Problem

Solve the given differential equation. $$x^{2} \f…

05:06

Problem 18 Medium Difficulty

Solve the given differential equation.
$$2 x y d y-\left(x^{2} e^{-y^{2} / x^{2}}+2 y^{2}\right) d x=0$$

Answer

$y^{2}=x^{2} \ln (\ln (C x))$

Related Courses

Calculus 2 / BC

Differential Equations and Linear Algebra

Chapter 1

First-Order Differential Equations

Section 8

Change of Variables

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Video Transcript

Hello. So today we're working on first order differentials, and ultimately, this is gonna boil down to a problem of separation of variables or B ah, split up energy to variable sets. And then we integrate with respect each variable. Um, and the reason that we know this is because we have a d y in the d X. All right, so what do we do? We given an expression that looks like this where we have an equation that set equal to zero. Well, our first thing that we want to do is sort of split up the parts so that we can reduce it down to two variables. So first, we're gonna want to add this complex expression to both sides. We have a brake line just how we differentiate steps within the problem. So that gives us two x y you I equals X squared e to the negative y squared over X squared plus two y squared DX. So next we can, uh, reduced down the expression but the differentials on one side and the only variable the others will divide by two x y and then by whatever DX, we're doing two things simultaneously here And that gives us Do I already X equals our x squared e It's the negative y squared over x squared What two y squared, all divided by to why accer two x life. Okay, well, no. From our textbook, we can recall that one way Teoh early Use a y ver expression is to equal to a new variable R b which has an X component within it. So the equals y over X or the other way to write that is why equals the time, Zach. So we can take this. Why expression and plug in for all of our wise. And then we can also do that with our, um the expression and plug that in when we have live. Rex. So doing this, we get the partial of, um, be of Akst times about about Deac equals X squared. Yeah, to the negative b squared, because why over act, um, squared. All right, we're just putting of the end for that expressions that becomes B squared, X squared over X squared. So I think to be squared, or if you just take it as an expression of, um, why over x squared, then that's the same thing is to be of X squared. So either way, it's the same thing if you just plug in for why or for the whole expression. Okay, so then we plug in for the second line. So we have vivax times act squared, all divided by two x squared and then put into that Why vivax times x. So just ah, short reminder that the reason why I did not as the vaccine because when we're taking the differential and I'll do this in green So the differential of this vivax times X we have to, uh, variables that are, um, that have x contained within them. So you have to dio of the differential breakdown by part Sophie left side equation times the derivative of the right side plus the derivative of the left side times the original with the right side. And that gives us, uh, well, now that we've done that, we don't need a do the and remind ourselves that it is a subset of acts or has subset of acts. So be plus e prime, and then we can take that and plug it in for that expression. So that's just a little reminder that we dio until we get to that point. So this is the same thing is ready TV over the x Times X plus de And then we can rewrite the expression above where we have X squared. Yeah, nothing b squared plus two b squared X squared Hold about it by two x squared the x The reason I'm not combining that denominators because we see that we have an X squared in both Art is the expression so I can cancel out and then we're left with e and I'm gonna be squared. What? To be squared all over too. The X equals R X TV over the X plus de So now we can attract the from both sides of the equation. And Teoh make it so, um sorry. Ah, so backtracking. I made a tiny mistake, so that's not squared. So it's always good, Teoh. We'll check yourself. You're solving these equations. So two x squared be, um And then there's no x there, So just sort of show you what? Where I made my mistake. So I have about my two y x here and then when I'm playing in for why it's two acts times p X and then we have the two x squared b on. We're getting that. So then I boil some of that. OK, so now that we've corrected that mistake, do you get that expression? Uh, the same have the same denominator so we can combine the two expressions to multiply that by to be over to be So then this boils down to X TV over DX. Um, And then when we combine these two expressions, we have to be squared in two b squared, and that's to B squared minus to B squared. So that cancels. We have e So the negative b squared over to be so this very complex expressions getting much simpler. And now we can separate variables such that we have. We can, uh, multiply the whole expression by to the over eat the negative b squared. Um, and I just I won't do this simultaneously this time. So we have that equals one. And then we have to be, um, he to the B squared. Because when you're dividing by a negative beast square, that's like saying one over one divided, divided by eat the V squared. So that's just equal to eat. of the B squared. Um, and then times are over the ec. So that may want to multiply the whole thing by the X over X, and that will give us to the B to the B squared equals, uh, one over X, the X. So now we've successfully separated are variables. We have these on one side X on the other, and we can integrate. Now, which the whole goal. Um, and I apologize. I left out my TV. There we go. Okay. So he can integrate. So the integration of the right side would be Oh, are only on the vax plus Ah, see some constant and then the adoration of the left side while we have two sets of the variables. So what do we do when we have that? Well, we can use a substitution or a u substitution. So you more equal b squared. The derivative of you would be to be bebe. And then this plugs in very nicely. We have are two vdv to be TV and then we have our b squared. So student Matt in, we have yeah to the u Do you you do you to you is are integral Well, sorry, just et you. Because the derivative of each of you is Are you dio so plugging in for you. So now we've successfully done both for immigration, but we want to put it back to our original format. So with our wives and our axes Sorry. So we wanna put it back into our original format with our wives and our axes. So first we have to reverse, solve and plug. Are you back in? So this becomes e. B squared equals natural. Lagerback's what I see. And then we can plug our, uh, wild racks squared back in, Harvey. And that's natural log max plus C. And, um, that's our final answer. Because we have successfully reversed integrated or river solved back and for what we substituted in for, um and ultimately, what you could dio is go back in and, uh, suffer why? And sort of winning. I realize that, Um, but you don't have to. It wasn't asked for in the problem. So this is indeed the final answer, and I'll just seem out so you can see everything we did to get to this point

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