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Complex algebra help

  1. Apr 16, 2007 #1
    1. The problem statement, all variables and given/known data

    I am told: [itex]{\frac {{\it du}}{{\it dx}}}=y[/itex] and [itex]{\frac {{\it du}}{{\it dy}}}=x[/itex]. Need to find u(x,y) which is a real valued function and prove the result.

    2. Relevant equations



    3. The attempt at a solution

    Well, I think the answer is of the form u(x,y) = xy + c because the answer makes sense but how should I go about proving it?

    Thanks
     
  2. jcsd
  3. Apr 16, 2007 #2

    quasar987

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    By integrating and reasoning logically.

    [tex]\frac{\partial u}{\partial x}=y[/tex]

    Therefor, integrating wrt x:

    [tex]\int\frac{\partial u}{\partial x}dx=\int ydx[/tex]

    [tex]u(x,y)= yx+\phi(y)[/tex]

    phi(y) is kind of like the constant of integration, but we reckon that in the most general case, it may actually be a function of y. Make sure you understand why.

    We can do the same with the other equation:

    [tex]\frac{\partial u}{\partial y}=x[/tex]

    [tex]u(x,y)=xy+\psi(x)[/tex]

    Now compare the two equations. What do you conclude about the forms of [itex]\phi(y)[/itex] and [itex]\psi(x)[/itex]?
     
  4. Apr 17, 2007 #3

    HallsofIvy

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    I would have done this slightly differently.
    You are given that
    [tex]\frac{\partial u}{\partial x}= y[/tex]
    so [itex]u(x,y)= xy+ \phi (y)[/itex]
    as quasar987 said.

    Now differentiate that with respect to y:
    [tex]\frac{\partial u}{\partial y}= x+ \frac{d\phi}{dy}[/tex]
    and that must be equal to x. What does that tell you about
    [tex]\frac{d\phi}{dy}[/tex]?

    (I said slightly differently!)
     
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