Prove that solution to Cauchy problem is even function.

Question

Please help me to solve this problem:

Differential equation $y'=f(x,y)$ has 2 properties:

1. For the right side the theorem about uniqueness and existence of solutions is true.
2. $f(-x,y) = -f(x, y)$.

I have to prove that the solution to Cauchy problem $y(0)=1$ is even function.

My ideas:

I know that the derivative of odd function is even and vice versa. So $y$ can not be odd. However, it is not enough - $y$ in theory may be neither odd nor even.

Can you please help me with the problem? Thanks a lot!

Answer 1

You can consider the function $z(x) := y(-x)$. It is not difficult to show that $z$ is a solution of the same Cauchy problem $$ z' = f(x,z), \qquad z(0) = 1. $$ Namely $$ z'(x) = -y'(-x) = -f(-x, y(-x)) = f(x,z(x)). $$ But, from uniqueness, it must be $z(x) = y(x)$, i.e. $y(-x) = y(x)$.