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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.

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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.

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  • $\begingroup$ This just gives us one solution. But in some cases there may be more than one solutions, which is not immediately obvious. Also, how does one know if the IVP is uniquely solvable? $\endgroup$ – sequence Nov 2 '18 at 14:36
  • $\begingroup$ @sequence: no, this gives us the general solution, which means all solutions will have this form. You just need to show that for any particular $(x_0, y_0)$ there is only one $C$ that satisfies $y_0=-\log(\cos(x_0+C))$ $\endgroup$ – Vasya Nov 2 '18 at 14:42
  • $\begingroup$ @Vasya Is this also because the solution is defined in all of $\mathbb{R}$? Otherwise wouldn't there be the possibility that there exist solutions outside of a certain interval? $\endgroup$ – sequence Nov 2 '18 at 14:52
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    $\begingroup$ @sequence: No, the solution defined on all reals is called global. General solution contains all solutions and any particular solution can be obtained from the general solution by choosing appropriate $C$ $\endgroup$ – Vasya Nov 2 '18 at 15:23

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