Convert $\frac{d^2y}{dx^2}+x^2y=0$ to Bessel equivalent and show that its solution is $\sqrt x(AJ_{1/4}+BJ_{-1/4})$ I have been following the thread " Convert Airy's Equation $y''-xy=0$  to Bessel equation $$t^2u''+tu'+(t^2-c^2)u$$ " but I can't join the dots to a solve similar equation $y''+x^2y=0$ so as to obtain a solution of the form $$\sqrt x\left(AJ_\frac{1}{4}+BJ_{-\frac{1}{4}}\right)$$
I actually get an equation that looks this way $$t^2\frac{du}{dt}+t\frac{du}{dt}+(t^2+\frac{5}{64})u$$ The above equation can not yield the desired solution. Please help me to clearly see this. Thank you.
 A: Perform the change of variables $y=\sqrt{x}f(x)$. Then, using the product rule, $f$ satisfies:
$$
\frac{x^2}{4}f''+\frac{x}{4}f'+(\frac{x^4}{4}-\frac{1}{16})f=0.
$$
Then make the change of variable $t=x^2/2$ and compute:
$$
\frac{df}{dx}=x\frac{df}{dt},
$$
and
$$\frac{d^2f}{dx^2}=\frac{df}{dt}+x^2\frac{d^2f}{dt^2}.
$$
Thus, the equation above becomes
$$
\frac{x^4}{4}f''+\frac{x^2}{2}f'+(\frac{x^4}{4}-\frac{1}{16})f=0,
$$
where the primes now denote derivatives w.r.t. the variable $t$.
Written in the variable $t$, the equation above becomes
$$
t^2f''+tf'+(t^2-\frac{1}{16})f=0.
$$
The solution of which is
$$
f=A J_{1/4}(t)+B J_{-1/4}(t).$$
Thus since $y=\sqrt{x}f(x^2/2)$, you get the desired solution.
A: Here are maple solutions to $y''(x)+x^2y(x)=0$ in terms of the Bessel functions
$$ y \left( x \right) ={c_1}\,\sqrt {x}
{{J_{1/4}}\left(\frac{{x}^{2}}{2}\right)}+{c_2}\,\sqrt {x} \left( 
{{J_{1/4}}\left(\frac{{x}^{2}}{2}\right)}-\sqrt {2}
{{J_{-1/4}}\left(\frac{{x}^{2}}{2}\right)} \right), $$
or you can have the form
$$ y \left( x \right) ={c_1}\,\sqrt {x}
{{J_{1/4}}\left(\frac{{x}^{2}}{2}\right)}+{c_2}\,\sqrt {x}\,
{{Y_{1/4}}\left(\frac{{x}^{2}}{2}\right)},$$
where $Y_{1/4}\left(\frac{{x}^{2}}{2}\right)$ is the Bessel function of the second kind.
