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) $
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) $
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.
$1=y\partial_yf(x,y)-x\partial_xf(x,y)$
$x\partial_xf(x,y)-y\partial_yf(x,y)=-1$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x$ , 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 $f(x,y)=F(y_0)-t=F(xy)-\ln x$