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Given DE is: $y''+ p(x)y'+ q(x)y = 0$

It's given in my book that if $p(x),q(x)$ are continuous on open interval I, they are linearly dependent if and only if their Wronskian W=0 at some point in the interval.

Also, by Abel's formula, W($y_1,y_2$)=$ce^{-\int p(x)dx}$

This is what I gathered/understood from these, but I'm not sure if they're right:

  1. If $y_1,y_2$ are linearly dependent, the Wronskian is identically zero.

  2. W can only be 0 everywhere or never 0.

  3. As a consequence of 2., if W is non zero anywhere, they're linearly independent and will be non-zero everywhere else too. If W is zero anywhere, it has to be identically zero(?) and they're dependent.

Conclusion#3 doesn't seem right to me. For example, $x^2$ and 1 are independent but their W is identically 0, so it says they must be dependent?

Where am I going wrong here?

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There are two points which are wrong:

First, the Wronskian of $x^2$ and $1$ is not identically $0$:

$$W(y_1, y_2)(x) = \begin{vmatrix} x^2 & 1 \\ 2x & 0 \end{vmatrix} = -2x.$$

Second: It is true for all differentiable functions $y_1$ and $y_2$ that: If $W(y_1, y_2) (x_0) \neq 0$ for at least one $x_0$, then $y_1$ and $y_2$ are linearly independent. But the backwards direction is not true in general, i.e., if $y_1$ and $y_2$ are linearly independent, then their Wronskian is not necessarily non-zero. Consider as a counterexample $y_1 = x^2/2$ and $y_2 = x\lvert x \rvert/2$. The backwards direction is true if you additionally assume that $y_1$ and $y_2$ are solutions of a linear second order differential equation.

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  • $\begingroup$ How does assuming that they're solutions of a 2nd order DE change anything? Also, since the first proof in my question has a 'if and only if', doesn't that mean the backwards direction is also true, i.e. if W=0, they are linearly dependent. So won't x^2 and x|x| be dependent? $\endgroup$ – user_9 Dec 30 '19 at 10:07
  • $\begingroup$ @user_9 In the first sentence above, which functions do you mean with "they"? The proof of the backwards direction needs this assumption since $x^2$ and $x \vert x \rvert$ are linearly independent ($x^2/(x \vert x \vert)$ is not a constant). $\endgroup$ – Jan Dec 30 '19 at 10:12
  • $\begingroup$ Ah, apologies. By 'they' in the first sentence, I meant y1, y2. Won't $x^2/x|x|$ being constant depend on the interval we choose though? $\endgroup$ – user_9 Dec 30 '19 at 12:32
  • $\begingroup$ @user_9 Yes, but you only need one counterexample. $\endgroup$ – Jan Dec 30 '19 at 14:03

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