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DAVID SHIROKOFF: Hi, everyone.
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Welcome back.
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So today, I'd like to tackle
a problem on direction fields.
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So given a
differential equation,
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y prime equals
negative y divided
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by x squared plus
y squared, we're
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asked to first sketch
the direction field.
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And then secondly,
given the curve
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that goes through
y(0) equal to 1,
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we're asked several
questions about it.
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Why is y of x greater than
0 for x greater than 0?
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And then, why is
y of x decreasing
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for x greater than 0?
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So I'll let you think about this
for a moment and I'll get back.
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Hi, everyone.
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Welcome back.
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OK, so to start
off this problem,
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we're asked to sketch
the direction field.
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Now when asked to sketch
a direction field,
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the first thing to do is to
look at a couple isoclines.
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So what is an isocline?
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Well, it's a curve
where the derivative y
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prime is equal to m,
which is some constant.
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So what are the isoclines for
this differential equation?
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Well, it would be minus
y divided by x squared
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plus y squared equals m.
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Now, of particular
interest, there's
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a very special isocline, which
is usually the easiest to plot.
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And that's a nullcline.
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And this is just a special
case where m is equal to 0.
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So what's the
nullcline for this ODE?
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Well, when m is equal to 0, the
only way that y prime can be 0
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is when y is 0.
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So the nullcline for
this ODE is y equals 0.
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Now, for more general
isoclines, we're
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left with this relation
between x and y.
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And we can massage
this expression
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to see exactly what
the isoclines are.
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So specifically, I'm
going to multiply through
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by x squared plus y squared,
and I'm going to divide by m.
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And we see that we have
a quadratic in x and y.
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And we have one linear term.
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So whenever we have
a relation like this
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and we want to understand
what it looks like,
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typically the approach is
just to complete the square.
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So I'm going to
bring the 1 over m y
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to the other side
of the equation.
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And I'm going to
combine it with y
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squared to complete the square.
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And when we do this, we
obtain the following equation.
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And we recognize this equation
as the equation for a circle.
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Specifically, it's
a circle that's
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centered at x is equal to 0
and y is equal to negative 1
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over 2m.
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And in addition, the
circle has a radius
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r squared is equal to 1
over 4m squared, which
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means its radius is 1 over 2
times the absolute value of m.
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OK, so why don't we take a look
and plot a couple isoclines?
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So as I mentioned before,
typically the first isocline
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that we should plot
is the nullcline,
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which is m is equal to 0.
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And we know that this
is y is equal to 0.
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So all along the
line y is equal to 0,
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I'm just going to draw dashes
that correspond to a slope of y
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is equal to 0.
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For the other isoclines, we just
have to pick some values of m
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and start plotting.
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So I'll take the value m
is equal to negative 1.
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And when m is equal
to negative 1,
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we have a circle which
is centered at 0 and 1/2
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with radius 1/2.
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So here's 1/2.
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Here's 1.
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And at every point
along the circle,
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we just draw a little
slope of negative 1.
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So at every point
along this curve,
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the solution has
slope negative 1.
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Now, in addition, we can also
get another circle, which
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is centered at negative 1/2.
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And this corresponds
to the isocline of m
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is equal to plus 1.
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And every point on this
circle has slope plus 1.
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Those should all be the same.
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And now we can pick
some other values of m.
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So for example, m is equal to 2.
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If m is equal to 2,
then we have a circle
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which is centered at
negative 1/4 and radius 1/4.
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So it might look
something like this.
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And m is equal to negative 2
might look something like this.
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So this is m is
equal to negative 2.
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This is m is equal to 2.
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And the slope on this curve
is going to be steeper.
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It's going to be negative 2.
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And the same goes
for this circle.
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And then lastly, I
can draw a sketch of m
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is equal to 3/4, which would
go up something like this.
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Sorry, m is equal
to negative 3/4.
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And that would have a slightly
weaker slope like this.
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So notice how the
collection of isoclines
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are a family of circles that
all are tangent to the origin.
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And in fact, if
we think about it,
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the nullcline, the
m equal 0 line,
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is actually, in
some sense, a limit
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circle where we take the
radius in the center going
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to infinity.
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So as the circles become
larger and larger,
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they tend to approach
this line y equals 0.
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So this concludes part one.
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So for part two, we're now
asked several questions
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about the curve which goes
through y(0) is equal to 1.
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So I'll just sketch what this
solution curve might look like.
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And it's going to start off
at a slope of negative 1.
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And then it's going to
hit these circles, which
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have a steeper slope.
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And then eventually
it's going to go
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through the circles
with a steeper slope
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and come back to the circle
that has slope negative 1.
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And eventually, decay outwards.
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So this is what the solution
curve might look like.
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I'm just sketching this to
set up parts 2a and 2b here.
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OK, so why is y of x
always bigger than 0
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for x greater than 0?
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Well, we see that
this solution curve
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stays in the upper half plane.
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And we note that y is equal
to 0, which is the nullcline,
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is actually a very
special curve.
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Not only is it the nullcline,
but it's actually a solution
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to the differential equation.
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So if we look back at the
differential equation,
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we see that y is equal
to 0 has zero derivative.
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And the right-hand side of
the differential equation
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is also 0.
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So this is not only a
nullcline, but it's a solution
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to the differential equation.
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Note that this is
extremely special.
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In no way does
every nullcline have
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to be a solution to a
differential equation.
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But in this case, we get lucky.
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Now we know from
the theory of ODEs
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that solution
curves can't cross.
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So a solution curve that
starts in the upper half plane
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can't cross another
solution, which in this case
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is the y equals 0 curve.
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Hence, it must be
bounded in the upper half
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plane for all x bigger than 0.
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The second part, why
is y of x decreasing
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for x greater than 0?
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Well, what we have
to look at-- so
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I'll just write in here part 2.
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We have to look at
the sign of y prime.
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Now y prime is equal
to negative y divided
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by x squared plus y squared.
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So as I've just
argued, a solution
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that starts in the
upper half plane
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and goes through the point
y of 0 is equal to 1,
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stays in the upper half plane
for all x bigger than 0.
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What this means is
that the solution curve
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has y bigger than 0 for
all x bigger than 0.
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So that means the numerator
is always positive.
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The denominator is
also always positive.
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So we have negative
a positive number
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divided by a positive number.
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And so this quantity is always
going to be less than 0.
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So for x bigger than
0, y prime is always
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going to be less than 0.
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Hence, the solution
that starts at y of 0
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is equal to 1 is going to
monotonically decay to y
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is equal to 0.
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OK?
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So this concludes the problem.
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Now just to recap,
we were given an ODE.
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The standard approach when
sketching direction fields
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is to pick a few nullclines.
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Or to pick a few isoclines.
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And specifically
pick the nullcline.
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Sketch the isoclines.
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And then, if you're asked
to plot any integral curves
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or solutions to the
differential equation,
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you just simply
connect the dots.
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OK, so I'd like
to conclude here,
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and I'll see you next time.