Finding a power series solution to a differential equation?

[Eclair_de_XII]

Homework Statement

"Find the recurrence relation in the power series solution for ##y''-xy'-y=0## centered about ##x_0=1##."

Homework Equations

##y=\sum_{n=0}^\infty a_nx^n##
Answer as given in book: ##(n+2)a_{n+2}-a_{n+1}-a_n=0##

The Attempt at a Solution

##y=\sum_{n=0}^\infty a_n(x-1)^n##
##xy'=x\sum_{n=0}^\infty na_n(x-1)^{n-1}=x\sum_{n=1}^\infty na_n(x-1)^{n-1}=x\sum_{n=0}^\infty (n+1)a_{n+1}n(x-1)^{n}##
##y''=\sum_{n=0}^\infty n(n-1)a_n(x-1)^{n-2}=\sum_{n=0}^\infty n(n-1)a_n(x-1)^{n-2}=\sum_{n=2}^\infty n(n-1)a_n(x-1)^{n-2}=\sum_{n=0}^\infty a_{n+2}(n+2)(n+1)(x-1)^{n}##

##y''-xy'-y=0=\sum_{n=0}^\infty[(n+2)(n+1)a_{n+2}-x(n+1)a_{n+1}-a_n](x-1)^n##

##(n+2)(n+1)a_{n+2}=x(n+1)a_{n+1}+a_n##

I don't know what I'm doing wrong. It's not matching the answer given in the back of the book.

 

[Eclair_de_XII]

Never mind, I found what I did wrong.

##xy'=x\sum_{n=0}^\infty na_n(x-1)^{n-1}=[1+(x-1)]⋅\sum_{n=0}^\infty na_n(x-1)^{n-1}=\sum_{n=1}^\infty na_n(x-1)^{n-1}+\sum_{n=0}^\infty na_n(x-1)^{n}=\sum_{n=0}^\infty (n+1)a_{n+1}(x-1)^{n}+\sum_{n=1}^\infty na_n(x-1)^{n}##