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Need help understanding an aspect of undetermined coeff's

  1. Jan 19, 2016 #1
    8TOXi9t.png

    imgur link: http://i.imgur.com/8TOXi9t.png

    I am comfortable with the need to multiply the polynomial in front of [itex]e^{2x}[/itex] by [itex]x^3[/itex], that makes perfect sense in terms of what the text has already said about how no term in the particular solution should duplicate a term in the complementary solution, and that itself is logical in terms of the linear independence of the solutions.

    What I don't fully understand is the need to multiply the polynomials in front of the trig functions by [itex]x[/itex].

    [itex](D+Ex)\cos{3x} + (F+Gx)\sin{3x}[/itex] already has non-duplicated terms in it, ie, [itex]Ex\cos{3x}[/itex] and [itex]Gx\sin{3x}[/itex].

    Why wouldn't it be sufficient to simply use [itex](D+Ex)\cos{3x} + (F+Gx)\sin{3x}[/itex] and assume that D and F will be zero or else absorbed by linearity into the undetermined constant coefficients of the corresponding trig functions in the complementary solution?

    I have of course tried this, and it turns out that indeed the [itex]x\cos{3x}[/itex] and [itex]x\sin{3x}[/itex] terms cancel out when substituted back into the original differential equation, which is

    FLRpzcr.png

    imgur link: http://i.imgur.com/FLRpzcr.png

    I've subbed the trial solution [itex](D+Ex)\cos{3x} + (F+Gx)\sin{3x}[/itex] into [itex](D^2+9)y[/itex] (the factor in the diff eq that produces the trig components of the RHS), and I can see how in this instance the [itex]x\cos{3x}[/itex] and [itex]x\sin{3x}[/itex] terms cancel out, leaving no [itex]x\sin{3x}[/itex] term.

    I have come to agree with the insufficiency of [itex](D+Ex)\cos{3x} + (F+Gx)\sin{3x}[/itex] in this particular case, but I don't know why or how this carries over to the general case.

    Why wouldn't [itex](D+Ex)\cos{3x} + (F+Gx)\sin{3x}[/itex] generally be okay in the particular solution when no [itex]x\sin{3x}[/itex] or [itex]x\cos{3x}[/itex] terms appear in the complementary solution?
     
  2. jcsd
  3. Jan 19, 2016 #2

    HallsofIvy

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    Because sin(3x) and cos(3x) are part of the solution to the homogeneous equation, "Ax sin(3x)+ Bx cos(3x)" would give a solution to an equation with non-homogeneous part "sin(3x)+ cos(3x)". In order to get a solution with non-homogeneous part "x sin(3x)+ x cos(3x)" you have to multiply by x yet again, using [itex]x^2 sin(3x)+ x^2 cos(3x)[/itex]. In general, if you have a non-homogenous part [tex]x^n sin(3x)[/tex] and/or [tex]x^n cos(3x)[/tex] you need to use [tex]Ax^{n+1} sin(3x)+ Bx^{n+1} cos(3x)[/tex].
     
    Last edited by a moderator: Jan 19, 2016
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