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Solve the differential equation.$$(\sec x) \frac{d y}{d x}=e^{y+\sin x}$$

$-y=\ln \left(c_{2}-e^{\sin x}\right)$

Calculus 1 / AB

Calculus 2 / BC

Chapter 7

Integrals and Transcendental Functions

Section 2

Exponential Change and Separable Differential Equations

Functions

Trig Integrals

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we want to solve this differential equation here. Well, let's go ahead and get all of our terms on the same side to remember on the right hand side that's really eat the White Times e to sign X. So we want to divide by E to the Y on each side. Let me just make one big fraction for it. We want to divide by E to the one on each side. And also we want to cancel e c connects. Angela would multiply, see connects. So that's going to give us e to the negative Y d y If we multiply the DX over and then one over seeking his co sign. So this is gonna be co sign e to the sine x d X Now we can go ahead, integrate each side. Ah, the left hand side. It looks like it will be negative e to the negative. Why? And I'll leave the constant for the right hand side and integrate this. We would do a u sub so we would do you is eager to sign X D'You use you to co sign X dx, and after you do that and interpret everything, we should just end it with V to sign X, plus their constancy. I don't call this C one because this is one that we should be able to just solve for why, pretty easily. So let's first go ahead and multiply. You decide by negative one there's gonna be eaten negative. Y is equal to e to the negative or negative e to the sine X. And then we multiply our constant by a negative, that still constant. So I'll just call the sea too. Then we're going to want to take a natural log on each side. And after we do that, we should end up. So it be negative. Why is equal to the natural log of? I'm gonna write this as C two minus e to these sine X and well, we just need to get rid of that negative mouse. Let's just won't quite each side by negative. And so this here would be our solution. So if you wanted, you could have also stopped uh, right here the other one that I boxed in black. It just kind of depends on how you want your solution to look

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