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Thread: Undetermined coefficients

  1. #1
    Junior Member
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    Undetermined coefficients

    $\displaystyle \frac{d^2x}{dt^2} + \omega^2x = F_0sin(\omega t), x(0) = 0 , x'(0) = 0$

    Because this problem is not in term of x and y. Then I assume x is t and y is x.

    For general solution,
    $\displaystyle x = x_c + x_p$

    $\displaystyle m^2 + \omega^2 = 0$

    $\displaystyle m = i\sqrt{\omega}$

    $\displaystyle x_c = e^t(c_1sin(\sqrt{\omega}t) + c_2cos(\sqrt{\omega}t))$

    And for particular solution,
    $\displaystyle x_p = Asin(\omega t)$

    $\displaystyle x'_p = Acos(\omega t)$

    $\displaystyle x''_p = -Asin(\omega t)$

    Substitute into $\displaystyle \frac{d^2x}{dt^2} + \omega^2x$ form,

    $\displaystyle -Asin(\omega t) + \omega^2Asin(\omega t) = F_0sin(\omega t)$

    Then, $\displaystyle A = \frac{F_0}{\omega^2 - 1}$

    Finally, $\displaystyle x = e^t(c_1sin(\sqrt{\omega}t) + c_2cos(\sqrt{\omega}t)) + \frac{F_0}{\omega^2 - 1}sin(\omega t)$

    My question is how x(0) = 0 and x'(0) = 0 involve with this problem?
    Thank you.

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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by noppawit View Post
    $\displaystyle \frac{d^2x}{dt^2} + \omega^2x = F_0sin(\omega t), x(0) = 0 , x'(0) = 0$

    Because this problem is not in term of x and y. Then I assume x is t and y is x.

    For general solution,
    $\displaystyle x = x_c + x_p$

    $\displaystyle m^2 + \omega^2 = 0$

    $\displaystyle m = i\sqrt{\omega}$

    $\displaystyle x_c = e^t(c_1sin(\sqrt{\omega}t) + c_2cos(\sqrt{\omega}t))$

    And for particular solution,
    $\displaystyle x_p = Asin(\omega t)$

    $\displaystyle x'_p = Acos(\omega t)$

    $\displaystyle x''_p = -Asin(\omega t)$

    Substitute into $\displaystyle \frac{d^2x}{dt^2} + \omega^2x$ form,

    $\displaystyle -Asin(\omega t) + \omega^2Asin(\omega t) = F_0sin(\omega t)$

    Then, $\displaystyle A = \frac{F_0}{\omega^2 - 1}$

    Finally, $\displaystyle x = e^t(c_1sin(\sqrt{\omega}t) + c_2cos(\sqrt{\omega}t)) + \frac{F_0}{\omega^2 - 1}sin(\omega t)$

    My question is how x(0) = 0 and x'(0) = 0 involve with this problem?
    Thank you.

    They are the initial conditions and determine the values of the arbitrary constants $\displaystyle c_1$ and $\displaystyle c_2$ (that is if the general solution was correct which it is not, the solution to the homogeneous equation is wrong ($\displaystyle m=\pm i \omega$) as is the particular integral)

    Spoiler:

    $\displaystyle x={\it k_1}\,\sin \left(t\,w\right)+{\it k_2}\,\cos \left(t\,w\right)-{{F_0\,t\,\cos \left(t\,w \right)}\over{2\,w}}$

    CB
    Last edited by CaptainBlack; Jul 10th 2009 at 02:08 AM.
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