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Math Help - help with ODE. *i got no solutions

  1. #1
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    help with ODE. *i got no solutions

    1) x'' + 2500x = 30 cos 45t where x(0) = 0; x''(0) = 0

    i worked it out as y = 0.03sin45t - 0.036 cos 45t.

    Yh = 0 and Yp = 0.03sin45t - 0.036 cos 45t.

    can anyone confirm this?

    2) x'' + 2x' + 26x = 82 cos 4t; x(0) = xa; x'(0) = 0
    for xa = 20; 0; 20
    somehow i worked it out as Yh = e^-t(cos5t) + (1/5)(e^-t(xa)(sin5t)
    Yp = (8/(-8+10i))cos4t + (8/(10+8i))sin4t

    i dun think it's right.

    really appreciate it if u guys can help. thanks!
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  2. #2
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    Quote Originally Posted by khairuldin View Post
    1) x'' + 2500x = 30 cos 45t where x(0) = 0; x''(0) = 0
    What is that ""? It looks like box to me. Is it a special symbol? I'm going to assume it was just a space and ignore it.

    i worked it out as y = 0.03sin45t - 0.036 cos 45t.

    Yh = 0 and Yp = 0.03sin45t - 0.036 cos 45t.

    can anyone confirm this?
    Do you mean that your "solution to the associated homogeneous equation" is Yh= 0? If so that is incorrect. Also it should be obvious to you that that equation does not satisfy either of your initial conditions! I suspect what you did was take the solution to the associated homogenous equation to be Yh= C cos(50 t)+ D sin(50 t) and then decide that C and D must to 0 to satisfy the initial conditions. You can't do that. You must find the entire solution first: Also your specific solution Yp is incorrect. It should be clear that since the differential equation has no "odd" derivatives, your solution cannot have a "sin(45 t)" term. Try just X(t)= A cos(45 t).

    Oh, and there is no "Y" in your problem. Your solution must be X(t), not Y(t).

    2) x'' + 2x' + 26x = 82 cos 4t; x(0) = xa; x'(0) = 0
    for xa = 20; 0; 20
    somehow i worked it out as Yh = e^-t(cos5t) + (1/5)(e^-t(xa)(sin5t)
    Yp = (8/(-8+10i))cos4t + (8/(10+8i))sin4t
    Again, you cannot determine the coeffients from the initial conditions until after you have the entire solution. Your particular solution does now involve both cos(4t) and sin(4t) but I have no idea how you could have gotten complex coefficients! I get the coefficient of cos(4x) to be 5 and the coefficient of sin(4x) to be 4!
    If you take x(t) (again, not Y!) to be A cos(4t)+ B sin(4t) then x'= -4A sin(4t)+ 4B cos(4t) and x"= -16A cos(4t)- 16B sin(4t). Putting those into the differential equation, -16A cos(4t)- 16B sin(4t)- 8A sin(4t)+ 8B cos(4t)+ 26A cos(4t)+ 26B sin(4t)= (-16A+ 8B+ 26A) cos(4t) + (-16B -8A+ 26B) sin(4t)= (10A+ 8B) cos(4t)+ (10B- 8A)sin(4t)= 82 cos(4t). In order for that to be true for all t, we must have 10A+ 8B= 82 and 10B- 8A= 0. I've been know to make arithmetic errors but I dont' believe that can give complex solutions for A and B!
    i dun think it's right.

    really appreciate it if u guys can help. thanks!
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