Proof involving Wronski determinant 
Let $x''+q(t)x=0$ with a continuous function $q:\mathbb R \to \mathbb R$
Let $t \mapsto x(t)$ and $t \mapsto y(t)$ be two solutions of the ODE. Their Wronski determinant is defined $W(t):=x(t)y'(t)-x'(t)y(t)$. The solutions $x(t), y(t)$ are linearly independent if $W(t) \neq 0 \ \forall t \in \mathbb R$.
a) Show that $W(t)$ is constant.

I show it by showing $W'(t)=0$:
$$W'(t)=x'y'+xy''-x''y-x'y'=xy''-x''y=x(-qy)-(-qx)y=0$$

b) Show that for linearly independent solutions $x(t),y(t)$ using a) that

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*i) $x(t_1)=0\Rightarrow x'(t_1)\neq 0$ and $y(t_1)\neq 0$

Assume that $y(t_1)=0$. Then $W(t_1)=x(t_1)y'(t_1)-x'(t_1)y(t_1)=0$
Assume that $x'(t_1)=0$. Then $W(t_1)=0$
Which are contradictions to linearly independent.
Why/How do I need a) here? a) Tells me that the Wronski determinant is $0$ everywhere but isn't it sufficient that $W(t_1)=0$?


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*ii) If $x(t_1)=x(t_2)=0$ and $x(t)\neq 0$ for $t \in (t_1,t_2)$, then $y(t)$ has exactly one root in $(t_1,t_2)$.


I don't know how to prove this. Looking for hints/solution.
Thanks in advance!
 A: The last question is the simple version of the Sturm-Picone comparison theorem.
If the signs of $y(t_{1,2})$ where the same, you would also get that the signs of $x'(t_{1,2})$ are the same as $W(t_{1,2})=-x'(t_{1,2})y(t_{1,2})$. This would mean that $x$ crosses the time line there in the same sense, which implies an additional root inside $(t_1,t_2)$ by the intermediate value theorem.
By contradiction, the signs of the $y$ values have to be different, implying a root in the interval.

On your other questions, yes, a) is more of a curiosity, as the general properties of the Wronskian only depend on the continuity of the coefficient functions (or even weaker assumptions). The solution of that part of the question shows that you understood the basic nature of the Wronskian.

As to uniqueness of the root of $y$, assume there were two. Use the just proven argument with switched functions to conclude that between the two roots of $y$ there must be a root of $x$. However that cannot exist by the assumption that $x$ has no roots inside $(t_1,t_2)$.
