Question

Find the eigenvalues and eigenfunctions of the given boundary value problem. Assume that all eigenvalues are real. \(y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad y^{\prime}(\pi)=0\)

Short answer

Answer: The eigenvalues for the given boundary value problem are \(\lambda_n = ((2n + 1)/2)^2\) and the corresponding eigenfunctions are \(y_n(x) = \sin(((2n + 1)/2)x)\), where \(n\) is an integer.

Step-by-step solution

Solve the differential equation

To solve the differential equation \(y^{\prime \prime}+\lambda y=0\), we'll first consider two cases: \(\lambda>0\) and \(\lambda<0\). For \(\lambda>0\), let \(\lambda = k^2\). The differential equation becomes: \(y^{\prime \prime} + k^2 y = 0\). The general solution is given by: \(y(x) = A\sin(kx) + B\cos(kx)\). For \(\lambda<0\), let \(\lambda = -k^2\). The differential equation becomes: \(y^{\prime \prime} - k^2 y = 0\). The general solution is given by: \(y(x) = A\sinh(kx) + B\cosh(kx)\).

Apply the boundary conditions

Using the boundary condition \(y(0) = 0\), we get: For \(\lambda > 0\): \(y(0) = A\sin(0) + B\cos(0) = 0\). Since \(\cos(0) = 1\), we have \(B = 0\). The equation becomes \(y(x) = A\sin(kx)\). For \(\lambda < 0\): \(y(0) = A\sinh(0) + B\cosh(0) = 0\). Since \(\cosh(0) = 1\), we have \(B = 0\). The equation becomes \(y(x) = A\sinh(kx)\). Now, we'll apply the second boundary condition: \(y^{\prime}(\pi) = 0\). For \(\lambda > 0\): The derivative of \(y(x)\) is \(y^{\prime}(x) = Ak\cos(kx)\). Using the boundary condition, \(Ak\cos(k\pi) = 0\). Since \(A\) and \(k\) cannot both be zero, we must have \(\cos(k\pi) = 0\). This condition is satisfied for \(k = (2n + 1)/2\), where \(n\) is an integer. Hence, the eigenvalues are \(\lambda_n = ((2n + 1)/2)^2\). For \(\lambda < 0\): The derivative of \(y(x)\) is \(y^{\prime}(x) = Ak\cosh(kx)\). Using the boundary condition, \(Ak\cosh(k\pi) = 0\). However, \(\cosh(kx)\) is always positive and non-zero, which means no negative eigenvalues can satisfy this condition.

Determine the eigenfunctions

For the eigenvalues \(\lambda_n = ((2n + 1)/2)^2\), the corresponding eigenfunctions are given by \(y_n(x) = A_n\sin(((2n + 1)/2)x)\), where \(A_n\) is an arbitrary constant. Eigenfunctions are determined up to a constant, so we can set \(A_n = 1\). Therefore, the eigenfunctions are \(y_n(x) = \sin(((2n + 1)/2)x)\). In conclusion, the eigenvalues for the given boundary value problem are \(\lambda_n = ((2n + 1)/2)^2\) and the corresponding eigenfunctions are \(y_n(x) = \sin(((2n + 1)/2)x)\).