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how could i solve the PDE (without boundary or other initial conditions)

$ 1= y\partial _{y}f(x,y) -x \partial _{x}f(x,y) $

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What do you mean by "solve"? Find all functions that solve it? Without conditions, the solution is not unique. –  joriki Apr 10 '12 at 10:32
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2 Answers 2

up vote 1 down vote accepted

One solution would be:

$f(x,y) = xy - log(x)$

But so would:

$f(x,y) = n\cdot xy - log(x)$

In general, you need a boundary condition to solve a first order PDE.

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ok , thanks is it possible to get ALL the solutions of $ f(x,y) $ into a closed expression :) –  Jose Garcia Apr 10 '12 at 10:33
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The equation is linear, so taking the difference between two solutions $f_1$ and $f_2$ you have that $y\partial_y(f_1 - f_2) - x\partial_x(f_1-f_2) = 0$, so every solution is a specific solution plus some homogeneous solution. The specific solution is given above by nbubis as $f_0 = \log(|x|)$. Any homogeneous solution has to be constant on the integral curves of $x\partial_x - y\partial_y$, and hence can be written as $g = g(xy,\mathop{sgn}(x))$. That is, as a function depending only on the product $xy$ and on the sign of the variable $x$. The second dependence is because outside $xy=0$... –  Willie Wong Apr 10 '12 at 11:29
    
... the level curves $\{xy = c\}$ (which are the integral curves of $x\partial_x - y\partial_y$) have two connected components, and the function is allowed to be different on the components. The two components can be distinguished by the sign of the $x$ variable there. –  Willie Wong Apr 10 '12 at 11:30
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This belongs to a PDE of the form http://eqworld.ipmnet.ru/en/solutions/fpde/fpde1211.pdf.

The general solution is $f(x,y)=C(xy)+\int\dfrac{dy}{y}=C(xy)+\ln y$ or $f(x,y)=C(xy)-\int\dfrac{dx}{x}=C(xy)-\ln x$ .

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