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Math Help - Solving for x ? [differential equation]

  1. #1
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    Solving for x ? [differential equation]

    Hello,

    I'm stuck with this equation and I can't figure out a way how to solve for t..
    So I got the solution of a differential equation(complex roots), and now I need to find the values of t for which y=3.5

    3.5 = 12e^(-αt)*cos(ω t) - 12α/ω*e^(-αt)*sin(ω t)

    α and ω are constants, I tried changing it to the exponential form, cosine form but it didn't help.
    In fact what I need is the time intervals for which y(t) > 3.5

    Any tips or advice how to do it ?

    Thanks !
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  2. #2
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    Exponential form should work...

    What is your original DE?
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  3. #3
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    d^2 y/dt^2 + 2α * dy/dt + (ω0)*y=0

    But with the exponential form, I get 6 *(e^complexNum + e^conjugate) ??
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  4. #4
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    Quote Originally Posted by Ihaz View Post
    d^2 y/dt^2 + 2α * dy/dt + (ω0)*y=0

    But with the exponential form, I get 6 *(e^complexNum + e^conjugate) ??
    Dear Ihaz,

    \frac{d^{2}y}{dt^{2}}+2a\frac{dy}{dt}+\omega_{0}y=  0

    The solution will be, y=Ae^{(-a+\omega)t}+Be^{(-a-\omega)t}~ where~\omega=\sqrt{a^2-\omega_{0}^{2}}~and~A,B:~arbitary~ constants

    So in this form using the boundry conditions you can find A and B. Can you give us more infromation about the question. What are the boundry conditions?
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  5. #5
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    I already found the solution of the D.E :
    12e^(-αt)*cos(ωn t) - 12α/ωn*e^(-αt)*sin(ωn t)

    Now what I need is at what values of time will this be equal to 3.5 ??
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