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A lot of what you'll learn in differential equations is

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really just different bags of tricks.

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And in this video I'll show you one of those tricks.

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And it's useful beyond this.

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Because it's always good when, if maybe one day, you become a

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mathematician or a physicist, and you

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have an unsolved problem.

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Some of these tricks that solved simpler problems back

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in your education might be a useful trick that solves some

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unsolved problems. So it's good to see it.

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And if you're taking differential equations, it

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might be on an exam.

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So it's good to learn.

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So we'll learn about integrating factors.

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So let's say, we have an equation that has this form.

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Let's say this is my differential equation.

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3xy-- I'm trying to write it neatly as possible-- plus y

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squared plus x squared plus xy times y prime is equal to 0.

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So, especially since we've covered this in recent videos,

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whenever you see an equation of this form where you have

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some function of xy, then you have another function of x and

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y times y prime equals 0, you said, oh, this looks like this

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could be an exact differential equation.

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And how do we test that?

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Well, we can take the partial derivative of this with

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respect to y, and we could call this

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function of x and y, M.

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So the partial of that, with respect to y, so M partial

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with respect to y, would be 3x plus 2y.

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And if this function right here, that expression right

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there, that's our function N, which is a

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function of x and y.

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We take the partial with respect to x, and we get that

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is equal to 2x plus y.

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And in order for this to have been an exact differential

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equation, the partial of this with respect to y would have

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to equal the partial of this with respect to x.

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But we see here, just by looking at these two, they

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don't equal each other.

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They're not equal.

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So, at least superficially, the way we looked at it just

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now, this is not an exact differential equation.

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But what if there were some factor, or I guess some

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function that we could multiply both sides of this

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equation by, that would make it an exact

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differential equation?

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So let's call that mu.

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So what I want to do is I want to multiply both sides of this

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equation by some function mu, and then see if I can solve

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for that function mu that would make it exact.

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So let's try to do that.

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So let's multiply both sides by mu.

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And just as a simplification, mu could be a

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function of x and y.

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It could be a function of x.

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It could be a function of just x.

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It could be function of just y.

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I'll assume it's just a function of x.

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You could assume it's just a function of y and

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try to solve it.

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Or you could just assume it's a function of x and y.

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If you assume it's a function of x and y, it becomes a lot

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harder to solve for.

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But that doesn't mean that there isn't one.

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So let's say that mu is a function of x.

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And I want to multiply it by both of these equations.

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So I get mu of x times 3xy plus y squared plus mu of x

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times x squared plus xy times y prime.

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And then, what's 0 times any function?

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Well, it's just going to be 0, right?

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0 times mu of x is just going to be 0.

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But I did multiply the right hand side times mu of x.

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And remember what we're doing.

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This mu of x is-- when we multiply it, the goal is,

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after multiplying both sides of the equation by it, we

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should have an exact equation.

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So now, if we consider this whole thing our new M, the

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partial derivative of this with respect to y should be

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equal to the partial derivative of this with

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respect to x.

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So what's the partial derivative of this with

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respect to y?

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Well, if we're taking the partial with respect to y

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here, mu of x, which is only a function of x, it's not a

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function of y, it's just a constant term, right?

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We take a partial with respect to y. x is just a constant, or

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a function of x can be viewed just as a constant.

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So the partial of this with respect to y is going to be

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equal to mu of x, you could just say, times 3x plus 2y.

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That's the partial of this with respect to y.

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And what's the partial of this with respect to x?

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Well, here, we'll use the product rule.

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So we'll take the derivative of the first expression with

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respect to x. mu of x is no longer a constant anymore,

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since we're taking the partial with respect to x.

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So the derivative of mu of x with respect to x.

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Well, that's just mu prime of x, mu prime, not U.

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mu prime of x.

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mu is the Greek letter.

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It's for the muh sound, but it looks a lot like a U.

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So mu prime of x times a second expression, x squared

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plus xy, plus just the first expression.

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This is just the product rule, mu of x.

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Times the derivative of the second expression

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with respect to x.

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So times-- ran out of space on that line-- 2x plus y.

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And now for this new equation, where I multiplied

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both sides by mu.

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In order for this to be exact, these two things have to be

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equal to each other.

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So let's just remember the big picture.

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We're kind of saying, this is going to be exact.

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And now, we're going to try to solve for mu.

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So let's see if we can do that.

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So let's see, on this side, we have mu of x times 3x plus 2y.

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And let's subtract this expression from both sides.

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So it's minus mu of x times 2x plus y.

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You'll see a lot of these differential equation problems

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that get kind of hairy.

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They're really just a lot of algebra.

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And that equals-- what do we have left?

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I'll write it in yellow.

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That equals-- I'm going to run out of space.

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I'm going to do it a little bit lower.

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That equals, just this term right here.

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That equals mu prime of x times x squared plus xy.

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And let's see, if we factor out a mu of x here, we get mu

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of x times 3x plus 2y minus 2x minus y is equal to mu prime

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of x, the derivative of mu with respect to x, times x

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squared plus xy.

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Now, we can simplify this.

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So we get mu of x times-- what is this-- 3x minus 2x is x.

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2y minus y, so x plus y, is equal to-- and I'm just going

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to simplify this side a little bit-- is equal

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to mu prime of x.

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Let's factor out an x here.

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And the reason why I'm doing that is because it seems like

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if I factor out an x here, I'll get an x plus y.

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So this is mu prime of x times x times x plus y.

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x times x plus y is x squared plus xy.

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So that's why I did it, and I have this x plus y on both

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sides equation, which I will now divide both sides by.

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So if you divide both sides by x plus y, we could maybe

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assume that it's not 0.

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That simplifies things pretty dramatically.

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We get mu of x is equal to mu prime of x times x.

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And now, just the way my brain works, I like to rewrite this

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expression just in our operator form, where instead

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of writing it mu prime of x, we could write

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that as d mu dx.

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So let's do that.

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So we could write mu of x is equal to d, the derivative of

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mu with respect to x, times x.

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And this is actually a separable differential

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equation in and of itself.

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It's kind of a sub-differential equation to

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solve our broader one.

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We're just trying to figure out the integrating factor

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right here.

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So let's divide both sides by x.

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So we get mu over x, this is just a separable equation now,

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is equal to d mu dx.

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And then, let's divide both sides by mu of x, and we get 1

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over x is equal to 1 over mu.

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That's mu of x, I'll just write 1 over mu right now, for

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simplicity, times d mu dx.

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I'm actually going to go horizontal right here.

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Multiply both sides by dx, you get 1 over x dx is equal to 1

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over mu of x d mu.

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Now, you could integrate both sides of this, and you'll get

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the natural log of the absolute value of x is equal

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to the natural log of the absolute value of mu, et

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cetera, et cetera.

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But it should be pretty clear from this that x is equal to

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mu, or mu is equal to x, right?

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They're identical.

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If you look at both sides of this equation there, you can

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just change x for mu, and it becomes the other side.

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So, this is obviously telling us that mu of x is equal to x.

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Or mu is equal to x.

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So we have our integrating factor.

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And if you want, you can take the antiderivative of both

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sides with the natural logs, and all of that.

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And you'll get the same answer.

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But this is just, by looking at it, by inspection, you know

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that mu is equal to x.

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Because both sides of this equation are

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completely the same.

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Anyway, we now have our integrating factor.

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But I am running out of time.

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So in the next video, we're now going to use this

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integrating factor.

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Multiply it times our original differential equation.

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Make it exact.

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And then solve it as an exact equation.

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I'll see you in the next video.

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