- #1
VinnyCee
- 489
- 0
The problem (#11, 5.2, boyce diprima):
[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?
[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?