In the previous problem, the solution of the eigenvalue problem is
A) $\lambda = n , \Theta = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 0,1,2 , \ldots$
B) $\lambda = n ^ { 2 } , \ominus = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 0,1,2 , \ldots$
C) $\lambda = n ^ { 2 } , \ominus = c _ { n } \cos ( n \theta ) + d _ { n } \sin ( n \theta ) , n = 1,2,3 , \ldots$
D) $\lambda = n ^ { 2 } , R = r ^ { n } , n = 0,1,2 , \ldots$
E) $\lambda = n , R = r ^ { n } , n = 0,1,2 , \ldots$

Question

Quizplus · Accepted Answer

The correct solution to the eigenvalue problem typically involves eigenvalues that are squared (e.g., $\lambda = n^2$) and eigenfunctions that are a combination of sine and cosine functions, indicating option B is correct. This is characteristic of solutions to differential equations in polar coordinates, where $\lambda$ often represents a squared quantity related to the separation constants, and the functions of $\theta$ are linear combinations of sine and cosine to satisfy periodic boundary conditions.

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 , \ldotsλ=n,Θ=cn​cos(nθ)+dn​sin(nθ),n=0,1,2,…

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