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Differential Equation

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Write down the complementary solution [tex]y_c[/tex] and the correct particular solution [tex]y_p[/tex] in terms of its undetermined coefficients.

    [tex]y^{'''}+y^{''}+3y^{'} -5y=x^2e^{x}+x^{2}e^{-x}cos(2x)+xe^{-x}sin(2x)[/tex]

    3. The attempt at a solution
    I can easily find [tex]y_c[/tex] but I dont understand [tex]y_p[/tex].

    The solution is:TRY:[tex](Ax^2+Bx+C)e^x+e^{-x}[(Dx^2+Ex+F)cos(2x)+(Gx^2+Hx+I)sin(2x)][/tex]. And then you would adjust the particular solution. How come The [tex]Gx^2+Hx+I[/tex] term is a quadratic? If I didnt know the solution I would have put is as Gx+H.

    If you had a good site explaining undetermined coefficients, that would be awesome, I dont really like the way the book explains it.

    Thanks alot.
  2. jcsd
  3. Mar 2, 2008 #2


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    Staff Emeritus
    Science Advisor

    Why aren't you asking about the "Dx2+ Ex + F" as well? The answer is exactly the same:
    Since e-x cos(2x) and e-xsin(2x) are already solutions to the homogeneous equation, using (Dx+ C)cos(2x) and (Gx+ Y)sin(2x) would only give you "e-xcos(2x)" and "e-xsin(2x)", not "xe-xcos(2x)" and "x-xsin(2x)". Whenever a function already satisfies the homogeneous equation, you need to multiply whatever you would "normally try" by x.

    (You don't actually need Dx2+ Ex+ F or Gx2+ Hx+ I. What you need is (Dx2+ Ex)e-xcos(2x) and (Gx2+ Hx)e-xsin(2x). Try those and see what happens.)
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