= Solution
The <diffusion generator> is $\mathcal L=\mu\partial_x+\tfrac12\sigma^2\partial_{xx}$. By the <Feynman-Kac formula>, the required <conditional characteristic function> solves
$$
F_t+\mu F_x+\tfrac12\sigma^2F_{xx}=0,\qquad F(T,x)=e^{iux}.
$$
Use <separation of variables> in the form $F(t,x)=e^{iux}h(t)$. Since $F_x=iuF$ and $F_{xx}=-u^2F$, the temporal factor satisfies the <ordinary differential equation>
$$
h'(t)+\left(iu\mu-\frac12\sigma^2u^2\right)h(t)=0,\qquad h(T)=1.
$$
Thus $h(t)=\exp[(iu\mu-\tfrac12\sigma^2u^2)(T-t)]$, and
$$
\boxed{\mathbb E[e^{iuX_T}\mid X_t=x]
=\exp\left(iu[x+\mu(T-t)]-\frac12u^2\sigma^2(T-t)\right).}
$$
This is bounded in absolute value by one, so it lies in the uniqueness class of the <Feynman-Kac formula>. As an independent check, integrating the <stochastic differential equation> gives $X_T=x+\mu(T-t)+\sigma(B_T-B_t)$. The independent <Brownian increments> have the <normal distribution> with mean zero and <variance> $T-t$, yielding exactly the same <characteristic function>. For $\sigma=0$, the <conditional distribution> is a point mass at $x+\mu(T-t)$, and the same boxed expression still applies.
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