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Reduction of order of an ODE

  1. Feb 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Use the method of reduction of order to find another independently linear solution y2(x) when given one solution.

    [tex] x^2y'' - x(x+2)y' + (x+2)y = 0 [/tex]

    [tex] y_1(x) = x [/tex]


    3. The attempt at a solution

    Hopefully y2(x) will take the form of v(x)y1(x) or I have no idea how to solve the ODE. I start by finding y'2(x) and y''2(x).

    [tex] y_2 = vx [/tex]
    [tex] y'_2= v + xv' [/tex]
    [tex] y''_2 = v' + v' + xv'' = 2v' + xv'' [/tex]

    I substitute the above into the equation:

    [tex] x^2(2v' + xv'') - x(x+2)(v + xv') + (x+2)vx = 0 [/tex]

    [tex] 2x^2v' + x^3v'' - x^2v - x^3v' - 2xv - 2x^2v' + x^2v + 2xv = 0 [/tex]

    And this is where I've gotten to. Everything cancels out and I can't see how I'll find my v...
     
  2. jcsd
  3. Feb 22, 2013 #2

    vela

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    You have two terms left over: ##x^3 (v'' - v')=0##.
     
  4. Feb 22, 2013 #3
    Oh, wow. I overlooked the same thing three times...

    Might be worth switching notation to avoid this from happening again.

    Thanks.
     
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