-1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.