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Find the series solution,power series

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dp182
#1
Jul3-11, 10:52 PM
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=[itex]\Sigma[/itex]anxn+r
y'=[itex]\Sigma[/itex](n+r)anxn+r-1
y''=[itex]\Sigma[/itex](n+r)(n+r-1)anxn+r-2

3. The attempt at a solution
after plugging into the solution I get
2[itex]\Sigma[/itex](n+r)(n+r-1)anxn+r-1-[itex]\Sigma[/itex](n+r)anxn+r+1-[itex]\Sigma[/itex](n+r)anxn+r-1-[itex]\Sigma[/itex]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[itex]\Sigma[/itex](n+r+1)(n+r)an+1xn+r-[itex]\Sigma[/itex](n+r-1)an-1xn+r-[itex]\Sigma[/itex](n+r+1)an+1xn+r-[itex]\Sigma[/itex]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
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vela
#2
Jul4-11, 02:17 AM
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The third sum should be positive. Explicitly writing in the limits, you have
[tex]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[/tex]
As you noted, only the first and third sums give you a [itex]x^{r-1}[/itex] term, and the second sum doesn't give you an [itex]x^r[/itex] 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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