PDE method of characteristics solving $e^{t^2}u_t+tu_x=0$ with $u(x,0)=x+2$ The equation is given as:
$$e^{t^2}u_t+tu_x=0$$ with $u(x,0)=x+2$
I've got $x=\frac{1}{2}e^{t^2}+x_0$ but I'm not sure where to go from there
 A: As you have done, the best way to solve the equation is to write it as
$$
t e^{-t^2} u_x + u_t = 0
$$
Let $u = u\big(x(\eta),t(\eta)\big)$, then
$$
\frac{d}{d \eta} u\big(x(\eta),t(\eta)\big) = u_x x' + u_t t' = 0
$$
means that
$$
t'(\eta) = 1, \qquad x'(\eta) = t e^{-t^2}, \qquad u'(\eta) = 0
$$
EDIT
The initial condition is given in the curve $t = 0$, this means we have $t(\eta=0) = 0$. 
Now,
$$
t'(\eta) = 1, \quad t(0) = 0 \,\Longrightarrow \,t(\eta) = \eta.
$$
Using this, we have for $x$
$$
\frac{d x}{d \eta} = \eta e^{-\eta^2},
$$
and then
$$
\int_0^\eta \frac{d x}{d \xi} d\xi = \int_0^\eta \xi e^{-\xi^2} d\xi \quad \Longrightarrow \quad x(\eta) - x(\eta = 0) = \frac{1}{2} \left(1 - e^{-\eta^2}\right)
$$
or
$$
x(\eta = 0) = x - \frac{1}{2} \left(1 - e^{-t^2}\right).
$$
Finally
$$
\frac{d u}{d \eta} = 0 \quad \Longrightarrow \quad u(\eta) - u(\eta = 0) = 0
$$
or
$$
u(\eta) = u(\eta = 0) = x(\eta = 0) + 2
$$
We now conclude that
$$
u(x,t) = x + 2 - \frac{1}{2} \left(1 - e^{-t^2}\right).
$$
