= Solution
{c}
The <function> $y_1=x$ solves the <linear ordinary differential equation>. Apply <reduction of order> by writing $y=xv(x)$. Substituting $y'=v+xv'$ and $y''=2v'+xv''$ and cancelling the terms in $v$ gives
$$
x^2(1-x)v''+(2x-3x^2)v'=0.
$$
For $w=v'$, this becomes a first-order <linear ordinary differential equation>:
$$
w'+\left(\frac2x+\frac1{x-1}\right)w=0,\qquad w=\frac{C}{x^2(x-1)}.
$$
A <partial fraction decomposition> provides the required <antiderivative>:
$$
\frac1{x^2(x-1)}=-\frac1x-\frac1{x^2}+\frac1{x-1},\qquad v=C_1+C_2\left(\log\frac{x-1}{x}+\frac1x\right).
$$
Therefore
$$
\boxed{y(x)=C_1x+C_2\left(1+x\log\frac{x-1}{x}\right),\qquad x>1.}
$$
The <logarithm> has positive argument on the required <interval>. The <Wronskian> of $x$ and $1+x\log((x-1)/x)$ is $1/(x-1)\ne0$, proving their <linear independence> and hence completeness of this <general solution>.
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