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

Answer 1

The correct option is D

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

Explanation for the correct option

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

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

integrating both sides

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

Let $e^x+1=t$

$e^x·dx=dt$

$$\begin{gathered} \Rightarrow \int \frac{1}{y^2}dx+\int 1dy=\int \left(\frac{1}{t}\right)dt \\ \Rightarrow \frac{y^{-1}}{-1}+y=\log \left(t\right)+c \end{gathered}$$

Put $t=1+e^x$

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

Since, the curve passes through the point $(0,1)$

Put $y=1,x=0$

$$\begin{gathered} \Rightarrow -\frac{1}{1}+1=\log (1+e^0)+c \\ \Rightarrow 0=\log \left(2\right)+c \\ \Rightarrow c=-\log \left(2\right) \end{gathered}$$

therefore, the equation becomes

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

Hence, option D is correct.