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

Answer 1

The correct option is D

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

Explanation for the correct option:

Find the required solution of given differential equation

Given, Differential equation $(1+e^{-x})(1+y^2)\frac{dy}{dx}=y^2$,which is passes through the point$(0,1)$.

We can write above equation as,

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

Let $e^x+1=u$

$\Rightarrow$ $e^x=\frac{du}{dx}$

$\Rightarrow$$e^{x}dx=du$

Substitute the value,

$$\begin{gathered} \Rightarrow \left(\frac{-1}{y}\right)+y=\int \frac{1}{u}du\mathbf{}\mathbf{[}\mathbf{\int }\frac{\mathbf{1}}{\mathbf{x}}\mathbf{dx}\mathbf{=}\mathbf{ln}\left(x\right)\mathbf{]} \\ \Rightarrow \left(\frac{-1}{y}\right)+y=\ln \left(u\right)+C \\ \Rightarrow \left(\frac{-1}{y}\right)+y=\ln \left(e^x+1\right)+C \end{gathered}$$

$\because$ It is passes through the point $\left(x,y\right)=(0,1)$

By substituting the value, we get

$$\begin{gathered} \Rightarrow -1+1=\ln \left(1+1\right)+C \\ \Rightarrow 0=\ln \left(2\right)+C \\ \Rightarrow C=-\ln \left(2\right) \\ \therefore \left(\frac{-1}{y}\right)+y=\ln \left(e^x+1\right)-\ln \left(2\right) \\ \Rightarrow y^2=1+y\ln \frac{\left(e^x+1\right)}{2} \\ \Rightarrow y^2=1+y{\log }_e\left(\frac{1+e^x}{2}\right) \end{gathered}$$ $\left[\ln \left(a\right)-\ln \left(b\right)=\ln \left(\frac{a}{b}\right)\right]$

Hence, option (D) is the correct answer.