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I know that if a set of functions are linearly dependent, then its Wronskian = 0 at all values of t in the interval.

So can you conclude that if Wronskian = 0 for all values of t in the interval, then the functions must be dependent?

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The answer is no. For instance, the functions $f_1(x) = x^2$ and $f_2(x) = x \cdot |x|$ are continuous with continuous derivatives, have a Wronskian that vanishes everywhere, but fail to be linearly dependent.

The Wronskian Wikipedia page has a good discussion about this. Note that if the set of functions considered is analytic, then their dependence over an interval is indeed equivalent to their having a Wronskian that is identically zero.

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The other direction does not hold. A counterexample is given by the functions $f_1,f_2:\Bbb R\to \Bbb R$ defined by \begin{align*} f_1(t)=\begin{cases} 0, &t\leq0\\ t^2, &t>0 \end{cases},\qquad f_2(t)=\begin{cases} t^2, &t\leq0\\ 0, &t>0 \end{cases}. \end{align*} The Wronkian is zero at every $t\in\Bbb R$ but the functions are linearly independent. Indeed, if $c_1f_1(t)+c_2f_2(t)=0$ for some constants $c_j\in\Bbb R$ and all $t\in\Bbb R$ we have \begin{align*} c_1&=c_1f_1(1)+c_2f_2(1)=0,\\ c_2&=c_1f_1(-1)+c_2f_2(-1)=0. \end{align*}

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Answer is no because for instance the functions \begin{equation} \begin{aligned} &\left\{\begin{array}{l} f(x)=0 \text { for } x<0 \\ f(x)=x^{2} \text { for } x \geqslant 0 \end{array}\right.\\ &\left\{\begin{array}{l} g(x)=0 \text { for } x>0 \\ g(x)=x^{2} \text { for } x \leq 0 \end{array}\right. \end{aligned} \end{equation} Functions have 0 wronksian determinant but apparently they are linearly independent.

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