JEE Main 20204 Sep, Evening shiftMathematicsMediumDifferential Equations163× since 2002The solution of the differential equation dydx−y+3xloge(y+3x)+3=0{{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0dxdy−loge(y+3x)y+3x+3=0 is: (where c is a constant of integration)AOption A. x−12(loge(y+3x))2=Cx - {1 \over 2}{\left( {{{\log }_e}\left( {y + 3x} \right)} \right)^2} = Cx−21(loge(y+3x))2=CBOption B. y+3x−12(logex)2=Cy + 3x - {1 \over 2}{\left( {{{\log }_e}x} \right)^2} = Cy+3x−21(logex)2=CCOption C. x – log_e(y+3x) = CDOption D. x – 2log_e(y+3x) = CCheckSkip · show solution
