# solving a PDE in 2 variables without boundary conditions

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

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
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

$1=y\partial_yf(x,y)-x\partial_xf(x,y)$

$x\partial_xf(x,y)-y\partial_yf(x,y)=-1$

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$
$\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=\dfrac{y_0}{x}$
$\dfrac{df}{dt}=-1$ , letting $z(0)=F(y_0)$ , we have $f(x,y)=F(y_0)-t=F(xy)-\ln x$