Solve the equation as a separable equation (x + xy^2) dx + e ^×^2 ydy =0 Write it as separable to identify the functions Use the corresponding formula to solve...

Solving separable equations Find the general solution of the following equations. Express the solution explicitly as a function of the independent variable. $$u^{\prime}(x)=e^{2 x-u}$$

We're gonna solve this equation D y over the X is equal to And then why X squared plus one. So we're gonna separate the east by bringing the d X up, and we're gonna divide by. Why? So we're gonna end up with one over Why d why we divide by why and X squared bought one DX. That's how we separate them. We integrates, they're gonna integrate this and that's the integral of one over. Why d Why is long why the integral of this is X cubed over three plus X, Let's see. And then what we're going to do is we're just gonna multiply everything by three. So I'm gonna get three long y Is he going X cubed plus three X and then again, plus a constant. And we're gonna bring everything to one side. So three lawn y minus X cubed minus three x minus. This constant C is equal to zero and we'll leave it like that

This question were asked to solve this X Times do you buy? The X is equal to you squared minus four. So in order to separate these variables, we have that D'You divided by you squared minus four. And that's gonna equal one overact. Yeah, if I bring again, I'm bringing this this whole piece to the other side. So I'm gonna multiplied by the DX and I'm gonna divide by the act that I'm bringing this You scared over by dividing by. But then I'm gonna integrate both of the the interval of one overact searches on exp But see, that's relatively easy. Ah, the integral of the U squared minus four is a little bit more interesting. So I'm not gonna go into All the details were how to do that, because it is fairly complicated, but it is an interval using partial fractions, and you could look up the solution of it online if you wish to. But essentially it's gonna become this and again. It's not the purpose of the video to go through all of that because it is a lengthy process and then have to solve this for you. So I'm gonna do this And then I'm just gonna put lawn of see there. And the only reason I'm gonna do that is because it's gonna gonna allow me to get the lawn to be removed. A long A C is also a constant. But what we can then do is make this four lawn of seat I'm X, which then gives me lawn of U minus two over U Plus two equals lawn of C X, the fourth power that then if I get rid of the lawns, allows me to have U minus two over U Plus two is equal to see Ex to the fourth. So that means that U minus two is equal to U Plus two multiplied by C X to the fourth power. And that gives me u minus two is equal to you Time see X to the fourth plus two times c x, the fourth and then I can bring the used together so I get u minus you to the c X. The fourth is equal to two times c x to the fourth plus two. By bringing that to over and then that means that you on the one minus c x to the fourth Power is going to equal do onto one plus back to the fourth power and then I divide so you becomes too onto one. Plus she x to the fourth power, all divided by one minus c x the fourth power.