solve-the-differential-equation-2-sqrtx-y-fracd-yd-x1-quad-x-y0-2

Question

Solve the differential equation.
$$2 \sqrt{x y} \frac{d y}{d x}=1, \quad x, y>0$$

Bobby Barnes · Accepted Answer

we want to solve this differential equation here, so this will end up being a separable equation, and it may help us see that if we were to divide each side by the square root of necks so that&#x27;s going to give us two square to buy d Y is equal to, and I&#x27;m going to rewrite this as X to the negative 1/2. I should still be over DX. But now let&#x27;s go ahead and multiply d X to the other side. I say multiply dx in quotes because we really can&#x27;t multiply. But it works out for other reasons. So, uh, DX on this side now we could go ahead and integrate each side and let me rewrite Screwed. Why? As why to the 1/2 power. So now we could go ahead and use powerful to integrate excite. So on the left, Where to? Why now this to the three house power that we want to multiply by 2/3 and this is going to be equal to so x two, the 1/2 power now and then we would multiply by. Two guys were multiplying by are really dividing by the new power and then we have our constant C. Right. So now let&#x27;s do a little bit of manipulations, toe, get some numbers to cancel out. So noticed that we can first let me expand screen and spewed all this over. Actually, um, it&#x27;s, um multiply we get or thirds why? 2 3/2 secret to X. I&#x27;m gonna rewrite that as square root of ex. Now, now, with this point, we can multiply each side by re force and move. It appears that&#x27;s gonna give us why is too re haps or why 23 house and then multiplying to buy 34 should give us three halfs square root of X and then we&#x27;re gonna have C times three halves. So that&#x27;s just gonna be another constant. So I&#x27;ll call this 1st 1 C one. So this one here will be see too. And now we would want to undo our power here. And we would do that by raising each side to the 2/3 power which is gonna give why is equal to 3/2 square root of X waas. See, too, All to the two herbs power. And so now, since we solved for why, let&#x27;s just go ahead and get rid of that C two and call that our new C. So this will be our solution to this differential equation.

Bobby Barnes · Answer

we want to solve this differential equation here. So the first thing we can do this get our variables on the same side or separate them, I should say, as our differentials. So it might be a good idea to abide this by square root of X on each side as well as multiply by D X. So that&#x27;s going to give. I&#x27;m gonna wreck assistant square root of two times the square root of Why do you Why is equal to? And I&#x27;ll rewrite one of the spirit of X as next to the negative 1/2 power like that. And then we have DX once we multiply that over. All right. No, we can integrate each side here, and since that&#x27;s weird to just a fact or it constantly just pull it out now, integrating each side, we would use power rule to do so. So let me rewrite this because that should be a 1/2 power. So power rule says add one to the power and then divide by it&#x27;s new power cells will be multiplied by 2/3 on the right side. We&#x27;re gonna do the same thing, so we add one to the power we get 1/2 divided by that and we end up with two. And on the right side. All right. I&#x27;ll are constant. C got C one, Let&#x27;s go ahead and Saul Or why so first noticed that we divide each side by to the twos council of each other. And then we would want to divide by Route three, the route to over three. So that would be Oh, why, to the three house is equal to three over route, too. Route, too. And I&#x27;m gonna be right next to the 1/2 of this. Where would Vex? And now, since we multiplied our constancy one by some other constants. We&#x27;ll just call that C two now we can raise each side to the 2/3 power. Do you get this to cancel out? And it should be. Why is equal to three square roots of two times the square root of X plus C two all raised to the 2/3 power. So this would be our solution. And if you wanted, you really could have left it. Ah, in this first step here as the square root of 2/3. Why? To the three, huh? Is equal to the square root of X plus e. It just depends on if you want to solve for why, explicitly or not.

