Decide if thc following differential cquation is exact , and if it is, find the gencral 32"y2 2cy" Hdlc (2yr" 52"y' _ 2sin %y) dy = 0...

Determine whether the given differential equation is exact. If it is exact, solve it. $$\left(x-y^{3}+y^{2} \sin x\right) d x=\left(3 x y^{2}+2 y \cos x\right) d y$$

All right. So for this problem let's first determine whether it's an exact differential equation or not before solving it. So to determine whether it's an exact differential equation would want to have the form M. D. X. Plus and D. Y. Equals to zero. And so let's go ahead and get this equation that we have to be in that form. And the way we'll do that as well bring all this junk to the right hand side. And so when we do that we have to Y I just wanna divide by X. Plus co sign three X. Dy DX equals two. Y. X squared negative plus four X cubed minus three. Y. Signed three X. Okay. I can multiply both sides by D. X. All right. And when I do we'll just have D Why the left hand side? And then I can bring this entire thing back to the right hand or left hand side. Sorry. And so then I'm going to have why divided by X squared -4 X. Cubed plus three. Y. sign three x. Uh huh. D X plus two Y minus one divided by X plus Co sign three X. D. Y. All right. And now we can go ahead and take the partial derivatives. This so what we're going to need to do is we're gonna take partial why of Y divided by X squared minus four X cubed plus three. Wise signed three X. And we have to take the partial derivative with respect to X. Off to why minus one divided by X plus. Co sign three X. So let's start off with the numerator are sorry? The the top partial derivative. And so what we get is one divided by X squared Um Plus three signed 3 x. And on the denominator we take the passenger with respect to X. We get one divided by X squared minus three. Sign three X. Okay. And so These do not equal one another. And what that means is we do not Mhm. Have and exact differential equation. Yeah.

Okay, so start this problem by determining whether it's a exact differential equation or not. And the way we go about doing that is we take the partial derivative of this and the partial derivative of this and see if they equal one another. So have partial Y of two, X, y squared minus three. And we'll have partial acts off two X squared y plus four. All right. On the left hand side, we take the partial derivative with respect to why we get four X. Y. And on the right hand side. And we take the partial respect to X. We get four X. Y. As well. So these equal one another. So we can make the conclusion that we have an exact french equation. And so what this allows us to do is it allows us to solve it like we would normally, So we would have the integral of two X. Why squared -3, the X. And the integral uh two X squared why? Plus four? Yeah, D. Y. All right. And so let's start off with the top integral. So that's gonna integrate too, um X squared y squared minus three X. And in the bottom we're going to have X squared y squared plus four Y. All right. So we need to take the union of these. So we'll have please. And well, that will be our solutions will have X squared y squared -3 x plus four y equals to see where C. Is some constant. And that's her answer.

So for this equation we're going to have we're going to write down that M is going to be co sign X y minus X y sine x y And the end is gonna be negative X squared sine x y And the reason why these parts are m and an is because these are the ones that are multiplied by the X ray here and D wired in for the m and n So when we take the derivative of n with respect two d y what we will get is that we will be getting uhm I'm going to just simplify it just to save some time But we will get to X sign x y you just, uh, reduce. We're here, X Why? And ah minus X squared times Why and and times co sign X y And for the when we take the derivative of de and by D X, we will get to be negative two X sign Sign X Y minus X squared. Why co sign co sign Thanks. Why and so just gonna make that s a little bit more like an s. So if you take a look, you can determine that this equation is there is identical to this one. So since oh, noes an accident since D M over de y is identical know that looks second X over de y is identical to de en over de acts, we can say it is exact.

Okay, so for this problem let's go ahead and take the partial derivatives. So we'll take partial. Why? Uh for t cubed y minus 15 T squared minus why? And partial t of teaching the 4th plus three Y squared minus T. And so we'll start off at the top one. We take the partial, We got 40 cubes minus one. And then on the bottom we also get 40 cubed minus one. So these two were equal to another. So what we can conclude is that we have an exact differential equation. Mhm. And now we can solve it like we would Usually, so we'll take the integral 40 cubes y on his 15 T squared minus Y. And will integrate with tea. And then we'll also be integrating t to the fourth plus three Y squared months T with respect to Y. Okay, so the top integral, let's go ahead and tackle that one first. So we're going to get you to the fourth Y-. Let's see 50. Or sorry, five two cubed minus T. Y. And then the bottom We'll get to the 4th y plus why cute minus T. Y. All right. And so let's go ahead and see what terms are gonna get into our final solution. So I have these two terms or this term. We'll also have this term appearing and then this term and this term. So we have four terms in total. So let's go ahead and put them all into our final solution. So I've teeth the fourth. Y plus Y cubed minus T Y -5 T. Let's see five T cubed. And this all equals to see where she is some constant and that's her answer.