
00:03:11,480 --> 00:03:15,150
So we end up with the value for
our constant of integration,

50
00:03:15,150 --> 00:03:16,570
c equals minus 1.

51
00:03:20,010 --> 00:03:27,440
And so the solution
to this problem is

52
00:03:27,440 --> 00:03:31,290
y of x equals 1 over 1 minus x.

53
00:03:31,290 --> 00:03:33,090
So if you examine
this expression,

54
00:03:33,090 --> 00:03:35,340
you see right away that
we have a problem for x

55
00:03:35,340 --> 00:03:37,390
equals to 1 because
at x equals to 1,

56
00:03:37,390 --> 00:03:40,440
we have 1 over 0 which means
that, then, the solution blows

57
00:03:40,440 --> 00:03:43,210
up and we have a
vertical asymptote.

58
00:03:43,210 --> 00:03:45,775
So let me draw this here.

59
00:03:52,580 --> 00:03:58,830
So we're going to have
an asymptote at x equals

60
00:03:58,830 --> 00:04:08,010
1 and a solution that passes
through our initial condition,

61
00:04:08,010 --> 00:04:13,670
y equals 1, going to infinity
when approaching x equals 1.

62
00:04:13,670 --> 00:04:18,070
But then on the right side
of the value x equals 1,

63
00:04:18,070 --> 00:04:21,339
we have another part of
the solution that goes

64
00:04:21,339 --> 00:04:23,610
to zero as x goes to infinity.

65
00:04:23,610 --> 00:04:27,300
And that diverges to minus
infinity when x approaches 1.

66
00:04:27,300 --> 00:04:31,240
So by convention, the solutions
of differential equations

67
00:04:31,240 --> 00:04:33,150
are defined on one
single interval.

68
00:04:33,150 --> 00:04:39,600
So we need here to realize
that the solution we had

69
00:04:39,600 --> 00:04:42,140
is basically two
parts, the parts

70
00:04:42,140 --> 00:04:43,950
on the left of the
asymptote and the part

71
00:04:43,950 --> 00:04:45,350
on the right of the asymptote.

72
00:05:01,910 --> 00:05:03,960
So the solution to
our initial value

73
00:05:03,960 --> 00:05:05,680
problems needs to
be the solution that

74
00:05:05,680 --> 00:05:08,290
passes through the imposed
initial condition, which

75
00:05:08,290 --> 00:05:10,270
was y of zero equals to 1.

76
00:05:10,270 --> 00:05:12,615
So it needs to be this solution.

77
00:05:17,270 --> 00:05:23,170
So now, if we move on to the
solution of the second part

78
00:05:23,170 --> 00:05:26,770
of the problem, b, we were asked
to find the general solution

79
00:05:26,770 --> 00:05:29,750
of the problem, which means
that we need to account now

80
00:05:29,750 --> 00:05:31,460
for all the
solutions, regardless

81
00:05:31,460 --> 00:05:32,910
of their initial condition.

82
00:05:32,910 --> 00:05:35,270
So we already answered
this partially

83
00:05:35,270 --> 00:05:38,260
during the solution
of part a where

84
00:05:38,260 --> 00:05:41,550
we solved using
indefinite integrals

85
00:05:41,550 --> 00:05:46,410
and arrived to the
solution minus 1

86
00:05:46,410 --> 00:05:48,540
over x plus c,
where here we had,

87
00:05:48,540 --> 00:05:51,420
basically, an undetermined
constant of integration.

88
00:05:51,420 --> 00:05:54,010
So this is one general solution.

89
00:05:54,010 --> 00:05:57,250
But remember that
we need to give all

90
00:05:57,250 --> 00:05:58,790
the solutions of the problem.

91
00:05:58,790 --> 00:06:00,760
So when we arrived
at the solution,

92
00:06:00,760 --> 00:06:03,760
we excluded the
solution y equals

93
00:06:03,760 --> 00:06:06,530
zero, which was basically
a lost solution because we

94
00:06:06,530 --> 00:06:10,870
had to impose the condition
y not equal to zero.

95
00:06:10,870 --> 00:06:15,310
So we need, when we give
the general solution

96
00:06:15,310 --> 00:06:20,010
to this differential equation,
to recover the lost solution.

97
00:06:20,010 --> 00:06:23,720
And then we basically have
one kind of solution, minus 1

98
00:06:23,720 --> 00:06:26,560
over x plus c, that
excludes y equals to zero

99
00:06:26,560 --> 00:06:28,020
and another kind
of solution that

100
00:06:28,020 --> 00:06:30,760
is simply the zero solution.

101
00:06:30,760 --> 00:06:33,280
So to summarize,
the important points

102
00:06:33,280 --> 00:06:35,820
of this problem is to remember
the separation of variables

103
00:06:35,820 --> 00:06:40,660
and how to use it and
the fact that using it

104
00:06:40,660 --> 00:06:44,544
imposes conditions that require
us to recover lost solutions

105
00:06:44,544 --> 00:06:45,960
at the end of the
problem if we're

106
00:06:45,960 --> 00:06:48,210
asked to give general solution.

107
00:06:48,210 --> 00:06:51,810
Another point to remember
is that even a simple ODE

108
00:06:51,810 --> 00:06:54,814
can lead to relatively
complex behavior which

109
00:06:54,814 --> 00:06:56,730
drives the presence of
this vertical asymptote

110
00:06:56,730 --> 00:07:00,330
that you need to then know
how to deal with and determine

111
00:07:00,330 --> 00:07:02,890
which part of the solutions
that you've obtained

112
00:07:02,890 --> 00:07:05,180
is the real solution to
the initial value problem

113
00:07:05,180 --> 00:07:06,610
that you're given.

114
00:07:06,610 --> 00:07:12,520
So I hope that you are
OK with this problem

115
00:07:12,520 --> 00:07:16,040
and you will use these
approaches many times

116
00:07:16,040 --> 00:07:17,750
for the rest of the course.