Consider the 3rd order ODE


$$\dddot{x}+A\ddot{x}-\dot{x}^{2}+x=0$$ where $\dot{x}\equiv \frac{dx}{dt},\ddot{x}\equiv \frac{d^{2}x}{dt^{2}}, etc$. $A$ is a constant.


If we multiply this equation by $\ddot{x}$ and integrate we can convert it into


$$\frac{1}{2}\ddot{x}^{2}-\frac{1}{3}\dot{x}^{3}+x\dot{x}=C+\int\_{0}^{t}(\dot{x}^{2}-A\ddot{x}^{2})ds$$


where $C$ is a constant and $t>0$. If $A\leq 0$ the integral diverges as $t\rightarrow \infty$ and at least one of $\dot{x},\ddot{x},x$ must also diverge. 


By numerically integrating it seems that for $0<A<1.98$ the solution also diverges very quickly. 


I have been wondering whether there is a way of proving this, i.e. that for all $A<A\*, A\*=1.97...$ (or some other constant $A\*>0$) the motion is unbounded (with the exception of $x(t)=0$). 



Is it possible to prove unboundedness of 3rd order ODE?
Actually, no matter what $A$ is, there will be nonzero solutions that will converge to zero, so you can't prove unboundedness, even though it may be true that the 'generic' solution is unbounded. Here is why: First, convert the system into a first order system in $\mathbb{R}^3$ by setting $x = x\_0$, $\dot x = x\_1$, and $\ddot x = x\_2$. Then the equation becomes the first order system
$$
\begin{aligned}
\dot x\_0 &= x\_1\,,\\
\dot x\_1 &= x\_2\,,\\
\dot x\_2 &= -x\_0 + {x\_1}^2 - A\,x\_2\,.
\end{aligned}
$$
This vector field (i.e., ODE) has one singular point, $(x\_0,x\_1,x\_2) = (0,0,0)$,
and its linearization at this point has the matrix
$$
\begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
-1 & 0 & -A
\end{pmatrix}
$$
The eigenvalues of this matrix are the roots of $\lambda^3+A\lambda^2 + 1 = 0$, and there is always at least one negative real root. Hence the stable manifold of $(0,0,0)$ has dimension at least $1$, so there will always be a nonzero solution that decays to zero (in infinite time).


