$Ty''+y'+1/Ty=0$の場合$(1/T+T)y+Ty''+y'-Ty=0$$(1/T+1+T)y+T(y''-y)+y'-y=0$$T(y''-y)+y'-y=T(y''-y')+(T+1)y'-y$より$(1/T+T)y+T(y''-y')+(T +1)y'=0$$\lbrace 1/T+T-(T+1)\rbrace y+T(y'-y)'+(T+1)(y'-y)=0$$(1/T-1)y+T(y'-y)'+(T+1)(y'-y)=0$
これで$y'-y$を変数で置く 。