The eigenvalue problem $y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0$ has the solution
A) $y = \sin ( n x ) , \lambda = n ^ { 2 } , n = 1,2,3 , \ldots$
B) $y = \cos ( n x ) , \lambda = n ^ { 2 } , n = 1,2,3 , \ldots$
C) $y = \sin ( n x ) , \lambda = n , n = 1,2,3 , \ldots$
D) $y = \cos ( n x ) , \lambda = n , n = 1,2,3 , \ldots$
E) none of the above

Question

Quizplus · Accepted Answer

The general solution to the differential equation $y'' + \lambda y = 0$ can be expressed in terms of sines and cosines. Given the boundary conditions $y(0) = 0$ and $y(\pi) = 0$, the cosine function does not satisfy these conditions for non-zero solutions, but the sine function does. Specifically, $y = \sin(nx)$ satisfies both boundary conditions if $n$ is an integer, and for the differential equation to hold, $\lambda$ must equal $n^2$, making option A correct.

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