I am now considering the behavior of a class of second order ODE as $x$ is large enough. $$y''+\alpha(x)y'+y=0 \qquad (1)$$ Clearly, when $\alpha$ is a small constant, the equation changes into the damped vibration equation which has infinite zero points. To be precisely, when $\alpha \in (0,2)$ is a constant, the characteristic function $\lambda^2+\alpha \lambda +1$ leads to the solution $$\phi=C_1\exp(-\alpha)\cos(\sqrt{4-\alpha}x+C_2\exp(-\alpha)\sin(\sqrt{4-\alpha}x)$$ which is of infinite zero points.
Now I wonder when $\lim_{x\to \infty}\alpha(x)=0$, say $\alpha = \frac{1}{x}$, does the solution of (1) has the infinite zero points as $x$ is large enough? (I tried to use the Sturm Comparison Theorem.)
For $\alpha>0$ the solution is a damped oscillation or simply an exponential, if critically overdamped. If damping goes to zero, oscillations of however small amplitude survive. These can be stopped by a $\delta$-kick inhomogenous part only (physical experience).
For $\alpha$ negative, damping is time reversed, oscillations increasing. If the critical damping treshold is passed, the solution is exploding exponentially. If the antidamping factor goes to zero, again, an infinite oscillation survives for all times.