Linear First Order

A first-order linear differential equation in standard form \frac{dy}{dx} + P(x)y = Q(x) is solved by multiplying through by an integrating factor \mu(x) = e^{\int P(x),dx}, which turns the left side into the derivative of a product. The central idea is that this factor is engineered precisely so that \frac{d}{dx}[\mu y] = \mu Q, collapsing two terms into one exact derivative. Here P(x) = \frac{2}{x}, so \int P,dx = \int \frac{2}{x},dx = 2\ln x = \ln x^2 for x > 0. Exponentiating gives \mu(x) = e^{\ln x^2} = x^2. Option e^{2x} would arise if P were the constant 2 rather than 2/x. Option \frac{1}{x^2} reverses the sign of the exponent, which would correspond to P = -2/x. Option 2\ln x is the integral of P but omits the essential exponentiation step, so it is not itself an integrating factor. This is exactly the standard linear-ODE technique emphasized in JEE Advanced. As a final consistency check, multiplying the equation by x^2 gives x^2 y' + 2x y = \cos x, and the left side is indeed \frac{d}{dx}(x^2 y), confirming that x^2 performs its intended role correctly.