Question

Which of the following gives the eigenvalues and eigenfunctions of the Sturm-Liouville problem (SLP) y''(x) + \lambda y(x) = 0, 0 < x < \pi/2 y(0) = 0, y'(\pi/2) = 0. (Hint: There is no eigenvalue for $\lambda \le 0$) $\lambda_n = (2n - 1)^2$, $y_n(x) = c_n \sin((2n-1)x)$, $c_n \ne 0$, $n = 1,2,...$ $\lambda_n = 4n^2$, $y_n(x) = c_n \cos(2nx)$, $c_n \ne 0$, $n = 0,1,2,...$ $\lambda_n = \frac{n^2}{4}$, $y_n(x) = c_n \sin(\frac{n}{2}x)$, $c_n \ne 0$, $n = 1,2,...$ $\lambda_n = (\frac{2n-1}{2})^2$, $y_n(x) = c_n \cos(\frac{2n-1}{2}x)$, $c_n \ne 0$, $n = 0,1,2,...$ $\lambda_n = (\frac{2n-1}{2})^2$, $y_n(x) = c_n \sin(\frac{2n-1}{2}x)$, $c_n \ne 0$, $n = 1,2,...$

          Which of the following gives the eigenvalues and eigenfunctions of the Sturm-Liouville
problem (SLP)
y''(x) + \lambda y(x) = 0, 0 < x < \pi/2
y(0) = 0, y'(\pi/2) = 0.
(Hint: There is no eigenvalue for $\lambda \le 0$)
$\lambda_n = (2n - 1)^2$, $y_n(x) = c_n \sin((2n-1)x)$, $c_n \ne 0$, $n = 1,2,...$
$\lambda_n = 4n^2$, $y_n(x) = c_n \cos(2nx)$, $c_n \ne 0$, $n = 0,1,2,...$
$\lambda_n = \frac{n^2}{4}$, $y_n(x) = c_n \sin(\frac{n}{2}x)$, $c_n \ne 0$, $n = 1,2,...$
$\lambda_n = (\frac{2n-1}{2})^2$, $y_n(x) = c_n \cos(\frac{2n-1}{2}x)$, $c_n \ne 0$, $n = 0,1,2,...$
$\lambda_n = (\frac{2n-1}{2})^2$, $y_n(x) = c_n \sin(\frac{2n-1}{2}x)$, $c_n \ne 0$, $n = 1,2,...$

Which of the following gives the eigenvalues and eigenfunctions of the Sturm-Liouville
problem (SLP)
y”(x) + λy(x) = 0, 0 < x < π/2
y(0) = 0, y'(π/2) = 0.
(Hint: There is no eigenvalue for λ≤ 0)
= (2n - 1)^2, yn(x) = cn sin((2n-1)x), cn 0, n = 1,2,...
= 4n^2, yn(x) = cn cos(2nx), cn 0, n = 0,1,2,...
= (n^2)/(4), yn(x) = cn sin((n)/(2)x), cn 0, n = 1,2,...
= ((2n-1)/(2))^2, yn(x) = cn cos((2n-1)/(2)x), cn 0, n = 0,1,2,...
= ((2n-1)/(2))^2, yn(x) = cn sin((2n-1)/(2)x), cn 0, n = 1,2,...

Added by Peter M.

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