The Solution of the Eigenvalue Problem from the Previous Problem $\lambda = n \pi / L , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots$

The solution of the eigenvalue problem from the previous problem is

A) $\lambda = n \pi / L , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldots$
B) $\lambda = n \pi / L , X ( x ) = \cos ( n \pi x / L ) , n = 0,1,2 , \ldots$
C) $\lambda = ( n - 1 / 2 ) \pi / L , X ( x ) = \sin ( ( n - 1 / 2 ) \pi x / L ) , n = 1,2,3 , \ldots$
D) $\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X ( x ) = \sin ( ( n - 1 / 2 ) \pi x / L ) , n = 1,2,3 , \ldots$
E) $\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X ( x ) = \cos ( ( n - 1 / 2 ) \pi x / L ) , n = 1,2,3 , \ldots$

Correct Answer:

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Access For Free