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Differential Equations question

2020 · 6 Sep · Shift 1 · Q21

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JEE Main
Mathematics
Differential Equations
2020 · 6 Sep · Shift 1 · Q21

Differential Equations question

2020 · 6 Sep · Shift 1 · Q21

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

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)

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

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