How can I solve $y'+\sin y'=x$?

Question

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.

Answer 1

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 $F$.