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

Question

Quizplus · Accepted Answer

The general solution to the differential equation $y'' + \lambda y = 0$ is $y = A\cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x)$. Given the boundary conditions $y(0) = 0$ and $y(\pi) = 0$, the cosine term must be zero because $\cos(0) = 1$, not satisfying the first boundary condition unless $A = 0$. For the sine term to satisfy both boundary conditions, $\sqrt{\lambda}$ must be an integer $n$, making $\lambda = n^2$, and $n$ must start from 1 to avoid the trivial solution, leading to $y = \sin(nx)$ for $n = 1,2,3,\ldots$.

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

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