Use characteristic curve method to solve the problem $u_t(x,t)+u_x(x,t)+u(x,t) = 0$ 
Use characteristic curve method to solve the problem:
   $$u_t(x,t)+u_x(x,t)+u(x,t) = 0 ~~\text{for } 0 < x < \infty, t > 0$$
   $$u(x,0) = \cos(2x)$$ $$u(0,t) = 1$$

Note that if I am allowed to use change of coordinates method (also a method in Strauss PDE book), I am able to solve this problem without much problem.
However, the question specifically asked me to use the characteristic curve method: My attempt trying to mimic some examples (those examples are not very similar to this question)
Let $u_x+u_t = -u$,  and we have $$\frac{dx}{ds} = 1, \frac{dt}{ds} = 1, \frac{du}{ds} = -u$$
Hence it follows that $\frac{dt}{dx} = 1$ and solving to get $c_1 =t-x$ where $c_1$ is a constant. Now $\frac{du}{dt} = -u$ solving to get $c_2 = e^t u$
where $c_1,c_2$ are both characteristic curves. Then I am quite lost as how should i continue using only the method of characteristics.
 A: As you have found yourself and from the other answers, along the characteristic curves you get for $z(s)=u(x(s),y(s))$
$$
x(s)=x_0+s,~ t(s)=t_0+s,~ z(s)=e^{-s}z_0\implies z_0=u(x_0,t_0)=e^{s}u(x_0+s,t_0+s)
$$
The characteristic curve through $(x,t)$ crosses the boundary of the first quadrant at $s=-\min(x,t)$, so that
\begin{align}
u(x,t)&=e^{-\min(x,t)}u\bigl(\max(0,x-t),\max(0,t-x)\bigr)
\\&=\begin{cases}
e^{-x}u(0,t-x)=e^{-x}&\text{ for }x\le t,\\
e^{-t}u(x-t,0)=e^{-t}\cos(2(x-t))&\text{for }x>t.
\end{cases}
\end{align}
A: $$
\frac{dx}{1}=\frac{dy}{1}= -\frac{du}{u}
$$
then follows
$$
\frac{dy}{dx}=1\Rightarrow x-y = C_1
$$
and
$$
dx = -\frac{du}{u}\Rightarrow e^x=\frac{C_2}{u}\Rightarrow e^x u = f(x-y) = \cos(2(x-y))
$$
hence
$$
u = e^{-x}\cos(2(x-y))
$$
or according to the right variables
$$
u(x,t) = e^t\cos(2(t-x)) =  e^t\cos(2(x-t)) 
$$
A: Incomplete answer:
If $\phi(x,t,u)=c_1$ and $\psi(x,t,u)=c_2$ are solutions to the characteristic equations, then $f(\phi,\psi)=0$ is the general solution to the PDE. As you have found, $\phi(x,t,u) = t-x$ and $\psi(x,t,u) = e^tu$, so the general implicit solution is $f(t-x,e^tu)=0$, for some function $f$. Rearranging, we can write $e^tu = g(t-x)$, so $u(x,t) = e^{-t}g(t-x)$ for some function $g$. To find $g$, we apply the Cauchy data to the general solution: $ u(x,0) = g(-x)=\cos(2x) $, so $g(x) = \cos(2x)$. Thus, $$ u(x,t) = e^{-t}\cos(2(t-x)), \quad x>t $$
