This is a problem from Braun's Differential Equations book. Let $p(t)$ and $q(t)$ continuous functions and $y_{1}$, $y_{2}$ a fundamental set of solutions of the ODE \begin{equation} y''+p(t)y'+q(t)y=0 \end{equation} in the interval $t\in(-\infty,\infty)$. Prove that there is only one zero of $y_{1}$ between two consequtive zeros of $y_{2}$. Hint: Differentiate the expression $y_{2}/y_{1}$ and use Rolle's theorem.
This is what I have. Let $a,b$ two consecutive zeros of $y_{2}$ with $a<b$. Since $y_{1}$, $y_{2}$ are a fundamental set of solutions, their Wronskian \begin{equation} W(y_{1},y_{2})=y_{1}y_{2}'-y_{1}'y_{2} \end{equation} is not zero for all $t$. This implies that neither $a$ or $b$ are zeros of $y_{1}$. Then, I tried to use the hint computing the derivative of $y_{2}/y_{1}$, \begin{equation} \dfrac{d}{dt}(y_{2}/y_{1})=W(y_{1},y_{2})/y_{1}^{2} \end{equation} but then I have no idea of what to do, thanks for the help.
We need to prove both the existence and uniqueness of a zero of $y_1$ between the consecutive zeroes $a,b$ of $y_2$. As you observed, $W$ vanishes nowhere.
For existence, assume for the sake of contradiction that $y_1$ does not vanish on $(a,b)$. Then $z:=y_2/y_1$ is differentiable there, and your calculation shows $z'=W/y_1^2$. By your observation that neither $a$ nor $b$ are zeros of $y_1$, we have $z(a)=z(b)=0$. By Rolle's theorem, there is some place in $(a,b)$ where $z'$, and thus $W$, vanishes. This contradicts the fact that $W$ never vanishes.
For uniqueness, assume $y_1$ has two roots on $(a,b)$, say $c$ and $d$. Apply the same argument above to the reciprocal $z^{-1}=y_1/y_2$ to conclude that $y_2$ must vanish somewhere in $(c,d)\subset(a,b)$, contradicting the fact that $a$ and $b$ are consecutive roots of $y_2$.