In constant accelaration and linear (1D) motion, we can show that relationship between velocity and position is quadratic (parabola) by
We can write $v$ in the form of $v=v(x(t))$ \begin{equation} a=\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \\ a=v\cdot\frac{dv}{dx}\\ \int a\text{ }dx = \int v\cdot\frac{dv}{dx} \text{ } dx\\ ax = \frac{v^2}{2} + C \end{equation} This relationship is not complicated to prove, it is also does not involve time relationship between velocity or position (that is we do not need to find v(t) and x(t).) Then, I think that it can be done in similar way under the condition constant jerk(and of course not 0).
My Attempt: Jerk is defined by $J=\frac{da}{dt}$
\begin{equation} J=\frac{da}{dt}=\frac{da}{dv} \cdot \frac{dv}{dt}\\ J=a \cdot \frac{da}{dv} \\ Jv+C_1 = \frac{a^2}{2}\\ a= \pm\sqrt{2Jv+C_2}\text{ }; C_2 = 2C_1 \end{equation} Acceleration is defined by $a=\frac{dv}{dt}$ \begin{equation} a=\frac{dv}{dt}=\frac{dv}{dx} \cdot \frac{dx}{dt}\\ a=v \cdot \frac{dv}{dx}\\ \pm\sqrt{2Jv+C_2}=v \cdot \frac{dv}{dx} \end{equation} Solve this DE and we will get \begin{equation} x=\pm \frac{(Jv-C_2)\sqrt{2Jv+C_2}}{3J^2}+C_3 \end{equation} This is where I start to think if there are any mistakes in my solution. Because the solution seems to be very ugly. But if the solution is correct, I would like to see an easier proof. So please verify this proof.