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Math Help - Power series question

  1. #1
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    Power series question

    Determine the recurrence relation for the coefficients of the power series solutions, and find explicitly the …first four non-zero terms in each of the two linear independent series solutions to:  y^{''} +xy^{'} +2y = 0 about Xo=0

    I found the recurrence relation  (n+1)(n+2)An+2 + (n+2)An) = 0

    I dont understand what it means "find explicitly" the first four non-zero terms.
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  2. #2
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    Quote Originally Posted by Oblivionwarrior View Post
    Determine the recurrence relation for the coefficients of the power series solutions, and find explicitly the …first four non-zero terms in each of the two linear independent series solutions to:  y^{''} +xy^{'} +2y = 0 about Xo=0
    Let y = \sum_{n=0}^{\infty}a_n x^n then it means,
    \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} + x\sum_{n=1}^{\infty} na_nx^{n-1} + 2\sum_{n=0}^{\infty} a_n x_n = 0
    This becomes,
    \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2} + \sum_{n=1}^{\infty} na_nx^n + 2\sum_{n=0}^{\infty}a_nx^n = 0
    Change index,
    \sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^n + \sum_{n=1}^{\infty}na_nx^n + 2\sum_{n=0}^{\infty}a_nx^n = 0
    Evaluate the first and third sum at n=0 and combine,
    2a_2 + 2a_0 + \sum_{n=1}^{\infty} [(n+2)(n+1)a_{n+2} + (n+2)a_n]x^n = 0
    But this is the same as writing,
    \sum_{n=0}^{\infty} [(n+2)(n+1)a_{n+2} + (n+2)a_n]x^n = 0

    The recurrence relation is therefore, (n+2)(n+1)a_{n+2} + (n+2)a_n = 0 \implies a_{n+2} = -\frac{a_n}{n+1} \text{ for }n\geq 0
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  3. #3
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    I have got that part, but I am not sure how they are getting these answers from the recurrence formula. These are the answers.

     Y1 = 1-x^{2} +\frac{1}{3}x^4 -\frac{1}{15}x^6+...
     Y2 = x -\frac{1}{2}x^3 +\frac{1}{8}x^5 - \frac{1}{48}x^7
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  4. #4
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    I'll do the even ones, y_1(x). Try and do the odd ones:

    a_2=-\frac{a_0}{1}

    a_4=-\frac{a_2}{3}

    a_6=-\frac{a_4}{5}

    \vdots

    a_{2k}=-\frac{a_{2k-1}}{2k-1}

    Now multiply all corresponding equations on each side:

    a_2 a_4 a_6\cdots a_{2k}=\frac{(-1)^k}{1\cdot 3\cdot 5\cdot\cdots (2k-1)} a_0 a_2 a_4\cdots a_{2k-2}

    Simplifying:

    a_{2k}=\frac{(-1)^k}{1\cdot 3\cdot 5\cdots (2k-1)} a_0

    Let a_0=1 and we can then apply the arbitrary constant to y_1(x). We then have:

    y_1(x)=1+\sum_{k=1}^{\infty}\frac{(-1)^k}{1\cdot 3\cdot 5\cdots (2k-1)} x^{2k}
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