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Complex Differntial Equation

  1. Aug 31, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\frac{dy}{dx}=e^{x+iy}[/tex]


    2. Relevant equations

    [tex]e^{x+iy}=e^{x}(\cos y+i\sin y)[/tex]

    3. The attempt at a solution
    [tex]\frac{dy}{dx}=e^{x}(\cos y+i\sin y)[/tex]
    [tex]\frac{dy}{\cos y+i\sin y}=e^{x}dx[/tex]

    I tried multiplying by [tex]\frac{\cos y-i\sin y}{\cos y-i\sin y}[/tex] and [tex]\frac{\cos y+i\sin y}{\cos y+i\sin y}[/tex] and tons of other bad math to try and eliminate i or get it so that I can put it with the constant. None of that led to any meaningful results. Complex analysis has never been my strong suit especially since I've never really been taught it.
     
  2. jcsd
  3. Aug 31, 2009 #2
    Suggestion:

    [tex]\frac{1}{\cos y + i \sin y}\cdot\frac{\cos y - i \sin y}{\cos y - i \sin y}=\frac{\cos y - i \sin y}{\cos^2 y + \sin^2 y}=\cos y - i \sin y[/tex]

    I assume the i can be taken out of the integral (after separating cos y and i sin y) since it is a constant?
     
  4. Aug 31, 2009 #3

    Hootenanny

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    It would however be much simpler to note that,

    [tex]\int\frac{dy}{e^{iy}} = \int e^x\;dx[/tex]

    [tex]\int e^{-iy}\;dy = \int e^x\;dx[/tex]

    [tex]\int \left\{\cos\left(-y\right) + i\sin\left(-y\right)\right\}\;dy = \int e^x\;dx[/tex]

    [tex]\int \left\{\cos\left(y\right) - i\sin\left(y\right)\right\}\;dy = \int e^x\;dx[/tex]

    which is of course the same result, but is a more straightforward approach.
     
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