# The solution curve of the differential equation,

Question:

The solution curve of the differential equation,

$\left(1+\mathrm{e}^{-\mathrm{x}}\right)\left(1+\mathrm{y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}^{2}$, which passes

through the point $(0,1)$, is :

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

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

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

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

Correct Option: 1

Solution:

$\left(1+\mathrm{e}^{-\mathrm{x}}\right)\left(1+\mathrm{y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}^{2}$

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

$\Rightarrow\left(y-\frac{1}{y}\right)=\ln \left(1+e^{x}\right)+c$

$\therefore \quad$ It passes through $(0,1) \Rightarrow \mathrm{c}=-\ell \mathrm{n} 2$

$\Rightarrow \quad \mathrm{y}^{2}=1+\mathrm{y} \ell \mathrm{n}\left(\frac{1+\mathrm{e}^{\mathrm{x}}}{2}\right)$