-1是特解
y=ux y=u+xu‘ y=u’‘+2u‘代入(1-x)y''+xy'-y=0
(1-x)(u’‘+2u‘)+x(u+xu‘)-ux=0
(1-x)u’‘+(x^2-2x+1+1)u‘=0
u'=ln(1-x)+x^2/2-x+C1
u=(x-1)ln(1-x)+(1-x)+x^2/2-x+C1x+C2
y=x[(x-1)ln(1-x)+(1-x)+x^2/2-x+C1x+C2]-1