WEBVTT

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>> Welcome to the Cypress
College Math Review

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on First-Order Linear
Differential Equations.

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A first-order linear
differential equation is one

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that can be written in the
form y prime plus some function

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of x times y equals
some other function

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of x. Your first step is
to write it in that form.

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The second step is to obtain

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and simplify the integrating
factor mu of x. Mu of x is equal

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to e to the power of the
antiderivative of p of x.

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So find mu of x, which
is e to the power

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of the antiderivative
of p and simplify.

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Leave off the constant of
integration and in this case,

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and simplify the result
that you get for mu.

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Third step, to solve the
differential equation y times mu

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is equal to the antiderivative
of q times mu.

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Let's first write its standard
form by dividing by 3x minus 1,

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so we have y prime minus 6 over
the quantity 3x minus 1 times y,

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is equal to negative 10 times
and we have 3x minus 1 quantity

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to the 1/3 power divided by
3x minus 1 to the 1st power,

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which gives us that quantity
to the negative 2/3 power.

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We find our integrating factor,
mu of x, so mu of x is equal

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to e to the power of an
antiderivative of the function p

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of x. And the function p
of x is this function here,

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including the negative.

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So we have e to the power of
the antiderivative of negative 6

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over the quantity 3x minus 1.

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Pull out the negative 6, we're
integrating 1 over 3x minus 1.

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So let's multiply by 3.

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And divide by 3.

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So we're integrating 1 over udu,
so we have natural log, yes.

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So we have e to the power
of negative 2 natural log

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of the absolute value of what
e was, which is 3x minus 1.

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And we leave off the
constant of integration

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for the integrating factor.

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Raise the negative 2
up for our natural log,

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so we get the absolute
value of 3x minus 1 raised

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to the negative 2 power

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and since we're squaring we can
leave off the absolute value.

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And we simply get quantity 3x
minus 1 to the negative 2 power.

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As our integrating factor.

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So, we now know that mu times y
is equal to the antiderivative

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of mu times q with respect to
x. So mu, which is 3x minus 1

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on a quantity to the
negative 2 power times y,

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is equal to the antiderivative
of mu,

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which is the quantity 3x minus
1 to the negative 2 power,

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times q which is
this function here,

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negative 10 times 3x minus
1 to the negative 2/3 power.

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We're integrating that
with respect to x.

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Alright, so over here
on the right hand side,

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we pull out the negative
10 and we're going

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to get integrating the
quantity 3x minus 1

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to the negative 8/3 power.

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So that would be our
u, u is 3x minus 1.

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So it would be 3x minus 1
to the negative 5/3 power,

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divided by 5/3, which
is the same thing

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as multiplying by negative 3/5.

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Because u is 3x minus 1, we
would have to divide by 3

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and then we add a
negative 10 out there.

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You don't leave off your
constant of integration

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in general, that was only
for the integrating factor.

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These 3 cancel, so do
the negatives, and the 10

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and the 5 reduce down to just 2.

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So, now we solve for y, so we
have y is equal to 3x minus 1

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to the positive 2 power, times
2, times quantity 3x minus 1

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to the negative 5/3 power
plus c. And distribute

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to get our answer of 2 times
the quantity 3x minus 1

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to the 1/3 power, plus c
times quantity 3x minus 1

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to the 2nd power.

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Let's write it in
standard form, so we divide

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by 1 plus x squared each term.

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So we have y prime minus 2x
over 1 plus x squared times y.

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And a profact routed
x square we're left

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with x squared plus 1.

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And the denominators the
same, x squared plus 1.

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So those cancel and
simply give us x squared.

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Alright, we have
the standard form.

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Let's the find the
integrating factor mu of x as e

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to the power of the function p

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So we have e to the power of,
don't forget the negative,

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to the antiderivative
of negative 2x

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over 1 plus x squared dx.

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So a u substitution
would help us here.

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U equals 1 plus x squared.

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So du is equal to 2x dx, ok.

