Prove that if $y'=f(x,y),y(x_0)=y_0$ is invariant under the transformation $(x,y) \mapsto (-x,-y)$, then $y(-x)=-y(x)$ provided the existence How to make a clear justification for the proposition:
If $y'=f(x,y),y(x_0)=y_0$ is invariant under the transformation $(x,y) \mapsto  (-x,-y)$, then $y(-x)=-y(x)$ provided the existence of solution.
For instance, $u(x)=-y(-x)$ can also satisfy the ODE $z'(x)=z(x)^2,~z(x_0)=z_0$, but how it is necessary whenever the ODE is invariant under $(x,y) \mapsto  (-x,-y)$?
 A: Assuming that the transformation is in the sense that the original initial condition $y(x_0)=y_0$ is transformed into $-y(-x_0)=-y_0$ the statement is not true. Consider the ODE $y'(x)=0, y(0)=1$ is invariant under the transformation, but its solution $y(x)=1$ is not an odd function. This is the case regardless of whether $y'(x)$ is transformed into $-y'(-x)$ or into
$$
  \frac{\partial}{\partial x} (-y)(-x) = -\frac{\partial}{\partial x} (y)(-x) = y'(-x),
$$
so the contra-example is valid under both interpretations of symmetry.
If the transformation was in the sense that the initial condition $y(x_0)=y_0$ is transformed into $-y(-x_0)=y_0$, then the statement would fail for other reasons as well. For example the ODE $y'(x)=y^2(x)$ and $y'(x)=y^3(x)$ (each of which is symmetric in one of the sense) has solution trajectories for each initial conditions that are not odd functions.
A: This statement is correct if the full IVP is symmetric in this sense.

*

*As the initial time is invariant under sign change, $x_0=0$ follows.

*As the initial value is invariant under sign change, $y_0=0$ follows.

In consequence, for $z(x)=-y(-x)$, it follows that
$$
z'(x)=y'(-x)=f(-x,y(-x))=f(-x,-z(x))=f(x,z(x))
$$
so that $z$ is another solution to the same IVP, so from uniqueness it follows that $z=y$ or $y(-x)=-y(x)$.
In the mentioned examples $y'=y^m$, the solution through the origin is always the zero-function, so that the symmetry is trivial.
As another example, the solution of $y'=x^2+y^2$ through the origin is a non-trivial odd function. The same goes for $y'=1+y^2$.
