The exercise is as follows: Find and show in the plane (x, y) characteristics of the following equation:
$$u_t+x^2u_x=u$$
Get the general solution and check for direct replacement.
I was able to calculate the general equation, but I can not show the characteristics in the plane (x, y) and do not know how to do the direct substitution.
My calculations:
First, let's find the general solution to:
$$ u_t+x^2u_x=u. $$ This means characteristic ordinary differential equations are given by: $$ \begin{align*} \dfrac{\partial t}{\partial s}(r,s) &= 1, \\ \dfrac{\partial x}{\partial s}(r,s) &= x^2, \\ \dfrac{\partial z}{\partial s}(r,s) &= z. \\ \end{align*} $$ Solving for each one, we have $$ \begin{align*} dt=ds &\implies t=s+c_1(r), \\ \frac{dx}{x^2}=ds &\implies -\frac{1}{x}=s+c_2(r), \\ \frac{dz}{z}=ds &\implies \ln z = s+c_3(r) \implies z(r,s)= C_3(r)e^{s},\mbox{ where } C_3(r) = e^{c_3(r)}. \\ \end{align*} $$ Since $s=t-c_1(r)$, we see that $-\frac{1}{x}=t-c_1(r)+c_2(r)$, or $\boxed{t=-\frac{1}{x}+x_0(r)}$, where $x_0(r)=c_1(r)-c_2(r)$.
So the general solution is $$ \begin{align*} z(r,s) &=u(x(r,s),t(r,s)) \\ &=u\left(x,-\frac{1}{x}+x_0(r) \right) \\ &=C_3(r) e^{-\frac{1}{x}-c_2(r)} \\ &= D(r)e^{-\frac{1}{x}+x_0(r)}, \mbox{ where }D(r)=C_3(r)e^{-c_1(r)}, \\ \end{align*} $$ and one could easily check that $\boxed{u\left(x, -\frac{1}{x}+x_0(r) \right) = D(r)e^{-1/x+x_0(r)}}$ satisfies $$ u_t+x^2u_x=u. $$
Finally, the characteristic curves on the $(x,t)$-plane are $t=-\frac{1}{x}+x_0(r)$.