Intuitive explanation of why some autonomous differential equations go to infinity in finite time

The point is that $dy/dx = y^p$ is equivalent to $dx/dy = y^{-p}$, i.e. instead of thinking of $y$ as the dependent variable and $x$ as independent, do the reverse. If you think of $x$ as position and $y$ as time, the velocity is $y^{-p}$. If $p > 1$, this goes to $0$ fast enough that the change in $x$ as $y$ goes from some finite positive value to $\infty$ is finite. Now change point of view again and it says that as $x$ goes to some finite value, $y$ goes to $\infty$.

Your intuition that a solution to a DE like this should grow quickly but finitely makes a lot of sense. One justification for this intuition is to look at the estimation Euler's method would give: entirely finite and defined for the whole real line. To fix this inaccurate intuition, consider the following improvement of Euler’s method: instead of increasing x a constant amount each time, only increase x far enough to let y double. Since y doubles with each jump, $y^n$ Increases by $2^n$, so the ratio of the size of the horizontal jump from one jump to the next decreases by a factor of $\frac {2}{2^n}$. since n>1, this ratio is less than one. As a result the x-position converges, so y is doubling with out bound but x converges.

There is a nice discussion of this problem here p. 423, where the authors show by example that what one expects is not necessarily what happens. Below is a sketch of their proof of a criterion which can be used to tell whether a solution will blow up in finite time. Namely, we have a

Theorem:

if $y'=f(y);\ y(0)=y_0;\ f(y)>0$ for all $y>y_0,$ then $y$ blows up at time $t_1$ if and only if $\int^{\infty}_{y_0}\frac{1}{f(y)}dy=t_1.$

For the proof, note that $\int^{y(t)}_{y_0}\frac{1}{f(u)}du=t$ whenever the integral is defined. Therefore, if $y$ satisfies $\underset{t\to t_1^-}\lim y(t)=\infty$ then $\underset{t\to t_1^-}\lim \int^{y(t)}_{y_0}\frac{1}{f(u)}du=\underset{t\to t_1^-}\lim t=t_1.$

On the other hand, if the integral converges to $t_1,$ then $t=\int^{y(t)}_{y_0}\frac{1}{f(u)}du<\int^{\infty}_{y_0}\frac{1}{f(u)}du=t_1$ so $t$ is bounded by $t_1$. Conclude by observing that

$\underset{t\to t_1^-}\lim \int^{y(t)}_{y_0}\frac{1}{f(u)}du=\underset{t\to t_1^-}\lim t=t_1=\int^{\infty}_{y_0}\frac{1}{f(u)}du$ so $\underset{t\to t_1^-}\lim y(t)=\infty.$