 Solution

The first iterate isOn the prescribed box, $F(x,y)=1+x^2+y^2$ is continuous, $|F|\le 102$, and $|F(x,y)-F(x,v)|\le 20|y-v|$. The initial value is distance nine from the nearer horizontal edge. Thus on $|x|\le h=9/102=3/34$, the Picard map keeps functions within the box: its values differ from one by at most $102h=9$.

Successive differences satisfy, by induction,The Weierstrass M-test proves uniform convergence on this local interval. The uniform limit is continuous, and the Lipschitz estimate allows passage to the limit inside the integral, giving the displayed integral equation. Differentiation gives $y^{\prime }=1+x^2+y^2$, $y(0)=1$. The same local Lipschitz bound proves local uniqueness.

Direct differentiation gives $u^{\prime }={\sec }^2(x+\pi /4)=1+u^2$, $u(0)=1$, and $u\to +\infty$ as $x\uparrow \pi /4$. On the common positive existence interval put $w=y-u$. It satisfies $w^{\prime }=(y+u)w+x^2$, $w(0)=0$, so the integrating-factor formula givesThus $y\ge u$. Let $x_{\ast }>0$ be the maximal forward existence endpoint. The solution is strictly increasing because $y^{\prime }\ge 1$. If $x_{\ast }$ were finite and $y$ remained bounded, its finite limit and the smooth right side would allow local continuation, contradicting maximality. Comparison with $u$ prevents existence through $\pi /4$. ConsequentlyThere is no contradiction with Picard's theorem, which gives local existence and uniqueness. Its box estimate gives only a short guaranteed interval, and the Lipschitz constant for $y^2$ is not globally bounded in $y$.