<s>[INST] Finding the general solution to $2u_x-3u_y+(U-x)=0$ The PDE I'm working on is: $$2u_x-3u_y+(U-x)=0.$$
Using the method of characteristics I obtained $c_1=2x+3y.$  Where I am stuck is on $c_2$;  currently I'm exploring $$\frac{dx}{2}=\frac{du}{u-x}.$$
My first instinct was to integrate as is and get the following: $$\frac{x}{2}+c_2=\log(u-x)$$ or $$x+c_2=2\log(u-x)$$ which becomes $$e^xe^{c_2}=(u-x)^2$$ Solving for $c_2$ we get $(u-x)^2e^{-x}=c_2$
Hence $F(2x+3y)=(u-x)^2e^{-x}.$
Am I on the right track for a general solution?
 [/INST] For the equation 
\begin{align}
2 u_{x} - 3 u_{y} + u - x = 0
\end{align}
it is seen that
\begin{align}
\frac{dx}{2} = \frac{dy}{-3} = \frac{du}{x-u}.
\end{align}
Solving for the $x-y$ equation, $2 dy + 3 dx = 0$, leads to $c_{1} = 3 x + 2 y$. Consider the second characteristic from 
\begin{align}
dy = \frac{3 \, du}{u-x}.
\end{align}
This leads to $c_{2} = y - 3 \ln(u-x)$, or
\begin{align}
u = x + e^{(y+c_{2})/3}.
\end{align}
Making appropriate connections of the characteristics $c_{1}$ and $c_{2}$, it can be seen that a solution is
\begin{align}
u(x,y) = e^{-(x + y/3)} + x -2.
\end{align}

Verification
From the solution proposed solution it can be seen that:
\begin{align}
u_{x} &= 1 - e^{-(x+y/3)} \\
u_{y} &= - \frac{1}{3} e^{-(x+y/3)} \\
\end{align}
and leads to
\begin{align}
2 u_{x} - 3 u_{y} &= 2 - e^{-(x+y/3)} \\
&= x - \left[ e^{-(x+y/3)} + x - 2 \right] \\
&= x-u
\end{align}
which becomes
\begin{align}
2 u_{x} - 3 u_{y} + u = x.
\end{align}
</s>