Show which of the following functional forms work or do not work as solutions to this differential equation (known as 'the wave equation'). Cheek the possible solution listed below and show mathematically how you know the solution works or not. E(x, t) = A cos(Bt) E(x, t) =A cos(Bt + C) E(x, t) = Ax^2t^2 E(x, t) = A cos (Bx + Ct) E(x, t) = A cos(Bx + Ct + D) For the ones that will work as a solution, figure out what constraints, if any, there are on the values for the constants A, B, C, and D. Note that when taking a partial derivative with respect to x or t, you only differentiate with respect to the one variable, ignoring the other as if it is a constant.

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moya1316 moya1316 · Answer 1 · 2020-01-16T10:14:55+01:00

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