1. Not finding help here? Sign up for a free 30min 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!

Find the power series in x for the general solution of?

  1. Jul 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the power series in x for the general solution of (1+x^2)y"+6xy'+6y=0.

    2. Relevant equations
    None.

    3. The attempt at a solution
    I got up to an+2=-an(n+3)/(n+1)
    for n=1, 2, 3, 4, 5, 6...
    a3=-2a1
    a4=0
    a5=3a1
    a6=0
    a7=-4a1
    a8=0
    The answer in the book says y=a0sigma from m=0 to infinity of (-1)^m * (2m+1)x2m+a1sigma from m=0 to infinity of (-1)^m * (m+1) x^(2m+1). How do I get the correct answer?
     
  2. jcsd
  3. Jul 28, 2016 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    You didn't explain how you got those results. One way to solve this kind of problem is to do manipulations with summations using the ##\Sigma## notation. Another way is to work out cases for specific powers of ##x## until you see how to express the result as a pattern using the ##\Sigma## notation. Which method do your class materials expect you to use ?

    Assume ## y = A_0 + A_1 x + A_2 x^2 + A_3 x^3 + .... = \Sigma_{i=0}^\infty A_i x^i##. To start things off, what is the summation that represents ##y'## ?
     
  4. Jul 30, 2016 #3
    Just like you mentioned, I got those results by finding y, y' and y'' and then plugging those into the differential equation and simplify. But at the end after finding all the values of a3, a4, a5, etc. I don't know how to get to the correct answer in the textbook.
     

    Attached Files:

    • de.jpg
      de.jpg
      File size:
      37.6 KB
      Views:
      39
  5. Jul 30, 2016 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    The picture of your work is upside down.
     
  6. Jul 31, 2016 #5
    Now I believe you can take a look at my work. Sorry about that.
     

    Attached Files:

    • de1.jpg
      de1.jpg
      File size:
      40.6 KB
      Views:
      36
  7. Aug 1, 2016 #6

    Stephen Tashi

    User Avatar
    Science Advisor

    After the step:

    ## \sum_{n=0}^\infty {(n+2)(n+1)A_{n+2}x^n} + \sum_{n=2}^\infty {(n)(n-1)A_nx^n}+ 6 \sum_{n=1}^\infty {(n)A_n x^n }+ 6 \sum_{n=0}^\infty A_n x^n = 0 ##

    I don't see how you concluded ##A_0 = 0 ##

    ## (2)(1) A_2 x^0 + (3)(2)A_3x^1 + \sum_{n=2}^\infty {(n+2)(n+1)A_{n+2}x^n}##
    ##+ \sum_{n=2}^\infty {(n)(n-1)A_nx^n}##
    ##+ (6)(1)A_1x^1 + 6 \sum_{n=2}^\infty {(n)A_n x^n }##
    ## + 6A_0 x^0 + 6A_1 x^1 + 6 \sum_{n=2}^\infty A_n x^n = 0 ##

    ## ( 2 A_2 + 6A_0)x^0 ##
    ## + (6A_3 + 12A_1)x^1##
    ## + \sum_{n=2}^{\infty} {[ (n+2)(n+1)A_{n+2} + (n)(n-1)A_n + 6nA_n + 6A_n] x^n} = 0 ##

    ## ( 2 A_2 + 6A_0)x^0##
    ## + (6A_3 + 12A_1)x^1##
    ## + \sum_{n=2}^{\infty} {[ n+2)(n+1)A_{n+2} + (n)(n-1)A_n + 6nA_n + 6A_n] x^n} = 0 ##

    ## ( 2 A_2 + 6A_0) ##
    ## + (6A_3 + 12A_1)x^1 ##
    ## + \sum_{n=2}^{\infty} {[ n+2)(n+1)A_{n+2} + ( (n)(n-1) + 6n + 6)A_n] x^n} = 0 ##
     
  8. Aug 2, 2016 #7
    Okay, so from where you left off above,
    2a2+6a0=0
    that means a2=0 and so does a0=0.
    And (6a3+12a1)x=0,
    that means a3=-2a1.
    Now what? How to find the answer from here?
     
  9. Aug 2, 2016 #8
    Look at what you have so far:

    $$y = a_0(1-3x^2...)+a_1(x-2x^3...)$$
     
  10. Aug 2, 2016 #9
    Okay, so I got y=a0(1-3x^2-5/3x^4+7/3x^6-3x^8)+a1(x-2x^3+3x^5-4x^7).
     
  11. Aug 2, 2016 #10
    The ##a_0## sum does not match the book answer, but the ##a_1## sum does.
     
  12. Aug 2, 2016 #11
    But how do I get to the answer in the book?
     

    Attached Files:

    • de2.jpg
      de2.jpg
      File size:
      26.3 KB
      Views:
      24
  13. Aug 2, 2016 #12
    Does the answer in the book satisfy the differential equation?
     
  14. Aug 2, 2016 #13
    To be honest, I'm not too sure...
     
  15. Aug 2, 2016 #14
    What do you get if you substitute their answer term for term?
     
  16. Aug 2, 2016 #15
    I still don't get it.
     
  17. Aug 2, 2016 #16
    Your methodology looks correct. That's the important part. If their answer is correct and you don't match it, you must have made a simple "arithmetic" error somewhere.
     
  18. Aug 2, 2016 #17
    After checking my work, I think I have made a stupid mistake.
    Now I got y=a0(1-3x^2-5x^4+7x^6-9x^8)+a1(x-2x^3+3x^5-4x^7). Does this match the answer in the book?
     
  19. Aug 2, 2016 #18

    Stephen Tashi

    User Avatar
    Science Advisor

    No, it doesn't imply a definite value for ##A_0## or ##A_2##. (After all, the solution to a differential equation should contain some arbitrary constants that are determined only when you are given some initial conditions.)
     
  20. Aug 2, 2016 #19

    Stephen Tashi

    User Avatar
    Science Advisor


    ##A_2 = - 3 A_0 ##

    ##A_3 = - 2 A_1 ##

    ## A_{n+2} = -\frac{n^2 + 5n + 6}{(n+2)(n+1)} A_n = -\frac{(n+2)(n+3)}{(n+2)(n+1) } A_n ##

    So we can take ##A_0## and ##A_1## as the arbitrary constants and express all the other ##A_i ## in terms of ##A_0 ## and ##A_1 ##.
     
  21. Aug 3, 2016 #20
    You should be able to determine this yourself. The answer does match.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find the power series in x for the general solution of?
Loading...