Differential Equation Containing Natural Log of Negative e

[chrisa88]

Hi I am working on a problem that ends up having the natural log of a negative e which I'm confused on how to find the explicit solution.

The Problem:
Find an explicit solution with C.
y'-e^{-y}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^{-y})=ln(sin(x) + C)
which could then be simplified to (if I'm not mistaken)
ln(-e^{-y})=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

 

[JJacquelin]

y'-exp(−y)cos(x)=0
y'exp(y)=cos(x)
Integrate...

 

[chrisa88]

So my mistake was in doing the integral of e^(-y) which yields -e^(-y), wow.. I kept thinking I had made some silly algebra error here haha.. Thank you very much for pointing out my trivial mistake!