JEE MAIN - Mathematics Hindi (2020 - 6th September Morning Slot - No. 2)
समीकरण का सामान्य समाधान
$$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $$ + xy$${{dy} \over {dx}}$$ = 0 है :
(जहाँ C एकीकरण की एक स्थिरांक है)
$$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $$ + xy$${{dy} \over {dx}}$$ = 0 है :
(जहाँ C एकीकरण की एक स्थिरांक है)
$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$
$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$
$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
Comments (0)
