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Partial differentail equation

  1. Dec 20, 2009 #1
    X*U'x+Y*U'y=U
    {U=1-x2}
    {y=1}

    -------------------------------------------

    dx/x=dy/y=du/u

    [tex]\int[/tex]dx/x=[tex]\int[/tex]dy/y

    ln|x|=ln|y|+ln|c1|
    x/y=c1

    [tex]\int[/tex]dx/x=[tex]\int[/tex]du/u

    ln|x|=ln|u|+ln|c2|
    x/u=c2

    [tex]\Phi[/tex](x/u , x/y)=0

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

    now i use the conditions
    {U=1-x2}
    {y=1}


    x/(1-x2)=[tex]\phi[/tex](x)=x/u

    and all i get is U=1-x2 but i already know that

    the correct answer is u=(y2 -x2)/y
     
  2. jcsd
  3. Dec 20, 2009 #2

    rock.freak667

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    Homework Helper

    Is your equation

    [tex]x \frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=1-x^2[/tex]

    or


    [tex]x \frac{\partiall u}{\partial x}+y\frac{\partial u}{\partial y}=0[/tex]

    with u(x,0)=1-x2 and u(0,y)=1 ?
     
  4. Dec 20, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    I interpret this to mean that U(x,y) satisfies
    [tex]]x\frac{\partial U}{\partial x}+ \frac{\partial U}{\partial y}= U[/tex]
    with the condition that [itex]U(x, 1)= 1- x^2[/itex]

    But then it is peculiar that there is no condition for x a constant.
     
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