Every solution of $x'' + x + x^3 = 0$ is set on $\mathbb R$. What happens with $x'' + x' + x + x^3 = 0$? Exercise :
Show that every solution of $x'' + x + x^3 = 0$ are set on $\mathbb R$. What happens with the solutions of $x'' + x' + x + x^3 =0 $ ?
Attempt :
It is : 
$$ x'' + x + x^3 = 0 \Leftrightarrow x'' = -(x+x^3) \Leftrightarrow x'' \cdot x' = -(x+x^3)x' \Rightarrow \int x'' \cdot x'dt = \int [ -(x+x^3)x' ]dt \Leftrightarrow \frac{1}{2}x'^2 + c_2 = -\frac{1}{4}x^2(x^2 + 2)  + c_1 \Leftrightarrow x'^2 = C - 2x^2 - x^4$$
So we conclude that : 
$$x'^2 = C - 2x^2 - x^4$$
and we can see that the right-handed side of this differential equation becomes negative for any $C$ when $x$ becomes large enough (due to $x^2$ and $x^4$), which means that any solution associated to $C$ gets "deleted" before it reaches such a point. Which leads us to the conclusion that the only global solution is $x' = 0$ . But how can I derive from that, that all the solutions of that equation are set on $\mathbb R$ ? Also, I cannot seem to grasp how to work on the second given equation. Any help would be appreciated !
 A: These equations are known as Duffing equations, which describes the oscillation of a non-linear... oscillator, of course; the second equation correspond to the damped case.
Your result do not show that the only solution is $x'=0$, but that the solution is bounded by $\pm(\sqrt{C}-1)$. And it makes sense, since this equation describes an oscillator. If $|x|<(\sqrt{C}-1)$ you ensure that your solution is real and bounded.
For the damped case, i.e., with the $x'$ term, we have
\begin{equation}
x'(x''+x'+x+x^3)=0
\end{equation}
\begin{equation}
\frac{d}{dt}\left( \frac{1}{2}(x')^2+\frac{x^2}{2}+\frac{x^4}{4}\right)=-(x')^2.
\end{equation}
The quantity between parenthesis corresponds to the total energy $E$ of the oscillator (it is the sum of the kinetic energy and the potential energy). In the first equation, you found that this total energy is $E=2C$, i.e., the energy remains constant. In the second equation,
\begin{equation}
E' = -(x')^2\leq0.
\end{equation}
The derivative of the total energy is negative, then our oscillator is loosing energy. It corresponds to the effect of the damping. From the definition of $E$, it must be positive (since it is the sum of some positive stuff). If its derivative is negative, we must expect that
\begin{equation}
\lim_{t\to \infty} (x')^2 + x^2+\frac{x^4}{2} = 0.
\end{equation}
Or, in words, the solution $x$ for the second equation is not only bounded, but also tends (oscillatorily) to $0$.
Sorry for the lack of mathematical rigour, but I'm used to see these equations from an applied point of view.
(The analysis of the second equation was based on https://en.wikipedia.org/wiki/Duffing_equation#Damped_oscillator).
A: As you showed, $x'^2 + 2 x^2 + x^4$ is constant for solutions of this differential equation.  Thus the curves $y^2 + 2 x^2 + x^4 = constant$ are trajectories, and they are all bounded.  But the only way for a solution to stop existing is for $x$ and/or $x'$ to be unbounded as $t$ approaches some finite value.  
