Wronskian determinant and Linear dependence I was trying to show that if functions $~f~$ and $~g~$ defined on interval $~I~$ are linearly dependent then the Wronskian determinant is zero.
Suppose $~f, g \in I~$ and $~f g~$ are linearly dependent, then $\forall x \in I$, there exist $c_1$ and $c_2$ where both are not zero such that:
$$c_1 f(x) + c_2 g(x) =0$$
$$\Leftrightarrow c_1 f'(x) + c_2 g'(x) =0~~~~~~~~\forall ~x \in I$$
Fix an arbitrary $x_0 \in I$
WLOG, suppose $c_1 \neq 0$
$$\Leftrightarrow f(x_0)= \dfrac{-c_2}{c_1}g(x_0)$$
Also $$f'(x_0) = \dfrac{-c_2}{c_1}g'(x_0)$$
Hence the determinant can be computed as $~fg'(x_0)-gf'(x_0)~$ and by substitution of $~f~$ and $~f'~$, we are able to get the determinant is zero for all arbitrary $~x~$. 
Hence the Wronskian determinant for $~f~$ and $~g~$ are zero?
 A: Let's do the claim in this way. First, I think you could able to see that if $W(f,g)\neq 0,~ \forall x\in I$ then $f$ and $g$ are linearly independent. We want to show that if $f$ and $g$ are linearly independent so for all $x\in I$, $$W(f,g)\neq 0$$ Assume $f,g$ are linearly independent on $I$ and for some $x_0\in I$, $$W(f,g)(x_0)=0$$ This yields that for some $c_1$ and $c_2$ (which are not both zero): $$c_1f(x_0)+c_2g(x_0)=0$$ so $$c_1f'(x_0)+c_2g'(x_0)=0$$ and so by defining $y=c_1f+c_2g$ we get $y(x_0)=0,~ y(x_2)=0$ In fact, we can see that $y$ is identically the zero function. Don't you think this violets the assumption? :-)
A: Your reasoning is pretty good, but I would suggest the following corrections: 


*

*"For all $x \in I$, there exist $c_1$ and $c_2$ where both are not zero such that...". The correct statement would be "at least one of them different from zero".

*You don't need to fix an arbitrary $x_0$. You could better say: "Assume $c_1 \neq 0$ (the same reasonning works assuming $c_2 \neq 0$), then for every $x$..."

