$y''+p(t)y'+q(t)y=0$ has finite number of zeros

Question

Given $y''+p(t)y'+q(t)y=0$ and $p,q$ are continuous on $\mathbb{R}$ and let $y$ be a non-trivial solution of the system.
Prove that in any finite interval $[a,b]$ exists at most a finite number of zeros $\{t_k\}_{k=1}^{n}\subset[a,b]$ such that $y(t_k)=0$ for every $k$

I've thought transforming into S-L form by multiply the equation with $e^{\int p(s)}$ and then I get :
$(y'e^{\int p(s)})'+e^{\int p(s)}q(t)y=0$ but I don't know how can I say something about the number of zeros from this equation. any hint?

Answer 1

Any solution with $y(c)=y'(c)=0$ is the zero function.

Thus any non-zero solution has only isolated roots (non-zero derivative at root and thus a root-free neighborhood).

There can only be a finite number of isolated roots in a finite interval.

Answer 2

Note that the zero solution is always a solution. Also note that any solution given a value and derivative in a point is unique. If for a given solution $y$ there is an infinite number of zeros in $[a, b]$, then there is a convergent subsequence of zeros in $[a, b]$, let's say this limit is $c$. What is the value and derivative of $y$ in $c$? If you can prove this means $y$ has to be the zero solution itself, this shows the only solution with an infinite number of zeros in $[a, b]$ is $0$.