Question

The eigenvalues of the boundary value problem \[ \frac{d^2y}{dx^2} + \lambda y = 0, x \in (0, \pi), \lambda > 0, \] \[ y(0) = 0, y(\pi) - \frac{dy}{dx}(\pi) = 0 \] are given by:

• A. \( \lambda = (n \pi)^2, \, n = 1, 2, 3, \dots \)
• B. \( \lambda = n^2, \, n = 1, 2, 3, \dots \)
• C. \( \lambda = k^2_{n}, \, \text{where} \, k_{n}, n = 1, 2, 3, \dots \, \text{are the roots of} \, k - \tan(k \pi) = 0 \)
• D. \( \lambda = k^2_{n}, \, \text{where} \, k_{n}, n = 1, 2, 3, \dots \, \text{are the roots of} \, k + \tan(k \pi) = 0 \)

Hint:

For boundary value problems involving trigonometric functions, the eigenvalues are found by applying boundary conditions that result in transcendental equations with roots corresponding to the eigenvalues.

Answer 1

The Correct Option is C

The given boundary value problem is a Sturm-Liouville problem. The general solution to the differential equation \( \frac{d^2y}{dx^2} + \lambda y = 0 \) is of the form: \[ y(x) = A \sin(\sqrt{\lambda}x) + B \cos(\sqrt{\lambda}x). \] Step 1: Apply the boundary conditions. - \( y(0) = 0 \) implies \( B = 0 \), so the solution becomes: \[ y(x) = A \sin(\sqrt{\lambda}x). \] - The second boundary condition is \( y(\pi) - \frac{dy}{dx}(\pi) = 0 \). Substituting the solution and its derivative at \( x = \pi \), we get: \[ A \sin(\sqrt{\lambda} \pi) - A \sqrt{\lambda} \cos(\sqrt{\lambda} \pi) = 0. \] This gives the condition: \[ \sin(\sqrt{\lambda} \pi) = \sqrt{\lambda} \cos(\sqrt{\lambda} \pi). \] Solving this equation for \( \lambda \), the eigenvalues are the square of the roots of \( k - \tan(k \pi) = 0 \), where \( k = \sqrt{\lambda} \). Thus, the correct eigenvalues are \( \lambda = k^2_{n} \), where \( k_{n} \) are the roots of the equation \( k - \tan(k \pi) = 0 \). Final Answer: (C) \( \lambda = k^2_{n}, \, \text{where} \, k_{n}, n = 1, 2, 3, \dots \, \text{are the roots of} \, k - \tan(k \pi) = 0 \).