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

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
$$
ρ(t)=\frac{ρ_0}{\sqrt{e^{-2t}(ρ_0^2+1)-ρ_0^2}}
$$
