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Let's have equation $$ y''(t) + \left(\frac{3}{16t^{2}} + \frac{a}{t} -\frac{b}{t^{\frac{5}{4}}}cos(2t)\right)y(t) = 0, \quad t \in (1, \infty), \quad a, b > 0 $$ How to derive explicit upper bound for the solution $y(t)$ of this equation for case $a << b, b >> 1$? Numerical simulations show that it is bounded from above as $|y(t)| < e^{\left(\frac{b}{\sqrt{a}}\right)^{n}}$, where $n$ is different for a different $a$ at constant $b$ (but approximately constant for different $b$ at constant $a$).

I can show that up to the moment of time for which $\frac{b}{t^{\frac{5}{4}}}$ is more than one (or before $\frac{a}{t}$ becomes larger than $\frac{b}{t^{\frac{5}{4}}}$), there is growth of solution, while after the moment of time $T$, $\frac{b}{T^{\frac{5}{4}}} = 1$, the growth terminates. Thus the solution is bounded from above. But I can't evaluate an explicit form of upper bound.

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