Let $y(x)$ be a solution of the differential equation $\left(1+e^x\right) y'+y e^x=1$. If $y(0)=2$, then which of the following statement(s) is (are) true?

(a) $y(-4)=0$
(b) $y(-2)=0$
(c) $y(x)$ has critical point in the interval $(-1,0)$
(d) $y(x)$ has no critical point in the interval $(-1,0)$.

(a), (c)

We have,

$\left(1+e^x\right) y'+y e^x=1$

$\Rightarrow y'+\frac{e^x}{1+e^x} y=\frac{1}{1+e^x}$            .....(i)

This is a linear differential equation with integrating factor

Integrating factor = $e^{\int \frac{e^x}{1+e^x} d x}=e^{\log \left(1+e^x\right)}=1+e^x$

Multiplying both sides of (i) by I.F. $=1+e^x$, we obtain

$y'\left(1+e^x\right)+y e^x=1$

Integrating both sides with respect to $x$, we obtain

$y\left(1+e^x\right)=x+C$            .....(ii)

It is given that $y(0)=2$ i.e. $y=2$ when $x=0$. Putting $x=0, y=2$ in (ii), we get

$4=0+C \Rightarrow C=4$

Putting $C=4$ in (ii), we obtain

$y\left(1+e^x\right)=x+4 \Rightarrow y=\frac{x+4}{1+e^x} \Rightarrow y(-4)=0$

Putting $y=\frac{x+4}{1+e^x}$ in (i), we obtain

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

$\Rightarrow y'(-1) =\frac{\left(1+e^{-1}\right)-3 e^{-1}}{\left(1+e^{-1}\right)^2}=\frac{e-2}{e\left(1+e^{-1}\right)^2}>0$ and $y'(0)=-\frac{1}{2}<0$

$\Rightarrow y'(x)=0$ for some $x \in(-1,0)$

$\Rightarrow y(x)$ has a critical point in the interval $(-1,0)$.

Hence, options (a) and (c) are true.