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Thread: Power Series Solution

  1. #1
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    Power Series Solution

    The given equation is $\displaystyle 2x^2y'' + (x^2 + 3x)y' - y = 0$



    1. Find two linearly independent solutions of the form $\displaystyle y = (x^r)\sum_{n=0}^{\infty} a_nx^{n}$. Using the recurrence relation.

    I changed the given y to $\displaystyle y = \sum_{n=0}^{\infty} a_nx^{n+r}$

    Then I found the first and second derivatives.

    $\displaystyle y' = \sum_{n=0}^{\infty} (n+r)a_nx^{n+r-1}$

    $\displaystyle y'' = \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r-2}$

    When I plug it in back into the original equation, I get:

    $\displaystyle 2 \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r}$ + $\displaystyle \sum_{n=0}^{\infty} (n+r)a_nx^{n+r+1}$ + $\displaystyle 3\sum_{n=0}^{\infty} (n+r)a_nx^{n+r}$ - $\displaystyle \sum_{n=0}^{\infty} a_nx^{n+r}$

    Ok, I understand up to this part. Now I'm just confused after this, but this is what I got. I took out $\displaystyle x^r$

    $\displaystyle x^r$$\displaystyle [\sum_{n=0}^{\infty} 2(n+r)(n+r-1)a_nx^{n}$ + $\displaystyle \sum_{n=0}^{\infty} (n+r)a_nx^{n+1}$ + $\displaystyle \sum_{n=0}^{\infty} 3(n+r)a_nx^{n}$ -$\displaystyle \sum_{n=0}^{\infty} a_nx^{n}]$

    Then I tried to get $\displaystyle a_0$ and changed some of the n values to $\displaystyle j $and $\displaystyle j+1$ similar to what my professor did.

    $\displaystyle x^r [(2r(r-1)+3r-1)a_0 $ + $\displaystyle \sum_{n=1}^{\infty} $$\displaystyle 2(j+r)(j+r-1)a_j + 3(j+r)a_j - a_j - (j-1+r)a_{j-1}]x^j$

    Then I tried to find the r by:

    $\displaystyle 2r^2 +r - 1 = 0$
    $\displaystyle (2r-1)(r+1)$

    I got r = 1/2 and r = -1.

    Is this correct so far? When my professor went over a homework question similar to this problem, he said that the r's would be -1 and -1/2. I just don't see where the -1/2 came from and even so. When I plug -1 , I get zero in the third term which ruins it unless that's ok?

    I know what to do pass this part, but it's very confusing. I spent more than 5 hours trying to do this and I know I'm doing something wrong or thinking of something else. Please help me lead to the right direction.
    Last edited by Kyeong; Jul 9th 2008 at 09:39 PM.
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  2. #2
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    What equation?
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  3. #3
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    Sorry, just added it.
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  4. #4
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    Quote Originally Posted by Kyeong View Post
    The given equation is $\displaystyle 2x^2y'' + (x^2 + 3x)y' - y = 0$
    The point $\displaystyle 0$ is a regular singular point of the differencial equation. Note, $\displaystyle \lim_{x\to 0}x\cdot \frac{x^2+3x}{x^2} =3$ and $\displaystyle \lim_{x\to 0}x^2\cdot \frac{-1}{x^2} = -1$. This means the indicial equation is $\displaystyle 2k(k-1) + 3k - 1 = 0$.
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  5. #5
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    Well, I believe I already got the indicial equation. It's near the end of the post, just in terms of r. First part of that long function.

    $\displaystyle x^r [(2r(r-1)+3r-1)a_0$
    ..........

    I got $\displaystyle r_1 = \frac {1} {2}$ and $\displaystyle r_2 = -1$.

    Then I plug in the r values and set up all the functions $\displaystyle \frac {a_{n-1}} {a_n}$. I did this for each r value and when r = -1, the third term is zero and ruins it and can't get the fourth non-zero term.

    Maybe I'm looking at this a different way?

    when $\displaystyle r = \frac {1} {2}$
    $\displaystyle a_j = \frac {-(j- \frac {1} {2})a_{j-1}} {2j^2 + 3j}$

    when $\displaystyle r = 1$
    $\displaystyle a_j = \frac {-(j- 2)a_{j-1}} {2j^2 - 3j}$

    Did I make a sign error or something?
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