Property of solution to initial value problem Let $f\colon [0,\infty) \to [0, \infty)$ be a continuous function such that $\int_0^\infty f(t) \, dt = \infty$. Assume that $y \colon [0, \infty) \to \mathbb R$ is a solution to the initial value problem
$$\begin{cases}y'' + f(t) y = 0 \\ y(0) = 1 \end{cases}. $$
I want to show that $y$ has infinitely many zeros that do not accumulate anywhere and at each zero the derivative of $y$ is non-vanishing. Furthermore, between the zeros $y$ is either positive and concave or negative and convex.
Thoughts so far: I could not come up with something smart that uses the integral assumption on $f$. Clearly the theorem holds for constant $f$, e.g. $f = 1$. Then the solutions are of the form $y(t) = A\sin(t) + B\cos(t)$ and satisfy the desired properties. Plotting the solution for other choices of $f$ give similar trigonometric patterns of the solution, but I could not make it further.
 A: 1. The set of zeros of $y$ has no upper bound.
Proof. Assume that the set of zeros of $y$ has an upper bound $a$. Then changing $y$ by $-y$ if necessary, we can assume that there is $x_0>0$ such that $y(t)>0$ for any $t\geq x_0$. Then for any $t\geq x_0$, we have $y''(t)=-f(t)y(t) \leq 0$. Thus, $y'(t)$ is decreasing when $t\geq x_0$.
If $y'(t)<0$ for some $t\geq x_0$, then we have $y(t)<0$ for some $t>x_0$. This is absurd. Hence, we must have $y'(t)\geq 0$ for all $t\geq x_0$. Then $y(t)$ is increasing when $t\geq x_0$, so $y(t)\geq y(x_0)>0$ for any $t\geq x_0$. If this happens, then we have for any $T>x_0$,
$$
\int_{x_0}^{T} f(t)dt = \int_{x_0}^{T} \frac{-y''(t)}{y(t)} dt\leq \int_{x_0}^{T} \frac{-y''(t)}{y(x_0)} dt=\frac{y'(x_0)-y'(T)}{y(x_0)}.
$$
Since $y'(t)$ is decreasing and $y'(t)\geq 0$ when $t\geq x_0$, the limit $\lim_{T\rightarrow\infty}y'(T)$ exists. Thus,
$$
\frac{y'(x_0)-y'(T)}{y(x_0)} \leq B<\infty.
$$
This contradicts $\int_0^{\infty} f(t)dt=\infty$. Therefore the set of zeros of $y$ does not have an upper bound. From this, we also obtain that $y$ must have infinitly many zeros.
2. The set of zeros of $y$ does not accumulate anywhere.
Suppose that the set of zeros accumulate somewhere in the positive reals. That is, there exists a sequence $z_n$ of zeros of $y$ so that $z_n\rightarrow z>0$ as $n\rightarrow\infty$. Then by continuity $y(z)=0$. Furthermore,
$$
y'(z)=\lim_{n\rightarrow\infty} \frac{y(z_n)-y(z)}{z_n-z} = 0.$$
Then this $y$ satisfies the differential equation  $y''(t)+f(t)y(t)=0$ and initial conditions $y(z)=y'(z)=0$. By the Existence and Uniqueness theorem, such $y$ must be identically $0$. This is a contradiction.
3. The zeros of $y$ are simple.
Suppose that some zero $z_1>0$ of $y$ satisfies $y'(z_1)=0$. Let
a solution $y_1(t)$ to the same differential equation such that $y$ and $y_1$ are linearly independent. Then the Wronskian $W(t)=y(t)y_1'(t)-y_1(t)y'(t)$ is nonvanishing everywhere. However, at $z_1$, we have
$$
W(z_1)=y(z_1)y_1'(z_1)-y_1(z_1)y'(z_1)=0. $$
This is a contradiction.
4. Between the zeros y is either positive and concave or negative and convex.
Let $0<z_1<z_2$ be zeros of $y$, and $y(t)\neq 0$ when $z_1<t<z_2$. We have either $y(t)>0$ or $y(t)<0$ on this interval. In the former case, we have for $z_1<t<z_2$,
$$y''(t)=-f(t)y(t) \leq 0.$$
In the latter case, we have for $z_1<t<z_2$,
$$
y''(t)=-f(t)y(t)\geq 0.$$
The statement therefore follows.
