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The fact that linear dependence guarantees that $W=0$ is easy to prove. What about the converse? According to Wikipedia its not necessarily true. But, what is the problem is this proof?

$$W=y_1y_2'-y_2y_1'=0\implies\frac{y_1y_2'-y_2y_1'}{y_1^2}=d(\frac{y_2}{y_1})=0\implies\frac{y_2}{y_1}=k$$

Which guarantees linear dependence. Why is this not what Wikipedia agrees with? Is the converse generally true as claimed in the proof above?

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    $\begingroup$ Thou shalt not divide by zero. $\endgroup$ – Angina Seng Apr 22 at 9:29
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That's fine if $y_1$ never vanishes. That's why your attempted proof doesn't work if $y_1(x)=x^2$ and $y_2(x)=|x|x$, which is the standard counterexample.

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  • $\begingroup$ Why does $x,2x$ work? $\endgroup$ – tatan Apr 22 at 9:37
  • $\begingroup$ I don't understand your question. $\endgroup$ – José Carlos Santos Apr 22 at 9:38
  • $\begingroup$ You are saying that $W=0$ implies linear dependence is fine if$ y_1 $never vanishes. If $y_1=x$ and $y_2=2x$, then $y_1$ vanishes but the $W=0$ and they are linearly dependent. $\endgroup$ – tatan Apr 22 at 9:41
  • $\begingroup$ What I wrote was that your proof doesn't work if $y_1$ has some zero. I did not write that what you were trying to prove is false in that case. $\endgroup$ – José Carlos Santos Apr 22 at 9:43
  • $\begingroup$ Okay, I understand:-) $\endgroup$ – tatan Apr 22 at 9:46
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Let us assume the functions $\ f_1,f_2,\cdots,f_n$ which are differentiable at least $\ (n-1)$ times in the interval $\ I$.

We now consider the equation $$\ c_1f_1+c_2f_2+\cdots+c_nf_n=0$$
Differentiating successively, we get $$\ \begin{align} &c_1f'_1+c_2f'_2+\cdots+c_nf'_n=0\\ &c_1f''_1+c_2f''_2+\cdots+c_nf''_n=0\\ &\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ &\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\ &c_1f_1^{(n-1)}+c_2'f_2^{(n-1)}+\cdots+c_nf_n^{(n-1)}=0\end{align} $$
Now if we consider these $\ n$ equations as a system of $\ n$ equations in $\ c_1,c_2,\cdots,c_n$ and if the determinant of the system does not vanish, the system will have no solution except the one with each of the $\ c_i'$s equal to $\ 0$.

Thus if the Wronskian $\ W\ne0$, then the functions $\ f_1,f_2,\cdots,f_n$ are linearly independent.

$\ \mathbf{Hence\; the\; non-vanishing\; of\; W\; is\; a\; sufficient\; condition\; that\; the \;functions\; are \;linearly\,independent.} $

Given two functions $\ f$ and $\ g$ that are differentiable on some interval $\ I$.
i) If $\ W(f,g)(x_0)\ne 0$ for some $\ x_0\in I$, then $\ f$ and $\ g$ are linearly independent $\ I$.
ii) If $\ f$ and $\ g$ are linearly dependent, then$\ W(f,g)(x)=0$ for all $\ x\in I$.

Be careful of the fact that it does not say that if $\ W(f,g)(x)=0$, then $\ f$ and $\ g$ are linearly dependent. In fact, it is possible for two linearly independent functions to have a zero Wronskian.

For example, if you take $\ f(x)=2x^2$ and $\ g(x)=x^4$, you will find that $\ W(f,g)(x)=4x^5\ne0\; \text{unless}\; x=0$. So $\ W$ is not identically $\ 0$ here. So $\ f$ and $\ g$ are linearly independent here.

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