There is an initial value problem:
$x' = (2 \sqrt{|x|} + x^2)(3 - t)$
$x(0) = 0$
Proof there is a solution going to infinity in finite time. Is there an instable, non-negative, global solution?
So I wanted to solve that by:
$x' = (2 \sqrt{|x|} + x^2)(3 - t) \iff \frac{dx}{dt} = (2 \sqrt{|x|} + x^2)(3 - t) \iff \frac{dx}{2 \sqrt{|x|} + x^2} = (3 - t) dt $
But it turns out it's not an option since the integral of $\frac{1}{2 \sqrt{|x|} + x^2}$ is something crazy. So there must be some other way to solve that by analyzing the initial equation.
If I had my general solution I would take "time" as a parameter $t \in [0, \infty)$. I would need to show that there exists a $t_0 \in [0, \infty )$ such that $x(t) \to \infty$ as $t \to t_0$.
I don't know how to do the other part of the exercise (with an instable, non-negative, global solution).
Edit 1 (Idea of @KBS): Changing variable $x = y^2$ I get:
$x = y^2$
$x' = 2y y'$
$x' = (2 \sqrt{|x|} + x^2)(3 - t) \iff 2y y' = (2 y + y^4)(3 - t) \iff y' = (1 + \frac{y^3}{2})(3 - t) \iff \frac{dy}{dt} = (1 + \frac{y^3}{2})(3 - t) \iff \frac{dy}{1 + \frac{y^3}{2}} = (3 - t)dt$
And again, I get some crazy integral of $\int \frac{1}{1 + \frac{y^3}{2}} dy$
For proving the existence of a finite escape time, consider the system
$$\dot{z}(t)=\left\{\begin{array}{rcl}3-t,&&\mathrm{if\ }t\in[0,t_1],\\ (3-t)(1+z(t)^2),&&\mathrm{if\ }t\in[t_1,3], \end{array}\right.$$ where $z(0)=0$ and $t_1=3-\sqrt{5}>0$.
I claim that $y(t)\ge z(t)$ for all $t$ (where I am considering that the functions may take $\infty$ as a value after possible escape times).
Note first that we have $$\dot{y}(t)-\dot{z}(t)=(3-t)y(t)^3/2\ge0$$ over $t\in[0,t_1]$ and therefore we have that $y(t)\ge z(t)$ over that same interval.
Evaluating $z(t)$ at $t=t_1$ yields $z(t_1)=2$ and we have that
$$\dot{y}(t)-\dot{z}(t)=(3-t)(y(t)^3/2-z(t)^2).$$
We have that $y(t_1)\ge 2$ and, therefore, $y(t_1)^3/2-z(t_1)^2\ge0$. This implies that $y(t)\ge z(t)$ also on that interval.
Solving for $z(t)$, we have that $z(t)=t^2/3-3t$ for all $t\in[0,t_1]$. For the second fragment, we note that the differential equation can be rewritten as
$$\dfrac{dz}{1+z^2}=(3-s)ds$$
where I am using the change of variables $s=t$ for technical reasons. Then we can integrate both sides as
$$\int_{z(t_1)}^{z(t)}\dfrac{dz}{1+z^2}=\int_{t_1}^t(3-s)ds$$
which is equivalent to
$$\left[\arctan(z)\right]_{z(t_1)}^{z(t)}=\left[-\dfrac{1}{2}s^2+3s\right]_{t_1}^{t}$$
and finally leads to
$$\arctan(z(t))=\arctan(2)-t^2/2+3t+t_1^2/2-3t_1,\ t\in[t_1,3]$$
where I have used $z(t_1)=2$.
We have a finite escape-time for $z$ if there is a $t_2>t_1$ such that the right hand side is equal to $\pi/2$. Solving numerically for that $t_2$ yields $t_2=0.9819$.
Since the original system is bounded from above by a system that has a finite escape-time, then the original system must have a finite escape-time which is less than $t_2$. Better approximations can be obtained by using tighter comparison systems.