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Power Series - Differential Equation (check my answer)

  1. Dec 7, 2009 #1
    Using the power series method to solve the differential equation
    y'+xy=0 when y(0)=1

    Write the solution in the form of a power series and then recognize what function it represents.
    ************************************

    My answer:

    [tex]\sum[/tex](-1)k*[(x2k)/(2k)*k!]

    Is my answer correct?
    Is it written in the form of a power series?
    What function does it represent?

    Thanks to anyone that helps!
     
  2. jcsd
  3. Dec 7, 2009 #2

    Dick

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    Yes, it's correct. And yes, it's a power series. It looks like exp(f(x)) to me. Can you guess what f(x) might be?
     
  4. Dec 8, 2009 #3
    I'd guess and say that f(x)=cos(x).
    So the function would be f(x)=x2cos(x).

    Please correct me if I am wrong.
    Thank you for helping thus far.
     
  5. Dec 8, 2009 #4

    HallsofIvy

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    Not quite. Your sum reduces to
    [tex]\sum(-1)^k \frac{\left(\frac{x^2}{2}\right)^2}{k!}[/tex]
    which is a "cosine" but it is
    [tex]cos(\frac{x}{2})[/tex]

    [tex]x^2cos(x)[/tex]
    would be
    [tex]x^2\sum(-1)^k \frac{x^k}{k!}= \sum (-1)^k \frac{x^{k+2}}{k!}[/tex]
     
  6. Dec 8, 2009 #5
    Ok -got it.
    A big "THANKKK YOUUU!" to all for all of your help!
     
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