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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.

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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 $\color{red}{1.27081}$.

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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 $\frac{\pi}2$.

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  • $\begingroup$ Simple and nice and $\to +1$ $\endgroup$ – Claude Leibovici Jun 2 at 13:27

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