Solution to differential equation a) How to solve, or at least to prove the existence of a solution to differential equation for given initial condition $y(s)=y_0>0$ and $y'(s)=y_1$, $s<0$,
$$y''+(2-n)\coth(t) y'=\frac{(n-1)\sinh(2y)}{2}, t<0.$$ Here $n$ is an integer $>2$.
b) Can the previous equation have two different solution (with different initial conditions) in $(-2,-1)$, such that one is bounded and the second is not bounded?
 A: The result you want is true because of existence and Uniqueness for second order non-linear ODE, for example see Boyce and Diprima. 
Writing $y^{\prime\prime}=f(t,y,y^\prime)$ and verifying that $f$, $f_y$, and $f_{y^\prime}$ are continuous you can guarantee existence and uniqueness in a small interval of the initial condition.  Here you need to see that the functions you have in the problem ($coth$ and $sinh$) are smooth.
A: $\coth(t)$ has a singularity at $t=0$, so the hypotheses of the existence and uniqueness theorems are not satisfied there.  In fact if $\lim_{t \to 0} y(t) = y_0$ and $\lim_{t \to 0} y'(t) = y_1$ exist, $y''(t) \sim  (n-2) y_1 t^{-1}$ as $t \to 0$.  If $y_1 \ne 0$, this is impossible, as $t^{-1}$ is not integrable at $0$.  So there are no solutions with such an initial condition.
EDIT: For some initial conditions at $t=-2$, the solution will "blow up" before $t=-1$. 
It suffices to prove, e.g., that on any interval $[-2,a)$ where the solution exists we have $\dfrac{dy}{dt} \ge y^2$ for $t \ge -2$ with $y(-2) > 1$, as then 
$$a - (-2) \le \int_{y(-2)}^{y(a)} \dfrac{dy}{y^2} < \int_{1}^\infty \dfrac{dy}{y^2} = 1$$   
Now note that if $f = \dfrac{dy}{dt} - y^2$ we have
$$ \dfrac{df}{dt}  = \dfrac{d^2y}{dt^2} - 2 y \dfrac{dy}{dt} = ((n-2) \coth(t) - 2 y)(y^2 + f) + \dfrac{n-1}{2} \sinh(2y) $$
Given $n$, there is some $Y$ such that for all $f \in [0,1]$, $t \in [-2,-1]$ and $y \ge Y$, 
the right side is positive.  So if $y(-2) > Y$ and $y'(-2) > y(-2)^2$, we will have $y' > y^2$ for $t \in [-2,a]$.
