Mht CetDifferential EquationsMHT CET 2026 11th April Morning ShiftMCQ+2 / -02026The differential equation representing the family of curves $x \sin x + y^3 = 4ax$ isA$\dfrac{dy}{dx} = \dfrac{y^3 + x^2 \cos x}{3xy^2}$B$\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3xy^2}$C$\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3xy}$D$\dfrac{dy}{dx} = \dfrac{y^3 - x^2 \cos x}{3x^2 y}$Check AnswerClear SelectionReveal AnswerShow Explanation
