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How can I solve this differential equation $$y'= \sin(x+y) ,\ \ y(0) = -\frac{\pi}{2}, -\infty < x < \infty$$

I tried to denote $z=x+y$ And I got unfamiliar integral.

Please help. Thanks

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    $\begingroup$ The fact that an integral is "unfamiliar" does not mean that it is wrong... Just become "familiar" with it! $\endgroup$ – Siminore Oct 29 '14 at 13:49
  • $\begingroup$ I tried this, but it looks worng for me.. $\endgroup$ – user2424684 Oct 29 '14 at 13:54
  • $\begingroup$ If $x+y(x)=z(x)$, then $y'=z'-1$, so that $z'-1=\sin z$. Now you should compute $$\int \frac{dr}{1+\sin r}.$$ $\endgroup$ – Siminore Oct 29 '14 at 13:55
  • $\begingroup$ I think you got an extra +1 in there, but you are on the right track. The solution is an arctangent, as you found. $\endgroup$ – GEdgar Oct 29 '14 at 13:58
  • $\begingroup$ Ok, I think I got it here But know I can't use y(0)=-pi/2 I getting zero denominator $\endgroup$ – user2424684 Oct 29 '14 at 14:00
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Here's a hint to solve the "unfamiliar" integral

$$ \begin{align} \int \frac{1}{1+\sin x}\, \mathrm{d}x &= \int \frac{1-\sin x}{1-\sin^2 x}\, \mathrm{d}x \\ &= \int \sec^2 x\, \mathrm{d}x + \int \frac{(-\sin x)}{\cos^2x} \mathrm{d}x \end{align} $$

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