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PROFESSOR: Welcome back
to recitation.

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Today we're going to work
on a problem involving

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differential equations.

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I'm going to read it to you,
give you a little hint, give

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you some time to work on it, and
then I'll be back and work

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it out for you.

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So, the problem is to find a
function y equals f of x that

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has the following two
properties: d squared y dx

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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, but it says the

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graph of the function in the
xy plane passes through the

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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

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buried 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 to work on

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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 solving the

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problem initially, give yourself
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 is try and figure

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out maybe what the first
derivative of the function y

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equals f of x is.

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I have its second derivative, so
in order to find the first

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derivative I want to find,
ultimately a function that

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when I take its derivative
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 would actually be to

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consider another function
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--

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d squared y dx squared.

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I'm going to say is the first
derivative of another

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function, we'll say dw dx.

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And the reason I'm going to do
that is so we're not too

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nervous about how 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, which is the

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first derivative of y with
respect to x, ultimately.

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Because its derivative is the
second derivative of y with

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respect to x.

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So let's see how to solve the
differential equation

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dw dx 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 do a

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separation of variables.

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So right away, maybe some of you
can see that this will be

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2x-- sorry--

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3x squared.

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But let's just double-check.

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So, separation of variables,
on this side we get the

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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 do 6x squared

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over 2 plus a constant.

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So that's 3x squared
plus a constant.

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I'm going to write c1 here
because we're went to need a

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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

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with respect 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 with

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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 and so now

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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 the problem without

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separating variables, because
I actually know what an

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antiderivative is of this, what
an antiderivative is of

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this, and then 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 c1 x plus c2.

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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 constant so its

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antiderivative is c1 times x.

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And then I have to add on
another constant because I

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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, does this

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really satisfy number one?

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So if we wanted to check that,
we just take two derivatives

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of this expression 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
number two?

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It says we have some initial
conditions here.

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The graph in the x y 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

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that x value 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, from what we've seen

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previously.

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In the lecture you saw examples,
at least certainly,

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where 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 of

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just the first derivative.

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So that's kind of, that's
why you see more initial

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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 c1 plus 1 times,

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or just 1, or c2 there--
sorry, there's no x

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there, so just c2.

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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

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times 1, plus c1.

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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

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them is very easy to solve
initially, then we can

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substitute.

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So what does this say?

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Well 3 plus c1 equals 0.

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So c1 is equal to minus
3, negative 3.

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If I plug in negative 3 for
c1, in this expression, in

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this equation up here, I get
negative 3 plus 1 plus c2 has

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to equal 1.

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If I subtract the 1's from both
sides I get negative 3

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plus c2 has to equal 0.

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So c2 has to actually equal 3.

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So the final, final answer is
evaluating, or plugging in the

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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 graph this on a

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calculator or computer and look
and see if it satisfies

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that it actually passes through
the point 1,1 and has

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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.

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