Bobby Barnes · Answer

I want to solve our differential equation here. So let&#x27;s go ahead and burst. Separate are variables. So we&#x27;re going to want to do by each side but square root of why. But I&#x27;m square root of were won over. Why on that side those cancel out now we can multiply each side by DX and I say multiply by DX and quotes. But after we do that, we should end up with X squared times. DX is equal to and let&#x27;s rewrite one over this world. Why, That should be why to the negative 1/2 power that we have a d y. Now we can integrate each side and each side will use powerful to solve. So there should be why to the 1/2 and then we divide by its new powers that should be multiplied by two. And this is going to be equal to so X cube now to the 1/3 plus our constant. And I&#x27;ll just call this one C one for now. Now we want to, Saul, were why? Well, we can divide everything by to start. So that gives us why to the 1/2 is equal to 16 x cubed loss of that C 1/2. That&#x27;s gonna be a new constantly just called C two. And at this point, you would want to square each side that would give us why is equal to one six x cubed plus C two squared. So since we solved for why, let&#x27;s go ahead and get rid of that, too. And that&#x27;s just our new constant. So our solution, it&#x27;s gonna be why is equal to 1 16 execute will see all squared.

Samuel Hannah · Answer

So for this problem, we have a first order differential equation, which I have free written on the top left of the screen that we would like to rewrite this equation in terms of homogeneous variables. Andi, Andi to vote in tomorrow Explicit form. We want to get everything in terms of why we will start this by looking at the school average. So if we factor out X squared underneath the square root we get, the square is equal to the square root of X squared, multiplied by the square root of one plus y squared over X squared. And then this is equal to X times the square root of one plus y squared over X. Glad now we take the positive square root of acts because by the question, X is always positive. So the right hand side of the differential equation now reads X squared. How does he square root of one plus y squared is X squared plus y squared all over X more and we can do one final rewriting by factoring out X squared in both the numerator and a denominator which gives us ey D X is equal to the square root of one plus y squared of the X squared plus y squared over X squared all over. Why over X this we see that we now have the differential equation in the form of being only dependent on why over acts. So it is in fact, homogeneous. We commend. Make the variable transformation off the equals y over X, or why equals v X equivalently. Using the second formulation, we can see that the want the X is equal to V plus X e v the X using the product role. If we substitute this into the left hand side of our equation onset of right hand side, we just substitute in the we get the differential equation being the plus X DVD X is equal to the square root of one plus V squared plus B squared over the we can make an additional simplification by subtracting V from both sides. But on the right hand side notice that if we was to split the fraction into its component forms, the second component V squared over V is simply the itself on V minus three vanishes. So the differential equation is simply x e v. The X is equal to the square root one plus v squared over the This is a separable differential equation. So we will separate the variables. We&#x27;re in V to the left hand side. Get that. The over the square root of one plus V squared BV is equal to one over x, the X and then to Seoul. So the we need to integrate both science. We will first note that the integral of one over X is simply the natural lager log of X, so that is quite simple. And to integrate the V variables, we will make the substitution off you equals one plus V squared. By differentiating this, we see that to be TV is equal to d you and baffle the integral transforms from the of of a square root of one plus V squared devi into the integral of one over to times square root of you. You. This is simple integration on how solution of the square root of you plus c and then we just we just substitute back into all the variables. Then this is equal to the square root of one plus b squared, plus the integration constant c. So we solved v integral on me nervous solution of the X integral. So the overall equation becomes the square root of one plus V squared. Plus the constancy is equal giving natural log of X. Now, by bringing the sea over to the right hand side and then square in both sides, we get the one plus V squared is equal to the square that the natural log of X plus k, where K is the constant equal to minus c and that&#x27;s all squared. So we can now simply bring the for want of a right hand side and get our final solution, which is V squared, is equal to the natural log of X plus K all squared minus one. So we now have our solution for they what? We&#x27;ve been asked to find a solution for my not be so We need to substitute variable the equals y of X back into this equation. They&#x27;re in this. We simply get the Y squared over. X squared is equal. Teoh natural log of X plus K all squared minus one and then multiple it multiplying through by X squared Received that This final solution is all I squared equals X squared all times the square of the natural, Other vex minus K plus K even minus one. This is the solution. The differential equation.

