Consider the following family of SDEs: $$ dX_t = \frac{1}{X_t^p} dt + dW_t $$ with $X_0 > 0$. For $p=1$, this is the 3-dimensional Bessel process, which stays positive for all time almost-surely.
Does this remain true for other values of $p$?
I was thinking that we could use the scale functions on the open interval $I = (0, \infty)$ to try and get this answer.
When $p\not=1$, the scale function $s_p(x)$ can be computed as $$ \begin{aligned} s_p(x) &= \int_c^x \exp\left( -\int_c^y 2 \frac{1}{z^p} dz \right) dy \\ &= \int_c^x \exp\left( \frac{2}{p-1} (y^{1-p} - c^{1-p}) \right) dy \end{aligned} $$
p<1
When $p=\frac{1}{2}$, we get $$ \begin{aligned} s_{1/2}(x) &= \int_c^x \exp\left( 4\sqrt{c}- 4\sqrt{y} \right) dy \\ &= \frac{1}{8}\Big(1 + 4\sqrt{c} - \exp(4\sqrt{c}-4\sqrt{x}) (1 + 4 \sqrt{x}) \Big) \end{aligned} $$ Since both $\lvert s_{1/2}(0) \rvert < \infty$ and $\lvert s_{1/2}(\infty)\rvert < \infty$, we conclude that the (nonzero) probability of hitting zero is $$ \frac{s_{1/2}(\infty) - s_{1/2}(X_0)}{s_{1/2}(\infty) - s_{1/2}(0)} = \exp(-4\sqrt{x}) (1 + 4 \sqrt{x}) $$ For $p \in (0, 1)$ in general, Mathematica shows $s_p(0)$ and $s_p(\infty)$ are both finite, which means the same conclusion applies for all these $p$s?
p>1
On the other hand, for $p=2$, we get? (using Mathematica) $$ \begin{aligned} s_{2}(x) &= \int_c^x \exp\left( \frac{2}{y}-\frac{2}{c} \right) dy \\ &= e^{-2/c} \left( -c e^{2/c} + e^{2/x} x + 2 E_i(2/c) - 2 E_i(2/x) \right) \end{aligned} $$ where $E_i$ denotes the exponential integral. Plugging in $0$ and $\infty$ again gives $$ \lvert s_2(0) \rvert = \infty, \quad \lvert s_2(\infty) \rvert = \infty $$ which implies that $X_t$ almost surely never hits $0$ and never diverges. I'm guessing that the same conclusion also holds for all $p>1$, although I wasn't able to verify this using Mathematica.
Is this approach correct?