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

$y = (\frac{1}{6}x^3 +C )^2$

Calculus 1 / AB

Calculus 2 / BC

Chapter 7

Integrals and Transcendental Functions

Section 2

Exponential Change and Separable Differential Equations

Functions

Trig Integrals

Campbell University

Baylor University

University of Michigan - Ann Arbor

University of Nottingham

Lectures

01:11

In mathematics, integratio…

06:55

In grammar, determiners ar…

07:09

Solve the given differenti…

11:49

03:53

Solve the differential equ…

03:18

03:07

00:38

08:19

03:21

01:29

04:08

I want to solve our differential equation here. So let's go ahead and burst. Separate are variables. So we're going to want to do by each side but square root of why. But I'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'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'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'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's go ahead and get rid of that, too. And that's just our new constant. So our solution, it's gonna be why is equal to 1 16 execute will see all squared.

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