Zeros of solution of a ODE 
Let $a$ be a real differentiable function $a(0)>0$ and $a(t)\geq 0$. Consider the solution $\phi$ of the differential equation
  $$x''+ ax=0$$
  with initial conditions $\phi(0)=1$ and $\phi'(0)=0$. Prove that there exist $t_0<0<t_1$ such that $\phi(t_0)=\phi(t_1)=0$.

I don't know exactly what to do with this problem. It is clear that $0$ is local maximum of $\phi$ because of the initial conditions. 
Any hint?
 A: Assume that $x(t)>0$ for all $t>0$. Then $x''(t)$ is negative and thus by mean value theorem also $x'(t)<0$. As consequently $x$ is concave and monotonously falling, you get that the graph of $x$ is below the tangent in any point, for example that $x(t)\le x(1)+x'(1)(t-1)$ which has a root $t_1>1$ at some finite time $$t_1=1+\dfrac{x(1)}{(-x'(1))}.$$ Which on the other hand requires $x(t_1)\le 0$, contradicting the assumption. In the same way you get the existence of a root for $t<0$.

See also the Storm-Picone comparison theorem. If $a(t)\ge m^2>0$ for all $0<t<\frac{\pi}m$, then that theorem guarantees a root of $x$ inside that interval. Which tells you that you need $a(t)\ll(\frac{\pi}t)^2$ to get the more interesting cases that do not look too much like harmonic oscillations. 
A: Preliminary Remarks:  We are given that
$\phi''(t) + a(t) \phi(t) = 0, \; t \in \Bbb R, \tag 1$
with
$\phi(0) = 1, \; \phi'(0) = 0; \tag 2$
and also that
$a(0) > 0, \; a(t) \ge 0, \; t \in \Bbb R, \tag 3$
with $a(t)$ differentiable; thus $a(t)$ is continuous; it follows then from the standard theory of ordinary differential equations, that $\phi(t)$, $\phi'(t)$, and $\phi''(t)$ are  continuous functions on $\Bbb R$; since from (1),
$\phi''(t) = -a(t) \phi(t), \tag 4$
we have, for $\tau > 0$,
$\phi'(\tau) = \phi'(\tau) - \phi'(0) = \displaystyle \int_0^\tau \phi''(s) \; ds = -\int_0^\tau a(s) \phi(s) \; ds; \tag 5$
now since $a(0) > 0$ and $\phi(0) = 1$, it follows by continuity that for sufficiently small $\tau$ the integrand on the right of (5) is positive, and hence that both
$\phi(\tau) > 0, \; \phi'(\tau) < 0; \tag 6$
in what follows, we shall call upon both (3) and (6).  End of Preliminary Remarks.
Now, if $t \ge \tau$, 
$\phi'(t) - \phi'(\tau) = \displaystyle \int_\tau^t \phi''(s) \; ds = -\int_\tau^t a(s) \phi(s) \; ds; \tag 7$
that is,
$\phi'(t) = \phi'(\tau) - \displaystyle \int_\tau^t a(s) \phi(s) \; ds; \tag 8$
if $\phi(t)$ has no zero $t_1 \ge \tau$, then since $\phi(\tau) > 0$, we must have $\phi(s) > 0$ for all $s \ge \tau$ (this follows of course from the intermediate value theorem) whence, since $a(s) \ge 0$,
$\displaystyle \int_\tau^t a(s) \phi(s) \; ds \ge 0, \tag 9$
which implies, via (6),
$\phi'(t) \le \phi'(\tau) < 0; \tag{10}$
for $t \ge \tau$,
$\phi(t) - \phi(\tau) = \displaystyle \int_\tau^t \phi'(s) \; ds, \tag{11}$
whence
$\phi(t) = \phi(\tau) +  \displaystyle \int_\tau^t \phi'(s) \; ds \le \phi(\tau) + \int_\tau^t \phi'(\tau) \; ds = \phi(\tau) + \phi'(\tau)(t - \tau); \tag{12}$
since
$\phi(\tau) + \phi'(\tau)(t - \tau) = 0 \tag{13}$
for
$t = t_1' =  \tau -\dfrac{\phi(\tau)}{\phi'(\tau)}, \tag{14}$
it follows that there exists $t = t_1$, $\tau \le t_1 \le t_1'$, with
$\phi(t_1) = 0. \tag{15}$
Essentially the same argument shows that there exists $t_0 < 0$ for which 
$\phi(t_0) = 0; \tag{16}$
one need merely show that there is a small $\tau < 0$ with
$\phi(\tau) > 0, \phi'(\tau) > 0, \tag{17}$
which follows again from the continuity of $\phi(t)$, $a(t)$ and the equation corresponding to (5):
$-\phi'(\tau) = \phi'(0) - \phi'(\tau) = \displaystyle \int_\tau^0 \phi''(s) \; ds = -\int_\tau^0 a(s) \phi(s) \; ds$
$\Longrightarrow \phi'(\tau) = \displaystyle \int_\tau^0 a(s) \phi(s) \; ds > 0; \tag{18}$
then for $t \le \tau$ (7) becomes
$\phi'(\tau) - \phi'(t) = \displaystyle \int_t^\tau \phi''(s) \; ds = -\int_t^\tau a(s) \phi(s) \; ds; \tag{19}$
which shows that $\phi'(t) \ge \phi'(\tau)$ when $t \le \tau$ and if we follow through the details we arrive at, instead of (12), 
$\phi(\tau) = \phi(t) +  \displaystyle \int_t^\tau \phi'(s) \; ds \ge \phi(t) + \int_t^\tau \phi'(\tau) \; ds = \phi(t) + \phi'(\tau)(\tau - t), \tag{20}$
whence
$\phi(t) \le \phi(\tau) - \phi'(\tau)(\tau - t), \tag{21}$
which shows that $\phi(t)$ will take the value zero for some $t_0 \ge t_0'$ where
$t_0' = \tau - \dfrac{\phi(\tau)}{\phi'(\tau)}. \tag{22}$