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So we pull out the negative,
so mu of x is equal to e

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to the power of negative

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and then I would
simply have 1 over u du.

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So we have e to the power

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of negative natural log absolute
value of 1 plus x squared.

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We leave off the
constant of integration.

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Since 1 plus x squared
is positive,

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we can leave off
the absent value.

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We bring the negative 1 up,
because of the natural log

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and we end up with
quantity 1 plus x squared

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to the negative 1 power.

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There's our integrating factor.

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Now we apply it.

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So mu times y is equal
to the antiderivative

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of mu times q integrated
with respect to x.

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So 1 plus x squared quantity

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to the negative 1
power times y is equal

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to the antiderivative
of that same thing.

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Times q which is
simply x squared.

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What I'm integrating is x
squared over x squared plus 1.

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To integrate that, I
would add and subtract 1.

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Just a trick for,
actually for dividing.

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Just a short hand way

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for actually doing
the long division.

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So I get 1 minus 1
over x squared plus 1.

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So we have x minus arc
tangent of x plus c

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and then we multiply both sides
by 1 plus x squared to solve

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for y. So y is equal to the
quantity 1 plus x squared

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to the positive 1 power
times the right hand side,

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which is x minus arc
tangent of x plus c

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and that gives us our answer.

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This equation is already in
standard form, so we straight

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to our integrating factor.

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So mu of x is equal to e to the
power of the antiderivative of p

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with respect to x. So
we have e to the power

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of the antiderivative
of the function tangent.

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And the antiderivative of
tangent is negative natural log

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of the absolute value of cosign.

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We leave off the
constant of integration.

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Bring the negative 1 up and
we have the absolute value

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of cosign raised to
the negative 1 power,

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which of course you're
going to write as secant.

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Now what about the
absolute value?

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You're actually going
to drop that also,

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so mu of x is simply secant.

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Why can you leave off
the absolute value?

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Because when you apply
an integrating factor,

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you're actually multiplying
both sides by it.

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When you multiply both
sides by a positive

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or multiply both sides by a
negative, it's really not going

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to make any difference is it?

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So if you have an absolute value
on your integrating factor,

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you can leave that off also.

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Alright, we have our integrating
factor, let's apply it.

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So we know that mu times y is
equal to the antiderivative

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of mu times q integrated with
respect to x. So we have secant

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of x times y is equal to the
antiderivative of secant of x,

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times cosign of x,
integrated with respect to x.

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So in the right hand side,
I'm only integrating 1.

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When I integrate 1 I get
x plus c. To solve for y,

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I multiply both sides by cosign.

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And there's our answer.

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This is already in standard
form, so we go straight

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to our integrating factor.

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Mu of x is equal to e to the
power of the antiderivative of p

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of x with respect to x.
This time that is 3x plus 1,

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quantity all over x. You
can split up the numerator

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but not a denominator.

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So that would be 3x over x
which is just 3 plus 1 over x.

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And we integrate that.

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We get 3x plus natural
log of the absolute value

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of x. Leave off the plus c. So
I have e to the power of a sum.

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So I can rewrite that as e to
the 3x times e to the power

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of natural log of the
absolute value of x,

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because of this property.

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And then the second factor can
be rewritten as absolute value

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of x and like we said
in the last problem,

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we leave off the absolute value.

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I'll put that out front.

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And our integrating factor
is x times e to the 3x.

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Now let's apply it.

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So we know that mu
times y is going

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to equal the antiderivative
of mu times q integrated

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with respect to x. So x times e

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to the 3x times y will
equal the antiderivative

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of that same quantity times q,
which was, e to the negative 3x

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over x, integrated with
respect to x. Once again,

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all we're integrating on the
right hand side is just 1.

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So we get, once again
x plus c. So y is equal

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to x plus c all over,
x e to the 3x.

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Which is our answer.

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Here's some extra
practice I encourage you

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to try these on your own.

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Pause the video,
do these problems.

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And then once you've completed
the problems, restart the video

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and check your answers.

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Here are the answers to
the two practice problems.