Properties of solutions to general ODE's For general homogenous ODE's, of the types:
$$
a(x)\frac{d^i f(x)}{dx^i}+b(x)\frac{d^{i-1} f(x)}{dx^{i-1}}+\cdots + g(x)\frac{d f(x)}{dx} = 0 \tag{1}
$$
where $i,$ order of the ODE is finite, if the coefficients are smooth bounded functions of $x,$ one can often assume(?) that $(1)$ admits solutions that carry over the same properties. 
First question: Is there any validity to the above's hand-wavy argument? Or instead, are the existence of solution for such ODE's, and the properties of solutions much more non-trivial questions? 
Second question: Granted we know the answer to the 1st question, will it hold if we set the order of the ODE to infinity (so $i\to \infty$)? In other not entirely equivalent words, whether having a finite or infinite order ODE can lead to completely different solutions?
 A: Let $a(x)=b(x) = \ldots = 0$ and suppose that $g(x)$ is the infinitely smooth, bounded function whose support is contained in $[-2,-1]$ (Google "Test Functions" to learn more).  Consider the once-differentiable function $$y = \begin{cases} 0 & x<0\\ x^2 & x \geq 0 \end{cases}.$$  Clearly, $g(x)y'=0$, since the supports of $g$ and $y'$ are disjoint.  So then, we have an equation which has infinitely smooth coefficients, but admits a once-differentiable solution.  So, we can conclude that every solution to the above ODE need not be infinitely smooth or bounded.  However, note that $y=c$ is an infinitely smooth, bounded solution. So, the problem admits infinitely smooth and bounded solutions, but not all of them need to be infinitely smooth and bounded.
For the second question, I can be sure that the solution is infinitely differentiable, since in order to satisfy the equation with $i \to \infty$, all derivatives must exist.  Hence, every solution to the infinite ODE must be smooth.  This holds even if the coefficients are just continuous.
