Let's have linear
$$ \tag 1 y''(t) + b(t)y(t) = 0, \quad t \in (t_{0}, \infty ) $$ Here $|b(t)|$ is monotonically decreased function of time, and $\lim_{t\to \infty }b(t) = \frac{a}{t} \to 0$.
How to prove that the solution of Eq. $(1)$ is bounded from above?
An edit. I've made the mistake. The correct question is about boundedness of $\frac{y(t)}{t^{\frac{1}{4}}}$. Instead of Eq. $(1)$ the following equation arises: $$ \tag 2 y''(t) + (k^{2} - ab(t))y(t) = 0, \quad b(t) \to \frac{f(t)}{\sqrt{t}}, $$ where $f(t)$ is bounded periodic function.
Numerical simulations show that $$ |y(t)| \leqslant \text{exp}\left[\alpha \times \frac{a}{\sqrt{k}}\right], \quad \alpha = O(1) $$ It seems that the proof of boundedness can be performed through manipulations with integral representation $$ y(t) = cos(kt)-a\int \limits_{t_{0}}^{t}sin(s - t)b(s)y(s)ds, $$ which is equal to Eq.$(2)$. But I can show that Eq. $(2)$ has bounded solutions only in case of $|b(s)| \leqslant \frac{1}{t^{d + 1}}, d > 0$, while in this case $b(s) \to \frac{f(t)}{\sqrt{t}}$.