Question #251534

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

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

∂x2 v2 ∂t2


1
Expert's answer
2021-10-18T11:14:42-0400
2y2x=Ak2sin(kxωt)2y2t=Aω2sin(kxωt)v=ωk\frac{\partial ^2y}{\partial ^2x}=-Ak^2 \sin(kx – ωt)\\ \frac{\partial ^2y}{\partial ^2t}=-Aω^2 \sin(kx – ωt)\\ v=\frac{\omega}{k}


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