- #1

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[tex](3\,-\,x^2)\,y''\,-\,(3\,x)\,y'\,-\,y\,=\,0[/tex]

I got the recursion formula as:

[tex]a_{n\,+\,2}\,=\,\frac{(n\,+\,1)}{3\,(n\,+\,2)}\,a_n[/tex]

Which give the following results:

[tex]\begin{flalign*}

a_2& = \frac{1}{6}\,a_n\,x^2&

a_3& = \frac{2}{9}\,a_n\,x^3&

a_4& = \frac{1}{4}\,a_n\,x^4&\\

a_5& = \frac{4}{15}\,a_n\,x^5&

a_6& = \frac{5}{18}\,a_n\,x^6&

a_7& = \frac{6}{21}\,a_n\,x^7&

\end{flalign*}[/tex]

When these are used, the answer does not match the book:

[tex]y(x)\,=\,a_0\,\left[1\,+\,\frac{x^2}{6}\,+\,\frac{x^4}{24}\,+\,\frac{5}{432}\,x^6\,+\,...\right]\,+\,a_1\,\left[x\,+\,\frac{2}{9}\,x^3\,+\,\frac{8}{135}\,x^5\,+\,\frac{16}{945}\,x^7\,+\,...\right][/tex]

What did I do wrong?