Missing one solution of a differential equation? Let $y(x)$ be a solution to the differential equation $(1+e^x)y'+ye^x=1$. If $y(0)=2$ then which of the following statements is (are) true. 
(A) $y(-4)=0$
(B) $y(-2)=0$
(C) $y(x)$ has a critical point in the interval $(-1,0)$
(D) $y(x)$ has no critical points in the interval $(-1,0)$
This is a question from IIT-JEE advanced. I have the general solution $y=\frac{x}{(1+e^x)} +c$
and if $y(0)=2$ then the specific solution is $y=\frac{x}{(1+e^x)} +2$
But the problem is only the D is true,then. But the answer sheet has A&C as correct answers. But I have checked many times and found no error in my approach. 
Maybe I am missing something. Need a hint.
 A: Hint: To rectify your mistake, note that if your equation were homogeneous, it would be exact and we would have
$$
g(x,y)=y+ye^x=c\implies y=\frac{c}{1+e^x}
$$
And for the particular solution, use variation of parameters and assume a solution of the form
$$
y(x)=\frac{c(x)}{1+e^x}
$$
Then the conditions of your ode require that 
$$
(1+e^x)y'+ye^x=1\implies c'(x)=1\implies c(x)=x+d
$$
and your solution is thus
$$
y(x)=\frac{x+d}{1+e^x}
$$
with $y(0)=2$ implying that $d=4$
$$
y(x)=\frac{x+4}{1+e^x}
$$
That $A$ is true is clear enough. That $C$ is true follows from the IVT applied to 
$$
1-3e^x-xe^x
$$
on the interval in question.
A: Notice that
$$
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}x}(1+\exp(x))=\exp(x)
\end{equation}
$$
so that
$$
\begin{equation}
(1+\exp(x))\frac{\mathrm{d}y}{\mathrm{d}x}+\exp(x)y(x)=(1+\exp(x))\frac{\mathrm{d}y}{\mathrm{d}x}+y(x)\frac{\mathrm{d}}{\mathrm{d}x}(1+\exp(x))
\end{equation}
$$
by the product rule your differential equation is then
$$
\begin{equation}
\frac{\mathrm{d}}{\mathrm{d}x}\left[(1+\exp(x))y(x)\right]=1
\end{equation}
$$
which can be integrated
$$
\begin{equation}
(1+\exp(x))y(x)=x+C
\end{equation}
$$
$$
\begin{equation}
y(x)=\frac{x+C}{1+\exp(x)}
\end{equation}
$$
Now for the initial condition
$$
\begin{equation}
y(0)=C/2=2
\end{equation}
$$
$$
\begin{equation}
C=4
\end{equation}
$$
$$
\begin{equation}
y(x)=\frac{x+4}{\exp(x)+1}
\end{equation}
$$
This solution clearly has a single zero at $x=-4$.
Solving for $y'$ in the differential equation
$$
\begin{equation}
\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1-y\exp{x}}{1+\exp{x}}
\end{equation}
$$
Substituting the IVP solution for $y(x)$.
$$
\begin{equation}
\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1-(\frac{x+4}{\exp(x)+1})\exp(x)}{1+\exp{x}}
\end{equation}
$$
$$
\begin{equation}
\frac{\mathrm{d}y}{\mathrm{d}x}=1-(x+3)\exp(x)
\end{equation}
$$
$$
\begin{equation}
\frac{\mathrm{d}y}{\mathrm{d}x}=0,\;\;\; for\;\;\; x=-0.79206
\end{equation}
$$
or rather
$$
\begin{equation}
\frac{\mathrm{d}y}{\mathrm{d}x}|_{x=-1}=1-2\exp(-1)=1-\frac{2}{e}>0
\end{equation}
$$
$$
\begin{equation}
\frac{\mathrm{d}y}{\mathrm{d}x}|_{x=0}=1-3=-2<0
\end{equation}
$$
Since we see that $y'(x)$ is a continuous function of $x$, there exists a $c\in(-1,0)$ for which $y'(c) = 0$.
