※ 引述《tratt692 (jech)》之銘言：
: https://i.imgur.com/Awy8CSq.jpg
: 想問這題該怎麼解
xy" - (x + 1)y' + y = 0
y = uexp(x)
y' = u'exp(x) + uexp(x)
y" = u"exp(x) + 2u'exp(x) + uexp(x)
xu" - (x + 1)u' + u + 2xu' - (x + 1)u + xu = 0
=> xu" + (x - 1)u' = 0
=> u' = cexp(-∫(1 - 1/x)dx)
= cxexp(-x)
=> u = c[-xexp(-x) - exp(-x)] + d
=> y = uexp(x) = -cx - c + dexp(x)
所以另一解 = -c(x + 1)