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Sin+cos solution undetermined coeff. de

  1. Jun 22, 2005 #1
    This looks like a very simple one, but i ran out of ideas. I do not get the answer and unlikely will get it without some help:

    y'' + 16y = 24 cos(4x)

    I found homogeneous solution:

    y = c1ei4 + c2e-i4

    I haven't re-written it using euler's formula yet.
    So, for particular solution the general form is:

    y = x(Acos(4x) + Bsin(4x))

    I differentiated it and plugged into the DE and here's what i have:

    -8Asin(4x) + 8Bcos(4x) + xsin(4x)*[16A - 16B] + xcos(4x)*[16B - 16A] = 24cos(4x);

    It seems like last two terms need to go away, but then A = B and to get rid of the sin A = 0! the answer is totally different though.
    Where am i going wrong?

    Thanks for any hints.

    EDIT: or would yp = x2(Acos(4x) + Bsin(4x))
    I'm confused... :confused: i can't work it out with x2 either ...
     
    Last edited: Jun 22, 2005
  2. jcsd
  3. Jun 23, 2005 #2
    I find the last two terms do cancel

    y = x(Acos(4x) + Bsin(4x))

    y' = Acos(4x) + Bsin(4x) + x (-4Asin(4x) + 4Bcos(4x))

    y" = -4Asin(4x) + 4Bcos(4x) -4Asin(4x) + 4Bcos(4x) + x (-16Acos(4x) - 16Bsin(4x))

    Putting back into the equation y" + 16y = 24cos(4x) gives

    -4Asin(4x) + 4Bcos(4x) -4Asin(4x) + 4Bcos(4x) + x (-16Acos(4x) - 16Bsin(4x))

    + 16 ( x(Acos(4x) + Bsin(4x)) ) = 24 cos(4x)

    Here we are left with

    -8Asin(4x) + 8Bcos(4x) = 24 cos(4x)

    so B = 3 and A = 0

    so a

    y(p) = 3x sin(4x)
     
  4. Jun 23, 2005 #3
    Thanks for reply.
    that was my "other" solution. Good thing it matched yours. Come to find out: the book answer was wrong :mad:
     
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