Find the series solution,power series

[dp182]

Homework Statement

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 a_n

Homework Equations

2xy''-x(x-1)y'-y=0 about x=0

assuming the solution has the form y=\Sigmaa_nx^n+r
y'=\Sigma(n+r)a_nx^n+r-1
y''=\Sigma(n+r)(n+r-1)a_nx^n+r-2

The Attempt at a Solution

after plugging into the solution I get
2\Sigma(n+r)(n+r-1)a_nx^n+r-1-\Sigma(n+r)a_nx^n+r+1-\Sigma(n+r)a_nx^n+r-1-\Sigmaa_nx^n+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)a_n+1x^n+r-\Sigma(n+r-1)a_n-1x^n+r-\Sigma(n+r+1)a_n+1x^n+r-\Sigmaa_nx^n+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 what's throwing me off any help would be great

 

[vela]

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}<br /> -\sum_{n=0}^\infty (n+r) a_n x^{n+r+1}<br /> +\sum_{n=0}^\infty (n+r)a_n x^{n+r-1}<br /> -\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.