= Solution
The index order in $M_\mu{}^\nu x^\mu$ fixes the rotation sign. With the entries specified in the paper, the <integral curves of a vector field> satisfy
$$
\frac{dx^1}{d\lambda}=A^1+x^2,\qquad
\frac{dx^2}{d\lambda}=A^2-x^1,\qquad
\frac{dx^j}{d\lambda}=A^j\quad(j\ge3).
$$
Shift to $u=x^1-A^2$, $v=x^2+A^1$. Then $\dot u=v$, $\dot v=-u$, so $u^2+v^2$ is constant. If $x(0)=x_0$, put $u_0=x_0^1-A^2$ and $v_0=x_0^2+A^1$. Solving this linear <ordinary differential equation> gives the full <local flow>:
$$
\boxed{\begin{aligned}
x^1(\lambda)&=A^2+u_0\cos\lambda+v_0\sin\lambda,\\
x^2(\lambda)&=-A^1+v_0\cos\lambda-u_0\sin\lambda,\\
x^j(\lambda)&=x_0^j+A^j\lambda\quad(j\ge3).
\end{aligned}}
$$
The planar motion is a circle centred at $(A^2,-A^1)$, traversed clockwise as $\lambda$ increases. A nonzero vector $(A^3,\ldots,A^n)$ produces a <helical orbit of a Euclidean Killing vector>; zero drift gives a circle. At zero radius the curves are straight lines in the remaining directions, or stationary points if that drift also vanishes. These formulas are defined for every real $\lambda$.
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