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

$(1+e^x)y' + y e^x = 1$

$y(0)=2$

$\text{To find: which statement(s) are true}$

$(1+e^x)y' + e^x y = 1 \iff \dfrac{d}{dx}\big[(1+e^x)y\big] = 1$

$(1+e^x)y = x + C$

$y(0)=2 \Rightarrow (1+1)\cdot 2 = C \Rightarrow C = 4$

$y(x) = \dfrac{x+4}{1+e^x}$

$y(-4) = \dfrac{0}{1+e^{-4}} = 0 \Rightarrow (1)\ \text{is true}$

$y(-2) = \dfrac{2}{1+e^{-2}} \neq 0 \Rightarrow (2)\ \text{is false}$

$y'(x) = \dfrac{1 - (x+3)e^x}{(1+e^x)^2}$

$y'=0 \iff (x+3)e^x = 1$

$g(x)=(x+3)e^x-1:\quad g(-1)=\dfrac{2}{e}-1<0,\quad g(0)=2>0$

$\Rightarrow \text{a critical point in } (-1,0):\ (3)\ \text{true},\ (4)\ \text{false}$

$\boxed{\text{Answer: statements (1) and (3)}}$