Question

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

Answer 1

allow me to complete this problem as far as I can, give me like 5 mins. Thanks!

Answer 2

\[\sum_{n=2}^{\infty}n(n-1)a_nx^{n-1}-\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum_{n=1}^\infty na_nx^{n-1}=0\]

Answer 3

\[\sum_{n=1}^{\infty}n(n+1)a_{n+1}x^n-\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^\infty (n+1)a_{n+1}x^n=0\]

Answer 4

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...
\[a_0+ \sum_{n=0}^\infty n(n+1)a_{n+1}x^n \]

Answer 5

no that would be
\[a_0+ \sum_{n=1}^\infty n(n+1)a_{n+1}x^n\]

Answer 6

cof of x^n (left)=cof of x^n (right)
don't use sigmas.

Answer 7

already I have showed you !
Take a look to my solution.

Answer 8

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.

Answer 9

(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)