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We know that

  1. if for functions $f$ and $g$, the Wronskian $W(f,g)(x_0)$ is nonzero for some $x_0$ in [a,b] then f and g are linearly independent on [a,b].
  2. If f and g are linearly dependent then the Wronskian is zero for all $x_0$ in [a,b].

My doubt is : If for some $x$ $W(f,g)(x)$ is zero, can we conclude that wronskian is identically zero as we know that wronskian is zero or never zero.

In one problem Wronskian $W$ was coming as $-x^2$ on $(\infty,-\infty)$. Since $W$ is $0$ for $x=0$ can we say wronskian is identically zero OR using point 1 we may conclude that we are getting more than one point where wronskian is not zero and hence functions are linearly independent.

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"Identically zero" means "equal to zero for all values of $x$".

The function $-x^2$ is not identically zero, because there are values of $x$ (such as $1,2,3,\dots$) for which it's nonzero.

Since the Wronskian of linearly dependent functions is identically zero, the functions whose Wronskian is $-x^2$ are not linearly dependent.


As an aside: there is a scenario in which $W$ is either always zero or never zero: it happens when the two functions are solutions of the ODE of the form $y''+p(x)y'+q(x)y=0$. For such solutions, the Wronskian satisfies the identity $W(t)=W(s)\exp\left(-\int_s^t p(x)\,ds\right)$ which implies that if $W$ is zero at some point, it is zero everywhere.

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  • $\begingroup$ Thanks a lot for clearing doubts. Now, the basic is very much clear to me. $\endgroup$ – monalisa Dec 23 '14 at 9:45

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