Consider the following PDE: $$xu_x+(y+1)u_y=u-1.$$
Using this formula: $$\frac{dx}{x}=\frac{dy}{y+1}=\frac{du}{u-1}.$$
This yields $c_1=\frac{y+1}{x}$ and $c_2=\frac{u-1}{x}.$
We have: $$F\left(\frac{y+1}{x}\right)=\frac{u-1}{x}.$$
Given the following Cauchy condition - $u(x,2x-1)=e^x$. This yields $$xF(2)+1=e^x.$$
Am I on right right track? I'm a little confused because we have an $F(2)$.
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$
$\dfrac{dy}{dt}=y+1$ , we have $y+1=y_0e^t=y_0x$
$\dfrac{du}{dt}=u-1$ , we have $u(x,y)=F(y_0)e^t+1=xF\left(\dfrac{y+1}{x}\right)+1$
$u(x,2x-1)=e^x$ :
$xF(2)+1=e^x$ , which is impossible.
$\therefore$ There is no solution.