Prove differential equation solution is even function Let $0\leq c&lt;1$,$u,v,w \in C^1(\mathbb{R})$. Suppose $u'=vw ,\; v'=-uw, \; w'=-c^2 uv$ and $u(0)=0, v(0)=1, w(0)=1$
Then, for all $x\in \mathbb{R} $,$v(x)=v(-x)$ and there is $x_0$ with $v(x_0)=0$

I know that $u^2+v^2=1,\; c^2 u^2+w^2=1$ and I think I use this property in order to prove $v$ is even and has zero point.
If $x\mapsto \bigl(u_0(x),v_0(x),w_0(x)\bigr)$ is the solution of the original IVP then the triple $x\mapsto \bigl(u_1(x),v_1(x),w_1(x)\bigr)$ defined by
$$u_1(x):=-u_0(-x),\quad v_1(x):=v_0(-x),\quad w_1(x):=w_0(-x)$$
is also a solution of the same IVP (check this!). The general existence and uniqueness theorem for systems of ODEs then implies that in fact
$$\bigl(u_0(x),v_0(x),w_0(x)\bigr)\equiv\bigl(u_1(x),v_1(x),w_1(x)\bigr)\ ;$$
in particular $x\mapsto v_0(x)$ is an even function.
First equality:
Multiply the first equation by $u$ and the second by $v$. Adding them together yields $uu' + v v' = 0$, which is equivalent to $\frac{1}{2} \frac{d}{dx}(u^2 + v^2) = 0$. Hence $u^2 + v^2 = C$. Plugging in the initial condition implies that $u^2 + v^2 = 2$.

Second equality:
Hint: Multiply first equation by $c^2 u$ and third equation by $w$.