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) $\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X ( x ) = \cos ( ( n - 1 / 2 ) \pi x / L ) , n = 1,2,3 , \ldots$

Question

Quizplus · Accepted Answer

The correct solution involves squaring the term $( ( n - 1 / 2 ) \pi / L )$ to get the eigenvalue $\lambda$, and using the sine function for $X(x)$ with $n = 1,2,3, \ldots$, which corresponds to the boundary conditions typically found in problems involving a string fixed at both ends or similar physical systems.

The Solution of the Eigenvalue Problem from the Previous Problem λ=nπ/L,X(x)=sin⁡(nπx/L),n=1,2,3,…\lambda = n \pi / L , X ( x ) = \sin ( n \pi x / L ) , n = 1,2,3 , \ldotsλ=nπ/L,X(x)=sin(nπx/L),n=1,2,3,…

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