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First order homogenous equations
I will now introduce you to the idea of a homogeneous differential equation. Homogeneous is the same word that we use for milk, when we say that the milk has been-- that all the fat clumps have been spread out. But the application here, at least I don't see the connection. Homogeneous differential equation. And even within differential equations, we'll learn later there's a different type of homogeneous differential equation. Those are called homogeneous linear differential equations, but they mean something actually quite different. But anyway, for this purpose, I'm going to show you homogeneous differential equations. And what we're dealing with are going to be first order equations. What does a homogeneous differential equation mean? Well, say I had just a regular first order differential equation that could be written like this. So dy dx is equal to some function of x and y. And let's say we try to do this, and it's not separable, and it's not exact. What we learn is that if it can be homogeneous, if this is a homogeneous differential equation, that we can make a variable substitution. And that variable substitution allows this equation to turn into a separable one. But before I need to show you that, I need to tell you, what does it mean to be homogeneous? Well, if I can algebraically manipulate this right side of this equation, so that I can actually rewrite it. Instead of a function x and y, if I could rewrite this differential equation so that dy dx is equal to some function, let's call that G, or we'll call it capital F. If I can rewrite it algebraically, so it's a function of y divided by x. Then I can make a variable substitution that makes it separable. So right now, it seems all confusing. Let me show you an example. And I'll just show you the examples, show you some items, and then we'll just do the substitutions. So let's say that my differential equation is the derivative of y with respect to x is equal to x plus y over x. And you can, if you'd like, you can try to make this a separable, but it's not that trivial to solve. Or at least, I'm looking at an inspection, and it doesn't seem that trivial to solve. And as we see right here, we have the derivative. It's equal to some function of x and y. And my question is to you, can I just algebraically rewrite this so it becomes a function of y over x? Well, sure, if we just divide both of these top terms by x. This is the same thing as x over x plus y over x. This equation is the same thing as dy over dx is equal to this. Which is same thing as rewriting this whole equation-- I'm going to switch colors arbitrarily-- as this, dy over dx is equal to x divided by x is equal to 1, if we assume x doesn't equal 0. Plus y over x. So you're probably wondering what did I mean by a function of y over x? Well, you can see it here. When I just algebraically manipulated this equation, I got 1 plus y over x. So if I said that y over x is equal to some third variable, this is just a function of that third variable. And actually, I'm going to do that right now. So let's make a substitution for y over x. Let's say that v-- and I'll do v in a different color-- let's say that v is equal to y over x. Or another way, if you just multiply both sides by x, you could write that y is equal to xv. And we're going to substitute v for y over x, but we're also going to have to substitute dy over dx. So let's figure out what that is in terms of the derivatives of v. So the derivative of y with respect to x is equal to-- what's the derivative of this with respect to x? Well, if we assume that v is also a function of x, then we're just going to use the product rule. So the derivative of x is 1 times v plus x times the derivative of v with respect to x. And now, we can substitute this and this back into this equation, and we get-- so dy over dx, that is equal to this. So we get v plus x dv dx, the derivative of v with respect to x, is equal to-- that's just the left hand side-- it's equal to 1 plus y over x. But we're making this substitution that v is equal to y over x. So we'll do 1 plus v. And now, this should be pretty straightforward. So let's see, we can subtract v from both sides of this equation. And then what do we have left? We have x dv dx is equal to 1. Let's divide both sides by x. And we get the derivative of v with respect to x is equal to 1 over x. It should maybe start becoming a little bit clearer what the solution here is, but let's just keep going forward. So if we multiply both sides by dx, we get dv is equal to 1 over x times dx. Now, we can take the antiderivative of both sides, integrate both sides. And we're left with v is equal to the natural log of the absolute value of x plus c. And we are kind of done, but it would be nice to get this solution in terms of just y and x, and not have this third variable v here. Because our original problem was just in terms of y and x. So let's do that. What was v? We made the substitution that v is equal to y over x. So let's reverse substitute it now, or unsubstitute it. So we get y over x is equal to the natural log of x plus c, some constant. Multiply both sides times x. And you get y is equal to x times the natural log of x plus c. And we're done. We solved that seemingly inseparable differential equation by recognizing that it was homogeneous, and making that variable substitution v is equal to y over x. That turned it into a separable equation in terms of v. And then we solved it. And then we unsubstituted it back. And we got the solution to the differential equation. You can verify it for yourself, that y is equal to the x natural log of the absolute value of x plus c. Oh, actually, I made a mistake. y over x is equal to the natural log of x plus c. If I multiply both sides of this equation times x, what's the solution? It's not just x natural log of x. I have to multiply this times x, too, right? Distributive property-- that was an amateur mistake. So the correct solution is y is equal to x natural log of the absolute value of x plus x times c. And if you want to figure out c, I would have to give you some initial conditions. And then you could solve for c. And that would be the particular solution, then, for this differential equation. In the next video, I'll just do a couple more of these problems. I'll see you then.