The solution of the eigenvalue problem $y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( 2 ) = 0$ is
A) $\lambda = n \pi / 2 , y = \cos ( n \pi x / 2 ) , n = 1,2,3 , \ldots$
B) $\lambda = ( n \pi / 2 ) ^ { 2 } , y = \cos ( n \pi x / 2 ) , n = 1,2,3 , \ldots$
C) $\lambda ^ { 2 } = n \pi / 2 , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots$
D) $\lambda = n \pi / 2 , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots$
E) $\lambda = ( n \pi / 2 ) ^ { 2 } , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots$

Question

Quizplus · Accepted Answer

The general solution to the differential equation $y'' + \lambda y = 0$ depends on the sign of $\lambda$. For positive $\lambda$, the solutions are of the form $y = A\cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x)$. Given the boundary conditions $y(0) = 0$ and $y(2) = 0$, the cosine term must be zero due to the first boundary condition, leaving $y = B\sin(\sqrt{\lambda}x)$. The second boundary condition requires that $\sqrt{\lambda}2 = n\pi$, where $n$ is a positive integer, leading to $\lambda = (n\pi/2)^2$. This matches choice E, where the solution is expressed in terms of sine, satisfying both boundary conditions.

The Solution of the Eigenvalue Problem y′′+λy=0,y(0)=0,y(2)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( 2 ) = 0y′′+λy=0,y(0)=0,y(2)=0

The Solution of the Eigenvalue Problem $y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( 2 ) = 0$