Initial value problem blowing up in finite time Lets consider the IVP $\dot x = x^{3} - e^{t^{2}}x^{2}$ with $(t,x) \in (0,\infty) \times \mathbb{R}$ and $x(0)=\xi$. I want to prove that when $\xi \in [0,1]$, then $\lim_{t\rightarrow \infty} x(t)=0$ and the solution is defined for all $t\geq 0$ and when $\xi\geq K$, with $K$ sufficiently big, $x(t)$ blows up in finite time. 
My main problem is that this is my first encounter with ODEs and is getting really hard for me to figure out the techniques for problem solving. Any help would be really appreciated.  
 A: In tackling such questions, it is quite useful to define sub- and supersolutions. Given an IVP $\dot x = f(x,t)$, $x(0) = \xi$, $(x,t) \in \mathbb{R}\times[0,\infty),$ a subsolution is a differentiable function $y$ with $\dot y < f(y,t)$ and $y(0) \leq \xi$. Then for all $t$, $y(t) < x(t)$. For a proof and examples, see Ordinary Differential Equations and Dynamical Systems (pdf), around page $24$. If you reverse all inequalities, $y$ is called a supersolution. The other thing to keep in mind is that solution curves can't cross each other by the Picard-Lindelöf uniqueness theorem.
Now, since $y = c$ is a supersolution for $c \in (0,1)$, and $x = 0$ is a solution, hence for $\xi \in (0,1)$, we know that $x(t) \in (0,1)$ for all $t$. This in turn implies $\dot x < 0$; therefore $a := \lim_{t \rightarrow 0} x(t)$ exists. Supposing $a \neq 0$, we quickly arrive at a contradiction: Supposing that for all $t$, $x(t) > a- \epsilon$, we have $\dot x(t) < -1$ for all $t > t_0$ for some large $t_0$, implying $x(t_0+2 \epsilon) < a-\epsilon$. Therefore, $\lim_{t \rightarrow 0} x(t) = 0$ for $\xi \in [0,1)$, and the claim for $\xi = 1$ follows because the flow is continuous since the ODE is Lipschitz-continuous.
The other claim can be proven by considering $y(t) = \frac{1}{\epsilon-t}$.
We have 
$$\dot y(t) = \frac{1}{(\epsilon-t)^2} < \frac{1}{(\epsilon-t)^2}\left(\frac{1}{(\epsilon-t)^2} - e^{t^2}\right) = f(y,t),$$
if we choose $\epsilon$ small enough that $\frac{1}{(\epsilon-t)^2} > e^{t^2}$ for all positive $t$. This is a subsolution blowing up in finite time, hence any solution with $\xi > y(0) = \frac{1}{\epsilon^2} =: K$ will blow up in finite time as well.
