# Wronskian identically zero or never zero?

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.
• Where is $r(x)$ :) ?,also since at $x = 0$, $p(x)$ is not continuous i thought it must be excluded in the solution but it is not excluding $0$ ? Oct 25 '17 at 2:51
• What's $r(x)$.? Oct 25 '17 at 2:53