= Solution
Put $v=e^S$ and $p=S'$. The <Liouville-Green exponential ansatz> turns the normal-form equation into the <Riccati equation>
$$
p'+p^2
=\frac14\left(1+\frac2x-\frac1{x^2}\right).
$$
Write
$$
p\sim b_0+\frac{b_1}{x}+\frac{b_2}{x^2}+\cdots.
$$
Matching the constant, $x^{-1}$, and $x^{-2}$ coefficients gives
$$
b_0^2=\frac14,\qquad
2b_0b_1=\frac12,\qquad
b_1^2+2b_0b_2-b_1=-\frac14.
$$
For the recessive branch,
$$
b_0=-\frac12,\qquad b_1=-\frac12,\qquad b_2=1,
$$
so
$$
S_-(x)
=-\frac x2-\frac12\log x-\frac1x+O(x^{-2}).
$$
Since $y=e^{x/2}x^{-1/2}v$, this gives
$$
y_-(x)
=x^{-1}\exp\left(-\frac1x+O(x^{-2})\right)
=\frac1x-\frac1{x^2}+O(x^{-3}).
$$
This agrees with the first two terms of part (a),
$$
U(x)\sim\frac1x-\frac1{x^2}+\frac2{x^3}-\cdots.
$$
Continuing the Riccati recursion reproduces the factorial asymptotic series, so $U$ is proportional to this recessive branch.
For the dominant branch,
$$
b_0=\frac12,\qquad b_1=\frac12,\qquad b_2=0,
$$
and in fact
$$
S_+(x)=\frac x2+\frac12\log x+\text{constant}
$$
solves the phase equation exactly. It follows that
$$
v_+(x)\asymp e^{x/2}x^{1/2},
\qquad
\boxed{y_+(x)\asymp e^x}.
$$
Indeed, $e^x$ is an exact second solution of the original differential equation.
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