1. ## series solution!

$\displaystyle y''(x) + \left[\frac{x^2+x}{x^2}\right]y'(x) -x^{-2}y=0$

$\displaystyle \sum_{i=0}^n(r+c)(r+c-1)a_rx^{r+c-2}+\sum_{i=0}^n(r+c)a_rx^{r+c-1}+\sum_{i=0}^n(r+c)a_rx^{r+c-2}-\sum_{i=0}^na_rx^{r+c-2}=0$

$\displaystyle \sum_{i=0}^n\left[(r+c)(r+c-1)+(r+c)-1\right]a_rx^{r+c-2}+\sum_{i=0}^n(r+c)a_rx^{r+c-1}=0$

$\displaystyle r=0$

$\displaystyle c(c-1)+c-1=0$
$\displaystyle c^2-c+c-1=0 , c=\pm1$

$\displaystyle \sum_{i=0}^n\left[(r+c)(r+c-1)+(r+c)-1\right]a_rx^{r+c-2}=-\sum_{i=0}^n(r+c)a_rx^{r+c-1}$

$\displaystyle \left[(r+c)(r+c-1)+(r+c)-1\right]a_{r+1}=-(r+c)a_r$

$\displaystyle r=r+1$

$\displaystyle \left[(r+1+c)(r+c)+(r+1+c)-1\right]a_{r+1}=-(r+1+c)a_r$

$\displaystyle a_{r+1}=\frac{-(r+1+c)a_r}{(r+1+c)(r+c)+(r+1+c)-1}$

r=0

$\displaystyle a_{1}=\frac{-(1+c)a_0}{(1+c)(c)+(1+c)-1}$

c=1

$\displaystyle a_{1}=\frac{-(2)a_0}{(2)(1)+(2)-1}$

r=1

$\displaystyle a_{2}=\frac{-(2+c)a_1}{(2+c)(1+c)+(2+c)-1}$

c=1

$\displaystyle a_{2}=\frac{-(3)a_1}{(3)(2)+(3)-1}$

pls having problem deriving the closed form of the series! help?

2. ## Re: series solution!

Hi !

The change of function z=xy leads to a simpler ODE with trivial solutions z=x-1 and z=exp(-x)