given $y_1 = x$ and the equation: $x^3y'''-3x^2y''+(6-x^2)xy'-(6-x^2)y = 0$
I determined $y_2 = ux$,
$y_2' = u+xu'$,
and $y_2'''= 3u''+xu'''$
substitution in the equation gives me: $x^4u'''-x^4u' = 0$
then, let $v=u'$, $v'=u''$, and $v''=u'''$ gives me $x^4v'' - x^4v = 0$
how do you go about solving after that? i believe it has to do with separable equations, but i can not figure it out. HELP PLEASE!!!