Homogeneous Equations

The right side is a ratio of homogeneous degree-two expressions, so dividing numerator and denominator by x^2 reveals dependence only on y/x, confirming the equation is homogeneous and inviting the substitution y = vx with \frac{dy}{dx} = v + x\frac{dv}{dx}. Rewriting, \frac{y^2}{xy - x^2} = \frac{v^2 x^2}{vx^2 - x^2} = \frac{v^2}{v - 1}. Thus v + x\frac{dv}{dx} = \frac{v^2}{v-1}. Subtracting v gives x\frac{dv}{dx} = \frac{v^2}{v-1} - v = \frac{v^2 - v(v-1)}{v-1} = \frac{v}{v-1}. Separating, \frac{v-1}{v},dv = \frac{dx}{x}, i.e. \left(1 - \frac{1}{v}\right)dv = \frac{dx}{x}. Integrating gives v - \ln|v| = \ln|x| + C_1. Substituting v = y/x gives \frac{y}{x} - \ln\left|\frac{y}{x}\right| = \ln|x| + C_1, that is \frac{y}{x} - \ln|y| + \ln|x| = \ln|x| + C_1, so the \ln|x| terms cancel and \frac{y}{x} - \ln|y| = C_1. Rearranging gives \ln|y| = \frac{y}{x} + C, writing C for -C_1. Option \ln|x| = \frac{y}{x} + C replaces \ln|y| with \ln|x| incorrectly. Option y = Cx ignores the nonlinear coupling. Option x = Cy^2 has no basis in the integration. This follows the JEE Advanced homogeneous reduction. As a final check, differentiating \ln|y| = \frac{y}{x} + C implicitly gives \frac{y'}{y} = \frac{xy' - y}{x^2}, and solving for y' reproduces \frac{dy}{dx} = \frac{y^2}{xy - x^2}, confirming the solution.