The solution of the eigenvalue problem in the previous problem is
A) $\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X = \sin ( ( n - 1 / 2 ) \pi x / L )$
B) $\sqrt { \lambda } / h = \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )$
C) $\lambda = ( n \pi / L ) ^ { 2 } , X = \sin ( n \pi x / L )$
D) $\sqrt { \lambda } / h = - \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )$
E) $\sqrt { \lambda } h = - \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )$

Question

Quizplus · Accepted Answer

The Answer of The solution of the eigenvalue problem in the previous problem is
A) $\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X = \sin ( ( n - 1 / 2 ) \pi x / L )$
B) $\sqrt { \lambda } / h = \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )$
C) $\lambda = ( n \pi / L ) ^ { 2 } , X = \sin ( n \pi x / L )$
D) $\sqrt { \lambda } / h = - \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )$
E) $\sqrt { \lambda } h = - \tan ( \sqrt { \lambda } L ) , X = \sin ( \sqrt { \lambda } x )$

The Solution of the Eigenvalue Problem in the Previous Problem λ=((n−1/2)π/L)2,X=sin⁡((n−1/2)πx/L)\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X = \sin ( ( n - 1 / 2 ) \pi x / L )λ=((n−1/2)π/L)2,X=sin((n−1/2)πx/L)

The Solution of the Eigenvalue Problem in the Previous Problem $\lambda = ( ( n - 1 / 2 ) \pi / L ) ^ { 2 } , X = \sin ( ( n - 1 / 2 ) \pi x / L )$