Non-zero solutions of Differential equations So I have here the following problem:
Let $f:I\rightarrow \mathbb{R}$, with $I$ an open interval be a non-zero solution of the equation $$y''+p(t)y'+q(t)y=0$$ with $p,q:I\rightarrow \mathbb{R}$ continuous functions. Prove that if $f(t_0)=0$ for some $t_0\in I$, then there is a $\delta \gt 0$ such that $f(t)\neq 0$ for all $t\neq t_0 \in ]t_0-\delta ,t_o+\delta[$. 
I'd like some advice on my attempt, whether it is right or wrong, and also any suggestion for a simpler proof. Here's what I tried:
As a consequence of the Existence and Uniqueness Theorem and from the fact that $f$ is a non-zero solution, we must have that $f'(t_o)\neq 0$ (otherwise we would have $f\equiv 0$).
Now, since $f$ is a solution of the differential equation, it has a second derivative, so $f':I\rightarrow \mathbb{R}$ must be continuous, so as $f'(t_o)\neq0$, there is a $\delta \gt 0$ such that $f'(t)\neq 0$ for all $t\in V=]t_o-\delta, t_o+\delta[$. So we should have $f'(t)\gt 0$ for all $t\in V$ or $f'(t)\lt 0$ for all $t\in V$, which means $f$ should be strictly increasing or strictly decreasing in $V$. Suppose, without loss of generality, that it is strictly increasing, so, given $t\in ]t_0-\delta , t_0 [$, $t\lt t_0$, which implies $f(t)\lt f(t_0) = 0$ (so $f(t)\neq 0$), analogously, given $t\in ]t_0,t_0+\delta [$, we must have $f(t)\gt f(t_0)=0$ (so $f(t)\neq 0$), which proves the result.

I think your reasoning is correct. Another usual way to answer the problem is the following : the negation of the conclusion is
$$(\forall \delta>0)(\exists t\in\left]t_0-\delta,t_0\right[\cup \left]t_0,t_0+\delta\right[)\, f(t)=0$$
This enables the construction of a sequence $(t_n)_{n\ge1}$ by $t_n\in\left]t_0-\frac{1}{n},t_0\right[\cup \left]t_0,t_0+\frac{1}{n}\right[$ and $f(t_n)=0$.
Now as $f$ is supposed differentiable at $t_0$, and as $\lim_{n\to\infty}t_n=t_0$, we have $0=\frac{f(t_n)-f(t_0)}{t_n-t_0}\xrightarrow[n\to\infty]{}f'(t_0)$, so $f'(t_0)$ must be $0$. But by uniqueness of solution, this leads to $f=0$, which contradicts the hypothesis. 
