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Math Help - Power Series Solution

  1. #1
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    Power Series Solution

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



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

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

    Then I found the first and second derivatives.

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

    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:

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

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

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

     x^r [(2r(r-1)+3r-1)a_0 + \sum_{n=1}^{\infty} 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:

     2r^2 +r - 1 = 0
     (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; July 9th 2008 at 10:39 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 2x^2y'' + (x^2 + 3x)y' - y = 0
    The point 0 is a regular singular point of the differencial equation. Note, \lim_{x\to 0}x\cdot \frac{x^2+3x}{x^2} =3 and \lim_{x\to 0}x^2\cdot \frac{-1}{x^2} = -1. This means the indicial equation is 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.

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

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

    Then I plug in the r values and set up all the functions \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 r = \frac {1} {2}
    a_j = \frac {-(j- \frac {1} {2})a_{j-1}} {2j^2 + 3j}

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

    Did I make a sign error or something?
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