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2nd order de

  1. Dec 9, 2009 #1
    1. The problem statement, all variables and given/known data

    [itex](x^2-1)y'' + 4xy' + 2y = 6x[/itex], given that [itex]y_1=\frac{1}{x-1}[/itex] and [itex]y_2=\frac{1}{x+1}[/itex].

    2. Relevant equations

    3. The attempt at a solution

    Since both solutions are given, the solution to the homogenous system is:
    [tex]y_h=C_1\frac{1}{x-1} + C_2\frac{1}{x+1}[/tex]
    And the solution to the original equation would be:
    [tex]y=C_1(x)\frac{1}{x-1} + C_2(x)\frac{1}{x+1}[/tex]

    To solve I use this system of equations:

    [tex]C_1'(x)\frac{1}{x-1} + C_2'(x)\frac{1}{x+1} = 0[/tex]
    [tex]-C_1'(x)\frac{1}{(x-1)^2} - C_2'(x)\frac{1}{(x+1)^2} = 6x[/tex]

    Somehow, all of this should end up being: [itex]y=\frac{C_1}{x-1} + \frac{C_2}{x+1} + x[/itex], according to the answers, but I just can't get there. Was there anything wrong in the systems of equations? Or is it me solving for the constants (I didn't write it here)?
  2. jcsd
  3. Dec 9, 2009 #2


    Staff: Mentor

    It's not clear in your OP, but apparently y1 and y2 are solutions to the homogeneous equation. If that's the case, then you should try yp = Ax + B for your particular solution.
  4. Dec 11, 2009 #3
    Yes, those are solutions to the homogenous equation. I'm trying to solve it using the variation method, using y = c1(x) y1 + c2(x) y2.
  5. Dec 11, 2009 #4


    User Avatar
    Science Advisor

    You could do that, but Mark44's suggestion for "undetermined coefficients" is much simpler.
  6. Dec 13, 2009 #5
    Ok, so let's assume I still want to solve it using the variation method, just to see that I'm doing it ok (because obviously I went wrong there).

    Is my system ok? Any recommendations on a preferred way to solve it?
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