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.