Question

If $y\left(x\right)$ satisfies equations $(1+e^x)\frac{dy}{dx}+y{e}^x=1$ and $y\left(1\right)=\frac{3}{1+e}$ then maximum value of $y\left(x\right)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

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

$I.F.=e^{\int (e^x/1+e^x)dx}=e^{\ln (1+e^x)}=1+e^x$

$\Rightarrow y(1+e^x)=\int (1+e^x)\frac{1}{(1+e^x)}dx\Rightarrow y(1+e^x)=x+c$
We know that,
$y\left(1\right)=\frac{3}{1+e}\Rightarrow \frac{3}{1+e}(1+e)=1+c\Rightarrow c=2\Rightarrow y(1+e^x)=2+x\Rightarrow y=\left(\frac{x+2}{1+e^x}\right)$

Finding maximum of $y$,
$y(1+e^x)=2+x\Rightarrow y^{\prime }(1+e^x)+y\left(e^x\right)=1$
Putting $y^{\prime }=0$
$\Rightarrow \left(\frac{x+2}{1+e^x}\right)e^x=1\Rightarrow x=e^{-x}-1\Rightarrow x=0$
Now,
$y^{\prime \prime }(1+e^x)+y^{\prime }\left(e^x\right)+y^{\prime }\left(e^x\right)+y\left(e^x\right)=0$
Putting $x=0,y^{\prime }=0$
$2y^{\prime \prime }+2=0\Rightarrow y^{\prime \prime }=-2$

As $y^{\prime \prime }<0$ at $x=0$
So $y$ is maximum
$y{|}_{\text{max}}=1$