Introduce the stretched coordinate
$$
\rho=\epsilon r,\qquad F(\rho)=f(r).
$$
The equation becomes
$$
F_{\rho\rho}+\frac{3}{2\rho}F_\rho+FF_\rho=0,
\qquad F\to1\quad(\rho\to\infty).
$$
Seek the three-term <outer expansion>
$$
F=1+\sqrt\epsilon F_1+\epsilon F_2+\cdots.
$$
At first order,
$$
\left(\rho^{3/2}e^\rho F_1'\right)'=0,
$$
so decay at infinity and matching give
$$
F_1=-\frac12E_{3/2}(\rho).
$$
At the next order,
$$
\left(\rho^{3/2}e^\rho F_2'\right)'
=\frac14E_{3/2}(\rho).
$$
Using the function $Y$ supplied in the question, the matched decaying solution is
$$
\boxed{
F(\rho)\sim
1-\frac{\sqrt\epsilon}{2}E_{3/2}(\rho)
+\frac{\epsilon}{4}
\left[-2\sqrt\pi\,E_{3/2}(\rho)+Y(\rho)\right]}.
$$
Here $Y$ is the stated particular solution normalized to decay at infinity.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 336 / 2 / 2 / Solution

{wiki}

An outer expansion describes the solution away from a thin or distant inner region and is matched to the inner expansion in their common limit.


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 Incoming links: Outer expansion