Answer:
E = A cos (Bx + Ct) and E = A cos (Bx + Ct + D)
Explanation:
The wave equation is
d²y / dx² = 1 /v² d²y / dt²
Where v is the speed of the wave
Let's review the solutions
In this case y = E
a) E = A sin Bt
This cannot be a solution because the part in x is missing
dE / dx = 0
b) E = A cos (Bt + c)
There is no solution missing the Part in x
dE / dx = 0
c) E = A x² t²
The parts are fine, but this solution is not an oscillating wave, so it is not an acceptable solution of the wave equation
d) E = A cos (Bx + Ct) and E = A cos (Bx + Ct + D)
Both are very similar
Let's make the derivatives
dE / dx = -A B sin (Bx + Ct + D)
d²E / dx² = -A B² cos (Bx + Ct + D)
dE / dt = - A C sin (Bx + Ct + D)
d²E / dt² = -A C² cos (Bx + Ct + D)
We substitute in the wave equation
-A B² cos (Bx + Ct + D) = 1 / v² (-A C² cos (Bx + Ct + D))
B² = 1 / v² (C²)
v² = (C / B)²
v = C / B
We see that the two equations can be a solution to the wave equation, the last one is the slightly more general solution
