The general solution of the differential equation $\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $ + xy${{dy} \over {dx}}$ = 0 is : (where C is a constant of integration)

Question

The general solution of the differential equation

$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $ + xy${{dy} \over {dx}}$ = 0 is :

(where C is a constant of integration)

$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$

$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$

$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$

$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$

Solution

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