Question about Different Types of First-Order PDEs I am currently working my way through a question, and wanted some verification for my work. The question reads:

In each of the following, $u(x,t)$ is a smooth solution to the PDE for $x\in\mathbb R$ and $t>0$, and:$$u(4,1)=1$$
  In each case, find a particular point $x_0\in\mathbb R$ for which we know the value of $u(x_0,t)$. 

$\boldsymbol{u_t+u_x=0}$: Using the Method of Characteristics, I found that $u(x,t) = g(x_0)$ where $x_0 = x-t$, so here $x_0=3$, and $u(3, 0) = 1$.
$\boldsymbol{u_t+uu_x=0}$: Using Method of Characteristics again I got that $x_0 = x-u(x,t) t$, so $x_0=3$ again and $u(3,0) = 1$. 
$\boldsymbol{u_t+t^2u_x=0}$: Using Method of Characteristics I got that $x_0 = x-\frac{t^3}{3}$, so $x_0 = \frac{11}{3}$ and so $u(\frac{11}{3},0)=1$. 
$\boldsymbol{u_t+u_x+u=0}$: Using Method of Characteristics I got that $u(x,t) = u(x-t,0)\cdot e^{-t}$, so thus $u(3,0) = e$.
Can someone verify my work? I'm not sure if this is 100 percent correct.
 A: $\boldsymbol{u_t+u_x=0}\quad$:  The general solution is $\quad u(x,t)=F(x-t)\quad$ where $F(X)$ is any differentiable function.
The specified condition $u(4,1)=1=F(4-1)=F(3)$ is not sufficient to derermine a unique fonction $F$, but a point on anyone of the functions is determined. Nevertheless it is sufficient to say that at point $(x_0,0)$ the condition is satisfied if $x_0-0=3$. Thus we know that $u=1$ at the particular point $(3,0)$.
$\boldsymbol{u_t+uu_x=0}\quad$: The general solution is $\quad u(x,t)=F(x-u\:t)\quad$ where $F(X)$ is any differentiable function.
The same reasoning as above applies with the condition $u(4,1)=1=F(4-1*1)=F(3)$. Thus, the particular point again is $(3,0)$ where we know that $u=1$.
$\boldsymbol{u_t+t^2u_x=0}\quad$: The general solution is $\quad u(x,t)=F\left(x-\frac{t^3}{3} \right)$
The same reasoning as above applies with the condition $u(4,1)=1=F\left(4-\frac{1^3}{3}\right)=F\left(\frac{11}{3}\right)$. At point $(x_0,0)$ the condition is satisfied if $x_0-\frac{0^3}{3}=\frac{11}{3}$. Thus we know that $u=1$ at the particular point $(\frac{11}{3},0)$.
$\boldsymbol{u_t+u_x+u=0}\quad$: The general solution is $\quad u(x,t)=e^{-t}F(x-t)\quad$. The same reasoning leads to $F(3)=e^1$. So, we know that $u=e$ at the particular point $(3,0)$.
All is in agreement with your own results.
