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Math Help - Initial Value Problem - Differential Equation!!!

  1. #1
    Junior Member
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    Initial Value Problem - Differential Equation!!!

    Hi, For the following problem I am having hard time trying to find the phase angle (alpha).

    How to find the phase angle,

    isnt that alpha = arctan(B/A) if A,B>0 ?


    Problem:

    Solve the initial value problem
    (where primes indicate derivatives with respect to t). Express your solution as the sum of two oscillations,

    x
    (t) = (49/27)*cos(6t - ) + (5/27)cos(3t)


    where =

    (Recall that must be between 0 and 2.)

    But the phase angle is not right.

    Please help,
    Many Thanks.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    There is no phase angle here as the coefficient of sin (6t) = 0

    x= Acos(6t)+Bsin(6t) + 5/27 cos(3t)

    x(0) = A + 5/27 = 0 A = -5/27

    x ' (0) = 6B = 0 B = 0

    x(t) = -5/27 cos(6t) +5/27cos(3t)
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  3. #3
    Junior Member
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    Thanks a lot.

    Actually, I misinterpreted A as 5/27 which made me write the angle as arctan((49/27)/(5/27)).
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