Solving a master equation with linear coefficients I have the following PDE:
$$
\partial_t P(x,y,t)=x\partial_xP(x,y,t)+(y-1)\partial_yP(x,y,t)+2P(x,y,t).
$$
Mathematica suggests that the solution is
$$
\dfrac{f((y-1)/x,t+\log x)}{x^2},
$$
where $f$ is an arbitrary function.
My question is: How to solve this equation? Or put it in other words, how can I arrive at the solution given by Mathematica?

$\partial_tP(x,y,t)=x\partial_xP(x,y,t)+(y-1)\partial_yP(x,y,t)+2P(x,y,t)$
$\partial_tP(x,y,t)-x\partial_xP(x,y,t)+(1-y)\partial_yP(x,y,t)=2P(x,y,t)$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dx}{ds}=-x$ , letting $x(0)=x_0$ , we have $x=x_0e^{-s}=x_0e^{-t}$
$\dfrac{dy}{ds}=1-y$ , letting $y(0)=y_0$ , we have $y=(y_0-1)e^{-s}+1=(y_0-1)e^{-t}+1$
$\dfrac{dP}{ds}=2P$ , letting $P(0)=f(x_0,y_0)$ , we have $P(x,y,t)=f(x_0,y_0)e^{2s}=f(xe^t,(y-1)e^t+1)e^{2t}$
