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This is the edited version of the original problem, hopefully presented in a clearer manner. (I have also renamed this post with a more befitting title)

Problem: $$y'(x) = 2\sin\left(\frac{y(x)}{2}\right)$$ subjected to boundary conditions: $\lim\limits_{x\to−\infty}y(x)=0$ and $\lim\limits_{x\to+\infty}y(x)=k$ for some constant $k$

Annoying bits:

After integration, I seem to get $y(x)=4\operatorname{arccot}(c\exp(x))$ which has limits $\lim\limits_{x\to−\infty}y(x)=2\pi$ and $\lim\limits_{x\to+\infty}y(x)=0$.

But then I can't fit the boundary conditions of the problem! (This is driving me insane!) Please help. Thanks in advance!

RESOLVED: I have stupidly left out a minus sign, should be $y(x)=4\operatorname{arccot}(c\exp(-x))$

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    $\csc\,x+\cot\,x=\cot\frac{x}{2}$2011-08-17
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    One other little question: Is it true then that there is no solution to the system if I impose boundary conditions $\lim_{x \to +\infty} y(x) = k$, where $k$ is some constant and $\lim_{x \to -\infty} y(x) = 0$? Because $y(x) = 4 \times arccot(c \times exp(x))$ thus $y(x) \rightarrow 0$ as $x \rightarrow +\infty$ while $y(x) \rightarrow \frac{\pi}{2}$ as $x \rightarrow -\infty$?2011-08-17
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    distressed, why do you have $4\mathrm{arccot}$ instead of $2\mathrm{arccos}$? (Why are you using $\times$ for multiplication and a vector norm out of nowhere?) Also, since $y$'s limits at $\pm\infty$ are already determined (and the second one is $\pi$ not $\pi/2$) independent of $c$ (unless it's $0$, which changes the $+\infty$ limit), you cannot get the boundary conditions you want. It sounds like there's something you're trying to achieve behind what you're showing us, but we can only help marginally because you're not letting us in on it.2011-08-17
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    @anon: Thanks for your comment :) Indeed, I should have a 2 instead of a 4, but why isn't the arccot, using the trig identity provided by J.M.? That is actually the entire question -- the 1st order D.E. and the 2 limits... I think I should probably edit my question to present it in a clearer manner...2011-08-17
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    distressed, sorry I meant $2\mathrm{arccot}$, that's a typo.2011-08-17
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    @anon: No worries. I just realized that one of the sources of confusion is that I did a change in variable mid way in my calculations hence the jumbled coefficients.2011-08-17

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Let's look at the differential equation $y' = 2\sin(y/2)$. This is autonomous, so

$$ \frac{dy}{2\sin(y/2)}=dx.$$

The antiderivative of $1/\sin(\cdot)$ is $\ln \tan( \cdot/2)$ (see W|A), so continuing

$$ \ln \tan (y/4)=x+C$$

$$y=4\arctan(Ae^x)+4\pi n$$

with $A=e^C$ and $n\in\mathbb{Z}$. The problem is you gave us the expression $\csc+\cot$ (which @J.M. gave an identity for), but this expression has no relation to your differential equation. Like I said, you had stuff going on behind the scenes but weren't completely forward so you weren't able to get the full help you needed. If you make an error and then give us the result of the error, we don't even know something went wrong!

Now we get $\lim_{x\to-\infty}y=4\pi n$, so we can choose $n=0$, and then $\lim_{x\to+\infty}y=\frac{\pi}{2}$, so $k$ can't be arbitrary. The issue here is that your differential equation is first-order, so a single initial condition suffices to determine the solution. Two conditions overdetermines it.

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    Thanks again :) I think I know where I went wrong in my calc. I left out a minus sign in my 1st attempt, which is why the limits are "mirrored". The "irrelevant expression" is actually the integral of csc($x$) (c.f. http://en.wikipedia.org/wiki/Lists_of_integrals#Trigonometric_functions) Sorry about the behind-the-scenes toils... :)2011-08-17
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    Oh, and there are solutions iff $k=2\pi$2011-08-17