I have an amazing exercice to do, and I share it with you. I have one answer, but I understand it only partially, and it sets an intermediary function, that makes the proof fuzzy.
Let $ q : \mathbb R^*_+ \rightarrow \mathbb R^*_+ $, such that $\int^{+\infty}_1 q(t) \, dt = + \infty $
Show that the solutions of this differential equation, nullify themselves infinity times:
$$y''(x) + q(x) y(x) = 0$$
P.S. maybe the expression to nullify themselves infinity times is not english... the idea is that : $| \{x\in \mathbb R\mid f(x)=0\}| =\infty$
My attempt :
I tackle the problem in this way : I assume by absurdo that the solution of the differential equation vanishes only a finite times. Thus, since one point (called $a$), the solution is of constant sign. WLOG the solution is positive since one point. Then we set the function : $$z(x) = - \frac {y'(x)}{y(x)} $$ we have that : $$ z'(x) = q(x) + z^2(x)$$ So $$ \forall x > a, z(x) = z(a) + \int_a^x z'(t) dt \geq z(a) + \int_a^x q(t) dt > + \infty$$ So $$ \exists b > a, z(b) > 0 $$ Thus finally : $$y'(b) = y(b) z(b) < 0.$$
You're on the right track!
Just change $z(a) + \int_a^x q(t) dt > + \infty$ to $z(a) + \int_a^x q(t) dt \xrightarrow [x \to +\infty]{} + \infty$ because $\int_a^x q(t) dt$ is finite. From this, obtain $\exists b > a, z(b) > 0$ (as you've done) and $$y'(b) = -y(b) z(b) < 0.$$Now, since $y''(x)=-q(x)y(x) \le 0$ for all $x > a$, $y'(x)$ is non-increasing and $y'(x) \le y'(b)$ for all $x \ge b$. Thus, $$\forall x \ge b,\; y(x)=y(b)+\int_b^x y'(t) dt \le y(b) + \int_b^x y'(b) dt = y(b) + (x-b)y'(b),$$ and taking $x > b - \frac {y(b)}{y'(b)}$, we get $y(x) < 0$, a contradiction.