How do I apply sturm comparison theorem to draw the conclusion? Let $y$ be a non-trivial solution of the boundary value problem, $$y"+xy=0;x\in [a,b];y(a)=y(b)=0.$$ Then


*

*$b>0$

*$y$ is monotone in $(a,b)$ if $a<0<b$

*$y'(a)=0$

*$y$ has infinitely many zeroes.


My attempt:-
If $p(x)$ is continuous  and $p(x)\leq 0$ on $(a,b)$, then any non-trivial solution of the differential equation $$y"+p(x)y=0$$ has atmost one zero in $(a,b)$ and is monotone in $(a,b).$

Using this theorem If $x\leq 0, $then $y$ has atmost one zero in
  $(a,b)$ and is monotone in $(a,b)$ $\tag{1}$

(1) Suppose on contrary $b\leq 0$. Then $y$ is monotone by $(1)$. Which is not possible by Rolle's theorem.
(4) If $x\ge 1$ then by Sturm comparison test $$y"+xy=0;x\in [a,b];y(a)=y(b)=0.$$ has infinitely many zeroes.  I am confused.
 A: I assume you have to decide if (1)-(4) are true or false. I don't understand how you are using Sturm comparison test.
First note that 3) is false. Indeed, if $y(a)=0$ and $y^{\prime}(a)=0$, then
writing $z_{1}=y$ and $z_{2}=y^{\prime}$, you have that $z=(z_{1},z_{2})$ is a
solution of a first order Cauchy problem
\begin{align*}
\frac{dz}{dx}  & =A(x)z\\
z(0)  & =(0,0)
\end{align*}
The only solution is $z=(0,0)$, which contradicts the fact that $y$ is non-trivial.
Next note that 4) is false. Assume there exist infinitely many $x_{n}\in(a,b)$
such that $y(x_{n})=0$. Since the sequence $x_{n}$ is bounded, you can find a
subsequence $x_{n_{k}}$ such that $x_{n_{k}}\rightarrow x_{0}\in\lbrack a,b]$ (are you allowed to use this fact?).
Also, you can assume that either $x_{n_{k}}$ is strictly increasing, that is
$x_{n_{k}}<x_{n_{k+1}}$ for all $k$ or strictly decreasing, that is,
$x_{n_{k}}>x_{n_{k+1}}$ for all $k$. Let's assume the first. Since
$y(x_{n_{k}})=y(x_{n_{k+1}})=0$, by the mean value theorem, there exists
$z_{k}\in(x_{n_{k}},x_{n_{k+1}})$ such that
$$
0=y(x_{n_{k+1}})-y(x_{n_{k}})=y^{\prime}(z_{k})(x_{n_{k+1}}-x_{n_{k}}),
$$
which implies that $y^{\prime}(z_{k})=0$. By the squeeze theorem,
$z_{k}\rightarrow x_{0}$. By continuity of $y$ and of its derivative,
$0=y(x_{n_{k}})\rightarrow y(x_{0})$ and $0=y^{\prime}(z_{k})\rightarrow
y^{\prime}(x_{0})=0$, and so $y(x_{0})=y^{\prime}(x_{0})=0$. But this is
impossible, since reasoning as before the Cauchy problem
\begin{align*}
\frac{dz}{dx}  & =A(x)z\\
z(x_{0})  & =(0,0)
\end{align*}
has only the zero solution, which again contradicts the fact that $y$ is non-trivial.
