Characteristics for PDE $u_x + 3u_y - (3x-y)u = -2 $ Question :

I am willing to solve the PDE :$$u_x(x,y) + 3u_y(x,y) - (3x-y)u(x,y) = -2 $$

Thoughts :
The characteristics are :
$$\frac{\mathrm{d}x}{1} = \frac{\mathrm{d}y}{3} \implies u_1 = 3x-y$$
I see now the similarity between the surface yielded by the characteristics and the coefficient of $u(x,y)$. Since we have both a non-zero coefficient of the $u(x,y)$ term and a constant on the right-hand side, I thought of a change of variables, but I couldn't find anything handy.
I am really troubled by such equation which involve both a non-zero $u$ coeffiecient and a constant (or expression) on the right hand side. How does one work with them, with characteristics or even not ?
I will be grateful for any explanations / elaborations.
 A: Method of characteristics simplifies the PDE, but does not necessarily solves them immediately. This is such a case. Let us consider the values of function $u$ along a single characteristic line and define:
$$f(t,x_0):=u(x_0+t,3t)$$
We have
\begin{align} f_t(t,x_0) &= u_x(x_0+t,3t)+3u_y(x_0+t,3t) = \\
&= \big(3(x_0+t) -3t\big)u(a+t,3t) - 2 = \\
&= 3x_0f(t,x_0)  -2 \end{align}
So we get an ODE ($x_0$ can be treated as a parameter): $$ f_t(t,x_0) = 3x_0f(t,x_0)  -2 $$
The similiarity of the equation of characteristic and the coefficient that you have notices makes it so that there's no free-standing $t$ on the right side, making the equation easy to solve. The solution is
$$ f(t,x_0) = \left\{\begin{array}{ll}\frac{2}{3x_0} + C(x_0) \exp\big(3x_0 t\big) & \text{for }x_0 \neq 0 \\ C(0) - 2t & \text{for }x_0 =0 \end{array}\right.$$
You have then
$$ u(x,y) = f(\frac{y}{3},x-\frac{y}{3}) = \left\{\begin{array}{ll}\frac{2}{3x-y} + C(x-\frac{y}{3}) \exp\big((x-\frac{y}{3}) y\big) & \text{for }x-\frac{y}{3} \neq 0 \\ C(0) - \frac{2y}{3} & \text{for }x-\frac{y}{3} =0 \end{array}\right.$$
You can find $C(x_0)$ from the initial conditions on $u$, if you are given any. Given $u(x,0)=u_0(x)$ the solution is
$$ u(x,y) = \left\{\begin{array}{ll}\frac{2}{3x-y}\Big(1-\exp\big((x-\frac{y}{3}) y\big)\Big) + u(x-\frac{y}{3},0) \exp\big((x-\frac{y}{3}) y\big) & \text{for }x-\frac{y}{3} \neq 0 \\ u(0,0) - \frac{2y}{3} & \text{for }x-\frac{y}{3} =0 \end{array}\right.$$
A: Set $u(x,y)=v(x,y-3x)$. For simplicity, denote the second argument position of $v$ as $z=y-3x$. Then the differential equation for $v$ is
$$
(v_x-3v_z)+3v_z+zv=-2\implies v_x(x,z)+zv(x,z)=-2
$$
This can now be solved along the lines $z=const.$ as
$$
(e^{zx}(zv+2))_x=0\implies v(x,z)=\frac{-2+\phi(z)e^{-zx}}z
$$
or
$$
u(x,y)=\frac{-2+\phi(y-3x)e^{-(y-3x)x}}{y-3x}
$$
A: Here is my solution:
$$x' = x + 3y \\ y' = y - 3x$$
Then, we have $$u_{x} = u_{x'} \cdot 1 + u_{y'} \cdot (-3) \\ u_{y} = u_{x'} \cdot 3 + u_{y'} \cdot 1$$
Then, $$u_{x} + 3u_{y} - (3x - y)u = -2 \\ u_{x'} - 3u_{y'} + 9u_{x'} + 3u_{y'} + y' \cdot u = -2 \\ 10u_{x'} + y' \cdot u = -2 \\ u_{x'} + \frac{y'}{10} \cdot u = -\frac{1}{5}$$
We multiply by the integrating factor $$e^{\frac{y'x'}{10}}$$ and we have
$$\Big(e^{\frac{y'x'}{10}} \cdot u\Big)_{x'} = -\frac{1}{5} \cdot e^{\frac{y'x'}{10}} \\ e^{\frac{y'x'}{10}} \cdot u = F(y') - \frac{1}{5} \int_{0}^{x'}{e^{\frac{y't}{10}}}~dt \\ e^{\frac{y'x'}{10}} \cdot u = F(y') - \frac{2}{y'} \cdot \Big(e^{\frac{y'x'}{10}} - 1\Big) \\ u(x',y') = \bigg(F(y') \cdot e^{-\frac{y'x'}{10}}\bigg) - \bigg(\frac{2}{y'}\bigg) + \bigg(\frac{2}{y'} \cdot e^{-\frac{y'x'}{10}}\bigg)$$
Finally,
$$u(x, y) = -\bigg(\frac{2}{y-3x}\bigg) + \bigg(F(y-3x) \cdot e^{-\frac{(y-3x)(x+3y)}{10}}\bigg) + \bigg(\frac{2}{y-3x} \cdot e^{-\frac{(y-3x)(x+3y)}{10}}\bigg)$$.
Reference guide: https://www.math.upenn.edu/~vedranso/Math425_Homework1_Solutions.pdf
PS: My answer is similar to LutzL's answer.
