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Math Help - Trig derivative questions.

  1. #1
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    Trig derivative questions.

    I\ have\ an\ equation\ and\ solution.\ I'm\ unclear\ on\ a\  couple\ of\ things.

    An\ analysis\ of\ the\ motion\ of\ a\ particle\ resulted\ in\ the\  differential\ equation

    \frac{d^2y}{dt^2}\ +\ b^2y\ =\ 0

     Show\ that\ y=a\ sin\ bx\ is\ a\ solution\ of\ this\  equation.

    Solution

    y=a\ sin\ bx

    \frac{dy}{dt}\ =\ ab\ cos\ bt\ Why\ has\ x\ become\ t\ and\ why\  is\ there\ another\ b\ in\ the\ equation?
    What\ is\ the\ rule\ for\ finding\ the\ derivative?

    \frac{d^2y}{dt^2}\ =\ -ab^2\ sin\ bt

    \frac{d^2y}{dt^2}\ =\ -b^2(a\ sin\ bt)

    \frac{d^2y}{dt^2}\ =\ -b^2y

    \frac{d^2y}{dt^2}+b^2y=-b^2y+b^2y=0

    y=a\ sin\ bt\ is\ the\ solution\ of \ the\ given\ differential\  equation.
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  2. #2
    Super Member Random Variable's Avatar
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    The particular solution should be  y = a \sin bt .

    Then using the chain rule,  \frac{dy}{dt} = a \cos bt \ \frac{d}{dt} \ (bt) = ab \cos bt

    and  \frac{d^{2}y}{dt^{2}} = -ab \sin bt \ \frac{d}{dt} (bt) = - ab^{2} \sin bt
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  3. #3
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    Quote Originally Posted by Random Variable View Post
    The particular solution should be  y = a \sin bt .

    Then using the chain rule,  \frac{dy}{dt} = a \cos bt \ \frac{d}{dt} \ (bt) = ab \cos bt

    and  \frac{d^{2}y}{dt^{2}} = -ab \sin bt \ \frac{d}{dt} (bt) = - ab^{2} \sin bt
    Why isn't the solution y=C_1cos(bt)+C_2sin(bt)?
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  4. #4
    Super Member Random Variable's Avatar
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    Quote Originally Posted by dwsmith View Post
    Why isn't the solution y=C_1cos(bt)+C_2sin(bt)?
    That's the general solution. What's given is a particular solution (when  C_{1}=0 and  C_{2} = a ).
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  5. #5
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    Quote Originally Posted by Random Variable View Post
    That's the general solution. What's given is a particular solution (when  C_{1}=0 and  C_{2} = a ).
    Ok
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