Suppose I have the equation
$$u_t+3t^2u-x = u^2, \;\; \; \; \; u(x,0)=u_0(x)$$
and then the goal is to sketch the characteristics of this equation.
Solving using the method of characteristics I let $$\begin{align} &x'(t) = 3t^2,\; \;x(0) = x_0\\&z'(t)=z^2, \; \; z(0) = u_0(x_0) \end{align}$$
which gives $$x=t^3+x_0, \; \; \, z = \frac{1}{C-t}, C \in \mathbb{R}.$$
We find $C$ from $z(0)=u_0(x_0) \implies u_0(x_0)=\frac1C$ which yields $$z(t) = \frac{u_0(x_0)}{1-u_0(x_0)t}.$$
Since we know $x_0 = x-t^3$, we end up with
$$u(x,t) = \frac{u_0(x-t^3)}{1-u_0(x-t^3)t}$$
When and how do I sketch the "characteristics" of this equation?