I am given the PDE:
$$u_x+u_y+u=e^{x+2y} \quad u(x,0)=0$$
I tried to do this using the method of the characteristics in the following way.
First I find the particular solution of this inhomogenous equation by substituting $u_p = Ae^{x+2y}$ giving me $A=\frac{1}{4}$ hence $$u_p=\frac{1}{4}e^{x+2y}$$
Now I try the homogenous equation
$$u_x+u_y+u=0$$
If I parameterize $x,y$ with $t$ and have now $u(x(t),y(t))$ I can write
$$\frac{du}{dt} = \frac{dx}{dt}\frac{\partial u}{\partial x} + \frac{dy}{dt}\frac{\partial u}{\partial y} $$
So equating like terms I have
$$\frac{dx}{dt}=1 \quad \frac{dy}{dt}= 1 \quad \frac{du}{dt}=-u$$
My initial conditions are $x(t=0)= x_0, y(t=0)=0$ and $u(t=0)=u(x(0),y(0))=u(x_0,0)=0$
So then I can solve these equations to get
$$x = t + x_0 \Rightarrow x_0=x-t \quad y = t \quad u = Ce^{-t}$$
Now I'm not 100% sure what $C$ should be but I assume it some function of the initial condition, i.e
$$u=F(x_0)e^{-t} \Rightarrow u=F(x-t)e^{-t} \Rightarrow u_0 = F(x_0)=0 ?$$
But the initial condition is $u_0 = 0$ so $u$ is $0$ at all $x$??
I'm getting stuck here but I followed similar method on other examples given and it works fine. Why is this initial condition $0$ giving me $u=0$, what am I missing?