I'm stuck with this question, can anyone help please?
Let $\phi(x)$ be the solution of the IVP
$y'(x)=1+y^4(x), x>0, y(0)=0.$
By considering an inequality $1+y^4 \geq \alpha+ \beta y^2$ with suitable $\alpha$ and $\beta$ show that $\phi(x) \to \infty$ as $x\to x_0$ for some $x_0$ such that
$0<x_0\leq \frac{1}{4}3^{3/4} \pi$.
A first, more simple estimate, has $y'\ge 1$ and thus $y(x)\ge x $ for $x\in[0,1]$.
Then $y'\ge y^4$ gives for $x\ge 1$ with $y(1)\ge 1$ $$ (y^{-3})'=-3y^{-4}y'\le -3\implies y(x)^{-3}\le y(1)^{-3}-3(x-1)\le 4-3x \\~\\ \implies y(x)\ge\frac1{\sqrt[3]{4-3x}} $$ so that $x_0\le\frac43$.
Seeing that $1.3333...<1.79032...$ this gives an even stricter bound than the proposed method.
Computing the exact solution, for instance using partial fraction decomposition after separation based on the factorization $$ 1+y^4=(1+y^2)^2-2y^2=(1+\sqrt2y+y^2)(1-\sqrt2y+y^2) $$ gives the implicit solution formula
\begin{multline} x = \frac1{4\sqrt2}\Bigl(-\log\bigl(y(x)^2 - \sqrt2 y(x) + 1\bigr) + \log\bigl(y(x)^2 + \sqrt2 y(x) + 1\bigr)\\ - 2 \arctan\bigl(1 - \sqrt2 y(x)\bigr) + 2 \arctan\bigl(1 + \sqrt2 y(x)\bigr)\Bigr) \end{multline}
which in inserting the limit $y\to\infty$ gives the exact value $x_0=\frac{\pi}{2\sqrt2}=1.11072...$