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Linearly Independent Solutions[help please]

  1. Sep 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Find two linearly independent solutions valid for x>0


    2. Relevant equations
    well first of all, all the solutions that i get to see in the forums is interms of the euler formula.. but in the course that I am taking(advanced engineering mathematics), it is expressed using a power formula... can anyone please teach me how to do this?..

    3. The attempt at a solution

    y = Summation of (a_n)(x^n) n=0 -> infinity
    y'' = Summation of (n)(n-1)(a_n)(x^[n-2]) n=0 -> infinity // so yeah just differentiation

    *substituting it to the equation

    = [ Summation of (n)(n-1)(a_n)(x^[n-1]) n=0 -> infinity ] + [Summation of (a_n)(x^n) n=0 -> infinity]

    *shifting the y equation to achieve same degree of x
    = [ Summation of (n)(n-1)(a_n)(x^[n-1]) n=0 -> infinity ] + [Summation of (a_[n-1] )(x^[n-1]) n=1 -> infinity]

    *getting the recurrence releationship

    a_2k = [-1]^k a_1 / 2k! [2k-1]!

    a_[2k+1] =[-1]^k a_1 / [2k+1]![2k]!

    *now substituting it to a power series formula

    y= a_0 + [ SUmmation of [a_2k] x^2k k=1 -> infinity ] + a_1 x + [ Summation of a_[2k+1] x^[2k+1] k=1 -> infinity ] +......

    *now after that I really don't know how to obtain the two linearly independent solutions.. can someone enlighten me?.. did I even do this right? =[

    any help will be very much appreciated....
    Last edited: Sep 9, 2008
  2. jcsd
  3. Sep 9, 2008 #2


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    Well, actually you can't get this "in terms of Euler's formula" if you are thinking of the formulas for solving "Euler type" equations because this is not. If the coefficient of y" were x2 then it would be.

    I'm afraid you aren't going to like the answer! x=0 is a singular point because the coefficient of y" becomes 0. Generally, speaking you CANNOT write both independent solutions as power series.

    You can use Frobenious' method: write
    [tex]y(x)= \sum_{n=0}^{\infty} a_n x^{n+ c}[/tex]
    where c is a constant which may not be positive or even an integer.
    Now we get
    [tex]y"= \sum_{n=0}^{\infty} (n+c)(n+c-1)a_n x^{n+c-1}[/tex]
    and the equation becomes
    [tex]y"+ y= \sum_{n=0}^{\infty} (n+c)(n+c-1)a_n x^{n+c-1}+ \sum_{n=0}^{\infty} a_n x^{n+ c}= 0[/itex]
    If we look for the lowest possible power of x, we see that taking n=0 in each sum gives xc-1 in the first sum and xc in the second. So the lowest possible power of x is xc-1 and its coefficient is c(c-1)a0. Since the sum itself is 0, all coefficients of powers of x must be 0 so we have c(c-1)a0= 0. Now there is an "ambiguity" in this: we could reduce c by an integer while absorbing that constant in n and get different coefficients giving the same sum. We avoid that ambiguity by requireing that the sum actually start with n= 0- that is, that a0 is not 0. Since c(c-1)a0= 0, we have the "indicial equation", c(c-1)= 0 so either c= 0 or c= 1. Taking c= 0, of course, is just a regular power series solution.

    Now for the part you won't like! Frobenius' theory shows that when the solutions of the indicial equation differ by an integer, the other solution cannot be written as a power series in that way. What does happen is that the second indpendent solution must be ln(x) times the first!
    Last edited: Sep 9, 2008
  4. Sep 9, 2008 #3
    i see.... hmmmm... now i get why I can't solve for anything... I guess I need to start all over again and restudy these type of problems and plugin "indicial equations."

    Thank you very much hallsofivy!

    oh my, I think I'm starting to get this. :) thank you very much! I'll try to post my solution when I get back from school.. thankiez again :)
    Last edited: Sep 9, 2008
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