Mht CetDifferential EquationsMHT CET 2026 20th April Evening ShiftMCQ+2 / -02026The general solution of the differential equation $\dfrac{dy}{dx} + \dfrac{y}{x} = x^2 + 5$ is ....A$\dfrac{x^4}{4} + \dfrac{5x^2}{2} - xy = c$B$\dfrac{x^4}{4} - \dfrac{5x^2}{2} - xy = c$C$\dfrac{x^4}{4} - \dfrac{5x^2}{2} + xy = c$D$\dfrac{x^4}{4} + \dfrac{5x^2}{2} + xy = c$Check AnswerClear SelectionReveal AnswerShow Explanation
