The equation $y'=1+y^4$ blows up in finite time Prove that the equation $$y'=1+y^4 \:, y(0)=0$$ has a maximal solution with a bounded domain. 
The existence and unicity of such a maximal solution is granted by Cauchy-Lipschitz theorem. 
I tried a proof by contradiction for the boundedness. One promising way would be to use the fundamental theorem of calculus and taking limits at infinity... 
Thanks for your suggestions. 
 A: Here is an indirect solution. If $y$ does not blow up in finite time, then the maximal solution is defined on all of $[0, \infty)$. We make the following two easy observations:


*

*Sine $y' \geq 1$, we get $y(t) \geq t$.

*Since $y' = 1 + y^{4} \geq 2y^{2}$, for any $0 < t_{0} < t_{1}$ we get
$$ \frac{y'}{y^{2}} \geq 2 \quad \Longrightarrow \quad \frac{1}{y(t_{0})} - \frac{1}{y(t_{1})} \geq 2(t_{1} - t_{0}). $$


Combining two observations, for any $0 < t_{0} < t_{1}$ we get
$$ \frac{1}{t_{0}} \geq \frac{1}{y(t_{0})} - \frac{1}{y(t_{1})} \geq 2 (t_{1} - t_{0}). $$
Then we get a contradiction by choosing large $t_{0}$ and $t_{1} = t_{0} + 1$.
A: Let $y(t)$ be the solution, so that we know that $y(0)=0$ and $y'(t)=1+y(t)^4$. Clearly the function is strictly increasing in its domain, so it is positive on the positive part of its domain.
We have $\frac{y'(t)}{1+y(t)^4}=1$. Integrating this from $0$ to $\tau$, we have $\int_0^\tau\frac{y'(t)}{1+y(t)^4}dt=\tau$, and the integral can be rewritten as $\int_0^{y(\tau)}\frac{d\xi}{1+\xi^4}$, and this can be computed, so we  can conclude that $$
\frac{-\tan ^{-1}\left(1-\sqrt{2} y(t)\right)+\tan
   ^{-1}\left(\sqrt{2} y(t)+1\right)+\tanh
   ^{-1}\left(\frac{\sqrt{2} y(t)}{y(t)^2+1}\right)}{2 \sqrt{2}}=\tau$$
for all $\tau$ in the positive part of the domain of $y$.
Now the function $$\frac{\tanh ^{-1}\left(\frac{\sqrt{2} t}{t^2+1}\right)-\tan
   ^{-1}\left(1-\sqrt{2} t\right)+\tan ^{-1}\left(\sqrt{2}
   t+1\right)}{2 \sqrt{2}}$$ is bounded. This means that the domain has to be bounded above. In fact, the upper limit of the domain has to be the limit value of this function, which is $\frac{\pi }{2 \sqrt{2}}$.
