I'm trying to find the flow line $\textbf{u}(t)$ of $\textbf{F}(x,y,z)=(2,-3y,z^3)$ that passes through a point $(a,b,c)$ at time $t=0$.
To solve this as I have been taught, I need to solve this system of differential equations:
$$\begin{cases} x'(t) = 2 \\ y'(t) = -3y(t) \\ z'(t) = {z(t)}^3 \end{cases} $$
Solving these, I find: $$\begin{cases} x(t) = 2t + C \\ y(t) = e^{-3t+K} \\ (z(t))^2=-\frac{1}{2t+J} \end{cases} $$
How would I deal with the exponent on $z(t)$?
So the equation is
$z'(t)=z(t)^3$
$\Rightarrow$ $\frac{dz}{dt}=z^3 $
$\Rightarrow$ $\frac{dz}{z^3}=dt$
$\Rightarrow$ $\int{\frac{dz}{z^3}}=\int{dt}$
$\Rightarrow$ $\frac{1}{-2z^2}=t+j$
$\Rightarrow$ $z(t)^2=-\frac{1}{2(t+j)}$ or $z(t)^2=-\frac{1}{2t+J}$ where $J=2j$.
Now $2t+J<0$ for you to get a sensible answer.
$\Rightarrow -2t-J>0$
$\Rightarrow z(t)=\frac{1}{\sqrt{-2t-J}}$