Maxwell's Equations in Vacuum: Constraints on Wave

[MisterX]

Homework Statement

Condensed/simplified problem statement

$\vec{E} = f_{y}(x-ct)\hat{y} + f_{z}(x-ct)\hat{z} \\ \vec{B} = g_{y}(x-ct)\hat{y} + g_{z}(x-ct)\hat{z} \\$

All the f and g functions go to zero as their parameters go to ±∞.

Show that g_y = f_z and g_z = -f_y

Homework Equations

$\nabla \cdot \vec{E} = 0 \\ \nabla \cdot \vec{B} = 0$
$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$
$\nabla \times \vec{B} = \frac{1}{c^{2}}\frac{\partial \vec{E}}{\partial t}$

The Attempt at a Solution

$\nabla \times \vec{E} = -\frac{\partial f_{z}(x-ct)}{\partial x}\hat{y} + \frac{\partial f_{y}(x-ct)}{\partial x}\hat{z} = -\frac{\partial \vec{B}}{\partial t} = \\-(g^{'}_{y}(x-ct)(-c)\hat{y} + g^{'}_{z}(x-ct)(-c)\hat{z}) = c(g^{'}_{y}(x-ct)\hat{y} + g^{'}_{z}(x-ct)\hat{z})$

The -c factors come from the chain rule when differentiating B with respect to t. So then I have the following.

$-\frac{\partial f_{z}(x-ct)}{\partial x} = -f^{'}_{z}(x-ct) = cg^{'}_{y}(x-ct) \\ \frac{\partial f_{y}(x-ct)}{\partial x} = f^{'}_{y}(x-ct) = cg^{'}_{z}(x-ct)$

The problem is to relate the functions directly and not by their derivatives. We are given that the functions all go to zero as the parameter goes to +/- infinity. So maybe it would be okay to just integrate without concern for a constant added on. But supposedly the c factor isn't there. The answer is supposed to have$g_{y} = -f_{z}$. How do I get there?

 

[fzero]

B and E do not have the same units, so f and g must be related with an appropriate factor of c. There's probably a typo in the problem. Your argument about the integration constants makes sense.