The solution curve of the differential equation, (1 + e-x)(1 + y2)${{dy} \over {dx}}$ = y2, which passes through the point (0, 1), is :

Question

The solution curve of the differential equation,

(1 + e^-x)(1 + y²)${{dy} \over {dx}}$ = y²,

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

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

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

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

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

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