The question involves finding the solution to a partial differential equation.
The general solution that I found was $u(x,t)=F(x^{2}-t^2 e^{u})$ and the initial condition is $u(x,0)=2\ln (x)$. The issue I've been having is trying to apply that data to the general form.
What I tried was plugging in the data to the general solution to get: $2\ln (x) = F(x^{2})$ From there I let $x^{2} = z$ and hence $x=\sqrt{z}$, which gives me: $2\ln(\sqrt z ) = F(z)$
and then I tried taking this and subbing it through to the original solution I had, which gives:
$u(x,t) = 2\ln (\sqrt{x^{2}-t^2e^{u}})$
The solution to the problem is $u(x,t) = \ln (\frac{x^{2}}{1+t^{2}})$, was wondering if anyone could help point out what i did wrong.
Original PDE is $\partial_t u + \big(\frac{t}{x}e^u\big)\partial_x u = 0$ (see problem statement).
If you use the logarithm laws, $a\ln b=\ln( b^a)$, you find that $F(x^2)=\ln(x^2)$, thus $F(z)=\ln z$ and with that $$ u=\ln(x^2-t^2e^u)\implies e^u=x^2-t^2e^u\implies e^u=\frac{x^2}{1+t^2}. $$