I am looking for the help with the solution to the following PDE (along with IC): $$ uu_x + u_t = u, u(x, 0) = 2x $$
Solution (without parameterization): Applying the method of characteristics one can get the following system of ODEs: $$\frac{dx}{u} = \frac{dt}{1} = \frac{du}{u}$$ Combaining first and third: $$\frac{dx}{u} =\frac{du}{u} \Rightarrow du = dx \Rightarrow u = x + C_1$$ Thus, $\boxed{C_1 = u - x}$
Combaining second and third: $$\frac{dt}{1} = \frac{du}{u} \Rightarrow \frac{du}{u} = dt \Rightarrow \ln(u) = t + C_2 \Rightarrow u = C_2e^{t}$$ Thus, $\boxed{C_2 = ue^{-t}}$
Taking into an account that, in general the arbitrary constant is an arbitrary function, we can write $C_2 = G(C_1)$:
$$ue^{-t} = G(u - x)$$
Using the IC one can substitute it into the equation above, getting the following: $$2xe^{0} = G(2x - x) \text{ or } G(x) = 2x$$ Nowing how the arbitrary function does work (give to the function the argument and it gives you back the argument multiplied by 2), the particular solution becomes: $$u = e^{t}(u-x)\cdot 2 \Rightarrow u = 2ue^{t} - 2xe^{t} \Rightarrow u(1-2e^{t}) = - 2xe^{t} \Rightarrow u(x,t) = \frac{2xe^{t}}{2e^{t} - 1}$$
The answer to this problem, unfortunately, $u(x,t) = \frac{2xe^{t}}{2e^{t} \boxed{+} 1}$. I do not get how in the denominator they got plus. Can someone explain it?
P.S. Sorry for my English, I am not native speaker.