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Math Help - second order initial value problem

  1. #1
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    Question second order initial value problem

    to solve this:
    y = (-1/20)*e^(-x)*cos(4*x) - (1/40)*e^(-x)*sin(4*x) + A*e^x + B*e^(-x)
    using
    y(0) = 1, y'(0) = 2

    i got:
    y = (-1/20)*e^(-x)*cos(4*x) - (1/40)*e^(-x)*sin(4*x) + 0+ 1

    which is not right. Whats wrong?
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  2. #2
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    Quote Originally Posted by taurus View Post
    to solve this:
    y = (-1/20)*e^(-x)*cos(4*x) - (1/40)*e^(-x)*sin(4*x) + A*e^x + B*e^(-x)
    using
    y(0) = 1, y'(0) = 2

    i got:
    y = (-1/20)*e^(-x)*cos(4*x) - (1/40)*e^(-x)*sin(4*x) + 0+ 1

    which is not right. Whats wrong?
    Are you trying to solve for the constants A and B? It might help if you give the DE.
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  3. #3
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    yes for A and B
    so i did:

    - for y(0) = 1 i get
    (-1/20) + A + B = 1 => A + B = (1/20)

    for y'(0) =2 i get
    -3Ae^(-3x) -3Be^(-3x) + x^2 -3e^(-3x)
    => -3A - 3B = 5

    then i got a = 0 and b = 1
    ?
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  4. #4
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    Quote Originally Posted by taurus View Post
    yes for A and B
    so i did:

    - for y(0) = 1 i get
    (-1/20) + A + B = 1 => A + B = (1/20) Mr F says: A + B = 1 + (1/20) = 21/20.

    for y'(0) =2 i get
    -3Ae^(-3x) -3Be^(-3x) + x^2 -3e^(-3x)
    => -3A - 3B = 5

    Mr F says: I have no idea why you have 3's instead of 4's in the exponentials, and I don't know where the trig bits have gone. I doubt -3 A - 3 B = 5 is correct. I suggest you go back to your expression for y and differentiate it carefully. Then substitute x = 0 and get a seond equation. I've probably slipped up but I get 2 = (-1/20)(-1) - (1/40)(1) + A - B etc. ..

    then i got a = 0 and b = 1
    ?
    I would suggest that greater care in doing the calculations rather than explicit help is what's need here.
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