Linear First Order

Before any technique applies, this equation must be put into standard linear form by dividing through by x, giving \frac{dy}{dx} + \frac{2}{x}y = x. Now P(x) = \frac{2}{x} and Q(x) = x. The integrating factor is \mu = e^{\int \frac{2}{x},dx} = e^{2\ln x} = x^2, which transforms the left side into the derivative of a product. Multiplying through by x^2 gives x^2\frac{dy}{dx} + 2xy = x^3, and the left side is exactly \frac{d}{dx}(x^2 y). Integrating both sides yields x^2 y = \int x^3,dx = \frac{x^4}{4} + C. Dividing by x^2 gives y = \frac{x^2}{4} + \frac{C}{x^2}. Option y = \frac{x^2}{2} + Cx uses an incorrect integrating factor. Option y = x^2 + \frac{C}{x} stems from misintegrating the right side. Option \frac{x^3}{4} + C forgets to divide by x^2 at the end. This is the canonical linear-equation method emphasized in JEE Advanced. As a final consistency check, substituting y = \frac{x^2}{4} into the original gives x\cdot\frac{x}{2} + 2\cdot\frac{x^2}{4} = \frac{x^2}{2} + \frac{x^2}{2} = x^2, confirming the particular part satisfies the equation.