1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: [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

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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


    User Avatar
    Homework Helper

    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

    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:

    → -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.


    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted