Coupled differential equations The question I have trouble with is from Intro to electrodynamics by Griffiths



I do not understand how the author obtained the general solution
\begin{gather}
y(t) = C_1 \cos (\omega t) + C_2 \sin (\omega t) + (E/B)t + C_3, \\
z(t) = C_2 \cos (\omega t) - C_1 \sin (\omega t) + C_4,
\end{gather}
and then found that
$$y(t) = \frac{E}{\omega B}(\omega t - \sin(\omega t)),$$
and the corresponding equation for $z(t)$, since plugging in $y(0) = z(0) = 0$ yielded $C_1 = C_2 = C_3 = C_4 = 0$ for me.
Thanks in advance.
 A: Let's define $\xi \equiv \dot{y} + {\rm i}\dot{z}$. Then

$$
\dot{\xi} + {\rm i}\omega\xi
=
{\rm i}\,{\omega E \over B}\,,
\quad
{{\rm d}\left({\rm e}^{{\rm i}\omega t}\xi\right)\over {\rm d}t}
=
{\rm i}\,{\omega E \over B}\,{\rm e}^{{\rm i}\omega t}\,,
\quad
{\rm e}^{{\rm i}\omega t}\xi
=
{\rm i}\,{\omega E \over B}\,
{{\rm e}^{{\rm i}\omega t} - 1 \over {\rm i}\omega} 
=
{E \over B}\,
\left({\rm e}^{{\rm i}\omega t} - 1\right) 
$$


$$
\dot{y} + {\rm i}\dot{z}
=
\xi
=
{E \over B}\,\left(1 - {\rm e}^{-{\rm i}\omega t}\right)\,,
\quad
y + {\rm i}z
=
{E \over B}\,
\left(t - {{\rm e}^{-{\rm i}\omega t} - 1 \over -{\rm i}\omega}\right)
$$


$$
y + {\rm i}z
=
{E \over \omega B}\,
\left[\omega t
+
{\rm i}\left(1 - {\rm e}^{-{\rm i}\omega t}\right)\right]
=
{E \over \omega B}\,\left\{\vphantom{\LARGE A}%
\left[\vphantom{\Large A}\omega t - \sin\left(\omega t\right)\right]
+
{\rm i}\left[1 - \cos\left(\omega t\right)\right]
\right\}
$$

$$\color{#ff0000}{\large%
y = {E \over \omega B}\,\left[\omega t - \sin\left(\omega t\right)\right]\,,\qquad
z = {E \over \omega B}\,\left[1 - \cos\left(\omega t\right)\right]}
$$
A: $$\begin{align}
&y(0) = C_1 + C_3 = 0 \\
&\dot{y}(0) = C_2\omega + E/B = 0 \\
&z(0) = C_2 + C_4 = 0 \\
&\dot{z}(0) = -C_1 = 0
\end{align}$$
$$\iff$$
$$\begin{align}
&C_1 = 0 \\
&C_2 = -\frac{E\omega}{B}\\
&C_3 = 0\\
&C_4 =\frac{E\omega}{B} \\
\end{align}$$
A: The author obtained the general solution by the method indicated in the footnote. $$y''=\omega z'$$ Differentiate to get $$y'''=\omega z''$$ Combine with $$z''=\omega((E/B)-y')$$ to get $$y'''=\omega((E/B)-y')$$ Now the equations are uncoupled; can you solve that last equation for $y$? and then work out $z$ from one of the original equations?
