The outer expansion behaves as $f\sim x^{-1}$ near zero. The terms $x$ and $\epsilon f$ become comparable when $x^2=O(\epsilon)$, so
$$
\boxed{x=\sqrt\epsilon X,
\qquad f=\epsilon^{-1/2}F(X).}
$$
The equation has the exact first integral
$$
xf+\frac\epsilon2f^2=x^2+1+2\epsilon,
$$
where the constant follows from $f(1)=2$. Thus
$$
XF+\frac12F^2=1+\epsilon(X^2+2).
$$
Writing $F=F_0+\epsilon F_1+\cdots$ and choosing the branch matching the positive outer solution gives
$$
F_0=\sqrt{X^2+2}-X,
\qquad
F_1=\sqrt{X^2+2}.
$$
Hence
$$
\boxed{f(x)\sim\epsilon^{-1/2}[\sqrt{X^2+2}-X]
+\epsilon^{1/2}\sqrt{X^2+2},
\qquad X=\frac x{\sqrt\epsilon}.}
$$
Its large-$X$ expansion matches the supplied outer series.


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 336 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-336/3/b/solution