My Math Forum Series solution near regular singular point

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 October 29th, 2012, 02:12 PM #1 Member   Joined: Dec 2010 Posts: 51 Thanks: 0 Series solution near regular singular point So you have $xy'' + y = 0$ and at $x= 0$ is a regular singular point. Therefore you plug in the generalized power series $y= \displaystyle\sum_{n=0}^{\infty} a_nx^{n+r}$ into the differential equation. You should get: $a_0x^r*(r-1)r + \displaystyle\sum_{n=1}^{\infty} $(n+r-1)(n+r)a_n + a_{n-1}$x^{n+r} = 0$ soo... the indicial equation is $x^r: \hspace{20 mm} (r-1)r= 0 \rightarrow r=0, r=1$ $x^{n+r}: \hspace{20 mm} a_n= \frac{-a_{n-1}}{(n+r-1)(n+r)} \hspace{20 mm} n=1,2,3,...$ Now, for my question: Why do I consider r=1, but not r=0? The back of the book only has $y_1= x^1$1 - \frac{x}{1!2!} + \frac{x^2}{2!3!} + \ldots + \frac{(-1)^nx^n}{n!(n+1)!}$$ as a solution. Why is r=0 not a solution? Is there a general rule on when to and when not to consider an indicial equation root? Also, what if I've gotten $r_1= r_2 = 0$? or $r_1= r_2 = k$ (where k is non-zero)? Thanks

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