Question

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

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

Answer 1

The correct option is D $y^2=1+y\left({\log }_e\left(\frac{1+e^x}{2}\right)\right)$

Given:
$(1+e^{-x})(1+y^2)\frac{dy}{dx}=y^2$
$\Rightarrow \frac{1+y^2}{y^2}dy=\frac{e^x}{e^x+1}dx$

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

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

$\Rightarrow \frac{-1}{y}+y=\ln |e^x+1|+c$
since it is passes through $(0,1)$
$\Rightarrow -1+1=\ln 2+c$
$\Rightarrow c=-\ln 2$
$\Rightarrow \frac{-1}{y}+y=\ln |e^x+1|-\ln 2$
$\Rightarrow y^2=1+y\left[\ln \left(\frac{e^x+1}{2}\right)\right]$