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Particular solution to cosine

  1. Mar 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Q'' + 100Q' +50000Q = 4000cos(100t)
    i found the general solution to be e-50t[Acos(50sqrt19)t +Bsin(50sqrt19)t]

    but i have a problem with the particular solution

    i tried Cei100t

    did i try the wrong expression? because when i compared coefficients, i found C=4000/40000 = 0.1
    but the answer given by my lecturer is 16/170 cos100t + 4/170sin100t

    idon't understand where the sin term comes from? when i try Cei100t, i should be getting just the cosine terms right? since my target is 4000cos100t. so how come there's a sin term?

    help appreciated!
     
  2. jcsd
  3. Mar 25, 2010 #2

    Mark44

    Staff: Mentor

    No, this isn't the general solution. It looks more like the solution to the homogeneous problem, namely Q'' + 100Q' + 50000Q = 0. I didn't check your solution, but you can by making sure that it satisfies Q'' + 100Q' + 50000Q = 0.

    BTW, the t factor should be inside the parentheses: e.g., Acos(50sqrt(19)t)
    For reasons too long to go into here, I would recommend that you try a particular solution of yp = Ccos(100t) + Dsin(100t).
     
  4. Mar 25, 2010 #3
    oh. i always thought that was called the general solution? where you let the characteristic equation = 0

    oh my mistake, t should be in the brackets yes

    thats the problem. i don't understand. in another question, y'' +4y = cos2x. i could try Z = Ae2ix, so that the real part would just be cos2x and i discard the imaginary part

    so why in this case i have to use Acosx + Bsinx ? also, how do i know when i have to use this instead of Z = Ae2ix?

    many thanks!
     
  5. Mar 25, 2010 #4

    Mark44

    Staff: Mentor

    The general solution consists of two parts: the solution to the homogeneous problem + the particular solution.

    For your second question, you're making two mistakes.
    1. In the homogeneous equation (y'' + 4y = 0), the characteristic equation has two solutions - r = 2i and r = -2i. Rather than have a homogeneous solution of y = Ae2ix + Be-2ix, it's much easier to use y = Acos(2x) + Bsin(2x).
    2. Since the nonhomogeneous equation is y'' + 4y = cos(2x), a particular solution involving e2ix or e-2ix isn't going to work. When a function is a solution of the homogeneous problem, it can't also be a solution of the corresponding nonhomogeneous problem.

    For your particular solution, try y = Cxcos(2x) + Dxsin(2x).
     
  6. Mar 25, 2010 #5
    ah i see... so as long as i see cosx or sinx, i should use Ccosx+Dsinx? when i see tan i should use method of variation of parameters?

    i went to look through my lecturer's notes again, and i realise what he did was to introduce a imaginary part in the exact same format as the real part

    i.e , if it was y" + y' + 4y = cosx
    he introduced a iv" + iv' +4iv = isinx

    then he combine to get z = y + iv. so he says y = Re(Z).
    and he trys eix or something like this

    so his method would work also?

    but if i use Ccosx+Dsinx, i don't have to worry about imaginary or real parts right?
     
  7. Mar 27, 2010 #6

    Mark44

    Staff: Mentor

    Yes, both methods work, but if you use the method I described, you don't have to be concerned with the imaginary parts.
     
  8. Mar 27, 2010 #7
    ah i see! thanks
     
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