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Let's do some more examples with exact

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differential equations.

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And I'm getting these problems from page 80 of my old college

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differential equations books.

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This is the fifth edition of Elementary Differential

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Equations by William Boyce and Richard DiPrima.

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I want to make sure they get credit, that I'm not making up

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these problems. I'm getting it from their book.

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Anyway, so I'm just going to give a bunch of equations.

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We have to figure out if they're exact, and if they are

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exact, we'll use what we know about exact differential

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equations to figure out their solutions.

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So the first one they have is, 2x plus 3, plus 2y minus 2,

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times y prime is equal to 0.

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So this is our M of x and y-- although, this is only a

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function of x-- and then this is our N, right?

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You could say that's M, or that's N.

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You could also say that, if this is exact-- well, first

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let's [? test ?]

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this exact, before we start talking about psi.

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

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The partial of M, with respect to y.

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Well, there's no y here, so it's 0.

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The rate of change that this changes with

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respect to y is 0.

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And what's the rate of change this changes,

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with respect to x?

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The partial of N, with respect to x is equal to-- well,

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there's no x here, right?

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So these are just constants from an x point of view, so

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this is all going to be 0.

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But we do see that they're both 0.

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So M sub y, or the partial with respect to y, is equal to

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the partial with respect to x.

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So this is exact.

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And actually, we don't have to use exact equations here.

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We'll do it, just so that we get used to it.

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But if you look here, you actually could have figured

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out that this is actually a separable equation.

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But anyway, this is exact.

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So knowing that it's exact, it tells us that there's some

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function psi, where psi is a function of x and y.

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Where psi sub x is equal to this function, is equal to 2x

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plus 3, and psi-- I shouldn't say sub x.

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I say the partial of psi, with respect to x.

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And the partial of psi, with respect to y, is equal to

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this, 2y minus 2.

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And if we can find our psi, we know that this is just the

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derivative of psi.

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Because we know that the derivative, with respect to x

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of psi, is equal to the partial of psi, with respect

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to x, plus the partial of psi, with respect

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to y, times y prime.

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So this is this just the same form as that.

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So if we can figure out y, then we can rewrite this

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equation as dx, the derivative of psi, with respect to x, is

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equal to 0.

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Let me switch colors, or it's going to get monotonous.

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This right here, if we can find a psi, where the partial

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with respect to x, is this, the partial with respect to y,

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is this, then this can be rewritten as this.

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And how do we know that?

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Because the derivative of psi, with respect to x, using the

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partial derivative chain rules, is this.

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This partial with respect to x, that's this.

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This partial with respect to y, is this, times y prime.

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So this is the whole point of exact equations.

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But anyway, so let's figure out what our psi is.

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Actually, before we figure out, if the derivative of psi,

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with respect to x, is 0, then if you integrate both sides,

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you just-- the solution of this equation is

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psi is equal to c.

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So using this information, if we can solve for psi, then we

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know that the solution of this differential equation is psi

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is equal to c.

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And then if we have some initial conditions, we could

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solve for c.

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So let's solve for psi.

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So let's integrate both sides of this equation,

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with respect to x.

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And then we get psi is equal to x squared plus 3x, plus

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some function of y.

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Let's call it h of y.

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And remember, normally when you take an antiderivative,

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you have just a plus c here, right?

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But you can kind of say we took an anti-partial

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derivative.

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So when you took a partial derivative, with respect to x,

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not only do you lose constants-- that's why we have

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a plus c, normally-- but you also lose anything that's a

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function of just y, and not x.

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So for example, take the partial derivative of this

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with respect to x, you're going to get this, right?

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Because the partial derivative of a function, purely of y,

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with respect to x, is going to be 0.

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So it will disappear.

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So anyway, we take the antiderivative of

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this, we get this.

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Now, we use this information.

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Well, we use this information.

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We take the partial of this expression, and we say, well,

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the partial of this expression, with respect to y,

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has to equal this, and then we can solve for h of y, then

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we'll be done.

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So let's do that.

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So the partial of psi, with respect to y, is equal to--

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well, that's going to be 0, 0, 0.

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This part is a function of x.

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If you take the partial with respect to y, it's 0, because

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these are constants, from a y point of view.

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So you're left with h prime of y.

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So we know that h prime of y, which is the partial of psi,

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with respect to y, is equal to this.

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So h prime of y is equal to 2y minus 2.

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And then if we wanted to figure out what h of y is, we

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get h of y-- just integrate both sides, with respect to

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y-- is equal to y squared plus-- sorry-- y

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squared minus 2y.

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Now, you could have a plus c there, but if you watched the

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previous example, you'll see that that c kind of merges

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with the other c, so you don't have to worry

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about it right now.

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So what is our psi function, as we know it now, not

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worrying about the plus c?

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It is psi of x and y is equal to x squared plus 3x, plus h

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of y-- which we figured out is this-- plus y

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squared, minus 2y.

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And we know a solution of our original differential equation

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is psi is equal to c.

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So the solution of our differential equation is this

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is equal to c.

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x squared plus 3x, plus y squared, minus

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2y is equal to c.

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If you had some additional conditions, you could test it.

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And I encourage you to test this out on this original

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equation, or I encourage you to take the derivative of psi,

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and prove to yourself that if you took the derivative of

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psi, with respect to x, here, implicitly, that you would get

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this differential equation.

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Anyway, let's do another one.

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Let's clear image.

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So the more examples you see, the better.

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So let's see, this one says 2x plus 4y, plus 2x minus 2y, y

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prime is equal to 0.

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

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So M, the partial of M, with respect to y-- this is 0-- so

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it's equal to 4.

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What's the partial of this, with respect to x, just this

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part right here?

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The partial of N, with respect to x, is 2.

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This is 0.

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So the partial of this, with respect to y, is different

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than the partial of N, with respect to x.

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So this is not exact.

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So we can't solve this using our exact methodology.

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So that was a fairly straightforward problem.

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Let's do another one.

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Let's see.

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I'm running out of time, so I want to do one that's not too

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complicated.

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Let's see, 3x squared minus 2xy-- actually, let me do this

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in the next problem.

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I don't want to rush these things.

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I will continue this in the next video.

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See you soon.

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