Patrick Vaughan · Answer

So today we&#x27;re given a question we&#x27;re asked to solve for it using our standard techniques. So this isn&#x27;t a very straight forward problem. And I just want to show you how you should be thinking about these types of problems. So I&#x27;m gonna start out using an incorrect method of how to solve this problem. Or at least it wasn&#x27;t a way that I found particularly useful using our typical y equals V X, um, solutions that our substitution that, um so to the left, you&#x27;ll see I have a skip to solution in a little box later on, I&#x27;m gonna edit in where you can the time to skip to when I start through the solution. But I just wanted to make it so that you can sort of see how we can start to think about these more complex problems. That&#x27;s where we&#x27;re looking at it and being like, Oh, well, I&#x27;m not seeing that this Methodist particularly working. This is how I&#x27;m gonna work. Maybe I should try something else. There&#x27;s something new, So I just want to sort of explain to you how I think about these problems when I get stuck or challenged and hopes that hopefully it will help you to try different things in a person here. Um, when you encounter things like this All right, So let&#x27;s recall that we have been typically taking our substitution as why equals r vivax time or written up another way. Are vivax equals? Why demanded by acts? So starting off with this problem that we want to isolate the wide prime so we can divide why, by both sides Or just most play the whole thing by one of her wife and I give bless our why prime or R d y Over the X equals our radical X squared plus y squared, divided by why minus X over y And then we take our live value where it&#x27;s being x, times X, and we substitute that infer wherever we have a wide so that he over the act time vivax time blacks equals radical X squared plus B squared X squared over all over. Why, um And then, uh, excellent. Why is the inverse of be so that&#x27;s one Overbey. Okay, so taking the derivative of the left side that makes it so that we have are the prime times acts, plus, uh, the which equals. And then on the right side will do. Ah, step will retake our X square out. We pull it out and hopes that this will help us. So that becomes X squared times one plus he squared over. Why? Minus what ever be, um, and then the left side. So we&#x27;ll have a brake line, and then we&#x27;ll have X times are devi over the act plus B equals X over y. So this right here becomes X over y because radical ax squared just acts. So again, I&#x27;m gonna have whatever be radical every time radical one plus squared minus one of her be and then also attracted me from both sides. And I&#x27;ll get TV over. The X equals well, this is radical. One plus b squared might be minus one minus. B squared over one. About it might be because to get that, uh, denominator and B, we multiply by the Overbey. Okay, so now we&#x27;re left with this complex expression on the right, and then we&#x27;re left with, um, access TV yet. So if you, uh, do our separation of variables, we get, um, TV times The over are very large denominator, which would be one plus B squared, minus B squared, minus one times like TV equals, um de X over X. And then you can try integrating mess. But I had difficulties. I couldn&#x27;t really figure out how to make this simpler. So I want you in culture, this type of situation, there&#x27;s to think that you could try the first is to check your assumptions. I didn&#x27;t start a problem that I was gonna be able to use that, um the substitution, nothing. And, uh, looks like that it didn&#x27;t really hope me because I could see integration part. It&#x27;s pretty this cold. I couldn&#x27;t really figure out how to go from there, so well, the right side pretty straightforward with left side. Um, that was a little more complex, and I didn&#x27;t like working with that. So one, uh, check your assumptions to you should, if your assumptions are correct, figure out a different way. Maybe try. Um, so see, it&#x27;s father methods in this. We&#x27;ll put a question mark better because we don&#x27;t know that&#x27;s something we have to investigate. Sometimes it will have other pre established methods for solving these types of problems. Sometimes it&#x27;s just you made the wrong assumption beginning problem. And in this case, I made the wrong assumption. Great being in my problem and I&#x27;ll show you what I mean. So I view this part and great silent create Green is good. Okay, So after a little bit of heaven on I decided to try you substitution where I said you equal to x squared plus y squared And then my d&#x27;you is to act Plus you Why times d y over the x So you might look at me wondering Well, how can I make that assumption? Well, we&#x27;re assuming that is our variable of interest or independent variable and that why is the output very well associated with facts? So that&#x27;s why I could do this. Okay, so now that we&#x27;ve established that way, can then rewrite this as the prime or the derivative of you equal to times X plus why and what&#x27;s another way to write? Do you everyday acts? Well, that&#x27;s why prime and something interesting happens if you use this, you substitution. So I&#x27;m gonna use red and under it highlights. So I have X squared And why squared? I squared y squared and then in my you if we divide by two. So we get you prime over to equals X plus. Why? Why prime? Well, if we look and if we look, we see that we have ah x y y prime we have Why comes like crap. And then if we took this, Max brought it over. So let me redirect this. Um, a little I&#x27;m equals radical. I could quitter plus wife minus X was the original expression and taking the original expression we can make it in. Why I just adding actual side explain. Plus why squared? And then all of a sudden, we see that we have I fly squared And then our x squared plus y squared this original expression that gives us you prime over to you equals radical. Well, what is you prime? Well, that&#x27;s just do you over the X. So do you Over the x times that to you equals radical you. Well, let&#x27;s separate power variables going into a little more space that gives us, uh, you to the negative 1/2 which is just one over radical. You do. You is equal to two e x. So now we can integrate. Well, that&#x27;s a pretty straightforward. Integral. On the right side, we just have to x plus C and on the left side. Well, what&#x27;s the in a role of you to the negative 1/2? That&#x27;s just you with 1/2 time here because between derivative of you to the 1/2 that two and 1/2 cancels out, we&#x27;re left with you to the next 1/2. Okay, so now let&#x27;s continue salt into the problem. What&#x27;s divide by two? Just get this in expression of you. So that gives us you the 1/2 equals plus C over to and then what was you? Let&#x27;s remind ourselves, growing up you was X squared plus y squared. So putting this back in for are you substitution? We get X squared plus y squared one equals X. What? See bottom by two and that&#x27;s your final answer. And now just seem out. There we go. And then, as a reminder, this not dead method on the left did not work and that the U substitution method was the one that ended up being successful

