The one-dimensional wave equation has the form
∂t2∂2u=c2⋅∂x2∂2u As you can see, you need to calculate the second partial derivatives of the specified function
u(x,t)=cos(αcx)⋅sin(αt) Then,
∂x∂u=∂x∂(cos(αcx)⋅sin(αt))=dxd(cos(αcx))⋅sin(αt)==−cα⋅sin(αcx)⋅sin(αt)→∂x2∂2u=∂x∂(∂x∂u)=∂x∂(−cα⋅sin(αcx)⋅sin(αt))==dxd(−cα⋅sin(αcx))⋅sin(αt)=−c2α2⋅cos(αcx)⋅sin(αt) Conclusion,
∂x2∂2u=−c2α2⋅cos(αcx)⋅sin(αt) Then,
∂t∂u=∂t∂(cos(αcx)⋅sin(αt))=cos(αcx)⋅dtd(sin(αt))==α⋅cos(αcx)⋅cos(αt)→∂t2∂2u=∂t∂(∂t∂u)=∂t∂(α⋅cos(αcx)⋅cos(αt))==cos(αcx)⋅dtd(α⋅cos(αt))=−α2⋅cos(αcx)⋅sin(αt) Conclusion,
∂t2∂2u=−α2⋅cos(αcx)⋅sin(αt)
Substitute all found derivatives in the specified one-dimensional wave equation equation:
∂t2∂2u=c2⋅∂x2∂2u→−α2⋅cos(αcx)⋅sin(αt)=c2⋅(−c2α2⋅cos(αcx)⋅sin(αt))→−α2⋅cos(αcx)⋅sin(αt)=−c4α2⋅cos(αcx)⋅sin(αt)→c4α2⋅cos(αcx)⋅sin(αt)−α2⋅cos(αcx)⋅sin(αt)=0→(c4−1)⋅α2⋅cos(αcx)⋅sin(αt)=0→c4−1=0→(c2−1)(c2+1)=0→(c−1)(c+2)(c2+1)=0→c−1=0orc+1=0orc2+1=0→c=1orc=−1orc∈∅ ANSWER
c=±1
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