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Solving an ODE using a power series

  1. Mar 31, 2008 #1
    1. The problem statement, all variables and given/known data

    solve the initial value problem:

    [tex] x(2-x)y'' - 6(x-1)y' - 4y = 0[/tex]
    [tex] y(1)=1 [/tex]
    [tex]y'(1) = 0 [/tex]

    hint: since the initial condition is given at [tex]x_0 = 1 [/tex], it is best to write the solution as a series centered at [tex]x_0 = 1 [/tex].

    2. Relevant equations
    I have attempted the question, but I got stuck in one of the steps. Here's what I have so far:

    assume the solution can be written as a power series centered about x = 1. Then the solution and its derivatives are:

    [tex] y(t) = \displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} [/tex]

    [tex] y'(t) = \displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^{n-1}} [/tex]

    [tex] y''(t) = \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}} [/tex]

    where the [tex] a_n [/tex] are constant coefficients.

    3. The attempt at a solution

    Substitute y, y', and y'' into the ODE in order to determine the [tex] a_n [/tex]:

    [tex]x(2-x)\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}-6(x-1)\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^{n-1}}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0[/tex]

    [tex](1-(x-1)^2)\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}-6(x-1)\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^{n-1}}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0[/tex]

    [tex]\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n ((x-1)^{n-2}- (x-1)^n)}-6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0[/tex]

    [tex]\displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^{n-2}}- \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^n} -6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0[/tex]

    now I need to shift indices to make all the exponents of (x-1) equal to n, and also to try to make all the summations start at n=0:

    [tex]\displaystyle\sum_{n=0}^{\infty}{(n+2)(n+1) a_{n+2} (x-1)^n}- \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^n} -6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0[/tex]

    I don't know how to proceed from here. I have 4 summations. Two of them start at n=0 and has the [tex](x-1)^n[/tex] factor. The other two summations start at n=2 or n=1. I can't get them to start at n=0 because if I try to do that, then I get:

    [tex]\displaystyle\sum_{n=0}^{\infty}{(n+2) (n+1) a_{n+2} (x-1)^{n+2}}[/tex]


    [tex]6\displaystyle\sum_{n=0}^{\infty}{(n+1) a_{n+1} (x-1)^{n+1}}[/tex]

    As you can see, these have the factors [tex] (x-1)^{n+2} [/tex] and [tex] (x-1)^{n+1} [/tex] instead of [tex] (x-1)^n [/tex]. So I don't know how to combine all this, since the summations do not start at the same number.

    I'm told that in the end, the answer to the coefficients should be:

    [tex]a_{2n} = (n+1)a_0 [/tex] and [tex]a_{2n+1} = \frac{2n+3}{3}a_1 [/tex]

    So I just don't know how to get from the point I'm stuck to this answer. I'm also quite confused as to how to solve the initial value problem, after obtaining the expected answer. The initial conditions are: [tex] y(1)=1 [/tex], [tex]y'(1) = 0 [/tex]. Using the initial condition y(1) = 1, I get:

    [tex] y(1)= \displaystyle\sum_{n=0}^{\infty}{a_n ((1)-1)^n} [/tex]
    [tex] 1 = \displaystyle\sum_{n=0}^{\infty}{a_n (0) [/tex]
    [tex] 1 = 0 [/tex]

    This does not make sense...what am I doing wrong?
  2. jcsd
  3. Apr 1, 2008 #2


    User Avatar
    Science Advisor

    After you have
    [tex]\displaystyle\sum_{n=0}^{\infty}{(n+2)(n+1) a_{n+2} (x-1)^n}- \displaystyle\sum_{n=2}^{\infty}{n (n-1) a_n (x-1)^n} -6\displaystyle\sum_{n=1}^{\infty}{n a_n (x-1)^n}-4\displaystyle\sum_{n=0}^{\infty}{a_n (x-1)^n} = 0[/tex]
    Do the n= 0 and n= 1 separately.

    When n= 0, the terms that start at n=0 give you (0+2)(0+1)a0+2- 4a0= 2a2- 4a0= 0 so a2= 2a0

    When n= 1, the terms that start at n= 1 give you (1+2)(1+1)a1+2- 6(1)a1- 4a1= 0 so 2a3= 10a1 and a3= 5a1.

    After that, you can use all the terms for n> 1.
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