Question: $xu_x-xyu_y-u=0$ for all $ (x,y)$
My attempt: Our characteristic curve is in the form of $\frac{dy}{dx}$. Since our $dy = -xy$ and $dx = x$ we have the following separable equation .
$\frac{dy}{dx} = \frac{-xy}{x}$
so that leaves us with
$\frac{dy}{dx} = -y$
$\frac{-1}{y}dy = dx$
$- \ln y = x+ C$
$ \ln y^{-1} = x+ C$
Taking the exponential, we have
$e^{ \ln y^{-1}}=e^{x+C}$
$y^{-1} = e^{x+C}$
$y = e^{-x-C}$
$ye^x = C $
Our characteristic lines and change of variables will be
$ W = ye^x \rightarrow \frac{w}{z}=e^x$
$ z = y \rightarrow y = z$
Taking partial derivatives of $W_x,W_y,Z_x,$ and $Z_y$, we have
$W_x = ye^x$
$W_y = e^x$
$Z_x = 0$
$Z_y=1$
$x[V_wye^x]-xy[V_we^x+V_z]-v=0$
$xye^xV_w-xyV_we^x-xyV_z-v=0$
$-xyV_z-v=0$
Dividing $-xy$ throughout the equation, we have
$V_z+\frac{1}{xy}v=0$
Ok. I am stuck. I feel like there are double logs if I choose $p(a) = \frac{1}{xy}$ that would be $v(a) = e^{\int \frac{1}{xy}} \rightarrow e^{\ln{x}\ln{y}} \rightarrow xy$ and that clearly doesn't make any sense. If I guess at the reverse product rule, I will have $[xy][V]$ or $ln[xy][V]$ which would probably be nonsense. So my question is ... can we find an integrating factor for this equation?
$V_z+\frac{1}{xy}v=0$
Edit: What if I try something like multiplying the $xy$ all the way through? Then I would have $xyV_z+v=0$. It might work since the reverse product rule gives us $[xy]v$ so if I leave the xy alone and deal with the V I would have $xyV_z$ and for leaving the v alone and dealing with the xy it would be zero because $V_z$ is the partial derivative in terms of Z and I have none. So assuming that's legal, I will have
$V{xy} = \int 0$
$V{xy} =F[w]$
$V{xy} =F[ye^x]$
ughhh but then I have to divide xy.. what the..!
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$
$\dfrac{dy}{dt}=-xy=-e^ty$ , letting $y(0)=y_0$ , we have $y=y_0e^{1-e^t}=y_0e^{1-x}$
$\dfrac{du}{dt}=u$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=f(y_0)e^t=f(e^{x-1}y)x=F(e^xy)x$