= Solution
The first iterate is
$$
\boxed{y_1(x)=1+2x+\frac{x^3}{3}.}
$$
On the prescribed box, $F(x,y)=1+x^2+y^2$ is continuous, $|F|\leq102$, and $|F(x,y)-F(x,v)|\leq20|y-v|$. The initial value is distance nine from the nearer horizontal edge. Thus on $|x|\leq h=9/102=3/34$, the Picard map keeps functions within the box: its values differ from one by at most $102h=9$.
Successive differences satisfy, by <induction>,
$$
|y_{n+1}(x)-y_n(x)|\leq102\frac{20^n|x|^{n+1}}{(n+1)!}.
$$
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 $y'=1+x^2+y^2$, $y(0)=1$. The same local Lipschitz bound proves local <uniqueness>.
Direct differentiation gives $u'=\sec^2(x+\pi/4)=1+u^2$, $u(0)=1$, and $u\to+\infty$ as $x\uparrow\pi/4$. On the common positive existence interval put $w=y-u$. It satisfies $w'=(y+u)w+x^2$, $w(0)=0$, so the integrating-factor formula gives
$$
w(x)=e^{\int_0^x(y+u)\,ds}\int_0^x t^2e^{-\int_0^t(y+u)\,ds}\,dt\geq0.
$$
Thus $y\geq u$. Let $x_*>0$ be the maximal forward existence endpoint. The solution is strictly increasing because $y'\geq1$. If $x_*$ 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 $\pi/4$. Consequently
$$
\boxed{0<x_*\leq\frac\pi4,\qquad\lim_{x\uparrow x_*}y(x)=+\infty.}
$$
There is \b[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 $y$.
Back to article page
