= Solution
{c}
For $y=x$, the first and second <derivatives> are one and zero, giving $-2x+2x=0$. To find an independent solution, use <reduction of order> and write $y=xv(x)$. Cancellation of the terms proportional to $v$ leaves
$$
x(1-x^2)v''+(2-4x^2)v'=0.
$$
Writing $w=v'$ and using <separation of variables> gives
$$
\frac{w'}w=-\frac2x+\frac{2x}{1-x^2},\qquad
w=\frac C{x^2(1-x^2)}.
$$
Since $1/[x^2(1-x^2)]=x^{-2}+(1-x^2)^{-1}$, integration yields the <general solution>
$$
\boxed{y(x)=C_1x+C_2\left(\frac x2\log\frac{1+x}{1-x}-1\right),\qquad0<x<1.}
$$
The second factor divided by $x$ has nonzero <derivative> $1/[x^2(1-x^2)]$, so the two solutions are <linearly independent>.
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