Question

Find the power series in x for the general solution of (1+x^2)y"+6xy'+6y=0.

Answer 1

@jim_thompson5910

Answer 2

Unfortunately I'm not that great with DE

Answer 3

Seek for \(y=\sum_{n=0} a_nx^n\)
\(y'=\sum_{n=0} na_n x^{n-1}~~~~~~~~~~~\rightarrow 6xy'= \sum_{n=0}6na_nx^n\)
\(y"=\sum_{n=0}n(n-1)a_nx^{n-2}\)
-->\((1+x^2)y"=\sum_{n=0}n(n-1)a_nx^{n-2}+\sum_{n=0}n(n-1) a_n x^n\)

Answer 4

change the first element of y" to \(x^n\) and add them all, you have
\[\sum_{n=0} ((n+1)(n+2)a_{n+2}+n(n-1)a_n+6na_n+a_n)x^n =0\]

Answer 5

that meant:

Answer 6

\(x^n\neq 0\), Hence, it =0 iff the coefficient =0, that is
\((n+1) (n+2)a_{n+2}+(n(n-1)+6n +1))a_n=0\)
simplify, you have \(a_{n+2}= -\dfrac{n^2+5n+1}{(n+1)(n+2)}a_n\)

Answer 7

Hope you can proceed from here.

Answer 8

@Kainui

Answer 9

I help can answer any questions you have about what @Loser66 wrote