Consider the equation $$y''- \dfrac{y'}{x} = 0~.$$ Solution is $$y=Cx^2 + d~.$$ The Wronskian of $x^2$ and $1$ turns out to be $-2x$ which is zero at $x = 0$ and non zero elsewhere.
But the Wronskian of solutions to an equation of type $$y'' + p(x) y' + q(x) y = 0$$ should be identically zero or never zero.
What am I missing?
When we discuss about solutions of the equation $y''+ p(x)y' + q(x)y = 0$, continuity of $p(x)$, $q(x)$ and $r(x)$ is initial assumption for finding solutions. Here clearly $p(x)=- \dfrac{1}{x}$ isn't continuous in $x=0$ and therefore our solutions $y(x)=Cx^2 + D$ is defined in every open interval not excluding $0$. Also Wronskian must be considered in such intervals.