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DiffEQ 2nd order series sol'n problem: (1 - x)y'' + y = 0, x0 = 0

  1. Jun 5, 2005 #1
    This problem is from section 5.2 in Boyce, DiPrima's Differential Equations 8th edition.

    [tex](1 - x)\,y''\,+\,y\,=\,0[/tex]

    I get:


    Which leads to one equation:



    and the recursion formula:


    Now I am totally lost, just like in this other thread, because I don't know how to move on from this step. Please help :confused:
  2. jcsd
  3. Jun 6, 2005 #2
    What now?

    Using the recursion formula above, i get the following:

    a_{(0)+2} =\frac{((0)+1)(0)a_{(0)+1}-a_0}{((0)+2)((0)+1)}\:\:\:\:\:\:\:
    a_2 =\frac{-a_0}{2*1}\\

    a_{(1)+2} =\frac{((1)+1)(1)a_{(1)+1}-a_{(1)}}{((1)+2)((1)+1)}\:\:\:\:\:\:\:
    a_3 =\frac{-a_0-a_1}{3*2}\\[/tex]

    [tex]a_{(2)+2} =\frac{((2)+1)(2)a_{(2)+1}-a_{(2)}}{((2)+2)((2)+1)}\:\:\:\:\:\:\:
    a_4 =\frac{-a_1-\frac{1}{2}a_0}{4*3}\\

    But I don't see much of a pattern. What am I doing wrong? Please help, thank you in advance.
    Last edited: Jun 6, 2005
  4. Jun 6, 2005 #3
    Almost done...

    After fixing the eroor in the first term of the recursion series above, I now set a0 = 1 and a1 = 0 to seperate the two-types of terms left in the formula:



    Now with a0 = 0 and a1 = 1:


    Now, factoring out the a0 and a1 terms from each:


    The book says this is the answer, but is there any way to generalize this equation more? Possibly by using a summation?
  5. Jun 6, 2005 #4


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    U could use factorials,but what would be the purpose?You found the solution,i say that's enough.

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