Solving $u_x+bu_y+cu=0$ I have the following problem: I solved the following partial differential equation by making changes of variables but I would like to know if it is possible to reach the same result by means of the characteristics method, however, I have no idea how to check my result . I really appreciate your time and help.
$u_x+bu_y+cu=0$
If I do the following variable changes $x'=x+by$ and $y'=bx-y$. By the chain rule,
$u_x=u_{x'}+bu_{y'}$
$u_y=bu_{x'}-u_{y'}$
And we have $(1^2+b^2)u_{x'}+cu=0$ which implies $u(x,y)=f(y)e^{-cx/(1^2+b^2)}$ and then $u(x,y)=f(bx-y)e^{-c(x+by)/(1+b^2)}$ where $f$ is a arbitrary differentiable function.
 A: $$u_x+bu_y=-cu$$
https://en.wikipedia.org/wiki/Method_of_characteristics
$$\begin{cases}
dx=ds\\
dy=b\:ds\\
du=-cu\:ds
\end{cases}\quad\implies\quad
\frac{dx}{1}=\frac{dy}{b}=\frac{du}{-cu}=ds$$
A first characteristic equation comes from solving $\quad dx=\frac{dy}{b}$
$$bx-y=c_1$$
A second characteristic equation comes from solving $\quad\frac{dx}{1}=\frac{du}{-cu}$
$$u\:e^{cx}=c_2$$
The general solution of the PDE expressed on the form of implicit equation $c_2=F(c_1)$ is :
$$u\:e^{cx}=F(bx-y)$$
$F$ is an arbitrary function (to be determined according to some boundary condition).
$$\boxed{u(x,y)=e^{-cx}F(bx-y)}$$
Checking :
$u_x=-ce^{-cx}F+be^{-cx}F'$
$u_y=-e^{-cx}F'$
$u_x+bu_y+cu=(-ce^{-cx}F+be^{-cx}F')+b(-e^{-cx}F')+c(e^{-cx}F)=0\quad$is OK.
A: Solution using the method of characteristics:
We need to solve $$u_{x}+bu_{y}+cu=0 \implies \underbrace{\left(-\frac{1}{c}\right)}_{m}u_{x}+\underbrace{\left(-\frac{b}{c}\right)}_{n}u_{y}-u=0\iff \boxed{mu_{x}+nu_{y}-u=0} $$
Now, we have a quasilinear partial differential equation and the relations between the differentials is $$\boxed{\frac{dx}{m}=\frac{dy}{n}=\frac{du}{u}} $$
We can pair the differentials in three ways:
$$ \frac{dy}{dx}=\frac{n}{m}, \quad \frac{du}{dx}=\frac{u}{m} \quad \text{and} \quad \frac{du}{dy}=\frac{u}{n} $$
Only two of these relations are independent. We focus on the first pair.
The first equation gives the characteristic curves in the $xy$-plane. This equation
is easily solved to give
$$\frac{dy}{dx}=\frac{n}{m}\implies \int dy=\frac{n}{m}\int dx\implies y=\frac{n}{m}x+c_{1}\implies \boxed{c_{1}=y-\frac{n}{m}x}$$
where $c_{1}$ is arbitrary constant.
The second equation can be solved to give $$ \frac{du}{dx}=\frac{u}{m} \implies m\int \frac{du}{u}=\int dx \implies \ln|u|=\frac{x+c_{2}}{m} \iff u=c_{3}e^{\frac{x}{m}} $$
The goal is to find the general solution to the differential equation. Since $u =
u(x, y)$, the integration “constant” is not really a constant, but is constant with
respect to $x$. It is in fact an arbitrary constant function. In fact, we could view it
as a function of $c_{1}$, the constant of integration in the first equation. Thus, we let
$c_{3} = F(c_{1})$ for $F$ and arbitrary function. Since $c_{1} = y − \frac{n}{m}x$, we can write the general solution of the differential equation as
$$u(x,y)=F\left(y-\frac{n}{m}x\right)e^{x}{m}\overbrace{=}^{m:=\frac{-1}{c}, n:=\frac{-b}{c}}F(y-bx)e^{-cx}$$
Therefore, we have
$$\boxed{u(x,y)=F(y-bx)e^{-cx}}$$
Solution using change of variables:
We need to solve $$u_{x}+bu_{y}+cu=0 $$
Let $\eta:=y-\frac{b}{a}x$ and $\xi:=x$, so we have $$\left\{\begin{aligned} u_{x}=\left(-\frac{b}{a}\right)u_{\eta}+u_{\xi}\\ u_{y}=u_{\eta} \end{aligned} \right.$$So, we have $$au_{\xi}+cu=0$$The solution is given by, $$u=f(\eta)e^{-\frac{c}{a}\xi}$$Finally, we have the solution by we problem $$\boxed{u(x,y)=f\left(y-\frac{b}{a}x\right)e^{-\frac{c}{a}x}}$$
Other changes variables:

*

*If you have $$\eta:=bx+ay$$ $$\xi:=bx-ay$$
Here, you can find $$\boxed{u(x,y)=f(bx-ay)e^{-\frac{c}{2ab}(bx+ay)}}$$

*If you have $$\eta:=ax+by$$ $$\xi:=bx-ay$$Here, you can find $$\boxed{u(x,y)=f(bx-ay)e^{-\frac{c}{a^{2}+b^{2}}(ax+by)}} $$
You can find the details this solutions here: solutions by @Kaster

Finally, you can find here more information about the method of characteristic: A. Salih. Department of Aerospace Engineering, Indian Institute of Space Science and Technology, Thiruvananthapuram 1 July 2016.
