# Clarification regarding the Wronskian in differential equations

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?

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.

• 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? Dec 30 '19 at 10:07
• @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).
– Jan
Dec 30 '19 at 10:12
• 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? Dec 30 '19 at 12:32
• @user_9 Yes, but you only need one counterexample.
– Jan
Dec 30 '19 at 14:03