Given the following IVP: $$y'=f(x,y)=4x^3 e^{-|x^2+y|}$$ $$y(x_0)=y_0$$
I need to show that the IVP has a single solution in $(-\infty,\infty)$, and that it is an even function.
I was able to show, after some struggles that $f(x,y)$ fulfills Lipschiz condition with respect to $y$ in every closed vertical infinite rectangle of the form $[a,b]\times(-\infty,\infty)$ (using showing it separately for the functions in the composition of $4x^3e^z$ with $|x^2+y|$), thus there is a single unique solution in $(-\infty,\infty)$.
Yet, I fail to show that the solution $u(x)$ is an even function.
The problem arises when I tried to show that the derivative of the solution, $u'(x)$, is odd, since $f(x,u)$ depends on $u(x)$ itself, which leads me to a cyclic argument.
Any hint would be appreciated.
Set $\tilde y(t)=y(-t)$. Then we see by inserting, that $\tilde y$ solves the same IVP as $y$. Since you have proven uniqueness, it follows that $\tilde y=y$. This proves that $y$ is even.