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

[VinnyCee]

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*}<br /> a_2&amp; = \frac{1}{6}\,a_n\,x^2&amp;<br /> a_3&amp; = \frac{2}{9}\,a_n\,x^3&amp;<br /> a_4&amp; = \frac{1}{4}\,a_n\,x^4&amp;\\<br /> a_5&amp; = \frac{4}{15}\,a_n\,x^5&amp;<br /> a_6&amp; = \frac{5}{18}\,a_n\,x^6&amp;<br /> a_7&amp; = \frac{6}{21}\,a_n\,x^7&amp;<br /> \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?

 

[Justin Lazear]

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

--J

 

[VinnyCee]

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

 

[Justin Lazear]

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