how to solve: y'(x)=e^(y/2)sinx ans: y=-2ln(1/2cosx+C)

This appeares to be a separable Differential Equation. Do you know how these generally play out? Like, what we should be looking to do?

nope, my teacher is awful/I slept through class. So, I'm looking through my book for relevant examples, but there's not many.

Ah. Okay. We can tell that it is "separable" when we have the ability to move all y terms to one-side and the x terms to the other. Here's a really general example that basically gives a framework for any Separable differential equation. \( \displaystyle y'(x) = G(y) F(x) \) F(x) a general function of x and G(y) a function of y We use a bit of a trick here, using the idea that: \( \displaystyle y' = \frac{\text{d}y}{\text{d}x} \) Now we rwrite: \( \displaystyle \frac{\text{d}y}{\text{d}x} = G(y) F(x) \) Let's multiply both sides by "dx" and divide both sides by G(y). \( \displaystyle \frac{1}{G(y)} \; \text{d}y = F(x) \; \text{d}x \) Friom here, our solution method is simply integratingth sides. This will usually give us a general solution (sometimes y won't be right there in front of us), so we may have to do a little further Algebra depending on the G(y).

In our case, we have: y'(x)=e^(y/2)sinx G(y) = e^(y/2), F(x) = sin x So it follows the same format as above. \( \displaystyle y'(x) = e^{y/2} \sin x \) <-- lets change out y' = dy/dx \( \displaystyle \frac{\text{d}y}{\text{d}x} = e^{y/2} \sin x \) <--- "separate" moving dx & e^(y/2) to the correct sides. \( \displaystyle e^{-y/2} \; \text{d}y = \sin x \; \text{d}x \) <-- Here, we can integrate both sides. \( \displaystyle \int e^{-y/2} \; \text{d}y = \int \sin x \; \text{d}x \)

So far, do you see how that works out here?

yes, I think I've got it I'll ask more question as I finish the problem.

Alright! :)

The situation with your teacher is unfortunate; from my experience, Differential Equations was a /lot/ of 'tricks' and things I would never have considered coming in from Calc I/II. The next topic following Separable Differential Equation is usually Linear DE, and one of the methods of solving them is using the property of derivatives that f' g + fg' = (fg)' and multiplying by a specific function to make that property work out / allow you to integrate both sides to break y out of the derivative.

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