Question

I cant get the recurrence relation to work at all, could someone give a brief summary on how to do 2y''+4xy=0 expansion is sumation A_n[x-x_0]^n centered at x_0= 0
i have absolutely no idea how to do this!

Answer 1

let y = a0 + a1x + a2 x^2 + a3 x^3 + ....

y'' = 2 a2 + 6 a3 x + 12 x^2 + ...

or written in summations:\[y=\sum_0a_n~x^n\]
\[y''=\sum_2a_n~n(n-1)x^{n-2}\]

plugging this in we get:
\[2\sum_2a_n~n(n-1)x^{n-2}+4x\sum_0a_n~x^n=0\]
and smoothing things thru
\[\sum_2~2a_n~n(n-1)x^{n-2}+\sum_04a_n~x^{n+1}=0\]

adjust the indexes/exponent to the lowest form

\[\sum_2~2a_n~n(n-1)x^{n-2}+\sum_{0+3}4a_{n-3}~x^{n-3+1}=0\]
\[\sum_2~2a_n~n(n-1)x^{n-2}+\sum_{3}4a_{n-3}~x^{n-2}=0\]
and pop out a term to line up the summation indexes
\[4a_2+\sum_3~2a_n~n(n-1)x^{n-2}+\sum_{3}4a_{n-3}~x^{n-2}=0\]

and combine the summations
\[4a_2+\sum_{3}2a_n~n(n-1)+4a_{n-3}~x^{n-2}=0\]
from this, we can determine that a2=0, and the coefficients need to go to zero as well
\[2a_n~n(n-1)+4a_{n-3}=0\]
solve for an
\[a_n=-\frac{2a_{n-3}}{n(n-1)}~:~n>2\]
start defining the terms
\[a_0=a_0, a_1=a_1, a_2=0\]
\[a_3=-\frac{2a_{0}}{3(2)}\]
\[a_4=-\frac{2a_{1}}{4(3)}\]
\[a_5=-\frac{2a_{2}}{5(4)}\]
\[a_6=-\frac{2a_{3}}{6(5)}\]
since all the a2 parts, and their related coeffs are zero, we can ignore those and define the rest in terms of a0 and a1