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Math Help - series solution!

  1. #1
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    series solution!

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

     \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

     \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

     r=0

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

     \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}

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

     r=r+1

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

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

    r=0

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

    c=1

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

    r=1

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

    c=1

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

    pls having problem deriving the closed form of the series! help?
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  2. #2
    Senior Member
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    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)
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