JEE Main 20206 Sep, Morning shiftMathematicsMediumDifferential Equations163× since 2002The general solution of the differential equation 1+x2+y2+x2y2\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}}1+x2+y2+x2y2 + xydydx{{dy} \over {dx}}dxdy = 0 is : (where C is a constant of integration)AOption A. 1+y2+1+x2=12loge(1+x2−11+x2+1)+C\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C1+y2+1+x2=21loge(1+x2+11+x2−1)+CBOption B. 1+y2−1+x2=12loge(1+x2−11+x2+1)+C\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C1+y2−1+x2=21loge(1+x2+11+x2−1)+CCOption C. 1+y2+1+x2=12loge(1+x2+11+x2−1)+C\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C1+y2+1+x2=21loge(1+x2−11+x2+1)+CDOption D. 1+y2−1+x2=12loge(1+x2+11+x2−1)+C\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C1+y2−1+x2=21loge(1+x2−11+x2+1)+CCheckSkip · show solution
