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PROFESSOR: Welcome
back to recitation.

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In this video what
we'd like to do

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is to solve a certain
differential equation.

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And I'm not even giving
you initial condition.

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So we just want to find a
y, such that x times dy/dx

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is equal to x squared plus
x times y squared plus 1.

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So I'll give you a
minute to think about it

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and then I'll be back.

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Welcome back.

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I'm going to use the technique
of separation of variables

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to solve this differential
equation problem.

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Hopefully you thought
about doing that as well.

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Because it's really
set up nicely

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for separation of variables.

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So let me let me first
get all of the values,

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or terms that involve y
on the left-hand side.

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And then I'm going to move
the dx to the right-hand side

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and get all the
terms that involve

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x to the right-hand side.

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So I'm going to
write dy divided by y

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squared plus 1 is equal to-- so
I'm going to multiply through

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by dx divided by x.

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So I'm going to get x
squared plus x over x dx.

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Now if x is 0, I
have a little problem

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but we're going to ignore
that for the moment.

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So I can rewrite
this as x plus 1 dx.

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And now I'm totally set
up with my next step

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in separation of variables and
so I can integrate both sides.

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So let's see what happens when
I integrate the left-hand side.

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OK?

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What is an antiderivative
to 1 over y squared plus 1?

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Well that's arctangent,
that's arctangent of y.

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So the derivative
of arctangent of y

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is 1 over y squared plus 1.

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And so on the left-hand
side I get arctan y.

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If you wrote tangent
to the minus 1 y,

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that's the same thing,
the inverse tangent of y,

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that's the same thing.

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And on the right-hand
side what do I get?

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Well this is a nice
easy thing to integrate.

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When I integrate x, I
get x squared over 2.

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And here I just get an x.

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And then I should
add in my constants.

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So x squared over 2
plus x plus a constant.

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So this is, so far I'm
doing everything OK.

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Now I need to figure out
how to isolate the y.

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Well arctan y is the inverse
of the tangent function.

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So if I want to isolate
y I have to take

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the tangent of both sides.

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So when I take tangent of
arctangent of y I just get y.

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Then over here, I get tangent of
x squared over 2 plus x plus C.

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And now because I didn't give
you any initial conditions,

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we can't say anything
about C. But if I gave you

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some initial conditions then we
could evaluate and solve for C.

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Now how do I go about checking
to make sure that this works?

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Well, what I do is actually
take the derivative.

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So you may want to
take the derivative

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of the right-hand
side, take dy/dx.

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Evaluate what that
is when you have y

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equals tangent of x squared
over 2 plus x plus C

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and see if you in fact get
the relationship you're

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supposed to get.

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But I'll stop there.