= Solution
The <Itô formula> verifies the <geometric Brownian motion> solution
$$
X_t=x\exp\!\left((b-\tfrac12\sigma^2)t+\sigma B_t\right).
$$
This is valid for negative $x$ as well as positive $x$; $x=0$ gives the zero solution. <Pathwise uniqueness> follows from the <Lipschitz continuity> of the linear coefficients. The <moment-generating function of a normal distribution> gives $\mathbb Ee^{k\sigma B_t}=e^{k^2\sigma^2t/2}$, and therefore
$$
\boxed{\mathbb E(X_t^k)=x^k\exp\!\left(\left(kb+\frac{k(k-1)}2\sigma^2\right)t\right)\qquad(k\geq1).}
$$
These <moments> are finite on every finite horizon. If the convention $0\in\mathbb N$ is used, the zeroth <moment> is $1$.
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