Answer to Question #49495 in Differential Equations for nwaar

Question

If w = sin(x+kt) + cos(2x+2kt).
Prove that ∂^2 w/∂t^2 = (k^2).∂^2 w/∂x^2

Answer 1

Solution:
∂ w/∂t = k*cos(x+kt)- 2k*sin(2x+2kt)
∂^2w/∂t^2 = -k^2 sin(x+kt) – 4k^2cos(2x + 2kt)
∂ w/∂x = cos(x+kt ) -2 sin(2x+2kt)
∂^2w/∂t^x = - sin(x+kt) – 4 cos(2x+ 2kt)


Let's calculate ∂^2 w/∂t^2 - (k^2).∂^2 w/∂x^2 =
= -k^2 sin(x+kt) – 4k^2 cos(2x + 2kt) – k^2 (-sin(x+kt) – 4 cos(2x + 2kt) =
= -k^2 sin(x+kt)– 4k^2 cos(2x + 2kt) + k^2 sin(x+kt) + 4k^2cos(2x+2kt) =0
So that’s true