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Use power series to solve the differential equation \[(x-1)y''+y'=0\] \[y=\sum_{n=0}^\infty a_nx^n\] \[y'=\sum_{n=1}^\infty na_nx^{n-1}\] \[y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}\] \[(x-1)\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}=0\]

Mathematics
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allow me to complete this problem as far as I can, give me like 5 mins. Thanks!
\[\sum_{n=2}^{\infty}n(n-1)a_nx^{n-1}-\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}=0\]
\[\sum_{n=1}^{\infty}n(n+1)a_{n+1}x^n-\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^\infty (n+1)a_{n+1}x^n=0\]

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Other answers:

ok so when we strip out a term on \[\sum_{n=1}^{\infty}n(n+1)a_{n+1}x^n\] for the sum to start at n=0 would I have... \[a_0+ \sum_{n=0}^\infty n(n+1)a_{n+1}x^n \]
no that would be \[a_0+ \sum_{n=1}^\infty n(n+1)a_{n+1}x^n\]
cof of x^n (left)=cof of x^n (right) don't use sigmas.
already I have showed you ! Take a look to my solution.
y=an y'=(n+1)an+1 xy'=nan y''=(n+2)(n+1)an+2 xy''=(n+1)nan+1 x2y=n(n-1)an and so on.
(n+1)nan+1-(n+2)(n+1)an+2+(n+1)an+1=0 an+2=(n+1)/(n+2) an+1 an+2=a1/(n+2) one answer is just a0(check it out) y=a0+a1sigma(n=1 to inf)(x^n/n)

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