Sturm Liouville Problem- Removing the Square from the Eigenvalue

Question

The Sturm Liouville problem

$y''=-\lambda y,$ $0<x<1,$ $y'(0)=0,$ $y(1)=0$

has eigenvalues $\lambda_n=(n\pi)^2, n = 1, 2, 3,\dots$, and eigenfunctions $y_n=sin(n\pi x), n =1,2,3,\dots$

Why are we removing the square in the eigenvalue from the formula for the eigenfunction?

Answer 1

The general sol'n is $c_1\cos(\sqrt\lambda x)+c_2\sin(\sqrt\lambda x)$, so the square in the solution will be removed by the square root.