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Solve the Differential Equation: (1 - x^2) y'' - (4x) y' + (6) y = 0

  1. Jun 6, 2005 #1
    Here is the problem (number 9 in 5.2 of Boyce, DiPrima 8th Edition Book):

    [tex](1 - x^2)\,y'' - (4x)\,y' + (6)\,y = 0,\,x_0 = 0[/tex]

    Here is what I have so far:

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

    Now, I took out the first two terms (n = 0 and n = 1) of the first sum to make it's index go to n = 2 in order to add all the sums together, right?

    [tex]2\,a_2 + 6\,a_3\,x[/tex]

    Continuing in that fashion to add the sums, i get this:

    [tex]2\,a_2\,+\,6\,a_3\,x\,-\,4\,a_1\,x\,+\,6\,a_0\,+\,6\,a_1\,x\,+\,\sum_{n = 2}^{\infty}\,\left[(n\,+\,2)\,(n\,+\,1)\,a_{n\,+\,2}\,-\,n\,(n\,-\,1)\,a_n\,-\,4\,n\,a_n\,+\,6\,a_n\right]\,x^n\,=\,0[/tex]

    This is where I am stuck. I am not sure if I pulled out those factors correctly or what, please help. Thank you.
     
  2. jcsd
  3. Jun 6, 2005 #2
    Whoops!

    I copied the problem wrong!

    It is supposed to be:

    [tex](1 + x^2)\,y'' - (4x)\,y' + (6)\,y = 0,\,x_0 = 0[/tex]

    And NOT:

    [tex](1 - x^2)\,y'' - (4x)\,y' + (6)\,y = 0,\,x_0 = 0[/tex]

    One little minus sign!

    The whole solution is worked out as the last problem here!
     
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