Question

Find the particular solution of the differential equation: \[ x^2 \frac{dy}{dx} - xy = x^2 \cos^2\left(\frac{y}{2x}\right), \] given that when \(x = 1\), \(y = \frac{\pi}{2}\).

Hint:

Linear differential equations can be solved using integrating factors. Always simplify terms and use initial conditions to find the constant of integration.

Answer 1

The differential equation provided is: \[ x^2 \frac{dy}{dx} - xy = x^2 \cos^2\left(\frac{y}{2x}\right). \] Rearranged, this becomes: \[ \frac{dy}{dx} - \frac{y}{x} = \cos^2\left(\frac{y}{2x}\right). \] This equation is of the linear form: \[ \frac{dy}{dx} + P(x)y = Q(x), \] with \(P(x) = -\frac{1}{x}\) and \(Q(x) = \cos^2\left(\frac{y}{2x}\right)\).

Step 1: Solve the associated homogeneous equation: \[ \frac{dy}{dx} - \frac{y}{x} = 0. \] Separating variables yields: \[ \frac{dy}{y} = \frac{dx}{x}. \] Integration of both sides gives: \[ \ln y = \ln x + C_1, \] where \(C_1\) is the integration constant. Simplifying results in the homogeneous solution: \[ y_h = C_1 x. \] 

Step 2: Employ an integrating factor to solve the non-homogeneous equation. The integrating factor (IF) is calculated as: \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln x} = \frac{1}{x}. \] Multiplying the original equation by \(\mu(x)\) produces: \[ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = \frac{\cos^2\left(\frac{y}{2x}\right)}{x}. \] 

This simplifies to: \[ \frac{d}{dx}\left(\frac{y}{x}\right) = \frac{\cos^2\left(\frac{y}{2x}\right)}{x}. \] Integrating both sides results in: \[ \frac{y}{x} = \int \frac{\cos^2\left(\frac{y}{2x}\right)}{x} \, dx + C_2. \] The constant \(C_2\) is determined using the initial condition \(x = 1\), \(y = \frac{\pi}{2}\). 
Further simplification or numerical methods may be required to compute the integral and obtain the particular solution.