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 given differential equation is: \[ x^2 \frac{dy}{dx} - xy = x^2 \cos^2\left(\frac{y}{2x}\right). \] 

Rearranging terms: \[ \frac{dy}{dx} - \frac{y}{x} = \cos^2\left(\frac{y}{2x}\right). \] This is a linear differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x), \] where \(P(x) = -\frac{1}{x}\) and \(Q(x) = \cos^2\left(\frac{y}{2x}\right)\). 

Step 1: Solve the homogeneous equation. The associated homogeneous equation is: \[ \frac{dy}{dx} - \frac{y}{x} = 0. \] 
Separating variables: \[ \frac{dy}{y} = \frac{dx}{x}. \] Integrating both sides: \[ \ln y = \ln x + C_1, \] where \(C_1\) is the constant of integration. Simplify: \[ y_h = C_1 x. \] 

Step 2: Solve the non-homogeneous equation using an integrating factor. The integrating factor (IF) is: \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln x} = \frac{1}{x}. \] 
Multiply through the original equation by \(\mu(x)\): \[ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = \frac{\cos^2\left(\frac{y}{2x}\right)}{x}. \] 

Simplify: \[ \frac{d}{dx}\left(\frac{y}{x}\right) = \frac{\cos^2\left(\frac{y}{2x}\right)}{x}. \] Integrating both sides: \[ \frac{y}{x} = \int \frac{\cos^2\left(\frac{y}{2x}\right)}{x} \, dx + C_2. \] Using the initial condition \(x = 1\), \(y = \frac{\pi}{2}\), we find \(C_2\). 
Solve further as needed. 

The particular solution will depend on further simplification or numerical methods to compute the integral.