Wronskian zero with linearly independent solutions

Question

Any ideas how to go about proving this?

Functions $\phi(x)$ and $\psi(x)$ are linearly independent on the interval $[\alpha,\beta]$, but their Wronskian $W(\phi,\psi)=0$ for some $x\in [\alpha, \beta]$.

Prove that exist at least two points $x_1,x_2 \in [\alpha, \beta]$ such that $\phi(x_1)=\psi′(x_2)=0$.

I've naturally started with considering the point at which the Wronskian vanishes, say $\phi(x_0)\psi'(x_0)-\phi'(x_0)\psi(x_0)=0$ and then took them to each side and arrive at the logs being equal when evaluated at x0 but I'm not sure if this makes sense and if it doesn, where to go from there?

I'm not sure how the linear independence comes in to play? Should I'm be attempting this at a higher level rather than constructively?

Any help would be greatly appreciated!

Answer 1

That's wrong. Consider $\phi(x) = x^2 + 10$ and $\psi(x) = \cos x+ 10$ on $[0,1]$. Then $\phi$ and $\psi$ are linearly independent, $\psi$ does not vanish on $[0,1]$, but $$ W(\phi, \psi)(0) = \phi(0)\psi'(0) - \phi'(0)\psi(0) = 10 \cdot 0 - 0 \cdot 10 = 0 $$

Answer 2

I assume that you left some important information out (since you've given it the differential equations tag), namely, that $\phi(x)$ and $\psi(x)$ are two linearly independent solutions of the same second order differential equation.

If that's indeed correct, all you have to do is look up Sturm's Separation Theorem, that should point you in the right direction.