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
 A: 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.
A: In your original path, you should, as you found out, apply some integration constant to indefinite integrals or compute all in definite integrals. Then you should arrive at
$$
(ρ^{-2})'=-2ρ^{-3}ρ'=-2(ρ^{-2}+1)\\
\log(ρ^{-2}+1)-\log(ρ_0^{-2}+1)=-2t
\\
\implies
(ρ^{-2}+1)=e^{-2t}(ρ_0^{-2}+1)
$$
which then can be transformed into the solution formula
$$
ρ(t)=\frac{ρ_0}{\sqrt{e^{-2t}(ρ_0^2+1)-ρ_0^2}}
$$
A: Thank you to all of you! I managed to get the solution that they asked me
\begin{align}
−\frac{1}{2}\frac{du}{dt} &= u+1 \\
\iff -\frac{1}{2} \int_{u_{0}}^{u}{\frac{1}{u+1}} &=\int dt \\
\iff \ln|u+1|-\ln|u_{0}+1| &= -2t \\
\iff u+1 &= e^{-2t}(u_{0}+1)
\end{align}
Thus, undoing $u:=\rho^2$, we obtain
$$\frac{1}{\rho^2}=e^{-2t} \left(\frac{1}{\rho_{0}^2}+1 \right)-1 \iff \rho=\frac{1}{\sqrt{e^{-2t}-1+\color{green}{e^{-2t}\frac{1}{\rho_{0}^2}}}}$$
Again, many thanks!
