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Variation of Parameters

  1. Mar 21, 2008 #1
    Solve by method of variation of parameters
    (x^2)y'' - (4x)y' + 6y = x^4*sinx (x > 0)

    Hey, I know how to solve problems using variation of parameters but only when the corresponding homogenous equation has constant coefficients...

    y'' - (4/x)y' + (6/x^2)y = 0.. the bit im confused about is how to obtain the fundamental solutions to this equation {y1, y2} when the coefficients are not constants. Any help would be appreciated.

  2. jcsd
  3. Mar 21, 2008 #2


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    Note that each term has the same "units", if you think of x as having units of length (so that a derivative removes one unit of length). Such differential equations have solutions of the form x^r. Plug this in and solve for r, and you'll quickly see why such solutions work. This is something you can just remember, although it would also fall out if you tried the Frobenius method.
  4. Mar 21, 2008 #3
    im not quite sure I understand where to plug in x^r .
  5. Mar 21, 2008 #4


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    I mean y(x)=xr is a solution to the homogenous differential equation for certain r. Plug in this y and see which r work.
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