I have a ODE system of the form $\dot{x} = f(x)$ with $x \in \mathbb{R}^3$. Now it is claimed that if $(x_1,x_2,x_3)$ is a solution to the system that also $(-x_1,-x_2,-x_3)$ is a solution.
How can I show that this holds, I'm namely not sure how to deal with this? I first thought that I would just show that $-f(x) = f(-x)$ but when computing I saw I wasn't getting anywhere.
ps. If you are wondering why I did not post the ODE system itself, it is because I want to try it for myself.
-edit-
Since I am not really getting response the ODE about which it is about it is
$$\dot{x}_1 = -a_1 x_1 + b_1 x_2 \\ \dot{x}_2 = -a_2 x_2 + b_2 x_1 x_3 \\ \dot{x}_3 = -a_3(x_3 - \lambda) - b_3 x_1 x_2 $$
-edit 2-
A friend of mine told me that if I would calculate the fixed points of the system with $(-x_1,-x_2,-x_3)$ that they give the same fixed points as $(x_1,x_2,x_3)$ and that it would imply that the solutions are the same? But I cannot see how this would imply that.
-edit 3- apparently there was a error in the exercise that I had. I had to proof that if $(x_1,x_2,x_3)$ then $(-x_1,-x_2,x_3)$ was also a solution and not $-x_3$. Subsituting then these variables then in the system delivers the same state space system.
If $x$ and $-x$ are both solutions, $\dot{x} = f(x)$ while $\dfrac{d}{dt}(-x) = -\dot{x} = f(-x)$, so it must be true that $f(-x) = -f(x)$.