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Math Help - Differential equations

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

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


    I do not understand why the 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 w is of the form  w = a + bi then the solution should be of the form
     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
    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 w is of the form  w = a + bi then the solution should be of the form
     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
    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?

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

    and why 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 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.,
     e^{a \pm i b} = e^a \cos b \pm e^a i \sin b.
    So if our solution has complex eigenvalues w_1,w_2 the solution
     x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}
    where w_1 = a+bi,\; w_2=a-bi becomes
     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
     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 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.,
     e^{a \pm i b} = e^a \cos b \pm e^a i \sin b.
    So if our solution has complex eigenvalues w_1,w_2 the solution
     x(t) = C_1 e^{w_1t} + C_2 e^{w_2t}
    where w_1 = a+bi,\; w_2=a-bi becomes
     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
     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 w_1 = \frac{-c}{2m} + wi and 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 w_1,w_2 are not correct. The correct one are the solutions to 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 w_1,w_2 are not correct. The correct one are the solutions to mw^2 + cw + k=0.
    This equation is correct
    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
    <br />
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<br />
.
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  10. #40
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    Quote Originally Posted by lvleph View Post
    The correct solution is
    <br />
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<br />
.

    What is: \frac{\sqrt{c^2 - 4mc}}{2m} t ???
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  11. #41
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    \frac{\sqrt{c^2 - 4mk}}{2m} t is the imaginary portion of w_1\cdot t and  \frac{-c}{2m} t is the real part of w_1\cdot t,w_2\cdot t.
    Last edited by lvleph; December 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
    \frac{\sqrt{c^2 - 4mc}}{2m} t is the imaginary portion of w_1\cdot t and  \frac{-c}{2m} t is the real part of 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 \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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