# Plotting a Function

How can one plot a three-dimensional plot of a differential equation system of the following to show the trajectories in mathematica:

$x' = yz,~ y' = -2xz,~ z' = xy$

I tried using VectorPlot3D and ParametricPlot3D

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ParametricPlot3D[] should be fine. What were you expecting that you didn't get? – J. M. Nov 11 '11 at 4:57
J.M.: I would like the trajectories to be displayed along the xyz axis. Parametric places it inside of a volume, (rectangle). – night owl Nov 11 '11 at 5:04
Oh, you want three-dimensional axes to be shown (somewhat like using Axes instead of Frame in the two-dimensional case)? – J. M. Nov 11 '11 at 5:08
As an alternative : Maple command DEplot3d – pedja Nov 11 '11 at 7:20
If you send me all necessary parameters I will try to plot your system... – pedja Nov 11 '11 at 14:27

How about like so, parameterizing initial conditions by spherical angles $\theta$ and $\phi$:

Traj[\[Theta]_, \[Phi]_, tmax_] :=
Module[{sol}, {sol} =
NDSolve[And @@ {x'[t] == y[t] z[t], y'[t] == -2 x[t] z[t],
z'[t] == x[t] y[t], x[0] == Sin[\[Theta]] Cos[\[Phi]],
y[0] == Sin[\[Theta]] Sin[\[Phi]], z[0] == Cos[\[Theta]]}, {x,
y, z}, {t, 0, tmax}];
sol]

Show[ParametricPlot3D[{x[t], y[t], z[t]} /. {Traj[Pi/5, Pi/3, 7],
Traj[3 Pi/5, Pi/3, 7], Traj[2 Pi/5, Pi/7, 7]}, {t, 0, 7},
PlotRange -> {{-1, 1}, {-1, 1}, {-1, 1}}, Evaluated -> True] /.
Line -> Tube, Graphics3D[{Sphere[{0, 0, 0}, 1]}]]

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@nightown Try this Show[ParametricPlot3D[{x[t], y[t], z[t]} /. {Traj[Pi/5, Pi/3, 7], Traj[3 Pi/5, Pi/3, 7], Traj[2 Pi/5, Pi/7, 7]}, {t, 0, 7}, PlotRange -> 1.4 {{-1, 1}, {-1, 1}, {-1, 1}}, Evaluated -> True] /. Line -> Tube, ParametricPlot3D[{Sin[\[Theta]] Cos[\[Phi]], Sin[\[Theta]] Sin[\[Phi]], Cos[\[Theta]]}, {\[Theta], 0, Pi}, {\[Phi], 0, 2 Pi}, PlotStyle -> None, MeshStyle -> LightGray], Boxed -> False, AxesOrigin -> {0, 0, 0}, Ticks -> None] – Sasha Nov 11 '11 at 18:44