= Solution
In position-velocity variables $(u,v)$, the scalar right-hand side is
$$
F(x,u,v)=\frac{v^2}{u}+2xv.
$$
Its only relevant singularity is $u=0$. The initial state $(x,u,v)=(0,1,1)$ is separated from it. For an explicit application of part (a), take $h=1$ and $k=1/2$ in the <L1 norm>. On the resulting <closed ball>, $1/2\leq u,v\leq3/2$, and $|x|\leq1$.
The <partial derivatives> obey
$$
|F_u|=\frac{v^2}{u^2}\leq9,\qquad
|F_v|=\left|\frac{2v}{u}+2x\right|\leq8.
$$
The ball is <convex>, so integrating these <partial derivatives> along a segment between two states gives the uniform <Lipschitz condition>
$$
|F(x,u,v)-F(x,\widetilde u,\widetilde v)|
\leq9\bigl(|u-\widetilde u|+|v-\widetilde v|\bigr).
$$
Continuity holds on the whole slab, and
$$
|F|\leq\frac{(3/2)^2}{1/2}+2\cdot1\cdot\frac32=\frac{15}2,
\qquad\|(v,F)\|_1\leq9.
$$
Thus $L=9$, $M\leq9$, and the choice $\boxed{\eta=1/20}$ satisfies the self-map and contraction bounds in part (a). This proves \b[a unique solution on $[-1/20,1/20]$], with $y$ bounded away from zero there.
There is also a useful exact representation. Where $y>0$, set $q=y'/y$, the <logarithmic derivative>. Differentiation cancels the quadratic term:
$$
q'=\frac{y''}{y}-\frac{(y')^2}{y^2}=2xq,\qquad q(0)=1.
$$
Solving this <first-order linear differential equation> gives $q=e^{x^2}$, and a second integration yields
$$
\boxed{y(x)=\exp\left(\int_0^x e^{t^2}\,dt\right).}
$$
This <function> is positive and smooth for every real $x$, satisfies both initial data, and verifies the original <ordinary differential equation> by direct differentiation. It therefore extends the unique local solution to the entire real line. The <logarithmic-derivative reduction of a second-order equation> explains why this apparently nonlinear example is explicitly integrable by <integration>.
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