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# First order homogeneous equations 2

## Video transcript

Let's do one more homogeneous differential equation, or first order homogeneous differential equation, to differentiate it from the homogeneous linear differential equations we'll do later. But anyway, the problem we have here. It's the derivative of y with respect to x is equal to-- that x looks like a y-- is equal to x squared plus 3y squared. I'm writing it a little bit small, so that I don't run out of space. Divided by 2xy. So in all of these homogeneous equations-- and obviously, we don't know it's homogeneous yet. So we have to try to write it as a function of y divided by x. So it looks like we could do that, if we divide the top and the bottom by x squared. So if we just multiply-- let me do it in a different color-- 1 over x squared, or x to the negative 2, over 1 over x squared. We're essentially just multiplying by 1. And then, that is equal to what? 1 plus 3y squared over x squared divided by 2-- if you divide x, divide by x squared, you just get a 1 over x-- so 2 times y over x. Or we could just rewrite the whole thing, and this is just equal to 1 plus 3y over x squared divided by 2 times y over x. So yes, this is a homogeneous equation. Because we were able to write it as a function of y divided by x. So now, we can do the substitution with v. And hopefully, this is starting to become a little bit of second nature to you. So we can make the substitution that v is equal to y over x, or another way of writing that is that y is equal to xv. And then, of course, the derivative of y with respect to x, or if we take the derivative with respect to x of both sides of this, that's equal to the derivative of x is 1 times v, this is just the product rule, plus x times the derivative of v with respect to x. And now, we can substitute the derivative of y with respect to x is just this. And then the right hand side of the equation is this. But we can substitute v for y over x. So let's do that. And so we get v plus x. Instead of dv [? dv ?] x, I'll write v prime for now, just so that I don't take up too much space. v prime is equal to 1 plus 3v squared, we're making the substitution v is equal to y over x. All of that over 2v. Now, let's see what we can do. This is where we just get our algebra hat on, and try to simplify until it's a separable equation in v. So let's do that. So let's multiply both sides of this equation by 2v. So we'll get 2v squared plus 2xv v prime-- 2v times x, yep, that's 2xv v prime-- is equal to 1 plus 3v squared. Now let's see, let's subtract 2v squared from both sides of this. And we will be left with 2xv v prime is equal to 1 plus-- let's see, we're subtracting 2v squared from both sides. So we're just left with a 1 plus v squared here, right? 3v squared minus 2v squared is just v squared. And let's see, we want it to be separable, so let's put all the v's on the left hand side. So we get 2xv v prime divided by 1 plus v squared is equal to 1. And let's divide both sides by x. So we get the x's on the other side. So then we get 2v-- and I'll now switch back to the other notation. Instead of v prime, I'll write dv dx. 2v times the derivative of v with respect to x divided by 1 plus v squared is equal to-- I'm dividing both sides by x, notice I didn't write the x on this side-- so that is equal to 1 over x. And then, if we just multiply both sides of this times dx, we've separated the two variables and we can integrate both sides. So let's do that. Let's go up here. I'll switch to a different color, so you know I'm working on a different column now. So multiply both sides by dx. I get 2v over 1 plus v squared dv is equal to 1 over x dx. And now we can just integrate both sides of this equation. This is a separable equation in terms of v and x. And what's the integral of this? At first, you might think, oh boy, this is complicated. This is difficult, maybe some type of trig function. But you'll see that it's kind of just the reverse chain rule. We have a function here, 1 plus v squared, an expression here. And we have its derivative sitting right there. So the antiderivative of this, and you can make a substitution if you like. You could say u is equal to 1 plus v squared, then du is equal to 2v dv. And then, well, you would end up saying that the antiderivative is just the natural log of u. Or, in this case, the antiderivative of this is just the natural log of 1 plus v squared. We don't even have to write an absolute value there. Because that's always going to be a positive value. So the natural log of 1 plus v squared. And I hope I didn't confuse you. That's how I think about it. I say, if I have an expression, and I have its derivative multiplied there, then I can just take the antiderivative of the whole expression. And I don't have to worry about what's inside of it. So if this was a 1 over an x, or 1 over u, it's just the natural log of it. So that's how I knew that this was the antiderivative. And if you don't believe me, use the chain rule and take the derivative of this, and you'll get this. And hopefully, it will make a little bit more sense. But anyway, that's the left hand side, and then that equals-- Well, this one's easy. That's the natural log, the absolute value of x. We could say, plus c, but just so that we can simplify it a little bit, an arbitrary constant c, we can really just write that as the natural log of the absolute value of some constant c. I mean, this is still some arbitrary constant c. So we can rewrite this whole equation as the natural log of 1 plus v squared is equal to-- when you add natural logs, you can essentially just multiply the two numbers that you're taking the natural log of-- the natural log of, we could say, the absolute value of cx. And so the natural log of this is equal to the natural log of this. So we could say that 1 plus v squared is equal to cx. And now we can unsubstitute it. So we know v is equal to y over x, so let's do that. So we get 1 plus y over x squared is equal to cx. Let me scroll this down a little bit. Let's multiply both sides of the equation times x squared. We could rewrite this as y squared over x squared. So we multiply both sides times x squared, you get x squared plus y squared is equal to cx to the third. And we're essentially done. If we want to put all of the variable terms on left hand side, we could say that this is equal to x squared plus y squared minus cx to the third is equal to 0. And this implicitly defined function, or curve, or however you want to call it, is the solution to our original homogeneous first order differential equation. So there you go. I will see you in the next video. And now, we're actually going to do something. We're going to start embarking on higher order differential equations. And actually, these are more useful, and in some ways, easier to do than the homogeneous and the exact equations that we've been doing so far. See you in the next video.