Subqueston (d):
$ \frac{dt}{1} = \frac{dx}{x^2} $
$ t = -x^{-1}+c $
so the general solution is $ u(t,x)=f(t+x^{-1})$.
$u(0,x)=f(x^{-1})=g(x)$
$f(y)=g(y^{-1})$
Since $y=x^{-1}$, so when $x>0$ we have $y>0$ as well.
$u(t,x)=f(t+x^{-1})$
We need $t+x^{-1}>0$
Since $x>0$,
therefore $tx+1>0$
so the domain be defined is $\{(t,x) | tx>-1 \}.