Answer
See below
Work Step by Step
Given $(1-x^2)y''-2xy'+2y=0$
We first compute the appropriate derivatives:
$y'=x(u'+u)\\
y''=x(u''+u'+u)$
Substitute:
$(1-x^2)(u''+u'+u)]-2x(u'+u)+2x=0$
which simplifies to
$x(1-x^2)u''-2u'(1-2x^2)=0$
or equivalently,
$w'x+w=0$
where $w=u'$
Separating the variables yields
$(1-2x^2)\frac{-2}{x(1-x^2)}=\frac{w}{w'}$
By integrating, we obtain
$\ln w=-2\ln x-\ln(1- x^2) +C\\
wx=k$
Therefore,
$u'=\frac{k}{x^2(1-x^2)}$
Integration by parts gives
$u=-\frac{k}{x}+\frac{1}{2}k\ln (\frac{1+x}{1-x})+C$
The general solution is
$y=\frac{1}{2}kx\ln (\frac{1+x}{1-x})-k+Cx$
