Question

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

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

Answer 1

Answer: (A)

$\left(1+ e ^{-x}\right)\left(1+ y ^{2}\right) \frac{ dy }{ dx }= y ^{2}$
$\Rightarrow \left(1+y^{-2}\right) d y=\left(\frac{e^{x}}{1+e^{x}}\right) d x$
$\Rightarrow \left(y-\frac{1}{y}\right)=\ln \left(1+e^{x}\right)+c$
$\therefore $ It passes through $(0,1)$
$\Rightarrow c=-\ln 2$
$\Rightarrow \,\,\,\ y^{2}=1+y \ln \left(\frac{1+e^{x}}{2}\right)$