= Solution
{c}
For $y=x$, the <derivatives> are $y'=1$ and $y''=0$, and substitution gives $2(x-1)-2(x-1)=0$.
Use <reduction of order>: set $y=xv$, so $y'=v+xv'$ and $y''=2v'+xv''$. Substitution and cancellation give
$$
x^2v''+2x^2v'=0.
$$
Since $x>0$, this reduces to $v''+2v'=0$, hence $v'=Ae^{-2x}$ and $v=C_1+C_2e^{-2x}$. The <general solution> is
$$
\boxed{y(x)=C_1x+C_2xe^{-2x},\qquad x>0.}
$$
The two solutions are <linearly independent>, since their ratio $e^{-2x}$ is nonconstant; equivalently, their <Wronskian> is $-2x^2e^{-2x}\ne0$.
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