I am dealing with following problem: $$ (3x^2+1)u_t + 2t u_x = 0, \quad u(x, 0) = e^{-\gamma x^2}$$
I started with stating the equations for the characteristics: $$\frac{dt}{ds} = 3x^2+1, $$ $$\frac{dx}{ds} = 2t,$$ $$\frac{du}{ds}=0,$$ and from the last equation I got $u = c_1$. Next, by dividing the first two equalities, I got $$ \frac{dx}{dt} = \frac{2t}{3x^2+1}$$ which is a separable ODE, giving me the implicit equation for the characteristics: $$x^3 + x - t^2 = c_2.$$ Following the method, $c_1 = u = F(c_2)$ and then it's easy to see that $u(x, t) = F(x^3+x-t^2)$ really is the general solution to the original PDE. My issue is now how to determine $F$ from the condition $u(x, 0) = F(x^3+x) = e^{-\gamma x^2}$. I can't really see how to find this undetermined function, since the degrees of the polynomials in the argument are different. Or did I make a different error that I didn't notice?