Given: $$\left(1+e^x\right) y^{\prime}+y e^x=1$$

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

$$\frac{d y}{d x}+\frac{e^x}{1+e^x} y=\frac{1}{1+e^x}$$, which is liner differential equation in $$y$$.

Comparing above equation with $$\frac{d y}{d x}+\mathrm{P} y=Q$$, we get $$Q$$, we get

$$\begin{aligned} \quad \mathrm{P} & =\frac{e^x}{1+e^x} \text { and } \mathrm{Q}=\frac{1}{1+e^x} \\ \text { So, I.F. } & =e^{\int \mathrm{P} \cdot d x} \\ \text { I.F. } & =e^{\int \frac{e^x}{1+e^x} d x} \\ \text { I.F. } & =e^{\ln \left(1+e^x\right)} \end{aligned}$$

$$\left\{\because \int \frac{f^{\prime}(x)}{f(x)} d x=\ln f(x)\right\}$$

$$\text { I.F }=1+e^x$$

So, solution of given differential equation is given by

$$y \text { (I.F.) }=\int \text { Q.(I.F.) } d x$$

$$\begin{aligned} y \cdot\left(1+e^x\right) & =\int \frac{1}{1+e^x} \cdot\left(1+e^x\right) d x \\ y\left(1+e^x\right) & =\int 1 d x \\ y\left(1+e^x\right) & =x+\mathrm{C} \\ \because \quad y(0) & =2 \\ \Rightarrow \quad 2\left(1+e^0\right) & =0+\mathrm{C} \\ c & =4 \end{aligned}$$

So, $$y(x)=\frac{x+4}{1+e^x}\quad \text{... (i)}$$

Put $$x=-4$$ in the equation, we get

$$y(-4)=0$$

Put $$x=-2$$ in the above equation (i), we get

$$y(-2)=\frac{2}{1+e^{-2}} \neq 0$$

For critical points, $$y^{\prime}=0$$

From e.q, (i), $$y\left(1+e^x\right)=x+4$$

Differentiating the above equation w.r.t. $$x$$, we get

$$\begin{aligned} y^{\prime}\left(1+e^x\right)+y\left(e^x\right) & =1 \\ 0\left(1+e^x\right)+y e^x & =1 \quad \left\{\because y^{\prime}=0\right\} \\ y e^x-1 & =0 \end{aligned}$$

$$\quad \begin{aligned} \text { Now, } \text { let } g(x) & =y e^x-1 \\ g(x) & =\frac{(x+4) e^x}{1+e^x}-1 \\ g(x) & =\frac{(x+3) e^x-1}{1+e^x} \end{aligned}$$

$$\text { Now, } \quad g(-1)=\frac{2 e^{-1}-1}{1+e^{-1}}=\frac{2-e}{1+e}<0 \quad \{\because e=2 \cdot 7 \cdot 8\}$$

$$\text { And, } \quad g(0)=\frac{3 e^0-1}{1+e^0}=1>0$$

So, there exists one value of $$x$$ in $$(-1,0)$$ for which

$$\begin{aligned} g(x) & =0 \\ \Rightarrow \quad y^{\prime} & =0 \end{aligned}$$

There exist a critical point of $$y(x)$$ in $$(-1,0)$$

Hint :

(i) Use solution of liner differential equation

$$\begin{aligned} \frac{d y}{d x}+\mathrm{P} y & =\mathrm{Q} \text { is given by } \\ y \text { (I.F.) } & =\int \mathrm{Q} \cdot \text { (I.F) } d x, \text { where I.F }=e^{\int p \cdot d x} \end{aligned}$$

(ii) For critical points, $$y^{\prime}=0$$