Question

Let y = y(x) be a solution of the differential equation (x cos x)dy +(xy sin x + y cos x – l)dx = 0, 0 < x < \(\frac{π}{2}\). If\(\frac{π}{3}y(\frac{π}{3})=\sqrt3\), then \(| \frac{π}{6 }y“( \frac{π}{6})+2 y'( \frac{π}{6})| \) is equal to______.

Answer 1

Correct Answer: 2

Given Differential Equation: \[ (x \cos x) \frac{dy}{dx} + \left(xy \sin x + y \cos x - 1\right) = 0, \quad 0 < x < \frac{\pi}{2} \]  
Rewriting the equation: \[ \frac{dy}{dx} + \frac{x \sin x + \cos x}{x \cos x}y = \frac{1}{x \cos x} \] 
Identifying the Integrating Factor (IF): \[ \text{IF} = x \sec x \] 
Multiplying through by the IF and solving the integral: \[ y \cdot x \sec x = \tan x + c \] 
Using the initial condition \(y\left(\frac{\pi}{3}\right) = \frac{3\sqrt{3}}{\pi}\): \[ \frac{\pi}{3} \sec \left(\frac{\pi}{3}\right) \cdot \frac{3\sqrt{3}}{\pi} = \sqrt{3} + c \implies c = \sqrt{3} \] 
Final solution: \[ y \cdot x \sec x = \tan x + \sqrt{3} \] 
Evaluating the expression: \[ \left|\text{Answer} \right| = 2 \]