Lorentz Force Law and cycloid motion

Question

The question I have is from Intro to Electrodynamics by Griffiths (Page 206, Example 5.2)

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When using the initial conditions $y(0) = z(0) = 0$, I get:

$y(0) = 0$ implies $C_3 = C_1 = 0$

and

$z(0) = 0$ implies $C_4 = C_2 = 0$.

However, how does this work with every $C_1, C_2, C_3, C_4 = 0$ ?

Also, how did the author arrive at the general solution that

$y(t) = C_1 \cos (\omega t) + C_2 \sin (\omega t) + (E/B)t + C_3$ \ $z(t) = C_2 \cos (\omega t) - C_1 \sin (\omega t) + C_4$

Answer 1

The general solution you mention is derived from solving a second order linear differential equation. You can differentiate $\ddot z=\omega (\frac EB-\dot y)$ to get $\ddot v_z=-\omega^2 \dot z$ where $v_z=\dot z$. This is a second order differential (linear) equation which has the general solution ( you can try out by using an exponential trial function) $$v_z=C_a \cos{\omega t}+C_b\sin {\omega t}$$ where the $C$ represents a constant (different from the final $C_1, C_2$ and so on). You can integrate this with respect to time to get the $z(t)$ and then apply similar treatment to the other equation ($\ddot y=\omega \dot z$) to find $y(t)$ and all the constants you mention will be the integraton constants and the arbitrary constants which are a part of the differential equation solution.

In general starting conditions, you can substitute $t=0$ in $y(t)$ and then get the value of $C_1+C_3$ and then substitute in $v_y(t)$ to get $C_1$ and so on till all the constants are determined. Hence, in a similar treatment to z direction, you can get all the constants in the general solution.