# Differential equation : u is a maximal solution, $u(t_0) = 0$, why $u'(t_0) \neq 0$?

Let $a : \mathbb{R} \rightarrow \mathbb{R}$, continuous, such as $a(t) > 0, \forall t \in \mathbb{R}$.

We consider the differential equation : $y''(t) = a(t) y(t)$.

We have to prove that if $u$ is a maximal solution defined on $I = \mathbb{R}$, $u$ is a solution not identically zero, and $u(t_0) = 0$ for a certain $t_0 \in \mathbb{R}$, then $u'(t_0) \neq 0$.

It's probably easy, but I'm a beginner and I don't see how to do it. Someone could help me ? Thank you in advance.

Suppose we had a maximal solution defined on $\Bbb R$ that had $y'(t_0) = y(t_0) = 0$. By "restarting the ODE at $t_0$", we notice that we have the initial value problem(IVP) \begin{align} y''(t) &= a(t) y(t) \quad t∈ [t_0,∞) \\ y(t_0) &= 0 \end{align} $0$ is a solution for the interval $[t_0,∞)$. By uniqueness of solutions, $y = 0$ for all $t≥t_0$.
By "time reversing" $w(t) := y(-t)$, we note that $w: [-t_0,∞) → \Bbb R$ solves the IVP \begin{align} w''(t) &= a(-t) w(t) \quad t∈ [-t_0,∞) \\ w(-t_0) &= 0\end{align} As before, $w = 0$ for all $t≥-t_0$, i.e. $y = 0$ for all $t≤t_0$. Hence $y$ is identically 0, contrary to assumption.
Remember that the only solution satisfying $y(t_0)=y'(t_0)=0$ is $y(t)\equiv 0$.