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Homework Help: Differential Equations

  1. Jan 7, 2006 #1
    Solve :

    [tex]y'' + y' - 12y = 4x^2[/tex]
    The complementary equation I get is [tex] y1 = C1 e^3x + C2 e^-4x [/tex]

    But how to solve for the trial solution?
    I do it in this way:

    [tex] f(x) = 4x^2 [/tex]
    [tex] y2 = D (Ax^2 + Bx + C )[/tex]......
    What I want to know is whether my y2 is correct.
     
    Last edited: Jan 7, 2006
  2. jcsd
  3. Jan 8, 2006 #2

    Tide

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    Are you familiar with the method of variation of parameters?
     
  4. Jan 8, 2006 #3

    TD

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    You can certainly use undetermined coefficients since your RHS is just a polynomial. Since its degree is 2, your suggestion for a solution should be a general second degree polynomial as well.

    Your y2 is fine, but it can be simplified a bit, the D isn't necessary. If you would work it out, you'd get ADx²+BDx+CD where 'AD', 'BD' and 'CD' are again just 3 constants so just using A, B and C is fine - then you have the most general second degree polynomial.

    Find its first and second derivative, plug it into your DE and identify coefficients to solve for A, B and C :smile:
     
  5. Jan 8, 2006 #4
    i think you should try variation of parameters as tide suggested
     
  6. Jan 8, 2006 #5
    What is variation of parameters??
     
  7. Jan 8, 2006 #6

    TD

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    You can google it, the method is described well in http://www.math.utah.edu/~gustafso/2250variation-of-parameters.pdf" [Broken].
    This is certainly a method worth learning since it applies more generally than the method of undetermined coefficients (which only works for a limited number of RHS functions)

    Just as a note: your method (undetermined coefficients) will work here as well!
     
    Last edited by a moderator: May 2, 2017
  8. Jan 8, 2006 #7
    Thanks a lot.
     
  9. Jan 8, 2006 #8

    TD

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    You're welcome, don't hesitate to ask for help if you're stuck :smile:
     
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