The problem (#11, 5.2, boyce diprima):
(3\,-\,x^2)\,y''\,-\,(3\,x)\,y'\,-\,y\,=\,0
I got the recursion formula as:
a_{n\,+\,2}\,=\,\frac{(n\,+\,1)}{3\,(n\,+\,2)}\,a_n
Which give the following results:
\begin{flalign*}
a_2& = \frac{1}{6}\,a_n\,x^2&
a_3& = \frac{2}{9}\,a_n\,x^3&...