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

$$e^{-y}+2 e^{\sqrt{x}}=C$$

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

Idaho State University

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I want to solve this differential equation here. Well, since it's separable, we probably want to move everything over on the same side. So what I'm gonna do is just divide each side by the square root of X and E to the Y. And doing that would give us so e to the negative wide on the left. And then we have, Do you why? And actually also multiply each side by the ex? I say multiply in quotes because we really can't multiply by our differential. But it works out for other reasons, right? But now we divide by square root so it would have e to the square of X over that's weird mix and then times DX. Now, at this point, this is a separable equation or we have it wrote out where we can integrate each side. So in the left hand side, this is just going to be negative e to the negative. Why and then on the right hand side looks like we'll need to do a U sub or that. So if you do that, you sub you is equal to the square root of X and do you is equal to one over the square root of X times 1/2 D X, and plug in all that. And we should come to the conclusion that this is two times he'd this word of ex Mostar constancy. Now this one seems nice. Or if we saw for why, let's just go ahead and do that. You could leave it at this point if you want to go. But let's just go ahead and keep on working. So almost by your side, by negative one that's going to give E to the negative. Why is he too negative too? So, actually, let's call this C one so negative to e to the square root of X plus our constancy to now because we just multiply that by a another constant. Now we want to take the natural log on each side for equation and let me actually move this up here that's gonna give negative why is equal to the natural log? Oh, C two minus two he to the square root. Oh, thanks. And then we would multiply each side by negative one and we'd get why is equal to negative natural log of so see too. But we just even a C now minus two E to the square root of X. So, like I was saying, you could either write your answer out like this here if you want it explicitly in terms of why Or you could also just have stopped at this part right here. But I also boxed in black, so just whichever one looks better to you.

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