The problem said "Determine the longest interval in which the given initial value problem is certain to have the solution.
given problem is $$x(x-15)y''+16xy'+4y=2, \quad y(14)=0, y'(14)= -1$$
so I change the equation to $$y''+\frac{16}{x-15}y'+\frac{4}{x(x-15)}y=\frac{2}{x(x-15)}$$
In this case the problem means that determine the wronskian of ODE? and if my thought is right, this equation is when $x\neq0,15$ the $p(x)$ and $q(x)$ is continuous. so the answer of problem is $ 0\lt x \lt 15$
Is my answer and thought is right?
This has nothing to do with the Wronskian $W[y_1,y_2]=y_1y_2'-y_1'y_2$.
What you need is the statement that on each interval $[a,b]$ where the coefficients in the second form/changed equation are continuous, the ODE is Lipschitz using the maxima of these continuous functions on a closed interval. Thus a solution exists and can be extended to any larger interval satisfying the same condition. In the union of all these intervals you can indeed cover the open interval $(0,15)$, which then is the domain of all maximal solutions to initial values inside this interval.