f \sim x^{- 1}

near zero. The terms

x

and

ϵ f

become comparable when

x^{2} = O (ϵ)

, so

x = ϵ X, f = ϵ^{- 1/2} F (X) .

x f + \frac{ϵ}{2} f^{2} = x^{2} + 1 + 2 ϵ,

where the constant follows from

f (1) = 2

. Thus

XF + \frac{1}{2} F^{2} = 1 + ϵ (X^{2} + 2) .

F = F_{0} + ϵ F_{1} + \dots

and choosing the branch matching the positive outer solution gives

F_{0} = X^{2} + 2 - X, F_{1} = X^{2} + 2 .

Hence

f (x) \sim ϵ^{- 1/2} [X^{2} + 2 - X] + ϵ^{1/2} X^{2} + 2, X = \frac{x}{ϵ} .

Its large-

X

expansion matches the supplied outer series.

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