Continuing Solutions of $\dot{x} = \frac{t^2x^5}{1 + x^2 + x^4}$ to entire number line Show that solutions of
\begin{equation}
    \dot{x} = \frac{t^2x^5}{1 + x^2 + x^4}
\end{equation}
can be continued to the whole real line.
I know that this ODE is seperable as follows
\begin{equation}
\frac{1+x^2+x^4}{x^5} dx = t^2 dt
\end{equation}
Thus, giving the solution
\begin{equation}
    -\frac{1}{4x^4} - \frac{1}{2x^2} + \ln(|x|) + C = \frac13 t^3
\end{equation}
However, forom here, it is not clear to me how any solution $x(t)$ can be continued to the entire real number line.
 A: By comparing the degrees, the right side of the equation is linearly bounded in the $x$-direction,
this can be made quantitative like
$$
|\dot x|\le t^2 |x| \implies |x(t)|\le|x(0)|e^{|t|^3/3}.
$$
At first this only holds on the domain where the solution $x$ exists. But as
this means that there can be no divergence to infinity at finite times, the domain can be extended without obstacle, so the maximal domain is the full real line.
A: While Lutz's answer is the most succinct, the approach in the question is valid, albeit more work.
It is clear from Picard-Lindelöf that local solutions exist and are unique.
Note that $x(t) = 0$ for all $t$ is a solution. In particular, if $x$ is a solution, and $x(t) \neq 0$ for some $t$, then $x(t) \neq 0$ for all $t$, and by continuity we see that if $x$ is a solution defined on some interval $I$ then exactly one of the following three cases holds: (i) $x(t) = 0$ for all $t \in I$, (ii) $x(t) > 0$ for all $t \in I$ and (iii) $x(t) < 0$ for all $t \in I$.
Since $f(x,t) = t^2 { x^5 \over 1 + x^2 + x^4}$ is odd in $x$, we see that if $x$ is a solution, then so is $-x$. Hence we can focus on Case (ii). All that needs to be done is to show that if there is a solution $x(t)>0$ for $t \in I$ then it is defined on $\mathbb{R}$ (or rather, it can be extended to).
The function $\phi(x) = -{1 \over 4x^4}-{1 \over 2 x^2} + \log x$ is defined for $t>0$, smooth and $\phi'(t)>0$ for all $t>0$, hence has a smooth inverse.
Define $y(t) = \phi^{-1}(\phi(x(t_0)+{1 \over 3} (t^3 - t_0^3))$, note that $y(t_0) = x(t_0)$ and
$\phi'(y(t)) y'(t) = t^2$, or $y'(t) = f(y(t),t)$, hence $y(t)=x(t) $ for $t \in I$. Hence $y$ is the extension we are looking for.
