Solve this PDE using the characteristic form
$\begin{equation} \frac{\partial w}{\partial t} -\frac{\partial w}{\partial x}=-w \\ w(0,t)=4e^{-3t} \end{equation}$
My attempt
We know $w(t)=w(x(t),t)$
Then by chain rule
$\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\times \frac{\partial x}{\partial t} + \frac{\partial w}{\partial t}$
This implies
$\begin{equation} \frac{\partial x}{\partial t}=-1 \\ \frac{\partial w}{\partial t}=-w \end{equation}$
Here, i'm stuck. Can someone help me?
You are not applying the method correctly. You are using $t$ to denote a coordinate in the plane and again for the parameter.
Write the equation in the form $1 \cdot w_x + (-1) \cdot w_t + (-1) \cdot w = 0$.
The characteristic curve $(x(s),t(s))$ then satisfies $x'(s) = 1$ and $t'(s) = -1$.
The solution $z(s)$ satisfies $z'(s) = z(s)$.
Solve these ODE with initial data $x(0) = 0$, $t(0) = t_0$, and $z(0) = w(0,t_0)$ to get \begin{align*} x(s) &= s \\ t(s) &= -s + t_0 \\ z(s) &= w(0,t_0)e^{s} = 4e^{-3t_0}e^{s}. \end{align*}
You can add the first two of these equations to find that the characteristic curve containing a point $(x,t)$ satisfies $t_0 = x+t$ and $s = x$. Thus $$\boxed{w(x,t) = z(s) = 4e^{-3(x+t)}e^{x} = 4e^{-2x - 3t}.}$$