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Help with interpretation of question. (Differential equations)

  1. Apr 26, 2012 #1
    http://dl.dropbox.com/u/33103477/Linear%20oscilator.png [Broken]

    I am having trouble understand the question, what I have done its solve the equation using the substitution [tex]x=e^{rx}[/tex]

    Then, I have the solution given by:

    [tex] x(t)=c_1 e^{t(\sqrt{\gamma^2 - \omega^2 })} + c_2e^{-t(\sqrt{\gamma^2 - \omega^2)}} [/tex]

    So at x(0)=0,

    [tex] c_1 + c_2 = 0[/tex]

    and, x'(0)=v
    [tex] v=c_2(\sqrt{\gamma^2 - \omega^2} - \omega) + c_1(\sqrt{\gamma^2 - \omega^2}+\omega) [/tex]

    Not quite sure how to proceed as if \omega is greater than \gamma I get imaginary values and the opposite gives real, but what does that mean ?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Apr 26, 2012 #2
    Assuming everything is correct and I see it is correct up to taking the derivative and plugging in zero (I was too lazy to take the derivative). If omega > gamma, you would have [itex]e^{ait}[/itex] where I am using a = radical.

    What is [itex]e^{ait}\mbox{?}[/itex] Can you rewrite e in this form as some other transcendental?

    If they are real, you just have a DE with of the form Ae+Be.
    Equal you just have a constant.

    Also, I see you didn't solve for you constant. You set it up but that was it.
    $$
    \begin{bmatrix}1&1&0\\
    \sqrt{\gamma^2-\omega^2}-\omega & \sqrt{\gamma^2-\omega^2}+\omega & v
    \end{bmatrix}
    $$
    When you solve this system, what do you get for [itex]c_1,c_2[/itex]?
     
    Last edited: Apr 26, 2012
  4. Apr 27, 2012 #3
    Okay so working out [tex] c_1 = -v/2\omega [/tex] and [tex] c_2 = v/2\omega [/tex] what does that mean for the formula I don't get what I'm suppose to interpret from these results ?
     
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