Behavior for small and large times of $\ddot{y}(t)=y^2(t), y(0)=y_0>0, \dot{y}(0)=0$. I am trying to understand the behavior of solutions for small and large times to the differential equation
$$\ddot{y}(t)=3y(t)^2, y(0)=y_0>0, \dot{y}(0)=0.$$
For small times we have that
$$\ddot{y}(t)=3y_0^2+6y_0(y-y_0)+O(y-y_0)^2$$
so we expect the solution to behave like a quadratic plus a positive  exponential. I'm not sure if the solution is defined for all times $t\ge 0$ or whether it blows up at some finite time $t$. We can integrate the equation of motion to get
$$\frac{1}{2}\dot{y}(t)^2=y(t)^3-y_0^3,$$
hence, assuming that $\dot{y}(t)\ge 0$ for all $t\ge 0$,
$$\int_{y_0}^{y(t)}\frac{dy}{\sqrt{y^3-y_0^3}}=\sqrt{2}t$$
though I'm not sure how to extract information out of this integral, which is not easily integrated. I thought about using the substitution $y(t)=y_0 e^{S(t)}$, $S(0)=0$ to obtain
$$\ddot{S}(t)+\dot{S}(t)^2=3y_0 e^{S(t)}$$
If we assume that the $\dot{S}^2$ dominates over $\ddot{S}$, we end up with
$$\dot{S}=\sqrt{3y_0}e^{S(t)},$$
i.e.
$$1-e^{-S(t)}=\sqrt{3y_0}t. $$
Thus,
$$S(t)=\log\left(\frac{1}{1-\sqrt{3y_0}t}\right)$$
giving us blow up at finite time, however $\ddot{S}\sim \dot{S}^2%$, violating our assumption so this approximation doesn't work. I would appreciate any ideas.
If we have that $y_0>1$, then $\ddot{y}(t)\ge 3y(t)$ and thus we should have $y(t)\ge y_0\cosh(\sqrt{3} t)$ and so we have super-exponential growth.
 A: Starting with the integrated equation of motion
$$\tag{1}
\frac{1}{2}\dot{y}(t)^2=y(t)^3-y_0^3,
$$
Rescale $t$ such that $t=s\sqrt{2}$, then we have
$$\tag{2}
\dot{y}(s)^2=4y(s)^3-4y_0^3,
$$
Looking at the DLMF page and the Mathworld page for Weierstrass elliptic functions, we see that $\wp(s)$ is a solution to (2), with $g_2=0$ and $g_3=4y_0^3$. Of interest to us are that the solutions are periodic, and have poles at finite times.

Let us take a different course and argue that the solutions have poles at finite times using the ansatz
$$\tag{3}
y(t)\sim A(t-t_0)^b \quad; \quad t\to t_0
$$
Substituting (3) into the DE $y''=3y^2$ we find that $A=2$ and $b=-2$, so the poles are second order. I'll write $z=t-t_0$ so poles occur at $z=0$. Using (3) we try an expansion of the form
$$\tag{4}
y(z) \sim 2z^{-2}+f(z) \quad ;\quad z \to 0
$$
Where $f\ll z^{-2}$ as $z \to 0$. Substituting (4) into the differential equation $\ddot{y}=3y^2$, we find that $f$ should satisfy
$$\tag{5}
f''=\frac{12 f}{z^2} + f^2 \\
\therefore z^2 f'' \sim 12 f \quad ;\quad z \to 0
$$
The asymptotic DE for $f$ is an Euler equation and may be solved directly. The solution is
$$\tag{6}
f(z) \sim C_1 z^{-3} +C_2 z^4
$$
We require $C_1=0$ because $f \ll z^{-2}$ as $z \to 0$. Since $C_2 z^4 \to 0$ as $z \to  0$ we have found the leading order term $\propto z^{-2}$. Substituting $y=2z^{-2}+C_2 z^4$ into (1) we find that $C_2=-y_0^3/10$. Conclusion: the pole structure is
$$\tag{7}
y(z) \sim \frac{2}{z^2} - \frac{y_0^3 z^4}{10} \quad ;\quad z \to0
$$

We can also estimate the time until blow up $t_0$ following the comment by @Hans Lundmark. Starting with the integral
$$\tag{8}
t\sqrt{2} = \frac{1}{\sqrt{y_0}}\int\limits_1^\eta \frac{du}{\sqrt{u^3-1}}
$$
I've rescaled variables in your integral $y=y_0 u$ and $\eta(t)=y(t)/y_0$. Write $\int\limits_1^\eta = \int\limits_1^\infty - \int\limits_\eta^\infty$. The first integral is a constant $K=\sqrt{4\pi}\frac{\Gamma(7/6)}{\Gamma(2/3)}\approx 2.4$. We need to study
$$\tag{9}
I(\eta)=\int\limits_\eta^\infty \frac{du}{\sqrt{u^3-1}}
$$
as $\eta\to\infty$. Integrating (9) by parts, we find
$$\tag{10}
I(\eta)\sim -\frac{1}{\sqrt{\eta}} \quad,\quad \eta\to\infty
$$
Using (8)
$$\tag{11}
t\sim \frac{1}{\sqrt{2y_0}}\left(K-\frac{1}{\sqrt{\eta}}\right) \quad,\quad \eta\to\infty
$$
Since $I\to 0$ as $\eta\to\infty$ we have
$$\tag{12}
t_0\approx \sqrt{\frac{2\pi}{y_0}}\frac{\Gamma(7/6)}{\Gamma(2/3)}
$$
