Estimate blow-up time. How am I supposed to find the blow-up time of this ODE solution?
$$y'=e^x + y^2 \qquad y(0)=0$$
The fact that it blows up it's granted by the fact that $y' \geq y^2$ which solution explodes. But how to estimate the time of explosion?
I suppose I should use Gronwall or things like this, but don't actually know.
Thanks in advance.
 A: Since $\forall x>0 \;\;y'(x)>0$, there exists an inverse function $x(y)$ defined on $[0,+\infty)$. Its derivative is
$$\tag{1}
x'(y)=\frac1{e^{x(y)}+y^2}.
$$
Integrating both sides of the equation (1), we obtain
$$
x(y)=\int_0^{y}\frac{dy}{e^{x(y)}+y^2}.
$$
Since $\forall x>0\;\; e^x>1$,
$$
\int_0^{y}\frac{dy}{e^{x(y)}+y^2}<\int_0^{y}\frac{dy}{1+y^2}
=\arctan y\Big|_0^{y}= \arctan y.
$$
Finally, $x(y)<\arctan y$ implies $\forall y>0\;\;x(y)<\frac{\pi}2$. Hence, the blow-up time is less than $\frac{\pi}2$.
A: If we set $y(x)=e^{x/2}f(e^{x/2})$ and $e^{x/2}=t$ we are left with
$$ e^{-x/2}f(e^{x/2})+f'(e^{x/2}) = 2 +  2\,f(e^{x/2})^2 $$
$$ \frac{f(t)}{t}+f'(t) = 2 + 2 f(t)^2,\qquad f(1)=0 \tag{A}$$
and $f(t)$ is greater than the solution of 
$$ g(t)+g'(t) = 2+2\,g(t)^2,\qquad g(1)=0 \tag{B}$$
which is a separable DE. The blow-up time of $g(t)$ is given by
$$ 1+\frac{\pi}{\sqrt{15}}+\frac{2}{\sqrt{15}}\arctan\frac{1}{\sqrt{15}} $$
hence the lifetime of $y(x)$ is bounded by 
$$ 2\log\left(1+\frac{\pi}{\sqrt{15}}+\frac{2}{\sqrt{15}}\arctan\frac{1}{\sqrt{15}}\right)<\color{red}{\frac{4}{3}}. $$
It is interesting to point out that $(A)$ can be solved in terms of a ratio of linear combinations of Bessel-J and -Y functions. The actual lifetime is twice the logarithm of the solution of $\frac{Y_0(2t)}{Y_1(2)}=\frac{J_0(2t)}{J_1(2)}$ which is closest to $2$, i.e. approximately $\color{red}{1.27081}$.
