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Power Series Solution

  1. Mar 1, 2012 #1

    Char. Limit

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    Gold Member

    1. The problem statement, all variables and given/known data
    I am trying to find the power series solution to y' = 4 x y + 2, with the initial condition of y(0)=1.

    2. Relevant equations

    3. The attempt at a solution

    Simple enough, I say, as I arrange the equation so I have 0 on one side. I get something like this:

    [tex]y' - 4 x y - 2 = 0[/tex]

    I then assume that [itex]y = \sum_{n=0}^\infty a_n x^n[/itex]. I also find that [itex]y' = \sum_{n=0}^\infty (n+1) a_{n+1} x^n[/itex] and I pick, for two, a series like [itex]\sum_{n=0}^\infty \frac{1}{2^n}[/itex]. Subbing this all in, I get:

    [tex]\sum_{n=0}^\infty \left(a_n - 4 \left(n+1\right) x a_{n+1} - \frac{1}{2^n}\right) x^n = 0[/tex]

    Or in other words...

    [tex]\left(a_n - 4 (n+1) x a_{n+1} - \frac{1}{2^n}\right) = 0[/tex]

    But this doesn't look right. There's an "x" in there that shouldn't be there. What's the best way to remove the x?
  2. jcsd
  3. Mar 1, 2012 #2


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    Hi Char. Limit! :smile:

    You needed to think ahead :wink:
    … you needed [itex]xy = \sum_{n=0}^\infty a_n x^{n+1}[/itex] :smile:

    (and then change the limits, of course)
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