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: $$\left|\text{Answer} \right| = 2$$