Solve Bernouilli differential equation I think that my question is dumb... but I am stuck in a step.
I need to solve this ode: $$\rho' = \rho(1+\rho^2)$$
I made the variable change $u:=\rho^{-2}$. Then, separating variables and replace $u$ by $\rho^{-2}$, i get $\rho=\frac{1}{\sqrt{e^{-2t}-1}}$.
But, when i check the solution, the result is: $\rho=\frac{1}{\sqrt{e^{-2t}-1+\color{red}{\frac{1}{\rho_{0}^2}e^{-2t}}}}$.
I think i forget in some integration step the term "$+C$"... I don't know
Thanks in advance

EDIT
Let $y = \rho$ for writing convenience. To begin with, notice that
\begin{align*}
y' = y(1+y^{2}) & \Longleftrightarrow \frac{y'}{y(1+y^{2})} = 1
\end{align*}
Then we have that
\begin{align*}
\frac{1}{y(1+y^{2})} = \frac{(1 + y^{2}) - y^{2}}{y(1+y^{2})} = \frac{1}{y} - \frac{y}{1+y^{2}}
\end{align*}
Finally, we obtain the solution by integrating both sides:
\begin{align*}
\int\frac{\mathrm{d}y}{y(1+y^{2})} = \int1\mathrm{d}x & \Longleftrightarrow \ln|y| - \frac{\ln(1+y^{2})}{2} = x + c\\\\
& \Longleftrightarrow \ln(y^{2}) - \ln(1+y^{2}) = 2x + k\\\\
& \Longleftrightarrow \ln\left(\frac{y^{2}}{1+y^{2}}\right) = 2x + k\\\\
& \Longleftrightarrow \frac{y^{2}}{1+y^{2}} = \exp(2x+k)\\\\
& \Longleftrightarrow y^{2}(1 - \exp(2x+k)) = \exp(2x+k)\\\\
& \Longleftrightarrow y^{2} = \frac{\exp(2x+k)}{1-\exp(2x+k)}\\\\
& \Longleftrightarrow y = \pm\left(\frac{\exp(2x+k)}{1-\exp(2x+k)}\right)^{1/2}
\end{align*}
In order to determine $k$, apply the initial condition $y(0) = y_{0}$.
Hopefully this helps.