Bobby Barnes · Answer

we want to Seoul this differential equation here. Well, first thing. Let&#x27;s go ahead and move all of our white terms up to look spell. We would want to divide each side by the square root of one minus y square on the left That&#x27;ll give us one over the square root of one minus y square. We&#x27;ll have d y. And then if we multiply the DX over, we get to X d X. Now let&#x27;s integrate. You decided this the left hand side and pretty sure that&#x27;s arc sine. So this is Sy inverse of why is equal to and then to integrate two ex Well, that&#x27;s going to be power rule. So we add one to the power it becomes like squared. Then we divide by its new power. So the two cancel out. So we just get X squared plus C Now, let&#x27;s just take sign on each side of this. Since in the problem, they tell us we&#x27;re restricted between why is even too negative one and one. We won&#x27;t need to put any extra restrictions on this. Why is he too, my abs sine of X squared plus seats of this here would be our solution

Problem

Solve the differential equation. $$\frac{d y}{d x…

Problem 9 Easy Difficulty

Solve the differential equation.
$$2 \sqrt{x y} \frac{d y}{d x}=1, \quad x, y>0$$

Answer

$$\frac{2}{3} y^{3 / 2}-x^{1 / 2}=C$$

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Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals

Grace H.

Anna Marie V.

Kayleah T.

Kristen K.

Integration Review - Intro

Definite vs Indefinite

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$$\frac{2}{3} y^{3 / 2}-x^{1 / 2}=C$$

University Calculus: Early Transcendentals

Grace H.

Anna Marie V.

Kayleah T.

Kristen K.

Integration Review - Intro

Definite vs Indefinite

Joel Hass, Christopher Heil, Przemyslaw BogackiUniversity Calculus: Early Transcendentals

Grace H.

Anna Marie V.

Kayleah T.

Kristen K.

Joel Hass, Christopher Heil, Przemyslaw Bogacki

University Calculus: Early Transcendentals