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For a set of functions $f_1,f_2,...f_n$, if their Wronskian determinant is identically zero $W(f_1,...f_n)(x) = 0$ for all $x$ in some interval $I$ we can't conlcude that these functions are linearly dependent.

But in case of differential equations, if $f_1,f_2,...f_n$ are solutions of some linear DE and the Wronskian is zero on some interval we say that the functions $f_1,f_2,...f_n$ are linearly dependent.

Is there some deep meaning why this is so?

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  • $\begingroup$ particular type means initial value problem? $\endgroup$ – Upstart Jul 29 '19 at 6:20
  • $\begingroup$ See en.wikipedia.org/wiki/Wronskian $\endgroup$ – Kavi Rama Murthy Jul 29 '19 at 6:34
  • $\begingroup$ What do you want to ask ? Pointout particularly where you stuck ? $\endgroup$ – nmasanta Jul 29 '19 at 7:43
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Have a look at the functions $f_1(x) = x^3$ and $f_2(x) = \vert x \vert^3$. Obviously, $f_1$ and $f_2$ are linearly independent. Moreover, we have $f_1'(x) = 3x^2$ and a little bit more calculus reveals that $f_2$ is differentiable with $f_2'(x) = 3 x \vert x \vert$. Furthermore,

$$W(f_1, f_2)(x) \equiv 0.$$

So what we have shown is, that in general the statement "$f_1, \dots, f_n$ linearly independent $\Longrightarrow$ $W(f_1, \dots, f_n) \neq 0$" is false. Since this if false, the contraposition "$W(f_1, \dots, f_n) \equiv 0 \Longrightarrow f_1, \dots, f_n$ linearly dependent" is also wrong, which explains, why we can't conclude the linear dependence in general.

In the case that $f_1, \dots, f_n$ solve an linear ODE and $W \equiv 0$ one can conclude that they are linearly dependent. A prove based on this additional assumption can be found e. g. here.

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From $W(f_1,...f_n)(x)=0$, only piecewise linear dependence follows. That means, the functions could be locally linear independent.

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