= Solution
Seek an order-one <outer expansion> as $\epsilon\to0^+$:
$$
y_{\rm out}(x)=y_0(x)+\epsilon y_1(x)+O(\epsilon^2).
$$
At leading order, $0=y_0-x^2$, so $y_0=x^2$. At order $\epsilon$,
$$
y_0''=y_1+2+y_0^2.
$$
Since $y_0''=2$, this gives $y_1=-x^4$. Therefore
$$
\boxed{y_{\rm out}(x)=x^2-\epsilon x^4+O(\epsilon^2).}
$$
This is an approximation away from any <boundary layer>; the small coefficient of the highest <derivative> means the reduced equation cannot generally impose both endpoint conditions.
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