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Thread: Trig derivative questions.

  1. #1
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    Trig derivative questions.

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

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

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

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

    $\displaystyle Solution$

    $\displaystyle y=a\ sin\ bx$

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

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

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

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

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

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

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

    and $\displaystyle \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 $\displaystyle y = a \sin bt $.

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

    and $\displaystyle \frac{d^{2}y}{dt^{2}} = -ab \sin bt \ \frac{d}{dt} (bt) = - ab^{2} \sin bt $
    Why isn't the solution $\displaystyle y=C_1cos(bt)+C_2sin(bt)$?
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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 $\displaystyle y=C_1cos(bt)+C_2sin(bt)$?
    That's the general solution. What's given is a particular solution (when $\displaystyle C_{1}=0 $ and $\displaystyle 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 $\displaystyle C_{1}=0 $ and $\displaystyle C_{2} = a $ ).
    Ok
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