JEE Main 2002Annual, Paper 1MathematicsEasyDifferential Equations163× since 2002The order and degree of the differential equation (1+3dydx)2/3=4d3ydx3\,{\left( {1 + 3{{dy} \over {dx}}} \right)^{2/3}} = 4{{{d^3}y} \over {d{x^3}}}(1+3dxdy)2/3=4dx3d3y areAOption A. (1,23)\left( {1,{2 \over 3}} \right)(1,32)BOption B. (3,1)(3, 1)(3,1)COption C. (3,3)(3,3)(3,3)DOption D. (1,2)(1,2)(1,2)CheckSkip · show solution
