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Thread: Differential equations

  1. #31
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    Your solution will be $\displaystyle x(t) = C_1 \cos w_1 t + C_2 \sin w_1 t$. You just set $\displaystyle w=w_1$. You could very well have set $\displaystyle w=w_2$ this would just change the values of $\displaystyle C_1,C_2$. The conventions is to set $\displaystyle w=w_1$, but it doesn't matter.
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  2. #32
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    Quote Originally Posted by lvleph View Post
    Your solution will be $\displaystyle x(t) = C_1 \cos w_1 t + C_2 \sin w_1 t$. You just set $\displaystyle w=w_1$. You could very well have set $\displaystyle w=w_2$ this would just change the values of $\displaystyle C_1,C_2$. The conventions is to set $\displaystyle w=w_1$, but it doesn't matter.

    The solution is:
    $\displaystyle x(t) = e^{-(\frac{c}{2m})t}(C_1coswt+C_2sinwt)$


    I do not understand why the $\displaystyle e^{-(\frac{c}{2m})t}$ ???
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  3. #33
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    I was just about to write this. I made a mistake. Say $\displaystyle w$ is of the form $\displaystyle w = a + bi$ then the solution should be of the form
    $\displaystyle x(t) = C_1 e^{at} \cos bt + C_2 e^{at} \sin bt$.
    So, in your case above you should actually have the solution
    $\displaystyle x(t) = C_1 e^{-\frac{c}{2m}t} \cos \frac{\sqrt{c^2 - 4mc}}{2m} t + C_2 e^{-\frac{c}{2m}t} \sin \frac{\sqrt{c^2 - 4mc}}{2m} t$.
    I apologize for the oversight.
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  4. #34
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    Quote Originally Posted by lvleph View Post
    I was just about to write this. I made a mistake. Say $\displaystyle w$ is of the form $\displaystyle w = a + bi$ then the solution should be of the form
    $\displaystyle x(t) = C_1 e^{at} \cos bt + C_2 e^{at} \sin bt$.
    So, in your case above you should actually have the solution
    $\displaystyle x(t) = C_1 e^{-\frac{c}{2m}t} \cos \frac{\sqrt{c^2 - 4mc}}{2m} t + C_2 e^{-\frac{c}{2m}t} \sin \frac{\sqrt{c^2 - 4mc}}{2m} t$.
    I apologize for the oversight.

    Why is the equation of the form?

    $\displaystyle x(t) = C_1 e^{at} \cos bt + C_2 e^{at} \sin bt$

    and why $\displaystyle w = \frac{\sqrt{4mk-c^2}}{2m}$ ?
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  5. #35
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    Like I said I made a mistake and $\displaystyle w = w_1$. Previously, I assumed that the eigenvalues were purely imaginary which leads to a different solutions. This was incorrect. The form of the solution comes from Euler's Formula, i.e.,
    $\displaystyle e^{a \pm i b} = e^a \cos b \pm e^a i \sin b$.
    So if our solution has complex eigenvalues $\displaystyle w_1,w_2$ the solution
    $\displaystyle x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}$
    where $\displaystyle w_1 = a+bi,\; w_2=a-bi$ becomes
    $\displaystyle x(t) = C_1\left(e^{at} \cos bt + e^{at} i \sin bt\right) + C_2\left(e^{at}\cos bt - e^{at} i \sin bt\right)$.
    After further simplification the solution becomes
    $\displaystyle x(t) = C_1 e^{at} \cos bt + C_2 e^{at} \sin bt$.
    Do you see why my previous solution works if the eigenvalues are purely imaginary?

    Again, I am sorry for the confusion.
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  6. #36
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    Quote Originally Posted by lvleph View Post
    Like I said I made a mistake and $\displaystyle w = w_1$. Previously, I assumed that the eigenvalues were purely imaginary which leads to a different solutions. This was incorrect. The form of the solution comes from Euler's Formula, i.e.,
    $\displaystyle e^{a \pm i b} = e^a \cos b \pm e^a i \sin b$.
    So if our solution has complex eigenvalues $\displaystyle w_1,w_2$ the solution
    $\displaystyle x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}$
    where $\displaystyle w_1 = a+bi,\; w_2=a-bi$ becomes
    $\displaystyle x(t) = C_1\left(e^{at} \cos bt + e^{at} i \sin bt\right) + C_2\left(e^{at}\cos bt - e^{at} i \sin bt\right)$.
    After further simplification the solution becomes
    $\displaystyle x(t) = C_1 e^{at} \cos bt + C_2 e^{at} \sin bt$.
    Do you see why my previous solution works if the eigenvalues are purely imaginary?

    Again, I am sorry for the confusion.


    Thank you. I still have a doubt, why $\displaystyle w_1 = \frac{-c}{2m} + wi$ and $\displaystyle w_2 = \frac{-c}{2m} - wi$

    I do not understand these two equations
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  7. #37
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    Sorry there is still confusion. Those formulations for $\displaystyle w_1,w_2$ are not correct. The correct one are the solutions to $\displaystyle mw^2 + cw + k=0$.
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  8. #38
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    Quote Originally Posted by lvleph View Post
    Sorry there is still confusion. Those formulations for $\displaystyle w_1,w_2$ are not correct. The correct one are the solutions to $\displaystyle mw^2 + cw + k=0$.
    This equation is correct
    $\displaystyle x(t) = e^{-(\frac{c}{2m})t}(C_1coswt+C_2sinwt)$
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  9. #39
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    The correct solution is
    $\displaystyle
    x(t) = C_1 e^{-\frac{c}{2m}t} \cos \frac{\sqrt{c^2 - 4mc}}{2m} t + C_2 e^{-\frac{c}{2m}t} \sin \frac{\sqrt{c^2 - 4mc}}{2m} t
    $.
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  10. #40
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    Quote Originally Posted by lvleph View Post
    The correct solution is
    $\displaystyle
    x(t) = C_1 e^{-\frac{c}{2m}t} \cos \frac{\sqrt{c^2 - 4mc}}{2m} t + C_2 e^{-\frac{c}{2m}t} \sin \frac{\sqrt{c^2 - 4mc}}{2m} t
    $.

    What is: $\displaystyle \frac{\sqrt{c^2 - 4mc}}{2m} t$ ???
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  11. #41
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    $\displaystyle \frac{\sqrt{c^2 - 4mk}}{2m} t$ is the imaginary portion of $\displaystyle w_1\cdot t$ and $\displaystyle \frac{-c}{2m} t$ is the real part of $\displaystyle w_1\cdot t,w_2\cdot t$.
    Last edited by lvleph; Dec 3rd 2009 at 08:22 AM. Reason: fixed typo
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  12. #42
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    Quote Originally Posted by lvleph View Post
    $\displaystyle \frac{\sqrt{c^2 - 4mc}}{2m} t$ is the imaginary portion of $\displaystyle w_1\cdot t$ and $\displaystyle \frac{-c}{2m} t$ is the real part of $\displaystyle w_1\cdot t,w_2\cdot t$.

    Thank you very much
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  13. #43
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    The imaginary part is not $\displaystyle \frac{\sqrt{c^2-4mk}}{2m}$ ???
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  14. #44
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    Sorry I had a typo.
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  15. #45
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    Quote Originally Posted by lvleph View Post
    Sorry I had a typo.
    Ok. Please look at this site, the imaginary part is different, you know why?
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