In which cases is the solution for $\frac{dx}{dt}=x-x^3+1$ an increasing function? I need to find the cases where the solution to the unidimensional ODE: $$\frac{dx}{dt}=x-x^3+1$$ is an increasing function.
My attempt was solving it like a separable ODE and then looking at how it behaves. But I couldn't do the integral, so I looked for it and it gives weird functions. Then I tried to see it like a Bernoulli equation (ignoring the $+1$) which gave me a cleaner result and an idea of the behavior of the solution but I don't know if my approach is correct. Could you please tell me if my approach is good or wether I can do something else? Thanks for the help.
 A: Let $p(x)=x-x^3+1$. As mentioned in the comments on the question, this real polynomial has a single real root that we'll call $x_c$. $p$ is negative for $x \lt x_c$ and positive for $x \gt x_x$.
Also notice that $p$ being a polynomial is locally Lipschitz for any $x \in \mathbb R$. According to Picard-Lindelöf theorem, the Initial Value Problem - IVP $x^\prime(t) = p(x(t))$, $x(t_0)=x_0$ has a unique maximal solution.
Now coming back to the question, which is essentially: for which $x_0$ is the maximal solution of the IVP increasing, notice that we must have $x_0 \lt x_c$ as otherwise $x^\prime(t_o) \lt 0$.
We have however to answer the question: is $x_0 \lt x_c$ a sufficient condition? The answer is positive. Let's see why. Suppose that the solution of the IVP with the initial condition $x(t_0) \lt x_c$ reaches the value $x_c$ in a finite time $t_c$. We would have at $t_c$, $x^\prime(t_c) = 0$ and the constant map equal to $x_c$ would be a solution. A contradiction with the uniqueness statement of Picard-Lindelöf theorem.
Conclusion: $x_0 \lt x_c$ is a necessary and sufficient condition for the solution to be increasing. In that case, it would be possible to prove that the solution is defined for $t \gt t_0$ as a solution that is only defined for finite times, blows up to infinite.
