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[Differential Equations] Verify by direct substitution that a given power series is a solution

  1. Nov 16, 2014 #1
    1. The problem statement, all variables and given/known data
    vTArFry.png

    2. Relevant equations
    All relevant equations are given above.

    3. The attempt at a solution
    I decided to make a substitution for x2 so that x2 = u. By doing that, I could make the k=n+1 and k=n-1 substitutions so I can keep the series in phase. I took y' and just plugged everything into the DE. In the 2xy part, I just pulled the x into y so that I get the x2n+1 term. However, I can't seem to combine the series in one without having a loose term included (the first term of the y' series). I get the combined series with the loose term, all equal to zero - and I don't know where to go from there.
     
  2. jcsd
  3. Nov 16, 2014 #2

    PeroK

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    I'm not sure why you would need ##u = x^2##. What did you get for ##y'##?
     
  4. Nov 16, 2014 #3

    Zondrina

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    The question is simply asking you to find ##\frac{d}{dx} [ \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} x^{2n} ]## and then use it along with ##y## to show the equation is satisfied.
     
  5. Nov 16, 2014 #4
    @PeroK

    I thought making that substitution would make the series easier to work with in terms of getting each series in phase. With that substitution, I get:

    y' = sum_n=1..inf 2n(-1)nx2n-1 / n! = sum_n=1..inf 2(-1)nun-1 / (n-1)!
    (canceled out the n's)

    and for 2xy (using the same substitution) I get:

    2xy = sum_n=0..inf 2(-1)nun+1 / n!

    But if I let k=n-1 for y' and k=n+1 for y, I can make both indices of summation start at the same number, k=1 if I shift the y' series up so that it starts at n=2. I get the following:

    y'+2xy=0
    → -2 + sum_k=1..inf 2(-1)k+1uk / (k)! + sum_k=1..inf 2(-1)k-1uk / (k-1)! = 0

    I can combine the two series, but after that, I'm not sure what to do.


    @Zondrina

    Yeah, that's why I'm not sure why this problem is causing difficulties for me. I was hoping it would be an easy substitute, but I keep running into the same issue above.
     
  6. Nov 16, 2014 #5

    PeroK

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    You got ##y'## right, but the you've gone astray with the pointless substitution ##u=x^2##.

    ##u^{n-1} = x^{2n-2} \ne x^{2n-1}##
     
  7. Nov 16, 2014 #6
    I see. Yeah, I tried to make that substitution because I needed a way to get the series in phase with each other / make their indices of summation the same and couldn't think of anything else. I see where I went wrong though.

    Anyway, I just decided to write out the terms and verify it that way. Thanks for the help!
     
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