MathematicsMedium38×since 2002Q3599The solution curve of the differential equation, (1 + e^-x)(1 + y²)dydx{{dy} \over {dx}}dxdy = y², which passes through the point (0, 1), is :Ay² + 1 = y(loge(1+e−x2)+2)\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)(loge(21+e−x)+2)By² + 1 = y(loge(1+ex2)+2)\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)(loge(21+ex)+2)Cy² = 1 + yloge(1+e−x2){y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}yloge(21+e−x)Dy² = 1 + yloge(1+ex2){y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}yloge(21+ex)Check answerSkip