Sketch characteristics of simple PDE 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?
 A: Probably they are typos in your question. You wrote :
$$u_t+3t^2u-x = u^2\tag 1$$
but you solve it as it was :
$$u_t+3t^2u_x = u^2\tag 2$$
If Eq.$(2)$ is the correct PDE you are right with the characteristic equations :
$$x-t^3=c_1 \tag 3$$
and
$$u=\frac{1}{c_2-t} \tag 4$$
The general solution of the PDE is $u=\frac{1}{F(c_1)-t}$
$$\boxed{u(x,t)=\frac{1}{F(x-t^3)-t}}\tag 5$$
where $F$ is an arbitrary function to be determined according to the specified condition.
$u(x,0)=u_0(x)=\frac{1}{F(x-0)-0}=\frac{1}{F(x)}$
$$F(x)=\frac{1}{u_0(x)}$$
The function $F$ is determined. We put it into the above general solution $(5)$ where the argument is $(x-t^3)$ thus $F(x-t^3)=\frac{1}{u_0(x-t^3)}$
$$u(x,t)=\frac{1}{\frac{1}{u_0(x-t^3)}-t}$$
$$\boxed{u(x,t)=\frac{u_0(x-t^3)}{1-t\: u_0(x-t^3)}}\tag 6$$
This is in agreement with your own result. So your calculus is correct for me.
In fact your question is : When and how do I sketch the "characteristics" of this equation?
For me the Eq.$(6)$ has no "characteristic" strictly speaking.
I suppose that it is asked to sketch the equations $(3)$ and $(4)$ which are characteristic equations of the PDE : i.e. the family of cubic curves for Eq.$(3)$ and the family of hyperbolic curves for Eq.$(4)$. But don't take it for granted.
