The solution of the eigenvalue problem $y ^ { \prime \prime } + \lambda y = 0 , y ^ { \prime } ( 0 ) = 0 , y ( 1 ) = 0$ is
A) $\lambda = n ^ { 2 } \pi ^ { 2 } / 4 , y = \cos ( n \pi x / 2 ) , n = 1,2,3 , \ldots$
B) $\lambda = n \pi / 2 , y = \cos ( n \pi x / 2 ) , n = 1,2,3 , \ldots$
C) $\lambda = n ^ { 2 } \pi ^ { 2 } / 4 , y = \sin ( n \pi x / 2 ) , n = 1,2,3 , \ldots$
D) $\lambda = ( 2 n - 1 ) \pi / 2 , y = \cos ( ( 2 n - 1 ) \pi x / 2 ) , n = 1,2,3 , \ldots$
E) $\lambda = ( 2 n - 1 ) ^ { 2 } \pi ^ { 2 } / 4 , y = \cos ( ( 2 n - 1 ) \pi x / 2 ) , n = 1,2,3 , \ldots$

Question

Quizplus · Accepted Answer

The differential equation $y'' + \lambda y = 0$ has solutions of the form $y = A\cos(\sqrt{\lambda}x) + B\sin(\sqrt{\lambda}x)$. Given the boundary conditions $y'(0) = 0$ and $y(1) = 0$, we find that $A = 0$ to satisfy $y'(0) = 0$ (since the derivative of $\cos$ at $0$ is $0$, and we need $B\sin(0) = 0$, which is automatically satisfied). For $y(1) = 0$, we require $\sin(\sqrt{\lambda}) = 0$, which gives $\sqrt{\lambda} = n\pi$, where $n$ is an integer. However, the condition $y(1) = 0$ with $A = 0$ leads us to consider only the odd multiples of $\pi/2$, hence $\sqrt{\lambda} = (2n-1)\pi/2$, and squaring both sides gives $\lambda = (2n-1)^2\pi^2/4$. The corresponding $y$ function that satisfies the initial conditions is $y = \cos((2n-1)\pi x/2)$.

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

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