Solving First Order PDE with Characteristics Method Let the first order PDE: $$u_t+au_x=b-cu$$
I tried Characterestics method:
$\frac{dt}{ds}=1$ and $\frac{dx}{ds}=a$ and $\frac{du}{ds}=b-cu$
Then we have if we suppose that $t(0)=0$, then we'll have $t=s$ and $x(t)=a.t+x(0)$
But I could not solve the last ode.
 A: Hint:
$$
\frac{du}{ds}+cu=e^{-cs}\frac{d}{ds}(ue^{cs})
$$
A: The "last equation" is
$\dfrac{du}{ds} = b - cu, \tag 1$
which may be written as
$\dfrac{du}{ds} + cu = b; \tag 2$
this is a first order, linear, constant-coefficient ODE, the solution of which is well-known; indeed, we may derive it in short order here; we multiply (2) through by $e^{cs}$:
$e^{cs}\dfrac{du}{ds} + ce^{cs}u = be^{cs}, \tag 3$
and note that
$\dfrac{d}{ds}(e^{cs}u) = e^{cs}\dfrac{du}{ds} + ce^{cs}u; \tag 4$
thus, (3) becomes
$\dfrac{d}{ds}(e^{cs}u) = be^{cs}, \tag 5$
which may readily be integrated 'twixt $0$ and $s$:
$e^{cs}u(s) - u(0) = \displaystyle \int_0^s \dfrac{d}{dr}(e^{cr}u(r)) \; dr = b \int_0^s e^{cr} \; dr = \dfrac{b}{c}(e^{cs} - 1); \tag 6$
thus,
$e^{cs}u(s) = u(0) + \dfrac{b}{c}(e^{cs} - 1), \tag 7$
and
$u(s) = u(0)e^{-cs} +  \dfrac{b}{c}(1 - e^{-cs}) = (u(0) - \dfrac{b}{c})e^{-cs} + \dfrac{b}{c}.  \tag 8$
We check:
differentiating (8),
$\dfrac{du(s)}{ds} = -c  (u(0) - \dfrac{b}{c})e^{-cs}  = -c((u(0) - \dfrac{b}{c})e^{-cs} + \dfrac{b}{c}) + b = -cu(s) + b, \tag 9 $
in light of (8).  This is then the required solution to (2).
A: $$u_t+au_x=b-cu$$
Charpit-Lagrange characteristic system : $\quad \frac{dt}{1}=\frac{dx}{a}=\frac{du}{b-cu}$ .
A first characteristic equation comes from $\frac{dt}{1}=\frac{dx}{a}$ :
$$x-at=C_1$$
A second characteristic equation comes from $\quad \frac{dt}{1}=\frac{du}{b-cu}\quad$  after solving :
$$\left(u-\frac{b}{c}\right)e^{c\,t}=C_2$$
The general solution of the PDE expressed on implicit form $C_2=F(C_1)$ is :
$$\left(u-\frac{b}{c}\right)e^{c\,t}=F(x-at)$$
with arbitrary function $F$.
Or equivalently on explicit form : $$\quad u(x,t)=\frac{b}{c}+e^{-c\,t}F(x-at)$$
The arbitrary function $F$ has to be determined according to some specified boundary and/or initial condition.
What is the condition which allows to determine the function $F$ ? 
In the wording of the question it is written : "Then we have if we suppose that $t(0)=0$ ". This is not clear. Is it a specification or is it a supposition ? Moreover there is no condition. Thus as the problem is presently raised they are an infinity of solutions :
$$\boxed{ u(x,t)=\frac{b}{c}+e^{-c\,t}F(x-at)\qquad\text{any function }F .}$$
If you put this result into the PDE you will see that it agrees.
