Zero's of non trivial solution of an second order ordinary differential equation What can we say about the zero's of  any non-trivial solution of the linear differential equation$$y^{''}+q(x)y=0$$ (where $q(x)$ is positive monotonically increasing continuous function of $x$). Can we say that it must have infinitely many zeros in $\mathbb{R}?$ For example $y^{''}+y=0$ has $sin(x)$ as a non trivial solution, which has infinitely many zero's. Please help me to prove the general result about infinitely many zero's. Thanks in advance.
 A: Let $0<c\le\sqrt{q(x)}$. The infinitesimal angle increment of the complexified phase vector $y'+icy$ is the imaginary part of
$$
\frac{d}{dt}Ln(y'+icy)=\frac{y''+icy'}{y'+icy}=\frac{(-qy+icy')(y'-icy)}{y'^2+c^2y^2}
\\
=\frac{-(q-c^2)yy'+ic(y'^2+qy^2)}{y'^2+c^2y^2}
$$
Thus the angular velocity around the origin is
$$
c+\frac{q-c^2}{y'^2+c^2y^2}cy^2\ge c,
$$
so there are infinitely many revolutions around the origin, each with time smaller $\frac{2\pi}c$, and each revolution gives two zeros of $y$.

Or, to state compatibility with the Sturm–Picone comparison theorem (where the second equation is $y''+c^2y=0$), each interval of length $\frac{\pi}c$ contains at least a half-revolution and thus at least one zero of $y$.
A: If $q(x)$ is a positive increasing function, then the Kneser oscillation criterion or can be used to prove that every solution has infinitely many zeros. Another possibility is to use Sturm comparison theorem and compare your equation with $$y''+\varepsilon y=0$$ for positive and sufficiently small $\varepsilon$. 
A: Consider the system
$$
y'=z,\,\, z'=-p(x)y
$$
Infinite zeros are guaranteed if we show that the vector
$$
\frac{1}{\sqrt{y^2+z^2}}(y,z)=(a,b)
$$
which lives on the unit circle, 
turns infinite times around the origin. 
But,
$$
(a,b)=(\cos\theta,\sin\theta)\Longrightarrow
(a',b')=\theta'(-\sin\theta,\cos\theta)\\ \Longrightarrow
\theta'=ab'-a'b=\cdots=\frac{y'z-yz'}{y^2+z^2}=\frac{(y')^2+q(x)y^2}{y^2+(y')^2}\ge \frac{(y')^2+q(0)y^2}{y^2+(y')^2}\ge\min\{1,q(0)\}>0.
$$
Thus $\theta(x)\to\infty$, as $x\to\infty$.
