= Solution
Let $\mathbf E=E\widehat{\mathbf x}$ and $\mathbf B=B\widehat{\mathbf y}$, and choose $\widehat{\mathbf z}=\widehat{\mathbf x}\times\widehat{\mathbf y}$. Part (a) gives $y=0$. With the origin at $\mathbf x_0$, the remaining Cartesian components are
$$
\ddot x=\frac{qE}{m}-\omega\dot z,
\qquad
\ddot z=\omega\dot x,
\qquad
\omega=\frac{qB}{m}.
$$
The initial conditions are $x=z=\dot x=\dot z=0$. Integrating the second equation gives $\dot z=\omega x$, so
$$
\ddot x+\omega^2x=\frac{qE}{m}.
$$
Solving this <forced harmonic oscillator> and then integrating $\dot z=\omega x$ yields
$$
\boxed{
\begin{aligned}
x(t)&=\frac{mE}{qB^2}\bigl(1-\cos\omega t\bigr),\\
y(t)&=0,\\
z(t)&=\frac EB\left(t-\frac{\sin\omega t}{\omega}\right).
\end{aligned}}
$$
\b[Thus the position vector is $\mathbf x_0+x(t)\widehat{\mathbf x}+z(t)\widehat{\mathbf z}$. The oscillation occurs at the signed <cyclotron frequency>, superposed on the usual <E-cross-B drift>.]
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