I have the following characteristic equation to solve:
$$ \frac{dt}{x^2} = \frac{dx}{-4xu} = \frac{du}{-4u^2} $$
I start with: \begin{align} \frac{dt}{x^2} &= \frac{dx}{-4xu} \\ 4udt &= -xdx \\ 4ut & = \frac{-1}{2}x^2 + A \end{align}
I then moved onto:
\begin{align} \frac{dx}{-4xu} & = \frac{du}{-4u^2} \\ \frac{dx}{x} & = \frac{du}{u} \\ \ln u & = \ln x + \ln B \\ B & = \frac{u}{x} \end{align}
I then said let $A = F(B)$ which then gives me: $$ 4ut + \frac{1}{2}x^2 = F(\frac{u}{x}) $$
However the solution given has: $$ x + \frac{4tu}{x} = F(\frac{u}{x}) $$
what am I doing wrong?
thanks.
None of the variables is constant along characteristics. Thus your first integration is wrong, $u$ is not constant. The second integration is correct.
If you insert $u=Bx$ into the first equation, you get $4Bdt=-dx$ so that $x=A-4Bt$.