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Partial Differential Equation

  1. Nov 21, 2006 #1
    x(δu/δx)-(1/2)y(δu/δy)=0

    By first looking for a separable solution of the form u(x, y)=X(x)Y(y), find the general solution of the equation given above.

    Determine the u(x,y) which satisfies the boundary condition u(1,y)=1+siny

    For the separable form I have u(x, y)=A(x^c)(y^2c), could someone please show me how to do the rest of it.

    Thank you.
     
  2. jcsd
  3. Nov 21, 2006 #2

    arildno

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    Write sin(y) as a power series.
     
  4. Nov 21, 2006 #3

    arildno

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    Each choice of c gives you a solution of your diff.eq.
    Since your diff.eq is linear, a sum of such solutions is also a solution of your diff.eq.
     
  5. Nov 21, 2006 #4
    How do I get from u(x, y)=A(x^c)(y^2c) to the general solution?

    Thank you.
     
  6. Nov 21, 2006 #5

    arildno

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    All right:
    A series solution of your diff.eq is:
    [tex]u(x,y)=\sum_{n=1}^{\infty}A_{n}x^{c_{n}}y^{2c_{n}}[/tex],
    whereby follows:
    [tex]u(1,y)=\sum_{n=1}^{\infty}A_{n}y^{2c_{n}}[/tex]
    and [itex]A_{n},c_{n}[/itex] are constants.

    Now, how can you fit this expression for u(1,y) to the given boundary condition?
     
  7. Nov 21, 2006 #6
    Please go on...
     
  8. Nov 21, 2006 #7

    arildno

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    Well, use my first hint in post 2.
     
  9. Nov 21, 2006 #8
    Ok if you could please show me the COMPLETE working I would really appreciate it... Thank you.
     
  10. Nov 21, 2006 #9
    Ok if you could please show me the COMPLETE working I would really appreciate it... Thank you.
     
  11. Nov 21, 2006 #10

    arildno

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    Do you know what a power series is? :confused:
     
  12. Nov 21, 2006 #11
    Yes lol, I just can't do this question, could you please show me the working for it... In fact can you do it yourself?
     
  13. Nov 21, 2006 #12

    HallsofIvy

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    If you are not capable of doing basic algebra, you should not be attempting partial differential equations!

    (Yes, I can do it myself! That's not really the point is it? You have been told exactly HOW to solve your equation, yet you have not even TRIED to apply what you have been told.)
     
  14. Nov 22, 2006 #13

    arildno

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    That's it. I'm out of here. It is long since I've met a more ungrateful and lazy f*ckhead on PF as you.
     
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