JEE Main 202118 Mar, Morning shiftMathematicsEasyDifferential Equations163× since 2002The differential equation satisfied by the system of parabolas y² = 4a(x + a) is :AOption A. y(dydx)2−2x(dydx)−y=0y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) - y = 0y(dxdy)2−2x(dxdy)−y=0BOption B. y(dydx)2−2x(dydx)+y=0y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) + y = 0y(dxdy)2−2x(dxdy)+y=0COption C. y(dydx)2+2x(dydx)−y=0y{\left( {{{dy} \over {dx}}} \right)^2} + 2x\left( {{{dy} \over {dx}}} \right) - y = 0y(dxdy)2+2x(dxdy)−y=0DOption D. y(dydx)+2x(dydx)−y=0y\left( {{{dy} \over {dx}}} \right) + 2x\left( {{{dy} \over {dx}}} \right) - y = 0y(dxdy)+2x(dxdy)−y=0CheckSkip · show solution
