Consider $$y'=e^y\sin x$$ Then $$y(x;C)=-\log (\cos x +C), \text{ with }C+\cos x > 0$$
How do we know that the above function represents all of the possible solutions of the ODE above and that each IVP of this ODE is uniquely solvable?
I'm reading the book Ordinary Differential Equations by Wolfgang Walter and am having some difficulty grasping parts of its contents, as it seems that I'm lacking parts of the background that this book seems to assume.
Update: The book has not introduced the Existence and Uniqueness Theorem at this point, and so one is supposed to deduce the above without using the theorem.
Suppose y is any solution to $$y′=e^y \sin x$$
Multiply both sides by $e^{-y}$ to get $$e^{-y}y' = \sin x$$
Integrate both sides and solve for $y$
You get the answer and that is it.