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Power Series Solution y' = xy

  1. Mar 5, 2014 #1
    1. The problem statement, all variables and given/known data
    $$y' = xy$$


    2. Relevant equations
    $$y = a_{0} + a_{1}x + a_{2}x^{2}+... = \sum\limits_{n=0}^∞ a_{n}x^{n}$$
    $$xy = a_{0}x + a_{1}x^{2} + a_{2}x^{3}+... = \sum\limits_{n=0}^∞ a_{n}x^{n+1}$$
    $$y' = a_{1} + 2a_{2}x + 3a_{3}x^{2}... = \sum\limits_{n=1}^∞ n a_{n}x^{n-1}$$


    3. The attempt at a solution

    $$\sum\limits_{n=1}^∞ n a_{n}x^{n-1} = \sum\limits_{n=0}^∞ a_{n}x^{n+1}$$
    $$\sum\limits_{n=1}^∞ n a_{n}x^{n-1} = \sum\limits_{n=2}^∞ a_{n-2}x^{n-1}$$
    $$\sum\limits_{n=2}^∞ n a_{n}x^{n-1} = a_{1}x^{0} + \sum\limits_{n=2}^∞ a_{n-2}x^{n-1}$$
    $$\sum\limits_{n=2}^∞ [n a_{n} - a_{n-2}]x^{n-1} = a_{1}x^{0}$$ (1)


    $$n a_{n} - a_{n-2} = 0$$ for all n = 2,3,4,...
    $$a_{1} = 0$$

    $$a_{0} = 2a_{2}$$
    let $$a_{0} = C $$
    $$a_{2} = \frac{C}{2} $$
    $$a_{1} = 3a_{3} = 0$$
    $$a_{2} = 4a_{4} $$
    $$a_{4} = \frac{C}{4*2} , a_{6} = \frac{C}{6*4*2}, a_{n} = \frac{C}{(2n)!!}$$
    I end up with the solution $$y = C*\sum\limits_{n=0}^∞ \frac{x^{2n}}{(2n)!!} $$

    What am I doing wrong? $$a_{1}$$ must be zero, correct?
     
    Last edited: Mar 5, 2014
  2. jcsd
  3. Mar 5, 2014 #2
    Please disregard, the solution is in fact $$ y = C*\sum\limits_{n=0}^∞ \frac{x^{2n}}{(2n)!!} $$
    $$ C*\sum\limits_{n=0}^∞ \frac{x^{2n}}{(2n)!!} \equiv C*e^{\frac{x^{2}}{2}} $$
     
    Last edited: Mar 5, 2014
  4. Mar 5, 2014 #3

    D H

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    Staff Emeritus
    Science Advisor

    That is correct.


    Your original series, the one with the those an coefficients, was of the form ##\sum_{n=0}^{\infty} a_n x^n##. Your new series is of the form ##\sum_{n=0}^{\infty} b_n x^{2n}##. All of your ##a_n## for odd n are zero.
     
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