Difficulty Understanding Conjugate points of a differential equation.

Question

I was struggling to understand this footnote from the book by Gelfand. I couldnot understand how he takes the solution of the differential equation outside the interval $[a,b]$ and also how will the solution of $(26)$ with initial condition $$h(a-\epsilon)=0$$ $$h'(a-\epsilon)=1$$ Cannot vanish on whole interval $[a,b]$

Answer 1

$P,Q$ are defined outside $[a,b]$ by the implicit assumptions of the cited footmark. Also, if $P$ is non-zero on $[a,b]$, then it is also non-zero on some open neighborhood of that interval. A linear ODE has solutions that can be extended to the full intervals over which the coefficients are continuous and the leading coefficient is non-zero.

This all makes it possible to take the initial condition at $a−ϵ$ for a sufficiently small positive $ϵ$.

For the overall claim, the conjugate point is a continuous function of the initial point. If it is greater than $b$ for initial point $a$, then that same is true for initial points in a neighborhood of $a$, simply by continuity.