Solve $e^yu_x-u_y+u=xe^y$
$u_x-\frac{u_y}{e^y}=x-\frac{u}{e^y}$
$\frac{b}{a}=\frac{dy}{dx}=-\frac{1}{e^y}$
$e^y=-x+c$,
$\eta=e^y+x,\xi=x$
$e^y=\eta-\xi$
$u(x,y)=w(\xi,\eta)=w_{\xi}+w_{\eta}-\frac{w_\eta e^y}{e^y}=\xi-\frac{w}{\eta-\xi}$
$w_{\xi}=\xi-\frac{w}{\eta-\xi}$
how can I continue?
This linear PDE can be solved by the method of characteristics, which reads:
$$ \frac{\mathrm{d}x}{e^y} = \frac{\mathrm{d}y}{-1} = \frac{\mathrm{d}u}{x e^y - u}. $$
For the first equality we have:
$$\mathrm{d}x + e^y \mathrm{d}y = 0,$$ so one of the characteristics may be defined as $\xi = x + e^y$. Use now, for example, the second and third terms taking into account the relation given by $\xi$ to come up with:
$$ \frac{\mathrm{d}y}{-1} = \frac{\mathrm{d}u}{(\xi - e^y) e^y - u} \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}y} =u - (\xi -e^y) \, e^y,$$ which is a 1st-order linear ODE for $u(y)$, which can be solved in terms of an integrating factor. With the help of Mathematica, we have:
$$u = e^y\, (e^y - y \, \xi) + e^y \, \eta,$$ put now $\eta$ as an arbitrary function of $\xi(x,y)$ to get the complete solution of the PDE as follows:
$$u = e^y\, \left(e^y - y \, (x+e^y) + \eta(x+e^y)\right).$$
Hope this helps.
I will double check on my calculations and edit if necessary.
Cheers!