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Differential Equations question

2020 · 3 Sep · Shift 1 · Q22

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JEE Main
Mathematics
Differential Equations
2020 · 3 Sep · Shift 1 · Q22

Differential Equations question

2020 · 3 Sep · Shift 1 · Q22

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

The solution curve of the differential equation, (1 + e-x)(1 + y2)

{{dy} \over {dx}}

= y2, which passes through the point (0, 1), is :

A
y2 + 1 = y $\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)$
B
y2 + 1 = y $\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)$
C
y2 = 1 + ${y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}$
D
y2 = 1 + ${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$

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