# Series Diff EQ problem: (3 - x^2) y'' - (3x) y' - y = 0

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& 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*}

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

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

What did I do wrong?

## Answers and Replies

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Your algebra is a little suspect. Try calculating those coefficients again.

--J

Incorrect recursion table is the trouble...

$$y(x)\,=\,\sum_{n\,=\,0}^{\infty}\,a_n\,x^n\,=\,a_0\,+\,a_1\,x\,+\,...$$

\begin{flalign*}a_2& = \frac{1}{6}\,a_0\,x^2&a_3& = \frac{2}{9}\,a_1\,x^3&a_4& = \frac{1}{24}\,a_0\,x^4&\\a_5& = \frac{8}{135}\,a_1\,x^5&a_6& = \frac{5}{432}\,a_0\,x^6&a_7& = \frac{16}{945}\,a_1\,x^7&\end{flalign*}

Also, you shouldn't include the xn in your coefficients. Remember that these are the coefficients of the powers of x! They don't include the power of x themselves. You must multiply them by the appropriate power of x to get your solution. Otherwise, it looks like you're set. Good job.

--J