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I will now introduce you to the idea of a homogeneous

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differential equation.

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Homogeneous is the same word that we use for milk, when we

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say that the milk has been-- that all the fat clumps have

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been spread out.

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But the application here, at least I don't see the

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connection.

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Homogeneous differential equation.

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And even within differential equations, we'll learn later

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there's a different type of homogeneous

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differential equation.

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Those are called homogeneous linear differential equations,

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but they mean something actually quite different.

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But anyway, for this purpose, I'm going to show you

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homogeneous differential equations.

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And what we're dealing with are going to

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be first order equations.

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What does a homogeneous differential equation mean?

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Well, say I had just a regular first order differential

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equation that could be written like this.

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So dy dx is equal to some function of x and y.

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And let's say we try to do this, and it's not separable,

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and it's not exact.

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What we learn is that if it can be homogeneous, if this is

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a homogeneous differential equation, that we can make a

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variable substitution.

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And that variable substitution allows this equation to turn

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into a separable one.

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But before I need to show you that, I need to tell you, what

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does it mean to be homogeneous?

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Well, if I can algebraically manipulate this right side of

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this equation, so that I can actually rewrite it.

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Instead of a function x and y, if I could rewrite this

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differential equation so that dy dx is equal to some

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function, let's call that G, or we'll call it capital F.

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If I can rewrite it algebraically, so it's a

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function of y divided by x.

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Then I can make a variable substitution

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that makes it separable.

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So right now, it seems all confusing.

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Let me show you an example.

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And I'll just show you the examples, show you some items,

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and then we'll just do the substitutions.

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So let's say that my differential equation is the

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derivative of y with respect to x is equal to

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x plus y over x.

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And you can, if you'd like, you can try to make this a

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separable, but it's not that trivial to solve.

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Or at least, I'm looking at an inspection, and it doesn't

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seem that trivial to solve.

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And as we see right here, we have the derivative.

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It's equal to some function of x and y.

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And my question is to you, can I just algebraically rewrite

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this so it becomes a function of y over x?

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Well, sure, if we just divide both of these top terms by x.

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This is the same thing as x over x plus y over x.

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This equation is the same thing as dy over

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dx is equal to this.

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Which is same thing as rewriting this whole

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equation-- I'm going to switch colors arbitrarily-- as this,

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dy over dx is equal to x divided by x is equal to 1, if

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we assume x doesn't equal 0.

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Plus y over x.

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So you're probably wondering what did I mean by a function

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of y over x?

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Well, you can see it here.

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When I just algebraically manipulated this equation, I

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got 1 plus y over x.

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So if I said that y over x is equal to some third variable,

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this is just a function of that third variable.

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And actually, I'm going to do that right now.

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So let's make a substitution for y over x.

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Let's say that v-- and I'll do v in a different color-- let's

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say that v is equal to y over x.

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Or another way, if you just multiply both sides by x, you

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could write that y is equal to xv.

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And we're going to substitute v for y over x, but we're also

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going to have to substitute dy over dx.

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So let's figure out what that is in terms of the

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derivatives of v.

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So the derivative of y with respect to x is equal to--

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what's the derivative of this with respect to x?

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Well, if we assume that v is also a function of x, then

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we're just going to use the product rule.

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So the derivative of x is 1 times v plus x times the

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derivative of v with respect to x.

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And now, we can substitute this and this back into this

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equation, and we get-- so dy over dx,

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that is equal to this.

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So we get v plus x dv dx, the derivative of v with respect

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to x, is equal to-- that's just the left hand side-- it's

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equal to 1 plus y over x.

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But we're making this substitution that v is equal

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to y over x.

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So we'll do 1 plus v.

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And now, this should be pretty straightforward.

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So let's see, we can subtract v from both

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sides of this equation.

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And then what do we have left?

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We have x dv dx is equal to 1.

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Let's divide both sides by x.

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And we get the derivative of v with respect to x is

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equal to 1 over x.

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It should maybe start becoming a little bit clearer what the

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solution here is, but let's just keep going forward.

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So if we multiply both sides by dx, we get dv is equal to 1

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over x times dx.

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Now, we can take the antiderivative of both sides,

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integrate both sides.

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And we're left with v is equal to the natural log of the

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absolute value of x plus c.

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And we are kind of done, but it would be nice to get this

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solution in terms of just y and x, and not have this third

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variable v here.

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Because our original problem was just in terms of y and x.

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So let's do that.

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What was v?

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We made the substitution that v is equal to y over x.

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So let's reverse substitute it now, or unsubstitute it.

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So we get y over x is equal to the natural log of x plus c,

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some constant.

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Multiply both sides times x.

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And you get y is equal to x times the natural

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log of x plus c.

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And we're done.

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We solved that seemingly inseparable differential

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equation by recognizing that it was homogeneous, and making

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that variable substitution v is equal to y over x.

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That turned it into a separable

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equation in terms of v.

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And then we solved it.

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And then we unsubstituted it back.

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And we got the solution to the differential equation.

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You can verify it for yourself, that y is equal to

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the x natural log of the absolute value of x plus c.

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Oh, actually, I made a mistake.

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y over x is equal to the natural log of x plus c.

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If I multiply both sides of this equation times x, what's

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the solution?

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It's not just x natural log of x.

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I have to multiply this times x, too, right?

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Distributive property-- that was an amateur mistake.

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So the correct solution is y is equal to x natural log of

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the absolute value of x plus x times c.

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And if you want to figure out c, I would have to give you

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some initial conditions.

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And then you could solve for c.

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And that would be the particular solution, then, for

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this differential equation.

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In the next video, I'll just do a couple more of these

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problems. I'll see you then.

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