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.

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.