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Using the D operator technique to solve a trig based DE

  1. Aug 4, 2012 #1
    hi everyone,

    im currently trying to teach myself the D operator technique, as opposed to the 'guess method' which i don't really like.

    I stumbled upon this on yahoo answers:

    (D^2 + 1)y = 4 cos x - sin x

    Find the complementary function by solving the auxiliary equation:
    m² + 1 = 0
    m² = -1
    m = ±i
    yᶜ = Asinx + Bcosx

    Find the particular integral by comparing coefficients:
    yᵖ = Cxsinx + Dxcosx
    yᵖ' = (C - Dx)sinx + (Cx + D)cosx
    yᵖ'' = (2C - Dx)cosx - (Cx + 2D)sinx
    yᵖ'' + yᵖ = 4cosx - sinx
    (2C - Dx)cosx - (Cx + 2D)sinx + Cxsinx + Dxcosx = 4cosx - sinx
    2Ccosx - 2Dsinx = 4cosx - sinx
    2C = 4
    C = 2
    2D = 1
    D = ½
    yᵖ = 2xsinx + xcosx / 2

    Find the general solution by combining these two parts:
    y = yᶜ + yᵖ
    y = Asinx + Bcosx + 2xsinx + xcosx / 2
    y = (2x + A)sinx + (x + B)cosx / 2

    That's all well and good, but from my understanding to solve a DE of the form
    y'' -2y' -3y = cos3x
    using the D operator technique I would say ok
    ypi=1/ (D2-2D-3) cos3x
    and then replace D2's with -(α)2 where α is 3 in this case
    and then proceed to solve by using D2 terms

    In the yahoo answers example are they using a different method because there are 2 trig functions on the RHS? I'm also unsure if they have used D as a constant and as the operator?

    Sorry for the long post, bottom line is I want to able to use the D operator technique to solve a DE with multiple trig terms on the RHS, and is there a way similar to how you would solve for one trig term, as I explained?

    Mitch
     
  2. jcsd
  3. Aug 5, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey mitch_1211.

    The idea will be to apply 1/P(D) (where P(D) is the operator definition on the LHS) to each term on the RHS. You have the property of linearity which means you can separate the terms and solve each one individually.

    Just to clarify, this means 1/P(D)[f(x) + g(x)] = [1/P(D)]f(x) + [1/P(D)]g(x)
     
  4. Aug 5, 2012 #3
    Oh I didn't even think of that! Thank you so much that makes things much easier
     
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