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Depending on the value of λ, we have three cases: Case 1: λ > 0 Let's assume λ = k² for some positive k. The given differential equation becomes: $$ y^{\prime \prime} + k^2 y = 0 $$ The general solution of the above differential equation is: $$ y(x) = A\cos{kx} + B\sin{kx} $$ Case 2: λ = 0 When λ = 0, the differential equation becomes: $$ y^{\prime \prime} = 0 $$ The general solution in this case is: $$ y(x) = Ax + B $$ Case 3: λ < 0 Let's assume λ = -k² for some positive k. The given differential equation becomes: $$ y^{\prime \prime} - k^2 y = 0 $$ The general solution of the above differential equation is: $$ y(x) = A\cosh{kx} + B\sinh{kx} $$

We will now use the boundary conditions \(y^{\prime}(0)=0\) and \(y(1)=0\) to find the possible eigenvalues (λ) and eigenfunctions (y(x)) for each case. Case 1: λ > 0 Applying the first boundary condition (y'(0) = 0), we get: $$ -kA\sin{0} + kB\cos{0} = 0 => kB = 0 $$ Since k is nonzero, this gives B = 0, and the solution becomes: $$ y(x) = A\cos{kx} $$ Applying the second boundary condition (y(1) = 0), we get: $$ A\cos{k} = 0 $$ For non-trivial solutions (y(x) ≠ 0), A ≠ 0 and \(\cos{k} = 0\). This implies that \(k = \dfrac{(2n-1)\pi}{2}\) for integers n. Therefore, λ = \(k^2 = \left(\frac{(2n-1)\pi}{2}\right)^2\) for integers n, and eigenfunctions are given by: $$ y_n(x) = A_n\cos{\left(\frac{(2n-1)\pi x}{2}\right)} $$ Case 2: λ = 0 In this case, the general solution is \(y(x) = Ax + B\). Applying the first boundary condition (y'(0) = 0), we get: $$ A = 0 $$ And the solution becomes: $$ y(x) = B $$ Applying the second boundary condition (y(1) = 0), we get: $$ y(1) = B = 0 $$ This leads to a trivial solution, meaning no eigenvalues or eigenfunctions for this case. Case 3: λ < 0 For the general solution \(y(x) = A\cosh{kx} + B\sinh{kx}\), applying the first boundary condition (y'(0) = 0), we get: $$ kA\sinh{0} + kB\cosh{0} = 0 => kB = 0 $$ Since k is nonzero, this gives B = 0, and the solution becomes: $$ y(x) = A\cosh{kx} $$ Applying the second boundary condition (y(1) = 0), we get: $$ A\cosh{k} = 0 $$ But \(\cosh{k}\) ≠ 0 for any k, so this case also leads to a trivial solution, meaning no eigenvalues or eigenfunctions for this case.

Based on the analysis of the three cases, we found eigenvalues and eigenfunctions only in the first case, where λ > 0. The eigenvalues and eigenfunctions for the given boundary value problem are: Eigenvalues: $$ \lambda_n = \left(\frac{(2n-1)\pi}{2}\right)^2, \quad n = 1, 2, 3, ... $$ Eigenfunctions: $$ y_n(x) = A_n\cos{\left(\frac{(2n-1)\pi x}{2}\right)}, \quad n = 1, 2, 3, ... $$