Solving this equation reducing it to Bessel Equation Question : Solve the differential equation  $ 4 \frac{d^2y}{dx^2}+9xy=0 $ by reducing it to Bessel equation.
What Have I tried so far?
I know that a standard Bessel equation is $$x^2 \frac{d^2y}{dx^2}+x\frac{dy}{dx}+(x^2-n^2)y=0$$
I tried using   $ t^2-n^2=9x$   to proceed with the problem but am reaching nowhere.
Kindly tell me how to approach this problem?
 A: First, to get $y'$ terms, we substitute with $s = x^n$. Then we have ($\mathrm d  \!*\!/\mathrm d\bullet$ is notated as $*_\bullet$)
$$y_x = nx^{n-1}\,y_s,\quad\text{and}$$
$$ y_{xx} = n(n-1)x^{n-2}\,y_s + (nx^{n-1})^2 y_{ss}.$$
Putting this into the orginal equation, 
$$4n(n-1)x^{n-2}\,y_s + 4(nx^{n-1})^2 y_{ss} + 9xy = 4n^2s^{\frac{2n-2}{n}} y_{ss} + 4n(n-1)s^{\frac{n-2}{n}}y_{s} + 9s^{\frac 1 n} y = 0.$$
Here, if we set $n = \frac 3 2$, the equation becomes 
$$9s^{\frac 2 3}y_{ss} + 3 s^{-\frac 1 3}y_s + 9s^{\frac 2 3}y = 0,\qquad \therefore 3sy_{ss} + y_s + 3sy = 0.$$
Now let $y = s^k u$ to get the form of the Bessel equation so that ($y_s = ks^{k-1}u + s^k u_s$ and $y_{ss} = k(k-1)s^{k-2}u + 2ks^{k-1}u_s + s^k u_{ss}$)
$$3s(k(k-1)s^{k-2}u + 2ks^{k-1}u_s + s^k u_{ss}) + (ks^{k-1}u + s^k u_s) + 3s\,s^{k}u = 0,$$ 
$$3s^{k+1} u_{ss} + (1+6k)s^k u_s + (3s^{k+1} + (3k^2 -2k)s^{k-1} )u = 0,$$
$$3s^{2} u_{ss} + (1+6k)s u_s + (3s^{2} + (3k^2 -2k) )u = 0.$$
Finally, choose $k$ so that $1+6k = 3$ ($k = 1/3$) and then $$s^2 u_{ss} + s\,u_s + \left(s^2 - \frac 1 9\right)u = 0.$$
A: See Airy function and its connection to Bessel functions. According to the cited formulas, you need to set
$$
y(x)=x^{1/2}f(x^{3/2}),
$$
then
$$
y'(x)=\frac12x^{-1/2}f(x^{3/2})+\frac32xf'(x^{3/2}),\\~\\ 
y''(x)=-\frac14x^{-3/2}f(x^{3/2})+\frac32f'(x^{3/2})+\frac94x^{3/2}f''(x^{3/2})=-\frac94x^{3/2}f(x^{3/2})
$$
so that with $t=x^{3/2}$
$$
t^2(f''(t)+f(t))=\frac19f(t)-\frac32tf'(t).
$$
