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PROFESSOR: Welcome back.

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In this session, we're going to
tackle initial value problem y

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dot plus t*y equals to t.

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And this initial
value problem is

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going to be subject to the
initial condition, y of 0

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equals to 3.

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We are going to use the
method of integration factor.

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And what I want you to use
is both definite integrals

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and indefinite integrals.

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So why don't you
take a few minutes

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to think about the problem?

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And we will be right back.

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Welcome back.

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So I hope that you worked out
the first part of the problem.

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So what are we going to
do to solve this ODE?

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First, we need to review what
is the method of integrating

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factor.

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So when we use the
integrating factor,

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basically, we're
trying to write down

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our ODE in a different form
by introducing a function u.

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And the goal is to find
the function u such

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that we can rewrite
this left-hand side

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as the derivative of
the product y dot u.

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So in this case, if we're
looking at identifying

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the function u that would
give us this form, y*u dot,

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we need to just basically
identify that this is u dot.

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And from previous
sessions, we saw

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that this would give us
classical solution that

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involves an exponential of the
integral of the right-hand side

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after dividing by u,
which gives us exponential

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of t squared over 2.

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So integrating factor is
just using a trick so that we

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simplify our left-hand side
and write it in the form

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of the derivative
of the product, y*u.

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So now, we identified
our integration factor.

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It's u equals exponential
of t squared over 2.

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Now, we can go back
to our equation.

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And I'm going to just
label it with a star here.

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So now, this equation
is written in this form.

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t.

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and the integral of just
basically derivative, just

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itself.

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So if we use first the approach
of definite integrals--

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actually, we're going
to switch the order,

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and I'm going to start with
indefinite integrals first.

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So using indefinite
integrals, we'll

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be integrating both sides.

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And on the left-hand side, we
would just be left with y*u,

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remembering that u is just
exponential of t squared over

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2.

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On the right-hand
side, we're just

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integrating t exponential
of t squared over 2.

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And here, you can recognize that
the derivative of exponential

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of t squared over 2 would
have a t in front of it.

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So this is actually a
very simple integration.

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But we are in the case of
a differential equation

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where we need a
constant of integration.

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And again, here, we would
end up with two constants

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of integration on both sides.

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But given that we are dealing
with a first-order differential

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equation, we can regroup
that in one constant.

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And then, we can just
find our solution

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by dividing this equation by u.

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And u, if you remember, is
just exponential of t squared

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over 2, so it's
equivalent to multiplying

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both sides by exponential
of minus t squared over 2.

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So I'm just going
to write it down.

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Then, it just gives
us exponential minus t

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squared over 2 multiplied by
exponential t squared over 2,

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which is 1 and c exponential
of minus t squared over 2.

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So that's our solution.

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But remember that we're trying
to solve an initial value

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problem they subject to
an initial condition.

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And our initial
condition is y of 0

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equals to 3, which
means that here, y of 0

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would give us exponential to
0, which is just a constant.

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So we end up with 3
equals to 1 plus c.

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Therefore, c is equal to 2.

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And the final form
of the solution

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would just be 1
plus 2 exponential

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of minus t squared over 2.

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So we started with the
indefinite integral.

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So what if we would do this
using definite integrals?

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So we don't need to
start from the beginning.

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We just need to take the
problem a few steps before

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when we integrated both
sides of the equation.

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And here, specify the
bounds of the integral.

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So how do we want to specify
the bounds of the integral?

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We're given an initial condition
that is at t equals to 0.

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So that's what we want here.

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And we're integrating
to the variable t.

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But one thing is
important when you

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do that is that you have
the variable t that is

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in the bounds of the integral.

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So we want the
integrand to not be

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written in terms of the
variable of integration.

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So it is very
important to change

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the label of your
variables in the integrand.

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That's how you proceed
for different integrals.

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So in this case then,
we end up with similar--

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so u dot y evaluated at t
minus u dot y evaluated at 0.

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And from our initial
conditions and the form of u,

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we know the value of this side,
of this term in the equation.

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And here, we're just
again recognizing

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that this is just the derivative
of exponential to s squared

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over 2 evaluated
between 0 and t.

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So here, if I just carry
on with the right-hand side

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and then go back to left-hand
side in the next step,

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we would end up here with
just t squared over 2 minus 1.

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Exponential of 0 is 1.

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And now, let's deal with
this left-hand side.

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So again, we end up with u*y
minus u dot y evaluated at 0.

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However, u of 0 is
just the function

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exponential to t squared over
2 evaluated at 0, which is 1.

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And y of 0, we have it because
that was our initial condition,

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and this is only 3.

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So we have u*y minus 1 dot 3,
which is just basically minus

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3.

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And on the right-hand
side, we have exponential

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of t squared over 2 minus 1,
which gives us exponential

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of t squared over 2 minus 1.

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Now bring in this minus
3 on the other side,

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you now have a plus 3,
all of this multiplied

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by 1 over u, which was our
exponential of minus t squared

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over 2.

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And therefore, we end up
with a similar solution

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that we had for the indefinite
integral, 3 minus 1 is just 2.

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So we have minus t
squared over 2 plus 1.

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So using both the definite
integrals approach

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and the indefinite integrals,
we recover the same results.

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And the main point
of this problem

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was really to practice
using the integration factor

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and practice using
both approaches

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with the definite and
indefinite integrals.

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So this ends the problem.

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And I'll see you next time.