On ODE with uniqueness solution Let $f:\mathbb{R}\times\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ continuous and Lipschitz. Let $\gamma(t)$ solution for the Cauchy Problem:
$$
\begin{cases}
(x,y)'=f(t,(x,y)) \\
(x,y)(0)=(7,-10)
\end{cases}
$$
I need to prove that, if $f(t,(x,0))=0$ for all $(t,(x,0))\in\mathbb{R}\times\mathbb{R}^{2}$, so $\gamma_2(t)\neq 5$ for all $t\in\mathbb{R}$, where $\gamma(t)=(\gamma_{1}(t),\gamma_{2}(t))$.
First, I know that my Cachy Problem has uniqueness of solution, for $f$ is Lipschitz. But the only I thing I can see some way to solve the problem is to prove that $\gamma_{2}(t)$ cannot reach positive numbers, since $\gamma(0)<0$. How can I do that?
 A: Suppose to the contrary that $\gamma_2$ takes value $5$ for some $T \ne 0$.  As the function $\gamma_2$ is continuous and $\gamma_2(0) < 0$ there is some $\tau \ne 0$ such that $\gamma_2(\tau) = 0$.
Denote $\xi := \gamma_1(\tau)$ and consider the initial value problem
\begin{equation*}
\tag{$*$}
\begin{cases}
(x, y)' = f(t, (x,y))
\\
x(\tau) = (\xi, 0).
\end{cases}
\end{equation*}
$\tau$ and $\xi$ are so chosen that $\gamma$ is a solution of $(*)$.  But the function $\tilde{\gamma} \colon \mathbb{R} \to \mathbb{R}^2$ defined as $\tilde{\gamma}(t) := (\xi, 0)$ for all $t \in \mathbb{R}$ satisfies $(*)$, too.  And this contradicts the uniqueness of a solution to $(*)$.  Incidentally, it is sufficient for $f$ to be continuous in $(t,x,y)$ and locally Lipschitz in $(x,y)$.
The uniqueness is essential here.  Indeed, let $f(t, (x,y)) := 3 y^{2/3}$.  Then $f(t,(x,0))=0$ for all $t \in \mathbb{R}$ and all $x \in \mathbb{R}$, but the solution $\check{\gamma} \colon \mathbb{R} \to \mathbb{R}^2$ given by $$
\check{\gamma}(t) := (7, (t - (-10)^{1/3})^3), \quad t \in \mathbb{R},
$$
serves as a counterexample.
