Answer to Question #251534 in Physics for jay

Question

Prove that y (x,t) = A sin(kx – ωt) is a solution of:

∂2y (x, t)  = 1 ∂2y (x, t)            

∂x2 v2 ∂t2

Answer 1

$$\frac{\partial ^2y}{\partial ^2x}=-Ak^2 \sin(kx – ωt)\\ \frac{\partial ^2y}{\partial ^2t}=-Aω^2 \sin(kx – ωt)\\ v=\frac{\omega}{k}$$