Solution: The given differential equation is \begin{equation}\label{eq:10Jan2025-1} \begin{cases} y''(x) + \lambda y(x) = 0 \\ y(0) = 0 \text{ and } y'(L) = 0. \end{cases} \end{equation} Consider the following cases.

Case 1: Let $\lambda = 0$. Then the equation \eqref{eq:10Jan2025-1} will become \[ y''(x) = 0 , \quad y(0) = 0 \text{ and } y'(L) = 0. \] The solution of the above equation is \[ y(x) = Ax + B \quad y(0) = 0 \text{ and } y'(L) = 0. \] With the two boundary conditions, we get $A = B = 0$. Thus, this can not be an eigenfunction.

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Case 2: Let $\lambda = -k^2$ for $k\neq 0$. The differential equation \eqref{eq:10Jan2025-1} reduces to \[ y''(x) - k^2 y(x) = 0 \implies y(x) = A e^{kx} + B e^{-kx}. \] Applying the boundary conditions, we get \begin{align*} a + b = 0 \text{ and } Ak - bk = 0 \implies A = B = 0. \end{align*} Thus, here also, we do not get any eigenfunction.

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Case 3: Let $\lambda = k^2,$ for $k\neq 0$. The differential equation will be \[ y''(x) + k^2 y(x) = 0 \implies y(x) = A\cos (kx) + B \sin (kx). \] Using the boundary conditions, we get \begin{align*} y(0) = 0 & \implies A = 0 \\ y'(L) = 0 & \implies -Bk \cos (kL) = 0. \end{align*} If $B = 0$, then again we will not get any eigenfunction, so we assume that $B\neq 0$. Thus, \begin{align*} \cos (kL) = 0 & \implies kL = (2n + 1) \frac{\pi}{2}, \ n \in \mathbb{Z} \\ & \implies k = (2n + 1) \frac{\pi }{2L}, \ n \in \mathbb{Z} . \end{align*} Therefore, the eigenvalues will be \[ \lambda _n = k^2 = (2n + 1)^2 \frac{\pi ^2}{4L^2}, \ n = 0,1,2, \dots \] and the corresponding eigenfunction will be \[ y_n(x) = B_n \sin \left( (2n + 1) \frac{\pi }{2L} \right), \ n \in \mathbb{Z} . \]