ODE solution - positivity and uniqueness I always see the theorem on the textbooks, but they only state the theorem and then give examples on finding the roots of the equation.
However, I would like to learn how it is proved that:

If $q(t)$ is non positive continuous function which is defined on $[0,\infty]$ and also $y_0$ and $y_1 $ are positive, then for $t>0$ solution of the problem:
  $$y''(t)+q(t)y(t)=0, y(0)=y_0, y'(0)=y_1$$
  is positive for all $t>0$ and it is unique.

 A: We begin by deriving a useful identity:
Since
$y''(t) + q(t)y(t) = 0, \tag{1}$
we have
$y''(t) = -q(t)y(t); \tag{2}$
upon multiplying (2) through by $y'(t)$ we obtain
$y''(t)y'(t) = -q(t)y(t)y'(t); \tag{3}$
noting that
$y''(t)y'(t) = \dfrac{1}{2}((y'(t))^2)', \tag{3}$
we write (3) as
$\dfrac{1}{2}((y'(t))^2)' = -q(t)y(t)y'(t); \tag{4}$
integrating (4) 'twixt $t_0$ and $t > t_0$ yields
$\dfrac{1}{2}((y'(t))^2 - (y'(t_0))^2) = \dfrac{1}{2}\int_{t_0}^t ((y'(s))^2)'  ds = -\int_{t_0}^t q(s)y(s)y'(s) ds. \tag{5}$
Now suppose that $y(t) = 0$ for some $t > t_0$; since we assume $y(t_0), y'(t_0) > 0$, there must be some $\tau$, $t_0 < \tau < t$, with $y'(\tau) = 0$; otherwise we would have
$y(t) = \int_{t_0}^t y'(s) ds + y(t_0) > y(t_0) > 0, \tag{6}$
since the integral appearing in this inequality is itself positive.  We choose the smallest such $\tau$.  Then by (5),
$0 > -\dfrac{1}{2}(y'(t_0))^2 = -\int_{t_0}^\tau q(s) y(s) y'(s) ds; \tag{7}$
however, the negative of the integral appearing on the right of (7) is non-negative, since $q(t) \le 0$ everywhere, and by our choice of $\tau$ and the initial conditions, $y(t), y'(t) \ge 0$ on $[t_0, \tau]$.  This contradiction implies $y(t) > 0$ for all $t \ge t_0$; taking $t_0 = 0$ then establishes the result in the present specific case.
The uniqueness of the solution follows from the standard theorems, since $q(t)y$, being continuous in $y$ and $t$ and Lipschitz continuous in $y$, satisfies the standard hypotheses these theorems require.
QED!!!
A: Here's an intuitive (not formal) proof of why the result is true. We can consider $y(t)$ the solution to be approximated by finite diferences taking intervals $\Delta t$, and then take the limit when $\Delta t \to 0$.
Since $q(t)$ is non positive we will have 
$$
y'(t) = - \int_{0}^t q(s)y(s) + y_1
$$
Iteratively we can use this expression and the fact that $y_1 >0$ to prove that $y'(t) >0$. For instance call $y(0) = y_0$ then by finite diferences: 
$$y'(1) = -q(0)y_0 +y_1 >0 $$
and similarly for all others. Then on constructing the solution by iterations:
$
y(t)= \int_{0}^t y'(s) ds + y_0 >0
$
About unicity, this comes about because in this iterating scheme once $y'(0) =y_1$ has been specified the differential equation gives us $y''$ at each iteration, which can be used to construct the whole of $y'(t)$ in a unique manner. Once we have $y'(t)$ we integrate it as above and produce a unique $y(t)$ that satisfies the given initial conditions and equation.
As I said this is not formal, is more heuristic, but I trust it can be made formal by looking with care at the details, the main idea however is correct I think.
