Suppose that $y_1$ and $y_2$ are a fundamental set of solutions, of the linear second order ODE $y''+p(t)y'+q(t)y=0$, on the interval $-\infty < t < \infty$. Show that there is one and only one zero of $y_1$ between consecutive zeros of $y_2$. Hint: differentiate the quantity $y_2/y_1$ and use Rolle's theorem.
I found an existing answer to this question to be unilluminating.
Here's how far I got.
Say, $a$ and $b$ are consecutive zeros of $y_2$. This would mean $\frac{y_2(a)}{y_1(a)} = \frac{y_2(b)}{y_1(b)}= 0$. Rolle's theorem now states that there exists $c \in (a,b)$ such that $\frac{\mathrm d y_2/y_1}{\mathrm{dt}}|_{t=c} = \frac{W[y_1,y_2](c)}{y_1^2(c)}=0$. The Wronskian being zero is in contradiction with $y_1$ and $y_2$ being a fundamental set of solutions. Hence, $\frac{W[y_1,y_2](t)}{y_1^2(t)}$ is nonzero for all $t\in (-\infty,\infty)$. Furtermore, the assumption we made in using Rolle's theorem is not correct.
Namely, $y_2(t)/y_1(t)$ is not continuous and therefore also not differentiable at some points the interval $(a,b)$. This is because the denumerator $y_1(t)$ has a zero in this interval.
Is my argument correct? And how should I now prove that that this zero of $y_1$ is unique?
Context: this is question 2.1.18 of Braun's Differential Equations and Their Applications, 4th edition.
Let $a$ and $b$ ($a<b$) be two consecutive zeros of $y_2$; namely $$ y_2(a)=y_2(b)=0. \tag1$$ Since $W(t)\neq0$, we can have $y_1(a)\neq0,y_1(b)\neq0$. Suppose that $y_1$ does not have zero in $(a,b)$. Since $W[y_1,y_2](t)\neq0$ for $t\in[a,b]$, we can assume $W[y_1,y_2](t)>0$ and hence we have $$ \frac{\mathrm d (y_2/y_1)}{\mathrm{dt}} = \frac{W[y_1,y_2](t)}{y_1^2}>0. $$ Namely $y_2/y_1$ is strictly increasing and hence $$ \frac{y_2(a)}{y_1(a)}<\frac{y_2(b)}{y_1(b)}. \tag2$$ Using (1) in (2), we have $0<0$ which is absurd. Thus $y_1$ has a zero in $(a,b)$. You can prove the uniqueness of zero of $y_1$.