What is the solution of the of the differential equation $x^{2}y''+xy'+\left(4x^{2}-\dfrac{9}{25}\right)y=0$ in terms of Bessel's polynomial of the form $y=AJ_{n}(x)+BJ_{-n}(x)$, where $A$ and $B$ are arbitrary constants?
How may the equation be transformed into standard form?
Standard trick $$ x\rightarrow \alpha x\\ y\rightarrow \beta y. $$
It will turn out you will not need to transform $y$. In any case $$ \bar{x}^2y'' + \bar{x}y' + \left(4\alpha^2\bar{x}^2 - \frac{9}{25}\right)y = 0 $$ let $\alpha = \frac{1}{2}$ we obtain $$ x^2y'' + xy' + \left(x^2 - \frac{9}{25}\right)y = 0 $$ which is bessel if we have the $x\rightarrow \frac{1}{2}x$ with the standard form being of the form $$ x^2y'' + xy' +\left(x^2-k^2\right)y = 0 $$ thus $k = \frac{3}{5}$ So your $n$ above is not integer, which I hope is ok for you to proceed..