= Solution
For $y_1=(1-x)^{-1}$, the <derivatives> are $y_1'=(1-x)^{-2}$ and $y_1''=2(1-x)^{-3}$. Consequently the left-hand side of the <ordinary differential equation> is
$$
\frac{2}{1-x}-\frac{4}{1-x}+\frac{2}{1-x}=0
$$
where $x\ne1$.
Use <reduction of order>, writing $y=u/(1-x)$. A direct application of the <product rule> gives
$$
(1-x)^2y''-4(1-x)y'+2y=(1-x)u''-2u'.
$$
For $v=u'$ the resulting <first-order ordinary differential equation> is $v'/v=2/(1-x)$ whenever $v\ne0$. <Integration> yields $v=C(1-x)^{-2}$ and $u=A+B/(1-x)$; the case $v=0$ is already included. Hence
$$
\boxed{y(x)=\frac{A}{1-x}+\frac{B}{(1-x)^2}}
$$
on either interval $x<1$ or $x>1$. The two displayed <solutions of a differential equation> are <linearly independent>, so they span the <solution space of a homogeneous linear differential equation> on either interval. The point $x=1$ is a <singular point of a differential equation> and cannot be crossed by a nonzero member of this family that remains finite there.
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