Question

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

Accepted Answer

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

Answer

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

Answer

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

Answer

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

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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