Solving $u_t+2u_x+3u=0$ with $u(x,t=0) = e^{-x^2/l^2}$ 
EDIT The issue is with the characteristic equations. 
It should be 
$$\frac{dt}{1} = \frac{dx}{2} =\frac{dw}{3u}$$
and not 
$$\frac{dt}{1} = \frac{dx}{2} =\frac{dw}{3}$$

Solving the following PDE $u_t+2u_x+3u=0$ with  $u(x,t=0) = e^{-x^2/l^2}$
So I've got my characteristics:
$$C_1 = x-2t$$
$$C_2 = 3x-2u$$
And I know the general form will be a function of the form 
$$f(x-2t, 3x-2u)$$
And we can do the following
$$3x-2u = f(x-2t) \Rightarrow u = \frac12\left(3x-f(x-2t)\right) $$
Now we know that:
$$u(x,t=0) = e^{-x^2/l^2} = \frac12\left(3x-f(x)\right)$$
and 
$$f(x) = 3x - 2e^{-x^2/l^2}\Rightarrow f(x-2t) = 3(x-2t) - 2e^{-(x-2t)^2/l^2} $$
Now the next step is where I'm having difficulties. Am I plugging f(x) back into here?
$$u = \frac12\left(3x-f(x-2t)\right)$$
I'd appreciate any hints
 A: $$u_t+2u_x=-3u$$
The system of characteristic differential equations can be summarized as :
$$\frac{dt}{1}=\frac{dx}{2}=\frac{du}{-3u}$$
The equation of a first family of characteristic curves comes from :
$$\frac{dt}{1}=\frac{dx}{2} \quad\to\quad x-2t=c_1$$
The equation of a second family of characteristic curves comes from :
$$\frac{dx}{2}=\frac{du}{-3u} \quad\to\quad ue^{\frac{3}{2}x}=c_2$$
The general solution of the PDE is, expressed on the form of implicit equation, is :
$$\Phi\left( (x-2t)\:,\:(ue^{\frac{3}{2}x})\right)=0$$
Where $\Phi$ is any differentiable function of two variables.
In this case, it is possible to separate $u$ to the explicit form :
$$ue^{\frac{3}{2}x}=f(x-2t) \quad\to\quad u= e^{-\frac{3}{2}x}f(x-2t)$$
where $f$ is any differentiable function.
With the condition :
$$u(x,0)=e^{-\frac{x^2}{L^2}}=e^{-\frac{3}{2}x}f(x-2*0)=e^{-\frac{3}{2}x}f(x)$$
The function $f$ is determined (doesn't matter the symbol of the variable)
$$f(X)=e^{\frac{3}{2}X}e^{-\frac{X^2}{L^2}}$$
In the above general solution we have $\quad X=x-2t\quad\to\quad f(x-2t)=e^{\frac{3}{2}(x-2t)}e^{-\frac{(x-2t)^2}{L^2}}\quad$ which leads to :
$$u(x,t)= e^{-\frac{3}{2}x} e^{\frac{3}{2}(x-2t)}e^{-\frac{(x-2t)^2}{L^2}}$$
$$u(x,t)= e^{-3t-\frac{(x-2t)^2}{L^2}}$$
A: This question can be solved without using the method of characteristics. Suppose we have $au_{t} + bu_{x} + cu = 0$ with initial value condition $u(x,0) = f(x)$. Then $a\frac{u_{t}}{u} + b\frac{u_{x}}{u} = -c \Rightarrow (log(u))_{t} + \frac{b}{a}(log(u))_{x} = -\frac{c}{a}$. Define $v = log(u)$, where our initial value condition is now $v(x,0)= log(f(x))$ - what we have is the transport equation for $v$, which has solution
$$ v(x,t) = log(f(x - \frac{b}{a}t)) + \int_{0}^{t}(-\frac{c}{a})ds = log(f(x - \frac{b}{a}t)) - \frac{c}{a}t$$
$$ \Rightarrow u(x,t) = exp(v(x,t)) = f(x - \frac{b}{a}t)exp(-\frac{c}{a}t)$$
See Evans' Partial Differential Equations, section 2.1 for a reference on the transport equation.
A: The simplest way to solve the problem is to 
consider the equation as a linear. (No need to hurry, the proven way to study first-order PDEs is first to study linear, then semilinear, and only then quasilinear equations; I'd recommend my book
on PDEs, in which I follow this approach.) 
There is a simple general result stating 
that if $(a,b) \ne (0,0),$ then the general solution of the linear PDE
$$
au_x + bu_y +cu=0
$$
with constant coefficients is given by
$$
u(x,y)=e^{-\lambda(ax+by)} f(bx-ay)
$$
where 
$$
\lambda=\frac c{a^2+b^2}
$$
and $f \in C^1(R)$ is a continuously differentiable function (it's a result worth remembering). In particular, the general solution of your equation is 
$$
u(t,x)=e^{-3t/5-6x/5}f(2t-x)
$$
where $f \in C^1(\mathbf R)$ is a continuously differentiable function. Now by the initial
condition,
$$
u(0,x)=e^{-6x/5}f(-x)=e^{-x^2/l^2},
$$
whence
$$
f(-x)=e^{-x^2/l^2}  e^{6x/5}
$$
whence
$$
f(x)=e^{-x^2/l^2}e^{-6x/5},
$$
since $f$ is defined everywhere on $\mathbf R.$ 
It then easily follows that the solution of the Cauchy
problem in question is 
$$
u(t,x)=e^{-3t} e^{-(2t-x)^2/l^2}.
$$
