I was trying to resolve the following differential equation: $$ y'=e^x(y+1)^2$$ where $y = y(x)$
and I start to resolve it using the following steps:
first we find the solution for when $(y+1)^2=0$ and we get $y=-1$
then we continue considering $y\neq-1$
\begin{align}
\frac{y'}{(y+1)^2} &= e^x \\
\int{\frac{y'}{(y+1)^2}}dx &= \int{e^x}dx\\
\end{align}
now I perform a u sub
\begin{align}
u &= y+1\\
du &= y'dx\\
dx &= \frac{du}{y'}
\end{align}
as such the left integral becomes
\begin{align}
\int{\frac{1}{u^2}}du &= \int{e^x}dx\\
\int{u^{-2}}du &= e^x+c_{1}\\
-\frac{1}{u}+c_{2} &= e^x+c_{1}\\
-\frac{1}{y+1}+c_{2} &= e^x+c_{1}\\
\end{align}
now we say that $c_{1}-c_{2}=c_{3}$
\begin{align}
-\frac{1}{y+1} &= e^x+c_{3}\\
-\frac{1}{e^x+c_{3}} &= y+1\\
y &= -1-\frac{1}{e^x+c_{3}}
\end{align}
for neatness purpose we consider $c_{3}=c$ and we include the solution we found at the beginning
$$y=-1-\frac{1}{e^x+c};y=-1$$
but my textbook gives another answer:
$$ y=-\frac{e^x+c+1}{e^x+c};y=-1$$
are my steps wrong or is the textbook solution wrong? because even wolfram gives the same answer as the textbook
