WEBVTT
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PROFESSOR: Welcome
back to recitation.
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Today we're going to work on a
problem involving differential
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equations.
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I'm going to read it to
you, give you a little hint,
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give you some time
to work on it,
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and then I'll be back
and work it out for you.
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So, the problem is to
find a function y equals
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f of x that has the following
two properties: d squared
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y dx squared is equal to 6x.
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And just to remind you what
this means, this is really,
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this is the second derivative
of y with respect to x.
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So, the second derivative with
respect to x should be 6x.
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And the second
condition's kind of long,
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but it says the graph of the
function in the xy-plane passes
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through the point (1, 1) with
a horizontal tangent there.
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So let me give you one hint.
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And that hint is that there are
some initial conditions buried
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in here.
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That's why we have
this condition.
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So I'm going to give
you a little bit of time
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to work on it and I'll be back
and I'll work it out with you.
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Welcome back.
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Hopefully you were
able to start at least
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solving the problem
initially, give yourself
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a little direction.
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So let's see how you did.
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OK, so the first thing
I would like to do
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is try and figure out maybe
what the first derivative
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of the function y
equals f of x is.
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I have its second
derivative, so in order
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to find the first
derivative I want
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to find, ultimately, a function
that when I take its derivative
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I get 6x.
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Right?
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So, we can think about this.
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Maybe the easiest
thing for us to do
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would actually be to
consider another function
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whose derivative is 6x.
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And we'll know that's dy dx.
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So I'm going to say this: dy--
sorry-- d squared y dx squared,
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I'm going to say is the first
derivative of another function,
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we'll say dw/dx.
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And the reason I'm
going to do that
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is so we're not too
nervous about how
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we solve this problem.
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So let's just assume that.
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So I'm introducing
another function w,
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which is the first
derivative of y with respect
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to x, ultimately.
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Because its derivative
is the second derivative
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of y with respect to x.
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So let's see how to solve the
differential equation dw/dx
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equals 6x.
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Now you may be able
to do that right away.
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You may see what
this is right away.
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If you're a little
nervous, we can
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do a separation of variables.
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So right away, maybe
some of you can
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see that this will be
2x-- sorry-- 3 x squared.
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But let's just double-check.
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So, separation of
variables, on this side
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we get the integral of dw.
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On this side we get
the integral of 6x dx.
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Here we get a w.
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And here, again we
get, we should really
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do 6 x squared over
2 plus a constant.
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So that's 3 x squared
plus a constant.
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I'm going to write
c_1 here because we're
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went to need a little
bit of information.
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We're going to need
another constant later.
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So we actually now know the
derivative of y with respect
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to x.
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What we found here, I'll
just write that in, is dy/dx.
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Again, let me remind you why.
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We had-- we were saying
the second derivative of y
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with respect to x we're
going to call dw/dx.
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So when I took an integral
of dw/dx, I got w.
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So that means I've taken
one antiderivative here
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and so now I have dy/dx.
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So this is dy/dx.
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I'm going to draw a line
and now we want to find y.
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And again we can use
separation of variables.
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Or this time I'm
just going to do
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the problem without
separating variables,
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because I actually know what an
antiderivative is of this, what
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an antiderivative
is of this, and then
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I need to add one more constant.
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So I can say that y is
definitely equal to x cubed
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plus c_1*x plus c_2.
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And let me just again,
let's see why that is.
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Right?
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This antiderivative
of this is x cubed.
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Antiderivative of
this is-- it's a
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constant so its
antiderivative is c_1 times x.
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And then I have to add
on another constant
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because I have a whole
family of possible solutions.
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So here I have y.
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And now I need to figure
out how to use number two.
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Maybe before you even go
on, you want to check,
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does this really
satisfy number one?
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So if we wanted to
check that, we just
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take two derivatives
of this expression
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on the right-hand side.
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Two derivatives of this is 0.
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Two derivative of this is 0.
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And two derivatives
of this is 6x.
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So I do indeed get what I want.
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So now we definitely
can go on to number 2.
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OK?
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So, what do we have?
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Number two, it says we have
some initial conditions here.
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The graph in the xy-plane
passes through the point (1, 1)
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and it has a horizontal
tangent there.
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Now what does that
actually mean?
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Let's think about that.
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That actually means that
two says f of 1 equals 1.
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Right?
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y is equal to f of x.
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So we can write
this also as f of x.
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So two, the first part
says that f of 1 is 1.
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And what does this
second condition?
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Let's check this
second condition.
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It says it has a
horizontal tangent there.
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Horizontal tangent means that
its derivative at that x-value
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is 0.
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So let me write down
that in a nice form.
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The derivative at that
x-value is equal to 0.
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This is a little
different, maybe,
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from what we've seen previously.
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In the lecture you saw examples,
at least-- certainly, where
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you had one initial condition.
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Here we need two
initial conditions.
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And you see why, is because we
actually have two constants.
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Where did that come from?
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It's because we started with
a second derivative instead
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of just the first derivative.
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So that's kind
of, that's why you
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see more initial conditions than
maybe you've seen previously.
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So, let's plug these in
and let's see what we get.
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If f of 1 equals 1, then
let's evaluate that.
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f of 1 is equal to 1 cubed
plus 1 times c_1 plus 1 times--
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or just 1, or c_2 there, sorry.
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There's no x there, so just c_2.
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And that all has to equal 1.
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And then let's look at
what the derivative is.
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The derivative is
still over here.
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So f prime at 1 is equal to 3
times 1 squared, so 3 times 1,
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plus c_1.
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And the condition says
that equals 0, equals 0.
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So we can read off.
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The nice thing is this is system
of equations but one of them
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is very easy to solve initially,
then we can substitute.
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So what does this say?
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Well 3 plus c_1 equals 0.
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So c_1 is equal to
minus 3, negative 3.
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If I plug in negative 3 for
c_1, in this expression,
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in this equation up here, I
get negative 3 plus 1 plus c_2
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has to equal 1.
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If I subtract the
1's from both sides
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I get negative 3 plus
c_2 has to equal 0.
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So c_2 has to actually equal 3.
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So the final, final
answer is evaluating,
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or plugging in the c1 and the
c2 in for the constants there.
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The final, final answer is
x cubed minus 3x plus 3.
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So we started with a
differential equation and two
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sets of initial conditions.
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And we came up with one
solution that satisfies that.
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Now you can look
at this, you could
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graph this on a
calculator or computer
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and look and see if it satisfies
that it actually passes
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through the point (1, 1) and
has a horizontal tangent there.
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But we know, based on our
work, that that actually
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should happen.
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So I think that's
where I'll stop.