# Find the series solution,power series

1. Jul 3, 2011

### dp182

1. The problem statement, all variables and given/known data
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

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

2. Jul 4, 2011

### vela

Staff Emeritus
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.