Find the recurrence equation for the coefficients of the series solution(in powers of $x$) of $y''-xy'+3y=0$ $,y(0)=1$ $,y'(0)=1$

and the first three nonzero terms. What is the radius of convergence of the series?

Here's my work

Let $y=\sum\limits_{n=0}^\infty a_nx^n$ ,

Then $y'=\sum\limits_{n=0}^\infty na_nx^{n-1}$

$y''=\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}$

$\therefore\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}-x\sum\limits_{n=0}^\infty na_nx^{n-1}+3\sum\limits_{n=0}^\infty a_nx^n=0$

$\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}-\sum\limits_{n=0}^\infty na_nx^n+\sum\limits_{n=0}^\infty3a_nx^n=0$

$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=0}^\infty(-na_n+3a_n)x^n=0$

change of index, let $n=m-2$ and $n=0$ means $m=2$

$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{m=2}^\infty(-(m-2)a_{m-2}+3a_{m-2})x^{m-2}=0$

now $n = m$, and sum up the two summations into one

$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=2}^\infty (-na_{n-2}+5a_{n-2})x^{n-2}=0$


$\therefore n(n-1)a_n-na_{n-2}+5a_{n-2}=0$

$a_n=\dfrac{(n-5)a_{n-2}}{n(n-1)}$ $,n\geq2$

Use the initial condition, I got $a_0=1$ and $a_1=0$

And all the coefficients $a_n$ with odd index are $0$. I cannot figure out the pattern for the coefficients $a_n$ with even index, and how do I calculate the radius of convergence?

Thanks for any help.


I made a big mistake in my original answer. ALso, I have not checked your recurrence relationship but it looks okay

Note that when $n=6$ $a_6 >0$. Hence all the $a_n$ for $n>6$ are positive. (This is really not needed but is useful to know)

Since only even terms remain, let $n=2k$ and let $$b_k = a_{2k}$$ Then $$ b_k = \frac{2 k -5}{2k (2k-1)} b_{k-1} $$ and the series is $$ \sum_{k=0}^\infty b_k \left(x^2\right)^k$$ Thus the $k$-th term is $$ c_k = b_k (x^2)^k$$

By ratio test $$ \frac{c_k}{c_{k-1}} = \frac{2 k -5}{2k (2k-1)} x^2 \to 0 \text{ as }\, k\to \infty ~(\forall x) $$

Hence the region of convergence is the entire real line.

  • $\begingroup$ Do I use the ratio test like this? $L = |x-0|\lim_{n\rightarrow\infty}(|\dfrac{a_{n+1}}{a_n}|)$ $\endgroup$ – user59036 Jan 20 '14 at 5:32
  • $\begingroup$ Can you give me a few minutes? $\endgroup$ – user44197 Jan 20 '14 at 5:37
  • $\begingroup$ When you do $\dfrac {c_k}{c_{k-1}}$, where are the $b_k$ and $b_{k-1}$ terms? $\endgroup$ – user59036 Jan 20 '14 at 23:50
  • $\begingroup$ They cancel to give ${2k-5}{2k(2k-1)}$ $\endgroup$ – user44197 Jan 22 '14 at 3:40

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