= Solution
For two solutions $y_1,y_2$, define the <Wronskian>
$$
W=y_1y_2'-y_1'y_2.
$$
Differentiate and substitute the <second-order linear differential equation>:
$$
W'=y_1y_2''-y_1''y_2=-p(y_1y_2'-y_1'y_2)=-pW.
$$
Integration proves the <Abel identity>, including the identically zero case:
$$
\boxed{W(x)=W(x_0)\exp\left[-\int_{x_0}^xp(\xi)\,d\xi\right].}
$$
For $p=-2/x$, this is $W=Cx^2$ on either interval excluding zero, since the integral uses $\log|x|$. The coefficient singularity at zero must not be ignored.
Multiply the particular equation by $x^2$ and substitute the <Frobenius solution> $y=x^\lambda\sum_{n\ge0}a_nx^n$. With $a_{-1}=a_{-2}=0$, equating coefficients gives
$$
\boxed{(n+\lambda)(n+\lambda-3)a_n=a_{n-2}.}
$$
The leading coefficient is nonzero, so the <indicial equation> is $\lambda(\lambda-3)=0$. Thus only $\lambda=0,3$ are possible, and the recurrences below construct both.
For $\lambda=0$, $a_1=0$, $a_2=-a_0/2$, and the $n=3$ recurrence reads $0=a_1=0$, leaving $a_3$ free. For all later indices,
$$
\boxed{a_{2k}=a_0\frac{1-2k}{(2k)!}\quad(k\ge0),\qquad
a_{2k+3}=\frac{6a_3(k+1)}{(2k+3)!}\quad(k\ge0),\qquad a_1=0.}
$$
This is an <undetermined coefficient at Frobenius resonance>; choosing $a_3=0$ selects a convenient even basis solution but is not forced by the differential equation.
For $\lambda=3$, all odd-indexed coefficients vanish and
$$
\boxed{a_{2k}=\frac{6a_0(k+1)}{(2k+3)!},\qquad a_{2k+1}=0\quad(k\ge0).}
$$
The factorial denominators show convergence for every finite $x$. Expanding the given <hyperbolic functions> directly yields
$$
\cosh x-x\sinh x=\sum_{k\ge0}\frac{1-2k}{(2k)!}x^{2k},
$$
$$
\sinh x-x\cosh x=-\sum_{k\ge0}\frac{2k+2}{(2k+3)!}x^{2k+3}.
$$
Consequently the exponent-zero family is $a_0(\cosh x-x\sinh x)-3a_3(\sinh x-x\cosh x)$, and the normalized exponent-three solution is $-3a_0(\sinh x-x\cosh x)$. This verifies the coefficient formulas and supplies two independent basis functions.
Finally $y_1'=-x\cosh x$ and $y_2'=-x\sinh x$, so
$$
W=(\cosh x-x\sinh x)(-x\sinh x)-(-x\cosh x)(\sinh x-x\cosh x)
=\boxed{-x^2}.
$$
This agrees with the Abel calculation. Its vanishing at zero does not contradict <independence> on regular intervals: the original equation is singular there.
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