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Math Help - Differential equations

  1. #16
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    Quote Originally Posted by lvleph View Post
    I guess I don't understand what you are asking. If we assume that -A\sin wt + B \cos wt \ne 0 this leads you to conclude the only time this is zero is when mw^2 + cw + k = 0. Therefore, the we must solve for the roots of the characteristic equatioin mw^2 + cw + k = 0. Again, I am not sure what you are having issues with, and so I am sorry I am not of much help.


    After finding the roots
    w = \dfrac{-c \pm \sqrt{c^2 - 4mk}}{2m}



    How do I determine the equation:
    x(t) = C_1 e^{w_1t} + C_2e^{w_2t}
    ???
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  2. #17
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    Refer to posts #9 and #11.
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  3. #18
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    Quote Originally Posted by lvleph View Post
    Refer to posts #9 and #11.
    Post #9 and #11 explain how to obtain C1 and C2. I wonder how after finding the roots get the equation
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  4. #19
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    Oh okay, sorry for the confusion. The characteristic polynomial comes from assuming a solution, i.e., we assume x(t) = e^{wt} and plug this into the originial DEQ. Thus we obtain the following
    m \frac{d^2x}{dt^2} \left[e^{wt}\right] + c \frac{dx}{dt}\left[e^{wt}\right] + k e^{wt}
    which leads us to
    (mw^2 + cw + k)(e^{wt}) = 0.
    This tells us the characteristic equation is
    mw^2 + cw + k = 0.
    When we find the roots to this equation it tells us that we have one solution containing w_1 and another solution containing w_2, i.e., we have one solution x_1(t) = C_1 e^{w_1t} and a second solution x_2 = C_2 e^{w_2 t}. However, since both these are solutions to the differential equation we know that a linear combination of the two is also a solution to the differential equation, and thus we obtain the general solution
    <br />
x(t) = x_1(t) + x_2(t) = C_1 e^{w_1t} + C_2e^{w_2t}.<br />
    I really hope I finally answered your question. I suppose after all this you will probably understand the problem more thoroughly.
    Last edited by lvleph; December 1st 2009 at 11:01 AM. Reason: Made a mistake with constants.
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  5. #20
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    Quote Originally Posted by lvleph View Post
    Oh okay, sorry for the confusion. The characteristic polynomial comes from assuming a solution, i.e., we assume x(t) = e^{wt} and plug this into the originial DEQ. Thus we obtain the following
    m \frac{d^2x}{dt^2} \left[e^{wt}\right] + k \frac{dx}{dt}\left[e^{wt}\right] + c e^{wt}
    which leads us to
    (mw^2 + kw + c)(e^{wt}) = 0.
    This tells us the characteristic equation is
    mw^2 + kw + c = 0.
    When we find the roots to this equation it tells us that we have one solution containing w_1 and another solution containing w_2, i.e., we have one solution x_1(t) = C_1 e^{w_1t} and a second solution x_2 = C_2 e^{w_2 t}. However, since both these are solutions to the differential equation we know that a linear combination of the two is also a solution to the differential equation, and thus we obtain the general solution
    <br />
x(t) = x_1(t) + x_2(t) = C_1 e^{w_1t} + C_2e^{w_2t}.<br />
    I really hope I finally answered your question. I suppose after all this you will probably understand the problem more thoroughly.



    Thank you helping me a lot.Sorry, but why we assume x(t) = e^{wt} ???


    The equation would be:
    m \frac{d^2x}{dt^2} \left[e^{wt}\right] + c \frac{dx}{dt}\left[e^{wt}\right] + k e^{wt} ???
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  6. #21
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    Your absolutely correct, I made a mistake and have corrected the post. We assume this is the solution, because we want
    m x''(t) + c x'(t) + k x(t) = 0.
    This will only occur if x'',x',x have the same form. Well, we know that the derivative of e^t is itself, so this clues us into the type of solution we need, i.e., e^{wt}.
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  7. #22
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    Quote Originally Posted by lvleph View Post
    Your absolutely correct, I made a mistake and have corrected the post. We assume this is the solution, because we want
    m x''(t) + c x'(t) + k x(t) = 0.
    This will only occur if x'',x',x have the same form. Well, we know that the derivative of e^t is itself, so this clues us into the type of solution we need, i.e., e^{wt}.


    Thank you very much. To determine C1 and C2 need to know the initial conditions? I know you already explained, the more I do not get the initial solutions in terms of f (x) and g (x)
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  8. #23
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    The f(x),g(x) are just general forms for the intial conditions. They can be constants or functions of x.
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  9. #24
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    Quote Originally Posted by lvleph View Post
    The f(x),g(x) are just general forms for the intial conditions. They can be constants or functions of x.


    The first I understood
    x(0) = C_1e^{w_1 \cdot 0} + C_2e^{w_2\cdot 0} = C_1 + C_2 = L OK

    But the second condition
    x_t(0) = w_1\cdot C_1 e^{w_1\cdot 0} + w_2 \cdot C_2 e^{w_2\cdot 0} = w_1\cdot C_1 + w_2 \cdot C_2 = 0


    How did you find this equation?
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  10. #25
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    Well, I was given an example where g(x) = 0 and so I took the derivative of  x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}
    Then I plugged in  t= 0, but I knew from that x(0) = 0. Putting this all together we could solve for C_1,C_2
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  11. #26
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    Quote Originally Posted by lvleph View Post
    Well, I was given an example where g(x) = 0 and so I took the derivative of  x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}
    Then I plugged in  t= 0, but I knew from that x(0) = 0. Putting this all together we could solve for C_1,C_2

    Thank you friend. You did x'(0), initial velocity, I correct ?
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  12. #27
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    Yes, but how to solve for C_1,C_2 will always depend on the problem and what is given for initial conditions. Typically you will be given x(0),x'(0)
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  13. #28
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    Quote Originally Posted by lvleph View Post
    Yes, but how to solve for C_1,C_2 will always depend on the problem and what is given for initial conditions. Typically you will be given x(0),x'(0)

    In the underdamped system, we have c^2-4mk<0, Complex Roots.

    How can we determine the equation of the position in function of time. x(t) ?
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  14. #29
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    The solution will be of the form
    x(t) = C_1 \cos wt + C_2 \sin wt,
    where w = \dfrac{-c+\sqrt{c^2 - 4mk}}{2m}. This comes from eulers formula e^{i\theta} = \cos \theta + i \sin \theta.
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  15. #30
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    Quote Originally Posted by lvleph View Post
    The solution will be of the form
    x(t) = C_1 \cos wt + C_2 \sin wt,
    where w = \dfrac{-c+\sqrt{c^2 - 4mk}}{2m}. This comes from eulers formula e^{i\theta} = \cos \theta + i \sin \theta.
    Ok. We have:
    w1 = \frac{-c+i}{2m}
    w2 = \frac{-c-i}{2m}

    \frac{-c}{2m}+or-wi


    I could not solve the equation
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