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Form of particular solution for y''-2y'+y=(e^2)/x

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the general solution for:
    y''-2y'+y=ex/x



    2. Relevant equations
    NONE - not an initial value problem


    3. The attempt at a solution
    Solve the homogeneous first:
    r2-2r+1=0
    r=1 as a double root

    So:

    y1=c1ex
    y2=c2xex

    ...but what in God's name is the form for the particular Y (based on the right side of the equation?). I suppose it's the "hard part" of this exercise but I've tried a couple and still don't see it.
     
    Last edited: Apr 21, 2010
  2. jcsd
  3. Apr 21, 2010 #2

    LCKurtz

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    Do you type that right and really mean the right side is

    [tex]\frac {e^2} x[/tex]

    If so, the method would be variation of parameters.
     
  4. Apr 21, 2010 #3
    Right side corrected, should be y''-2y'+y=(ex)/x
     
  5. Apr 21, 2010 #4

    LCKurtz

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