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?


1 Answer 1


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.

  • $\begingroup$ Okay, that clears it up i guess. Thanks:) $\endgroup$
    – Arkonaire
    Oct 7, 2017 at 19:17
  • $\begingroup$ 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 $0$ ? $\endgroup$
    – BAYMAX
    Oct 25, 2017 at 2:51
  • $\begingroup$ What's $r(x)$.? $\endgroup$
    – Nosrati
    Oct 25, 2017 at 2:53

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