$\coth(t)$ 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.

Moreover, it appears that for at least some initial conditions with $y_1 = 0$, solutions will exist but will not be unique. I tried (with $n=3$) numerical solutions with different initial conditions at $t = -1$, and plotted the results as an animation. As $t \to 0-$, these solutions appear to approach the $y'=0$ axis. For some initial points farther away from the origin the solutions appear to "blow up" in finite time.

alt text http://www.math.ubc.ca/%7Eisrael/problems/deanim.gif

$\coth(t)$ 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 $t \in [-2,a]$.

$\coth(t)$ 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.

$\coth(t)$ 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.

Moreover, it appears that for at least some initial conditions with $y_1 = 0$, solutions will exist but will not be unique. I tried (with $n=3$) numerical solutions with different initial conditions at $t = -1$, and plotted the results as an animation. As $t \to 0-$, these solutions appear to approach the $y'=0$ axis. For some initial points farther away from the origin the solutions appear to "blow up" in finite time.

alt text http://www.math.ubc.ca/%7Eisrael/problems/deanim.gif