Does the solution of differential equations exist for all times? I was trying to analyse the differential equation $\frac{dx}{dt} = -x + x^{3} , x(0) = x_{0}$,After solving I got the solution as $- \ln(|x|) + \frac{1}{2}\ln|x^2 - 1 | = t + (-\ln|x_{0}| + \frac{1}{2}\ln|x_{0}^2 - 1|)$,now by doing computational analysis I see that solution $x(t) $ diverges to infinity when the initial condition is $>1$ or $<1$ and the solution decreases and attains a steady stste when $x_0{ \in [-1,1]}$. 
But how do I prove analytically that when $x_{0} > 1$,$x_{0}<-1$,the solution diverges to infinity!
and attains a steady state when $-1<x(0)<1$?
And I think for negative time $t$,the solution exist for finitely many $t$,how can i intuitively think of this?
Any help is great!
 A: So you have
\begin{align}
\ln\left|\frac{x^2-1}{x^2}\right|=2t+\ln\left|\frac{x^2_0-1}{x^2_0}\right|
\end{align}
then by exponentiation we have that
\begin{align}
\frac{|x^2-1|}{x^2} = \frac{|x^2_0-1|}{x^2_0}e^{2t}.
\end{align}
Suppose $x(t)$ is a solution such that $x_0>1$, then by the uniqueness theorem we see that $x(t)>1$ for all $t$ since $x(t) =1$ is a solution. In particular, we get that
\begin{align}
\frac{x^2-1}{x^2} = \frac{x_0^2-1}{x_0^2}e^{2t}\ \implies x(t) = \frac{x_0}{\sqrt{x_0^2-(x_0^2-1)e^{2t}}}
\end{align}
which blows up in finite positive time. 
In the case $x(t)$ is a solution such that $-1<x_0<1$, then we see that $-1<x(t)<1$. In particular, we have
\begin{align}
\frac{1-x^2}{x^2} = \frac{1-x_0^2}{x_0^2}e^{2t}  \ \implies x(t) = \frac{x_0}{\sqrt{x_0^2+(1-x_0^2)e^{2t}}} 
\end{align}
which is defined for all time where $\lim_{t\rightarrow -\infty} x(t) = \operatorname{sgn}x_0$ and $\lim_{t\rightarrow \infty}x(t) = 0$. Also, note that $0$ is a steady state solution, which mean either $0<x(t)<1$ or $-1<x(t)<0$. 
Finally, the analysis for $x_0<-1$ is similar.
A: You already have the solution. Just think, what happens when $x_0 \in \{\pm 1 \}$?. Also, remember that $\ln|x| \to -\infty$ when $x \to 0$. 
A: You can see this behavior by analyzing $\frac{dx}{dt}$ as a function of $x$. If $x < -1$ or $x > 1$, then the function $x$ is either decreasing or increasing without bound. If $-1\leq x \leq 1$, then $x$ cannot diverge since it will only increase or decrease with bounded derivative. So we simply must restrict $x$ from being outside of $[-1,1]$ if we don't want it to diverge. 
