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

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 $y = A\cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x)$. The boundary conditions $y(0) = 0$ and $y(1) = 0$ imply that $A = 0$ and $\sin(\sqrt{\lambda}) = 0$, respectively. The latter condition is satisfied for $\sqrt{\lambda} = n\pi$, leading to $\lambda = n^2\pi^2$. The sine function is used because the cosine function does not satisfy the boundary condition at $x = 1$ for any $n$ except $n = 0$, which leads to the trivial solution. Therefore, the correct solution involves sine functions and $\lambda = n^2\pi^2$, where $n$ is a positive integer.

The Solution of the Eigenvalue Problem y′′+λy=0,y(0)=0,y(1)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( 1 ) = 0y′′+λy=0,y(0)=0,y(1)=0

The Solution of the Eigenvalue Problem $y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( 1 ) = 0$