Stuck trying to solve a PDE by method of characteristics I've been trying to solve the inhomogeneous PDE IVP $u_t+c u_x = e^{2x}$; $u(x,0)=f(x)$, but got stuck. Would appreciate some help.
Here's what I did by trying to use the method of characteristics:
$\frac{dt}{ds}=1$, $\frac{dx}{ds}=c$, $\frac{dz}{ds}=e^{2x}$, where $s$ is the parametrization variable for a characteristic curve lying in the surface $u(x,t)$.
So I get $t(s) = s+C_1$, $x(s) = cs+C_2$, $z(s) = e^{2x}+C_3$. We can thus see that $ct-x = C_1-C_2 = D_1$, where $C_i$ and $D_i$ are some constants.
But here's where I'm stuck since I have no idea what should be done next. That is, how can we now relate $ct-x$ to $z(x,t)$ to build the surface?
I was thinking that, maybe, do something like this: $z-te^{2x}=C_3-C_1=D_2$, so $z(x,t) = D_2 + te^{2(ct-D_1)}$ (after substituting). But is this the solution then and what's remaining is to apply the initial condition?
$z(x,t) = u(x,t)$, $u(x,0) = D_2 = f(x)$, so $u(x,t) = f(x) + te^{2(ct-D_1)}$. Unfortunately, this does not match the solution from WolframAlpha.
 A: The answer to the question raised by "sequence" is given in the comments. So, my answer is only a different form (but equivalent) of the method of characteristics with advantage of clearness :
$$u_t+cu_x=e^{2x}$$
The characteristic equations are : $\quad \frac{dt}{1}=\frac{dx}{c}=\frac{du}{e^{2x}}$
From $\frac{dt}{1}=\frac{dx}{c}$ , the first characteristic curve :
$$x-ct=c_1$$
From $\frac{dx}{c}=\frac{du}{e^{2x}} \quad\to\quad du-\frac{1}{c}e^{2x}dx=0$ , the second characteristic curve :
$$u-\frac{1}{2c}e^{2x}  =c_2$$
Thus, the general solution of the PDE, expressed on implicit form, is :
$$\Phi\left((x-ct)\:,\:(u-\frac{1}{2c}e^{2x})\right)=0$$
where $\Phi$ is any differentiable function of two variables.
Solving for the second variable leads to the explicit form : $\quad (u-\frac{1}{2c}e^{2x})=F(x-ct)$
$$u=\frac{1}{2c}e^{2x}+F(x-ct)$$
where $F$ is any differentiable function.
The condition : $u(x,0)=f(x)$ implies $\frac{1}{2c}e^{2x}+F(x-0)=f(x) \quad\to\quad F(x)=f(x)-\frac{1}{2c}e^{2x}$
$ F(x-ct)=f(x-ct)-\frac{1}{2c}e^{2(x-ct)}$ 
$u=\frac{1}{2c}e^{2x}+F(x-ct)=\frac{1}{2c}e^{2x}+f(x-ct)-\frac{1}{2c}e^{2(x-ct)}$
$$u(x,t)=f(x-ct)+\frac{1}{2c}e^{2x}(1-e^{-2ct})$$
