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Thread: 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 $\displaystyle -A\sin wt + B \cos wt \ne 0$ this leads you to conclude the only time this is zero is when $\displaystyle mw^2 + cw + k = 0.$ Therefore, the we must solve for the roots of the characteristic equatioin $\displaystyle mw^2 + cw + k = 0.$ 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
    $\displaystyle w = \dfrac{-c \pm \sqrt{c^2 - 4mk}}{2m}$



    How do I determine the equation:
    $\displaystyle x(t) = C_1 e^{w_1t} + C_2e^{w_2t}$
    ???
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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 $\displaystyle x(t) = e^{wt}$ and plug this into the originial DEQ. Thus we obtain the following
    $\displaystyle m \frac{d^2x}{dt^2} \left[e^{wt}\right] + c \frac{dx}{dt}\left[e^{wt}\right] + k e^{wt}$
    which leads us to
    $\displaystyle (mw^2 + cw + k)(e^{wt}) = 0.$
    This tells us the characteristic equation is
    $\displaystyle mw^2 + cw + k = 0.$
    When we find the roots to this equation it tells us that we have one solution containing $\displaystyle w_1$ and another solution containing $\displaystyle w_2$, i.e., we have one solution $\displaystyle x_1(t) = C_1 e^{w_1t}$ and a second solution $\displaystyle x_2 = C_2 e^{w_2 t}.$ 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
    $\displaystyle
    x(t) = x_1(t) + x_2(t) = C_1 e^{w_1t} + C_2e^{w_2t}.
    $
    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; Dec 1st 2009 at 10:01 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 $\displaystyle x(t) = e^{wt}$ and plug this into the originial DEQ. Thus we obtain the following
    $\displaystyle m \frac{d^2x}{dt^2} \left[e^{wt}\right] + k \frac{dx}{dt}\left[e^{wt}\right] + c e^{wt}$
    which leads us to
    $\displaystyle (mw^2 + kw + c)(e^{wt}) = 0.$
    This tells us the characteristic equation is
    $\displaystyle mw^2 + kw + c = 0.$
    When we find the roots to this equation it tells us that we have one solution containing $\displaystyle w_1$ and another solution containing $\displaystyle w_2$, i.e., we have one solution $\displaystyle x_1(t) = C_1 e^{w_1t}$ and a second solution $\displaystyle x_2 = C_2 e^{w_2 t}.$ 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
    $\displaystyle
    x(t) = x_1(t) + x_2(t) = C_1 e^{w_1t} + C_2e^{w_2t}.
    $
    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 $\displaystyle x(t) = e^{wt}$ ???


    The equation would be:
    $\displaystyle m \frac{d^2x}{dt^2} \left[e^{wt}\right] + c \frac{dx}{dt}\left[e^{wt}\right] + k e^{wt}$ ???
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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
    $\displaystyle m x''(t) + c x'(t) + k x(t) = 0.$
    This will only occur if $\displaystyle x'',x',x$ have the same form. Well, we know that the derivative of $\displaystyle e^t$ is itself, so this clues us into the type of solution we need, i.e., $\displaystyle e^{wt}$.
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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
    $\displaystyle m x''(t) + c x'(t) + k x(t) = 0.$
    This will only occur if $\displaystyle x'',x',x$ have the same form. Well, we know that the derivative of $\displaystyle e^t$ is itself, so this clues us into the type of solution we need, i.e., $\displaystyle e^{wt}$.


    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 $\displaystyle f(x),g(x)$ are just general forms for the intial conditions. They can be constants or functions of $\displaystyle x$.
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  9. #24
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    Quote Originally Posted by lvleph View Post
    The $\displaystyle f(x),g(x)$ are just general forms for the intial conditions. They can be constants or functions of $\displaystyle x$.


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

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


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

    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 $\displaystyle C_1,C_2$ will always depend on the problem and what is given for initial conditions. Typically you will be given $\displaystyle x(0),x'(0)$
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  13. #28
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    Quote Originally Posted by lvleph View Post
    Yes, but how to solve for $\displaystyle C_1,C_2$ will always depend on the problem and what is given for initial conditions. Typically you will be given $\displaystyle x(0),x'(0)$

    In the underdamped system, we have $\displaystyle c^2-4mk<0$, 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
    $\displaystyle x(t) = C_1 \cos wt + C_2 \sin wt$,
    where $\displaystyle w = \dfrac{-c+\sqrt{c^2 - 4mk}}{2m}$. This comes from eulers formula $\displaystyle e^{i\theta} = \cos \theta + i \sin \theta$.
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  15. #30
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    Quote Originally Posted by lvleph View Post
    The solution will be of the form
    $\displaystyle x(t) = C_1 \cos wt + C_2 \sin wt$,
    where $\displaystyle w = \dfrac{-c+\sqrt{c^2 - 4mk}}{2m}$. This comes from eulers formula $\displaystyle e^{i\theta} = \cos \theta + i \sin \theta$.
    Ok. We have:
    $\displaystyle w1 = \frac{-c+i}{2m}$
    $\displaystyle w2 = \frac{-c-i}{2m}$

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


    I could not solve the equation
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