Find a solution $g(x)$ that satisfies the PDE $u_x +3u_y-u = 1$. Question: What form must $g(x)$ have in order that the following problem have a solution? $u_x+3u_y-u=1,u(x,3x)=g(x)$. If $g(x)$ has the required form, will there be more than one solution?
My attempt:
$u_x+3u_y-u=1$
First we need the slope and the characteristic lines.
The characteristic line is in the form of $bx-ay =d$ for the equation $au_x+bu_y+cu=f(x,y)$
Since $ b = 3$ and $ a = 1$, our slope is $\frac{3}{1} \rightarrow 3$
Our characteristic lines are
$3x-y = w$ and $y = z$
and using the change of variables, we have 
$ x = \frac{1}{3}w+\frac{1}{3}z$ and $ z =y$
Using the chain rule, we have
$V_wW_x+V_zZ_x + 3(V_wW_y+V_zZ_y ) - v= 1$ 
and taking the partial derivatives, we have
$ W_x = 3, W_y = -1, Z_x = 0$ and $Z_y =1$ 
Our equation becomes
$V_z - v= 1$ 
$\frac{dv}{dz}-v=1$
$l(z) = e^{-\frac{1}{3}z}$
so multiplying the integrating factor we have
$e^{-\frac{1}{3}z}V_z - e^{-\frac{1}{3}z}v= e^{-\frac{1}{3}z}$ 
Taking the reverse product rule, we have
$v(e^{\frac{-1}{3}z})= \int \frac{1}{3}e^{\frac{-1}{3}z} $
Integrating with respect to z, we have
$v(e^{\frac{-1}{3}z})= -e^{\frac{-1}{3}z} +C(w)$
Dividing $(e^{\frac{-1}{3}z})$, we have 
$v= -1 +C(w)e^{\frac{1}{3}z}$
Since $ z = y$
$v= -1 +C(w)e^{\frac{1}{3}y}$
and $y = 3x$
$v= -1 +C(w)e^{\frac{1}{3}3x}$
$v= -1 +C(w)e^{x}$
So I am stuck at the $w$ part. The only thing that I can think of is to substitute $ w =3x-y$ and I am not sure what to do. The answer to do is $ g(x) = -1+2e^x$ and I almost have it except I don't know where the 2 comes from. 
$v= -1 +C(3x-y)e^{x}$
and from the condition $u(x,3x) = g(x)$
$ v = -1+C(3x-3x)e^{x}$
$v= =1+C(0)e^{x}$
what the!!
Edit: I was stressing out for no reason. It turns out that as long as $C(0) =k$ or we are in $C^1$ the solution of $g(x)$ has to be $ 1 +ke^x$. Now I feel like a dummy. 
 A: I would approach this differently. First, $u\equiv -1$ is a particular solution of the inhomogeneous equation, so we only need the general solution for homogeneous one, 
$u_x +3u_y-u = 0$. Take exponential ansatz $u=C e^v$ and simplify to $v_x+3v_y - 1=0$. Again, a particular solution is easy to find: $v(x,y)=x$. The general solution of $v_x+3v_y=0$ is any function that is constant on lines of slope $3$, which is $f(y-3x)$. 
Collect together: $x+f(y-3x)$ solves $v_x+3v_y - 1=0$, hence $u(x,y)=C e^{x+f(y-3x)} -1$ solves the original equation. Tidy up: $$u(x,y) = e^x h(y-3x)-1,\quad h\text{ arbitrary}\tag{1}$$
which is the same what you found.
Since $u(x,3x) = e^x h(0)-1$, you need some pretty special values on that line to have a solution. But with these special values you have lots of solutions, since $h(t)$ for $t\ne 0$ can be anything. 

In the special case $f(x,3x)=-1+2e^x$ we see that $h(0)=2$. This is the only constraint on $h$. You can take $h(t)\equiv 2$, or $h(t)=2+t$, or $h(t)=2e^{t}$... all these give different solutions of the PDE according to (1).
