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PROFESSOR: Hi everyone.

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Welcome back.

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So in this problem,
I'd like to take

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a look at autonomous
equations and phase lines.

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And specifically, we're going
to take a look at the simple

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equation x dot
equals a*x plus 1,

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which models births and death
rates and a fixed replenishment

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rate for a population.

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So in this case, the
variable a represents births

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minus deaths in a population.

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And 1 represents the constant
input of new creatures.

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So for part A, we're
asked to find intervals

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of the variable
a which determine

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the long-time stability
of the population.

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And then secondly,
for each typical case

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of a in the first
part of the problem,

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we're asked to
sketch a phase line

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and then also to sketch the
solution x of t versus t.

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So I'll let you take a
look at this problem,

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and I'll be back in a moment.

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Hi Everyone.

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Welcome back.

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So the problem we're looking
at is x dot equals a*x.

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And as mentioned in the
title of the problem,

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this is an autonomous
equation which

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means that the right-hand side
is not a function of time.

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The right-hand side
only depends on x.

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And for these types of
problems, the critical points

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or the points at which
the right-hand side vanish

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tend to determine the
long-time characteristics

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of the solution.

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So we can kind of get a
handle on what intervals of a

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determine the long-time
stability by sketching what

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x dot versus x might look like.

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And we see that
if a is positive,

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this represents a line
with a positive slope a.

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So on the right-hand side of
the intercept with the x-axis,

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x dot is positive, which means
that when the line goes through

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the point x dot equals 0, the
term a*x plus 1 is going to be

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positive.

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I'll just point out that
this point right here

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is when x dot is equal to 0.

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And this happens when x is
equal to negative 1 over a.

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So this point right here
is negative 1 over a.

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And when x is below
negative 1 over a,

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x dot is negative, which
means that solutions

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will want to move away from
the point negative 1 over a.

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What other qualitative
behavior could we have?

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Well, when a is below 0,
when a is less than 0,

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the slope of this curve
is going to be negative.

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So I'll plot again
x dot versus x.

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And again, the intercept is
going to be at negative 1

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over a.

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However, when x is above
the critical point,

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x dot is negative.

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So solutions will want to tend
to move back towards negative 1

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over a.

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And when x is below
the critical point,

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we see that x dot is in
the upper plane, which

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means it's positive.

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So solutions will want to grow.

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So we have two cases right now.

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One is when a is bigger than 0.

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One is when a is less than 0.

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And then, of course,
there's going

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to be an intermediate point,
which is when a is equal to 0.

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And in this case, the
curve x dot versus x,

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it's just going to be a straight
line hat goes through 1.

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So this gives us three
regions that will determine

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the long-time behavior.

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So the three regions are
a equals 0, a less than 0,

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and then a bigger than 0.

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So this concludes part
A. And for part B,

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we're asked to investigate the
behavior of each of these three

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regions by constructing
a phase line,

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and then, secondly, by
sketching a solution

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or several solutions.

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So for part B, let's take
a look at the case when

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a is bigger than 0 first.

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And I'm just going to
pick one value of a.

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I'll just pick a is equal to 1
to work things out concretely.

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And we're asked to
sketch a phase line.

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So plot the phase line x here.

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And we know from the ODE
that the critical point

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occurs at negative 1 over a.

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So in this case, there's going
to be one critical point,

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and it's going to
be at negative 1.

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Now when x is bigger
than negative 1,

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we've already seen that x
dot is going to be positive.

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So when x is bigger than
negative 1, in this region,

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x dot is positive.

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So above negative 1,
x dot is positive,

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which means that if
a solution starts

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at a point above
negative 1, it's going

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to constantly increase forever.

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And likewise, if it's
below negative 1,

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it's going to decrease forever.

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So this is x dot is
negative in this region.

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And of course, x dot is 0
at the critical point, which

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means if we start at
the critical point,

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we stay there for all time.

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So let's sketch some solutions.

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So whenever I have to sketch
solutions for a phase line,

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the first solution
I always plot is

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going to be the critical point.

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So note that x equals negative
1 solves the differential

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equation.

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So x equals a critical point
always solves the differential

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equation.

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So that means when x
is equal to negative 1,

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we solve the
differential equation.

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When x is bigger
than negative 1,

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we attain exponential growth.

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And when x is less
than negative 1,

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we have exponential growth
in the opposite direction.

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So this is sometimes said to
be an unstable critical point,

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because any small perturbation
away from negative 1

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will make the solution diverge.

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Also note that once I
have one of these curves,

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I can generate all the others by
just picking it up and shifting

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it over.

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And the reason this works is
because the original equation

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is autonomous, meaning that
the right-hand side didn't

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depend on time.

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So whenever you have
an autonomous equation,

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when you sketch
one solution, you

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can always get the
others by just picking--

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or you can always get
a family of others

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by picking up that solution and
just shifting it to the right

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and to the left.

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Now when a is below
0, I'll just pick

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the value of a
equals negative 1,

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we have a critical point at 1.

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And in this case,
the arrows are going

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to point towards 1 like this.

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And again, I can
sketch some curves.

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So I'll first draw the
solution when x is equal to 1.

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And in this case,
solutions are going

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to converge towards
the critical point.

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And in this case, we say that
the critical point x equals 1

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is stable.

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And that's because any small
perturbation in the solution

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will eventually just come back
to the critical point at x

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equals 1.

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And now lastly, we
have the third case

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which is when a equals 0.

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And in this case, there
are no critical points.

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In fact, x dot was equal to 1.

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So x dot is positive
for all values of x.

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So our phase line is
just a line with arrows.

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And in this case, if I were
to sketch some solutions,

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I'll just draw my axes, x and
t, there are no critical points.

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X is just increasing.

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So the solutions are going
to look something like this.

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They're just going to be a
family of straight lines.

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So we've just sketched
some phase lines,

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and we sketched several
solutions for each phase line.

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We've seen how the critical
points and the arrows

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around the critical points
affect the long time

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behavior of the solutions.

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And this is
typically the process

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that you go through when
looking at autonomous equations.

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You always first, A, find
the critical points, then B,

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find the regions between
the critical points

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and figure out if x dot is
either positive or negative.

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And then that automatically
gives you the phase line.

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Once you have the phase line,
you can always just sketch

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a family of solutions.

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So I'd like to conclude here.

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And I'll see you next time.