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First order linear partial differential equation

  1. Feb 26, 2009 #1
    Do these equations have two general solutions!?

    e.g. z_x + z_y -z = 0

    Using the method of characteristics


    Therefore dx/1=dy/1=dz/z

    Taking first two terms: x = y + A
    *Taking last two terms: z = Be^y
    So general solution is z = f(x-y)e^y

    BUT if we took first and last terms: z=Be^x
  2. jcsd
  3. Feb 26, 2009 #2


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    Neither of those is the "general" solution. z= f(x-y)ex+ g(x-y)ey is the general solution.
  4. Feb 27, 2009 #3
    You are quite the genius! Thanks
  5. Apr 2, 2009 #4
    I disagree - first order PDE's don't have two arbitrary functions in their solutions!

    Actually, they're both right.

    First solution [tex]z = e^xf(x-y)[/tex] second solution [tex]z = e^y g(x-y)[/tex]. Since [tex]f[/tex] is arbitrary the set [tex]f(x-y) = e^{-(x-y)} g(x-y)[/tex] and the first becomes the second.
  6. Apr 3, 2009 #5


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    What, you mean I'm NOT a genius?
  7. Apr 3, 2009 #6
    I've never met you so I really don't know :rofl:
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