blowup in finite time of $\frac{dx}{dt}=1+x^{10}$ I feel like this is extremely silly question. It is a question from Strogatz' Nonlinear Dynamics and Chaos
I want to show that $\frac{dx}{dt}=1+x^{10}$ blows up in finite based on the fact that  $\frac{dx}{dt}=1+x^2$ blows up in finite time.
Well clearly, $1+x^2 < 1+x^{10}$ but then I realized that this is only true for $x > 1$ (in absolute value)
So what if $x<1$?
Is there a nice and elegant way to do this problem without explicitly solving for a solution?
 A: You can calculate that $x$ gets to say $1.1$ in finite time--because $dx/dt \ge 1$ for all $x$, and then go from there.
A: I am working on Strogatz as well and stumbled upon this question. I fleshed out the solution proposed by Michael E2 in the second comment on the question, figuring someone following in my footsteps would be curious to see it:
Let $f(t)$ be the solution to the differential equation $\dot{x}=1+x^{10}$. Because $1+x^{10} > 0$ for all $x$, we know that $f'(t)$ is strictly positive and $f(t)$ is strictly monotone increasing and therefore invertible. Thus, we can calculate the time to reach infinity given we start at a value $x=a$ as follows:
$$ f^{-1}(\infty)-f^{-1}(a) = f^{-1}(x)\Big|_a^\infty $$
Now, because $f$ is the solution to a differential equation, we know that it is differentiable on the relevant domain and the above discussion implies that it is never zero. Therefore, we can use the well known result about derivatives of inverse functions to observe that:
\begin{align*}
    f^{-1}(x)\Big|_a^\infty &= \int_a^\infty \left( f^{-1}(x)\right)' \ dx\\
    &= \int_a^\infty \frac{1}{f'(f^{-1}(x))} \ dx \\
    &= \int_a^\infty \frac{1}{1+f(f^{-1}(x))^{10}} \ dx \tag{ \(f'(t)=1+f(t)^{10}\)  }\\
    &= \int_a^\infty \frac{1}{1+x^{10}} \ dx \\
\end{align*}
Let $a \geq 1$ then we have:
$$\int_a^\infty \frac{1}{1+x^{10}} \leq \int_a^\infty \frac{1}{1+x^{2}}< \infty $$
