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In the Previous Problem, the Solution of the Eigenvalue Problem λ=n,Θ=cncos(nθ)+dnsin(nθ),n=0,1,2,\lambda = n , \Theta = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 0,1,2 , \ldots

Question 21

Multiple Choice

In the previous problem, the solution of the eigenvalue problem is


A) λ=n,Θ=cncos(nθ) +dnsin(nθ) ,n=0,1,2,\lambda = n , \Theta = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 0,1,2 , \ldots
B) λ=n2,=cncos(nθ) +dnsin(nθ) ,n=0,1,2,\lambda = n ^ { 2 } , \ominus = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 0,1,2 , \ldots
C) λ=n2,=cncos(nθ) +dnsin(nθ) ,n=1,2,3,\lambda = n ^ { 2 } , \ominus = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 1,2,3 , \ldots
D) λ=n2,R=rn,n=0,1,2,\lambda = n ^ { 2 } , R = r ^ { n } , n = 0,1,2 , \ldots
E) λ=n,R=rn,n=0,1,2,\lambda = n , R = r ^ { n } , n = 0,1,2 , \ldots

Correct Answer:

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