# Showing that all solutions to the ODE $x''=4x^3$ cease to exist after a finite time

For fixed C, once x is large enough y is bounded from below by $x^2$, so the time it takes for $x$ to go from $n$ to $n+1$ is at most $n^{-2}$. This is summable over $n$, therefore there is a time $T$ such that $x\to\infty$ as $t\to T^-$. The same if they are negative.

Now you have to show that the solution eventually leaves a finite region so you can apply this observation which requires x large enough. If they go in the negative direction a similar argument works.

If $C\ne 0$ you can show that $\|(x',y')\|$ is bounded from below on the entire curve so it has to go away.

Also, notice that not all solutions cease to exist after a finite time. Taking $C=0$, you have $y=2x^2$ for $x<0$ and $y=-2x^2$ for $x>0$ which approximate the fixed point at $t\to\infty$. More explicitly, $x=\pm \sqrt 2 (t-t_0)^{-1} $ are solutions that are defined for all positive times.

Since $$y=\frac{dx}{dt}\\ t=\int_{x(0)}^\infty \frac{dx}y\\ =\int_{x(0)}^\infty \frac{dx}{\sqrt{2C+2x^4}}$$ and you need to show this is finite.