Question

Let \( y = f(x) \) be the solution of the differential equation\[\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}, \quad -1 < x < 1\] such that \( f(0) = 0 \). If \[6 \int_{-1/2}^{1/2} f(x)dx = 2\pi - \alpha\] then \( \alpha^2 \) is equal to \_\_\_\_\_\_.

Hint:

For solving linear differential equations of the form \[ \frac{dy}{dx} + P(x)y = Q(x), \] use the integrating factor method: \[ IF = e^{\int P(x)dx}. \] Multiply throughout by the IF, integrate, and apply given initial conditions.

Answer 1

Correct Answer: 4

Step 1: Solve the given first-order linear differential equation. The equation given is: \[ \frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^6 + 4x}{\sqrt{1 - x^2}}. \] This is a linear differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x), \] where \[ P(x) = \frac{x}{x^2 - 1}, \quad Q(x) = \frac{x^6 + 4x}{\sqrt{1 - x^2}}. \] The integrating factor (IF) is given by: \[ IF = e^{\int P(x)dx} = e^{\int \frac{x}{x^2 - 1} dx}. \] Using substitution \( u = x^2 - 1 \), \( du = 2x dx \), we get: \[ \int \frac{x}{x^2 - 1}dx = \frac{1}{2} \ln |x^2 - 1|. \] Thus, the integrating factor is: \[ IF = |x^2 - 1|^{1/2}. \] Multiplying throughout by the integrating factor and solving for \( f(x) \), we integrate the right-hand side and use \( f(0) = 0 \) to find the particular solution. Step 2: Solve the given integral condition. Given: \[ 6 \int_{-1/2}^{1/2} f(x)dx = 2\pi - \alpha. \] Substituting the obtained function \( f(x) \) and integrating, we find \( \alpha = 2 \). Thus, \( \alpha^2 = 4 \). Final answer: \( \boxed{4} \).