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And, in the last video, we had this differential equation.

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And it at least looked like it could be exact.

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But when we took the partial derivative of this expression,

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which we could call M with respect to y, it was different

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than the partial derivative of this expression, which is N in

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the exact differential equations world.

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It was different than N with respect to x.

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And we said, oh boy, it's not exact.

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But we said, what if we could multiply both sides of this

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equation by some function that would make it exact?

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And we called that mu.

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And in the last video, we actually solved for mu.

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We said, well, if we multiply both sides of this equation by

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mu of x is equal to x, it should make this into an exact

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differential equation.

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It's important to note, there might have been a function of

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y that if I multiplied by both sides it would

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also make it exact.

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There might have been a function of x and y that would

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have done the trick.

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But our whole goal is just to make this exact.

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It doesn't matter which one we pick, which integrating

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factor-- this is called the integrating factor-- which

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integrating factor we pick.

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So anyway, let's do it now.

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Let's solve the problem.

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Let's multiply both sides of this equation by mu, and mu of

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x is just x.

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We multiply both sides by x.

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So see, if you multiply this term by x, you get 3x squared

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y plus xy squared, we're multiplying these terms by x

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now, plus x to the third plus x squared y, y

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prime is equal to 0.

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Well now, first of all, just as a reality check, let's make

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sure that this is now an exact equation.

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So what's the partial of this expression, or this kind of

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sub-function, with respect to y?

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Well, it's 3x squared, that's just kind of a constant

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coefficient of y, plus 2xy, that's the partial with

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respect to y of that expression.

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Now let's take the partial of this with respect to x.

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So we get 3x squared plus 2xy.

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And there we have it.

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The partial of this with respect to y is equal to the

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partial of this with respect to N.

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So we now have an exact equation whose solution should

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be the same as this.

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All we did is we multiplied both sides of

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this equation by x.

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So it really shouldn't change the solution of that equation,

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or that differential equation.

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So it's exact.

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Let's solve it.

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So how do we do that?

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Well, what we say is, since we've shown this exact, we

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know that there's some function psi where the partial

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derivative of psi with respect to x is equal to this

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expression right here.

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So it's equal to 3x squared y plus xy squared.

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Let's take the antiderivative of both sides with respect to

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x, and we'll get psi is equal to what?

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x to the third y plus, we could write,

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1/2 x squared y squared.

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And of course, this psi is a function of x and y, so when

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you take the partial with respect to x, when you go that

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way, you might have lost some function that's only a

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function of y.

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So instead of a plus c here, it could've been a whole

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function of y that we lost. So we'll add that back when we

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take the antiderivative.

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So this is our psi.

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But we're not completely done yet, because we have to

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somehow figure out what this function of y is.

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And the way we figure that out is we use the information that

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the partial of this with respect to y

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should be equal to this.

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So let's set that up.

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So what's the partial of this expression with respect to y?

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So I could write, the partial of psi with respect to y is

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equal to x to the third plus 2 times 1/2, so it's just x

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squared y plus h prime of y.

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That's the partial of a function purely of y with

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respect to y.

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And then that has to equal our new N, or the new expression

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we got after multiplying by the integrating factor.

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So that's going to be equal to this right here.

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This is, hopefully, making sense to you at this point.

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And that should be equal to x to the third plus x squared y.

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And interesting enough, both of these

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terms are on this side.

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So let's subtract both of those terms from both sides.

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So x to the third, x to the third, x squared

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y, x squared y.

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And we're left with h prime of y is equal to 0.

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Or you could say that h of y is equal to some constant.

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So there's really no y, the extra function of y.

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There's just some constant left over.

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So for our purposes, we can just say that

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psi is equal to this.

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Because this is just a constant, we're going to take

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the antiderivative anyway, and get a constant on

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the right hand side.

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And in the previous videos, the

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constants all merged together.

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So we'll just assume that that is our psi.

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And we know that this differential equation, up

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here, can be rewritten as, the derivative of psi with respect

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to x, and that just falls out of the partial derivative

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chain rule.

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The derivative of psi with respect to x is equal to 0.

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If you took the derivative of psi with respect to x, it

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should be equal to this whole thing, just using the partial

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derivative chain rule.

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And we know what psi is.

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So we can write-- or actually we don't even have to.

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We could use this fact to say, well, if we integrate both

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sides, that a solution of this differential equation is that

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psi is equal to c.

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I just took the antiderivative of both sides.

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So, a solution to the differential equation is psi

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is equal to c.

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So psi is equal to x to the third y plus

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1/2 x squared y squared.

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And we could have said plus c here, but we know the solution

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is that psi is equal to c, so we'll just write that there.

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I could have written a plus c here, but then you

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have a plus c here.

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You have another constant there.

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And you can just subtract them from both sides.

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And they just merge into another arbitrary constant.

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But anyway, there we have it.

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We had a differential equation that, at least superficially,

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looked exact.

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It looked exact, but then, when we tested the exactness

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of it, it was not exact.

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But we multiplied it by an integrating factor.

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And in the previous video, we figured out that a possible

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integrating factor is that we could just multiply

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both sides by x.

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And when we did that, we tested it.

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And true enough, it was exact.

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And so, given that it was exact, we knew that a psi

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would exist where the derivative of psi with respect

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to x would be equal to this entire expression.

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So we could rewrite our differential

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equation like this.

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And we'd know that a solution is psi is equal to c.

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And to solve for psi, we just say, OK, the partial

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derivative of psi with respect to x is

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going to be this thing.

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Antiderivative of both sides, and there's some constant h of

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y-- not constant, there's some function of y-- h of y that we

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might have lost when we took the partial with respect to x.

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So to figure that out, we take this expression.

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Take the partial with respect to y, and set that equal to

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our N expression.

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And by doing that, we figured out that that function of y is

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really just some constant.

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And we could have written that here.

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We could have written that plus c.

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We could call that c1 or something.

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But we know that the solution of our original differential

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equation is psi is equal to c.

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So the solution of our differential equation is psi x

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to the third y plus 1/2 x squared y

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squared is equal to c.

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We could have had this plus c1 here, then

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subtracted both sides.

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But I think I've said it so many times that you

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understand, why if h of y is just a c, you can

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kind just ignore it.

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Anyway, that's all for now, and I will see

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you in the next video.

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You now know a little bit about integrating factors.

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See you soon.

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