It's given differential equation. How do I determine the solution curve of the begining values $x(0) = 0$,$p(0) = 1$ in the $(x, p)$ plane? It's given differential equation

$\dot{x}=p$

$\dot{p}=-x^3+x$

How do I determine the solution curve of the begining values $x(0) = 0$,$p(0) = 1$ in the $(x, p)$ plane?
As
$$
0=\dot x\dot p - \dot p\dot x=p\dot p+(x^3-x)\dot x
$$
solutions satisfy
$$
2p^2+(x^2-1)^2=C
$$
and for the given initial values $C=2·1^2+(0-1)^2=3$.
As
$$
0=\dot x\dot p - \dot p\dot x=p\dot p+(x^3-x)\dot x
$$
solutions satisfy
$$
2p^2+(x^2-1)^2=C
$$
and for the given initial values $C=2·1^2+(0-1)^2=3$.