Solved

In the Previous Problem, If We Also Require That Θ\Theta Be Bounded Everywhere, the Solution of the Eigenvalue Problem Is

Question 7

Multiple Choice

In the previous problem, if we also require that Θ\Theta be bounded everywhere, the solution of the eigenvalue problem is


A) λ=n(n+1) ,Θ=Pn(sin⁡θ) ,n=1,2,3,…\lambda = n ( n + 1 ) , \Theta = P _ { n } ( \sin \theta ) , n = 1,2,3 , \ldots
B) λ=n(n−1) ,Θ=sin⁡(nθ) ,n=1,2,3,…\lambda = n ( n - 1 ) , \Theta = \sin ( n \theta ) , n = 1,2,3 , \ldots
C) λ=n(n−1) ,Θ=cos⁡(nθ) ,n=1,2,3,…\lambda = n ( n - 1 ) , \Theta = \cos ( n \theta ) , n = 1,2,3 , \ldots
D) λ=n(n+1) ,Θ=Pn(cos⁡θ) ,n=1,2,3,…\lambda = n ( n + 1 ) , \Theta = P _ { n } ( \cos \theta ) , n = 1,2,3 , \ldots
E) λ=n(n−1) ,Θ=Pn(sin⁡θ) ,n=1,2,3,…\lambda = n ( n - 1 ) , \Theta = P _ { n } ( \sin \theta ) , n = 1,2,3 , \ldots

Correct Answer:

verifed

Verified

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

Related Questions