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 \}.