I have no idea how to start with this. I tried taking the derivative and making use of $\cos{y'}^2+\sin{y'}^2=1$ but it seems to get even more confusing. Of course we can replace $y'$ with $f$ since $y$ does not appear but that's as far as I can get.
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How far have you gotten in diff eq? – Rushabh Mehta Sep 10 '18 at 03:31
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I've taken a full course on ODEs and another one on PDEs. – John Katsantas Sep 10 '18 at 03:34
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Your observation that this is $f(x) + \sin(f(x)) = x$ is about as far as I would expect can be done on this problem, as I don't think you can invert the function $f + \sin(f)$. This is now no longer a question about ODEs or PDEs. – Michael Sep 10 '18 at 03:36
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This was a question in a previous exam on the course I'm having. So there has to be a way to solve this . – John Katsantas Sep 10 '18 at 03:37
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1@JohnKatsantas : You seem to have a lot of faith in exams! – Michael Sep 10 '18 at 03:39
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Haha, I'm supposed to have in my teacher. Hopefully this doesn't drop in the exam today. – John Katsantas Sep 10 '18 at 03:40
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It doesn't seem that you can get some elementary function. Write $F(t) = t+\sin t$, then it is
$$ F(y') = x\Rightarrow y' = F^{-1}(x) \Rightarrow y = \int F^{-1}(x) dx.$$
It seems that's the best you have, see here for $F$.