The solution curve of the differential equation,

Question:

The solution curve of the differential equation,

$\left(1+e^{-x}\right)\left(1+y^{2}\right) \frac{d y}{d x}=y^{2}$, which passes through the point $(0,1)$, is :

  1. (1) $y^{2}+1=y\left(\log _{e}\left(\frac{1+e^{-x}}{2}\right)+2\right)$

  2. (2) $y^{2}+1=y\left(\log _{e}\left(\frac{1+e^{x}}{2}\right)+2\right)$

  3. (3) $y^{2}=1+y \log _{e}\left(\frac{1+e^{x}}{2}\right)$

  4. (4) $y^{2}=1+y \log _{e}\left(\frac{1+e^{-x}}{2}\right)$


Correct Option: , 3

Solution:

$\int\left(\frac{y^{2}+1}{y^{2}}\right) d y=\int \frac{e^{x} d x}{e^{x}+1}$

$\Rightarrow y-\frac{1}{y}=\log _{e}\left|e^{x}+1\right|+c$

$\because$ Passes through $(0,1)$.

$\therefore c=-\log _{e} 2$

$\Rightarrow y^{2}-1=y \log _{e}\left(\frac{e^{x}+1}{2}\right)$

$\Rightarrow y^{2}=1+y \log _{e}\left(\frac{e^{x}+1}{2}\right)$

 

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