The solution curve of the differential equation,(1+e−x)(1+y2)dydx=y2, which passes through the point (0, 1) is

Given differential equation (1+e−x)(1+y2)dydx=y2 ⇒ ∫1+y2y2dy=∫ex1+exdx ⇒ (−1y+y)=loge(1+ex)+C,Which passes through (0,1),So −1+1=loge2+C⇒C=−loge2 So, equation of required curve is y2=1+yloge(1+ex2)Hence, option © is correct.

The solution curve of the differential equation,(1+e−x)(1+y2)dydx=y2, which passes through the point (0, 1) is

Given differential equation (1+e−x)(1+y2)dydx=y2 ⇒ ∫1+y2y2dy=∫ex1+exdx ⇒ (−1y+y)=loge(1+ex)+C,Which passes through (0,1),So −1+1=loge2+C⇒C=−loge2 So, equation of required curve is y2=1+yloge(1+ex2)Hence, option © is correct.

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The solution curve of the differential equation, $(1 + e^{- x}) (1 + y^{2}) \frac{d y}{d x} = y^{2},$ which passes through the point (0, 1) is

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$y^{2} = 1 + y \log_{e} (\frac{1 + e^{- x}}{2})$

$y^{2} + 1 = y (\log_{e} (\frac{1 + e^{x}}{2}) + 2)$

$y^{2} + 1 = y (\log_{e} (\frac{1 + e^{- x}}{2}) + 2)$

$y^{2} = 1 + y \log_{e} (\frac{1 + e^{x}}{2})$

answer is C.

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Detailed Solution

Given differential equation
$(1 + e^{- x}) (1 + y^{2}) \frac{d y}{d x} = y^{2} \Rightarrow \int \frac{1 + y^{2}}{y^{2}} d y = \int \frac{e^{x}}{1 + e^{x}} d x \Rightarrow (- \frac{1}{y} + y) = \log_{e} (1 + e^{x}) + C,$
Which passes through (0,1),
So $- 1 + 1 = \log_{e} 2 + C \Rightarrow C = - \log_{e} 2$
So, equation of required curve is
$y^{2} = 1 + y \log_{e} (\frac{1 + e^{x}}{2})$
Hence, option © is correct.