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Math Help - Having trouble with a trig under. coeffecient

  1. #1
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    Having trouble with a trig under. coeffecient

    im having trouble doing a trig coeffecient.

    t'' + 2t' +6x = 10 cos 3t

    i got the homogenius side, but now im having trouble with the right. Any help would be appreciated.

    *edit*

    I did get it to -3 A cos (3t) - 3Bsin (3t) - 6A sin (3t) + 6b cos (3t) = 10 cos 3t
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  2. #2
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    Quote Originally Posted by neven87 View Post
    im having trouble doing a trig coeffecient.

    t'' + 2t' +6x = 10 cos 3t

    i got the homogenius side, but now im having trouble with the right. Any help would be appreciated.

    *edit*

    I did get it to -3 A cos (3t) - 3Bsin (3t) - 6A sin (3t) + 6b cos (3t) = 10 cos 3t
    Something is really messed up in this equation. Are we solving for t(x) or x(t)? Is it possibly supposed to be
    x'' + 2x' +6x = 10 cos 3t ?

    -Dan
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  3. #3
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    well actually equals d2x/dt2= t'' and dx/dt

    dont know why i used t <shrug>
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by neven87 View Post
    well actually equals d2x/dt2= t'' and dx/dt

    dont know why i used t <shrug>
    Okay, so it's
    x^{\prime \prime} + 2x^{\prime} + 6x = 10 cos(3t)

    You said you have the homogeneous solution, so I'm not going to do that. I will simply say that the homogeneous solution contains no terms like Asin(3t) or Bcos(3t).

    We need a particular solution. Given the form of the RHS I'm going to guess a particular solution
    x_p(t) = Asin(3t) + Bcos(3t)

    x_p^{\prime}(t) = 3Acos(3t) - 3Bsin(3t)

    x_p^{\prime \prime} = -9Asin(3t) - 9Bcos(3t)

    Inserting these into the differential equation gives:
    (-9Asin(3t) - 9Bcos(3t)) + 2(3Acos(3t) - 3Bsin(3t)) + 6(Asin(3t) + Bcos(3t))  = 10cos(3t)

    (-9A - 6B + 6A)sin(3t) + (-9B + 6A + 6B)cos(3t) = 10cos(3t)

    The only way for this to be true for all t is for the coefficients of the trig functions to be the same on both sides of the equation. Thus:
    -9A - 6B + 6A = 0
    and
    -9B + 6A + 6B = 10

    The first equation says
    -3A - 6B = 0 ==> B = -\frac{1}{2}A

    Inserting this into the second equation gives:
    -9 \cdot -\frac{1}{2}A + 6A + 6 \cdot -\frac{1}{2}A = 10 <-- Multiply both sides by 2

    9A + 12A - 6A = 20

    15A = 20

    A = \frac{20}{15} = \frac{4}{3}

    Thus
    B = -\frac{1}{2} \cdot \frac{4}{3} = -\frac{2}{3}

    So your particular solution is:
    x_p(t) = \frac{4}{3}sin(3t) - \frac{2}{3}cos(3t)

    -Dan
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  5. #5
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    Ah, made a mistake after the subing, thanks.
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