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PROFESSOR: Welcome back
to recitation.

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In this video what we'd like
to do is to solve a certain

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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 is

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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 and

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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 to

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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 for

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separation of variables.

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So let me let me first get all
of the values, or terms that

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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 and

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get all the terms that involve
x to the right-hand side.

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So I'm going to write dy
divided by y squared

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plus 1 is equal to--

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so I'm going to multiply through
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 but we're going to

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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 in

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separation of variables and so
I can integrate both sides.

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So let's see what happens
when I integrate

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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 is 1 over y

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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, that's the same

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thing, the inverse tangent of
y, 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 the tangent of

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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 your I get tangent
of x squared over 2 plus x

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plus C.

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

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

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conditions then we could
evaluate and solve for C. Now

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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 of the right-hand

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side, take dy dx.

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Evaluate what that is when you
have y equals tangent of x

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squared over 2 plus x plus C and
see if you in fact get the

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relationship you're
supposed to get.

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But I'll stop there.

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