= Solution
With no electric field, the <Lorentz force> is $F=q\dot r\times B\hat z$. The equation from <Newton's second law> and its first integral are
$$
m\ddot r=qB\dot r\times\hat z,\qquad
\frac{d}{dt}\bigl(\dot r-\omega r\times\hat z\bigr)=0,
\qquad \boxed{\omega=\frac{qB}{m}}.
$$
The <cyclotron frequency> is $|\omega|$; the sign of $\omega$ specifies the sense of rotation. Since $\hat x\times\hat z=-\hat y$, the initial data give
$$
\boxed{c=(u+a\omega)\hat y+v\hat z.}
$$
The component equations are $\dot x=\omega y$, $\dot y+\omega x=u+a\omega$ and $\dot z=v$. For $\omega\ne0$, differentiating the first gives
$$
\ddot x+\omega^2x=\omega(u+a\omega),\qquad x(0)=a,\quad\dot x(0)=0.
$$
Solving this <linear differential equation> and then using $y=\dot x/\omega$ gives
$$
\boxed{r(t)=\left[a+\frac u\omega(1-\cos\omega t)\right]\hat x
+\frac u\omega\sin\omega t\,\hat y+vt\,\hat z.}
$$
The perpendicular projection is a circle of radius $|u/\omega|$ centred at $(a+u/\omega,0)$, the <guiding centre>. The velocity parallel to the <magnetic field> stays constant, producing <helical motion in a uniform magnetic field>.
When $a\omega+u=0$, this simplifies to
$$
\boxed{x=a\cos\omega t,\qquad y=-a\sin\omega t,\qquad z=vt.}
$$
Thus the trajectory is a <helix> about the $z$ axis, with radius $|a|$ and axial advance $2\pi v/|\omega|$ per complete revolution in increasing time. Its projection moves clockwise when viewed from positive $z$ for $\omega>0$. If $v=0$ it is a circle; if $a=0$ it is a straight line along the axis.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-4-magnetic-helix.png]
{title=Helical trajectory about the magnetic-field axis when a omega plus u is zero}
For $\omega=0$, the <Lorentz force> vanishes and the solution is instead $r(t)=a\hat x+ut\hat y+vt\hat z$, also the limit of the general formula. The special condition then forces $u=0$.
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