- #1

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**The Problem:**

Find an explicit solution with C.

y'-e[itex]^{-y}[/itex]cos(x)=0

**My Conclusion:**

First of all, I'm confused how I should solve this explicitly if I'm not given an initial condition. I'm assuming that is why they said "with C", however I'm now in the conundrum of how to solve this with just y on the left side since I have reached the following:

ln(-e[itex]^{-y}[/itex])=ln(sin(x) + C)

which could then be simplified to (if I'm not mistaken)

ln(-e[itex]^{-y}[/itex])=c*ln(sin(x))

BUT from my research and trials I have found that the natural log of a negative number, even e, is undefined. So how would I get this simplified to solving for y?

Thank you very much!

Chris