# find the series solution,power series

by dp182
Tags: around a point, ode, power series, series solution
 P: 22 1. The problem statement, all variables and given/known data 2xy''-x(x-1)y'-y=0 about x=0 what are the roots of the indicial equation and for the roots find the recurrence relation that defines the the coef an 2. Relevant equations 2xy''-x(x-1)y'-y=0 about x=0 assuming the solution has the form y=$\Sigma$anxn+r y'=$\Sigma$(n+r)anxn+r-1 y''=$\Sigma$(n+r)(n+r-1)anxn+r-2 3. The attempt at a solution after plugging into the solution I get 2$\Sigma$(n+r)(n+r-1)anxn+r-1-$\Sigma$(n+r)anxn+r+1-$\Sigma$(n+r)anxn+r-1-$\Sigma$anxn+r then I attempt to make all the x's the same and and make the sigma's equal so after doing that I get 2$\Sigma$(n+r+1)(n+r)an+1xn+r-$\Sigma$(n+r-1)an-1xn+r-$\Sigma$(n+r+1)an+1xn+r-$\Sigma$anxn+r I know that I need to replace the 0 under the sigma's with a (-1) on terms 1,3 but term 2 is whats throwing me off any help would be great
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,378 The third sum should be positive. Explicitly writing in the limits, you have $$2\sum_{n=0}^\infty (n+r)(n+r-1)a_n x^{n+r-1} -\sum_{n=0}^\infty (n+r) a_n x^{n+r+1} +\sum_{n=0}^\infty (n+r)a_n x^{n+r-1} -\sum_{n=0}^\infty a_n x^{n+r} = 0$$ As you noted, only the first and third sums give you a $x^{r-1}$ term, and the second sum doesn't give you an $x^r$ term. Separating those terms out, you have \begin{eqnarray*} &&[2r(r-1) + r]a_0x^{r-1} + \\ &&[2(r+1)r a_1 + (r+1)a_1 - a_0]x^r + \\ &&2\sum_{n=1}^\infty (n+r+1)(n+r)a_{n+1}x^{n+r} -\sum_{n=1}^\infty (n+r-1)a_{n-1}x^{n+r} +\sum_{n=1}^\infty (n+r+1)a_{n+1}x^{n+r} -\sum_{n=1}^\infty a_n x^{n+r} = 0 \end{eqnarray*} By assumption, a0 isn't equal to 0, so you must have 2r(r-1)+r=0. That's your indicial equation.

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