Prove that an initial value problem has more than 1 solution

Question

let $y' = \sin{\sqrt{|y|}}$, I need to prove that there is more than one solution with the initial value $y(0) = 0$, and I was given guidance to use $y = z^2$, reach $z'=f(x, z)$ and use the existence and uniqueness theorem to find a solution for $z'=f(x, z)$, with it I'm suppose to prove the existence of a second solution.

the first solution is easy to find $y \equiv 0$ is a solution. But I'm not sure how to find the second even with the guidence, here is what I tried:

Using the guidance I say $$ y = z^2 \Longrightarrow y' = 2z \cdot z' \Longrightarrow 2z\cdot z' = \sin{\sqrt{|z^2|}} = \sin{z} \Longrightarrow z' = \frac{\sin{z}}{2z} $$

Because $\frac{\sin{z}}{2z}$ is continuous and $\frac{\partial}{dz}\frac{\sin{z}}{2z}$ is also continuous when $z\neq 0$ from the existence and uniqueness theorem I get that there is a solution $u(x)$ for $z' = \frac{\sin{z}}{2z}$ and a point $(x_0, z_0)$ ($z_0 \neq 0$), so I create a new function $v(x)$ where $v(0) = 0$ and $v(x) = u^2(x)$ for $x\neq 0$, $v$ is a solution for $y' = \sin{\sqrt{|y|}}$, and it upholds the initial value. Because it doesn't uphold the existence and uniqueness theorem 's requirements there is no contradictions with having 2 solution to the same initial value problem

Though my solution seems good to me I feel like something is missing, but I can't put a finger on it

EDIT:

I'm was told to work with $y = z^2$, I understand that the solution to the IVP should look like $$ v(x) = \left\{ \begin{array}{ll} 0 & \mbox{if } x \leq 0 \\ u^2(x) & \mbox{if } x > 0 \end{array} \right. $$ where $u(x)$ is a solution for $z' = \frac{\sin{z}}{2z}$. $u(x)\not\equiv 0$ because $0$ isn't a solution for $z$, so $v(x) \not\equiv 0$ but I don't know how I can prove that $v(x)$ is derivable

Answer 1

Based on $$ z'=\frac{\sin(|z|)}{2z} $$ for $z\ne 0$, you can consider the ODEs $$ z'=\pm f(z)~~\text{ where }~~f(z)=\begin{cases}\frac{\sin(z)}{2z},&z\ne 0\\\frac12,&z=0\end{cases} $$ As you observed, $f(z)$ has a nice power series expansion, so satisfies the conditions of the existence-and-uniqueness theorem.

Solutions of $z'=f(z)$ give solutions of the original ODE whenever $z\ge0$, and those of $z'=-f(z)$ whenever $z\le0$. Both variants give a solution each for $x\ge 0$ that passes through $z(0)=0$.