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Math Help - I need to solve this D. E. (( plz help )

  1. #1
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    I need to solve this D. E. (( plz help )

    Attached Thumbnails Attached Thumbnails I need to solve this  D. E. (( plz help )-untitled-1.jpg  
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by amro05 View Post
    Are you sure? Where did this ODE come from? It clearly has no real solutions and I suspect the complex solutions are all constant derivative solutions.

    CB
    Last edited by CaptainBlack; June 28th 2009 at 09:30 AM.
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  3. #3
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    I get y(t)=c_1-t\text{W}(-1)

    where W is the Lambert W-function. May I ask what's wrong with a complex solution?
    Last edited by CaptainBlack; June 28th 2009 at 09:30 AM. Reason: edited rather than quoted
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by shawsend View Post
    I get y(t)=c_1-t\text{W}(-1)

    where W is the Lambert W-function. May I ask what's wrong with a complex solution?
    Except the Lambert-W is mulitivalued (and W(-1) is complex for all branches IIRC), you do not know which branch you are on without an initial condition, and not all initial conditions give solutions.
    CB
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  5. #5
    MHF Contributor chisigma's Avatar
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    The equation...

    e^{z}=z (1)

    ... in the complex domain is equilvalent to a couple of equations in real domain. Setting  z= \sigma + i\cdot \omega the (1) became...

    e^{\sigma} \cdot \cos \omega = \sigma

    e^{\sigma}\cdot \sin \omega = \omega (2)

    If \sigma^{*} + i\cdot \omega^{*} is a solution of (2) then the proposed D.E. becomes...

     y^{'}= \sigma^{*} + i\cdot \omega^{*} (3)

    ... and its solution is...

     y^{'}= \{\sigma^{*} + i\cdot \omega^{*}\} \cdot x + c (4)

    The question is now: does (2) have solutions?...

    Kind regards

    \chi \sigma
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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
    Except the Lambert-W is mulitivalued (and W(-1) is complex for all branches IIRC), you do not know which branch you are on without an initial condition, and not all initial conditions give solutions.
    CB
    If you might, which initial conditions do not give solutions?
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    Are you sure? Where did this ODE come from? It clearly has no real solutions and I suspect the complex solutions are all constant derivative solutions.

    CB
    let meput it this way
    this ODE came from the solution of

    the last one gives this ODE


    the last one is to hard to solve so I put it = 0 < jest hopping to find same idees
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  8. #8
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by chisigma View Post
    The equation...

    e^{z}=z (1)

    ... in the complex domain is equilvalent to a couple of equations in real domain. Setting  z= \sigma + i\cdot \omega the (1) became...

    e^{\sigma} \cdot \cos \omega = \sigma

    e^{\sigma}\cdot \sin \omega = \omega (2)

    ...

    ... the question is now: does (1) have solutions?...
    The answers is done by the follwing lemma of complex variable functions theory...

    Given the complex variable function...

    w = f(z) = c_{1} z + c_{2} z^{2} + c_{3} z^{3} + ... , c_{1} \ne 0 (1)

    ... holomorphic in a circle |z|< \epsilon , its inverse function is done by the Taylor expansion...

    z= f^{-1} (w) = d_{1} w + d_{2} w^{2} + d_{3} w^{3} + ... ,  d_{n} = \frac{1}{n!} \lim _{z \rightarrow 0} \frac {d^{n-1}}{dz^{n-1}} \{\frac{z}{f(z)}\}^{n} (2)

    Now we use this lemma to find the solution z=z(w) of the trascendental equation...

     w= z\cdot e^{-z} (3)

    With the formulas (2) we obtain the Taylor expansion...

     z= \sum_{n=1}^{\infty} \frac {n^{n-1}}{n!} w^{n} (4)

    ... which converges only for w=0. But our equation was...

    z\cdot e^{-z}=1 (5)

    ... and then it has been demonstrated it has no solutions...

    Kind regards

    \chi \sigma
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