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For $0<\delta <1$, the bracket in the explicit solution is positive and increasing. Its integral is finite on every finite interval because $F$ is continuous and strictly positive there. The same is immediate from the exponential formula for $\delta =1$. Neither solution can reach zero on a finite interval. Hence both cases have global existence almost surely.

For $\delta >1$, put $q=\delta -1>0$ and $K=x_0^{-q}/q$. The explosion threshold for a Bernoulli stochastic differential equation isIf the threshold is reached at a finite time, $F_{\tau }$ is positive and finite, and $X_t\to \infty$ as $t\uparrow \tau$.

When $\alpha =0$, $F=1$ and $\tau =K$ deterministically. When $\alpha \ne 0$, choose a constant $c$ large enough that $e^{qc}>K$. The eventhas positive probability: the first part is a Gaussian tail event, the second has positive probability by the Brownian reflection principle, and they are independent. On this event $\alpha B_s-{\alpha }^{2}s/2>c$ throughout $[1,2]$, so ${\int }_1^2{F}_s^{q}ds>e^{qc}>K$. Thus $\mathbb{P}(\tau \le 2)>0$ and global existence is not almost sure. We concludeThis argument asserts positive explosion probability for $\alpha \ne 0$, not certain explosion. Indeed $B_t/t\to 0$ almost surely makes ${\int }_0^{\infty }{F}_s^{q}ds$ finite in that case; that observation does not remove the positive chance that its value exceeds the threshold.