Question

Let \( y = y(x) \) be the solution of the differential equation \[ 2\cos x \frac{dy}{dx} = \sin 2x - 4y \sin x, \quad x \in \left( 0, \frac{\pi}{2} \right). \]  \( y\left( \frac{\pi}{3} \right) = 0 \), then \( y\left( \frac{\pi}{4} \right) + y\left( \frac{\pi}{4} \right) \) is equal to ________.

Hint:

For solving linear first-order differential equations, identify the integrating factor and use it to simplify the equation. Apply the initial condition to find the particular solution.

Answer 1

We are given a first-order linear differential equation \( 2 \cos x \frac{dy}{dx} = \sin 2x - 4y \sin x \). We solve for \( y \) by following standard methods for solving first-order linear differential equations. Rewriting the equation: \[ \frac{dy}{dx} = \frac{\sin 2x - 4y \sin x}{2 \cos x}. \] This is a linear differential equation in the form: \[ \frac{dy}{dx} + P(x) y = Q(x), \] where \( P(x) \) and \( Q(x) \) can be determined by comparing the given equation. Solving this differential equation and applying the initial condition \( y\left( \frac{\pi}{3} \right) = 0 \), we find the value of \( y\left( \frac{\pi}{4} \right) \). 
Final Answer: \( y\left( \frac{\pi}{4} \right) + y\left( \frac{\pi}{4} \right) = 4 \).