Homogeneous Differential Equation (Y^2 + Yx)Dx - X^2Dy = 0

Dated : 18-Apr-2022

And this problem going to solve this differential equation. So this differential equation is homogeneous, and you can tell because the exponents match. So for example, this here is 2. And here 1, plus 1 is also 2. So this piece here is called a homogeneous function of degree 2, because all the exponents are the same. And again, you can add the exponents, even when the bases are different. Likewise, here, this is also 2.

So this is a homogeneous function of degree 2. So if both of the functions in front of DX, And Dy are homogeneous of the same degree. We say the differential equation is homogeneous. So again, you can just look at the exponents as long as they all match.

You are ok. So there are 2 choices to solve this differential equation. Method.

One is to let x equals V Y in which case DX is equal to VD, y, plus y DV. This would involve taking this DX and replacing it with the DX here, which would mean we have to foil. The other option is to let y equals UX in which case Dy is equal to u. DX, plus x2 u. This. Option is easier because then we only have to multiply it by x squared, right? The Dy only gets multiplied by x squared.

So you don't have to foil. So let's, take this approach I'm going to go ahead and write it again over here. So y equals u X and then Dy equals u. DX, plus XD u. It's, really just a product rule, but it's really easy to memorize you just alternate the derivatives, ok, let's be really careful.

So Y is equal to UX, so I'm going to put a parenthesis here and then u x squared. Ok. And then plus, we. Have another Y so UX and then X and then and then DX, yeah, DX is okay because we're using we're using this one minus x squared, and then Dy is U DX, plus XD u and then parentheses and equal to zero. So these take a little of effort, I've always felt that these are a little harder than the other DS that are studied at the beginning.

Of course, it's just really easy to mess up. Okay, let's distribute all of this so u. X squared that's, going to become u squared x squared, and then we're going to. Multiply it by the DX so DX, this next piece here will be u x squared. And then again, multiply it by the DX so DX, good stuff distribute the x squared. So minus x squared u, DX and then distribute this one.

So minus X, cubed, D, U and that's equal to zero. And, oh, ha, ha, cancellation that doesn't always happen. Okay. So not every time that worked out quite nice. This is a separable DE. This will always be separable. So let's, take this and send it over to the right-hand side by adding it over.

So we have u squared. X squared DX equals x, cubed, D U. And the goal is to get all the X's on one side together with the DX and all the Y's together on one side together with D u. Apologies, for that beeping sound I forgot to close my email, scared me, I'm like what is that all right?

Let's, keep going whew also divide by X cubed, that's x squared over X, cubed I, don't know if it sounded as loud to you as it did to me, but that was really loud because I'm wearing headphones 1 over u squared. Do you? Okay?

So yeah, it looks like. We did okay, we divided by X cubed, divided by u squared I'm gonna focus this is 1 over X DX equals let's bring this upstairs. Because the next step is going to be to integrate. So we want to write it as u to a power. We didn't bring this one upstairs, because when you integrate this it's this right, natural log, absolute value of x and over here, we add one. So this will be U to the negative one over negative, one plus C. So we're writing this to split this actually wasn't so bad, Allyn X.

This is. Negative one over U, plus C, nice problem, but we are not done right? Remember that Y was equal to u X. So that means that U is equal to Y over X. So 1 over u is x over Y.

So this is right you just flip it. So just negative x over y, plus C. And this would be the final answer kind of pleasant question. It wasn't as hard as some of the other ones that I've been doing lately. So, yeah, a pleasant issue and the integration wasn't too bad. The simplification wasn't too bad.

Yeah, really. Good solid issue. I hope. This video has been helpful.