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Nonhomogeneous problems

  1. Oct 17, 2006 #1
    I was wondering if anyone could check my work on this to make sure I'm doing this right for finding a particular solution to y''' + 3y'' + 3y' + y = e^(-x) + 1 + x. First I split the problem into 2 halfs y_p1 and y_p2.

    y_p1 = Ce^(-x)
    -Ce^(-x) + 3Ce^(-x) - 3Ce^(-x) + Ce^(-x) = e^(-x)
    0*Ce^(-x) = e^(-x)
    y_p1 = 0

    y_p2 = C1 + (C2)x
    0 + 3(0) + 3(C2) + C1 + (C2)x = 1 + x
    get like terms together so...
    (C2)x = x and 3(C2) + C1 = 1 therefore C2 = 1 and C1 = -2
    y_p2 = -2 + x

    particular solution = y_p1 + y_p2 = -2 + x
    and then the general solution to the problem would be:
    C1e^(-x) + c2(x)e^(-x) + C3(x^2)e^(-x)
     
  2. jcsd
  3. Oct 17, 2006 #2

    HallsofIvy

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    e-x, xe-x and x2e-x are solutions to the homogeneous solution so of course y_p1= e-x gives you nothing. Try y_p1= x3e-x.
     
  4. Oct 17, 2006 #3
    If I use (x^3)(e^-x) I get y_p = (1/6)(x^3)e^(-x) - 2 + x does that sound right?
     
  5. Oct 18, 2006 #4

    HallsofIvy

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    Yes, that is correct.
     
  6. Oct 18, 2006 #5
    Thanks... I understand why now.
     
  7. Oct 21, 2006 #6
    i find in this situation variation of parameters works better than undetermined coefficients, when you get to having to guess the form of somin like x^3e^-x i just freak out :P
     
  8. Oct 21, 2006 #7

    HallsofIvy

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    I don't see why you would have to "guess". Obviously, something of the form e-x is called for but e-x, xe-x, and x2e-x are already "taken". x3e-x is obviously "next in line".
     
  9. Oct 23, 2006 #8
    ok fair point :P im lazy though, i freak out in exams and make mistakes, variation of parameters, for me, leaves less room for error as its more methodical
     
  10. Nov 1, 2006 #9
    it involves less work in alot of cases as well
     
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