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Partial Differentiation

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    The function f(x,y) satisfies the d.e.

    [tex] y{\partial f \over \partial x} + x{\partial f \over \partial y} = 0 [/tex]

    By changing to new vars [tex]u = x^2-y^2[/tex] and [tex]v=2xy[/tex] show that f is a function of [tex]x^2-y^2[/tex] only.

    2. Relevant equations

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

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

    3. The attempt at a solution

    [tex] y (\frac{\partial u}{\partial x}\frac{\partial }{\partial u}+\frac{\partial v}{\partial x}\frac{\partial }{\partial v}) g + x (\frac{\partial u}{\partial y}\frac{\partial }{\partial u}+\frac{\partial v}{\partial y}\frac{\partial }{\partial v}) g = y{\partial f \over \partial x} + x{\partial f \over \partial y} = 0 = 0[/tex]

    [tex]\left(2y^2+2x^2\right)\frac{\partial g}{\partial v}=x\frac{\partial f}{\partial y}+y\frac{\partial f}{\partial x}=0[/tex]

    I did that and so

    [tex] \frac{\partial g}{\partial v}=0 [/tex]

    How does this confirm g=g(u)? Is is because the slope of g with respect to v is 0? The so called function of v is just a constant or 0?
     
  2. jcsd
  3. Sep 22, 2009 #2

    LCKurtz

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    Gold Member

    I didn't check your arithmetic, but to answer your question, if g is a function of u and v and gv = 0, then yes, that does show g is only a function of u.
     
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