Solution of equation $y''+x^2y=0$ Does equation $y''+x^2y=0,$ where $y$ is function of $x$ have explicit solution? Perhaps with some conditions or in special case? I came across that when we have $x$ instead of $x^2,$ solutions are connected with some Airy functions. 
 A: $\newcommand{\d}{\mathrm{d}}$
As pointed out in the comments, understanding differential equations like yours leads to the parabolic cylinder functions. In particular, the standard form
$$\frac{\d^2 y}{\d x^2}+(\tfrac{1}{4}x^2-a)y=0$$
(DLMF 12.2.3)
admits standard solution
$$y(x)=c_+W(a,x)+c_-W(a,-x)$$
where $W(a,x)$ is chosen to satisfy the initial conditions
$$\begin{align}W\left(a,0\right)&=2^{-\frac{3}{4}}\left|\frac{\Gamma\left(\tfrac{1}{4}+\tfrac{%
1}{2}ia\right)}{\Gamma\left(\tfrac{3}{4}+\tfrac{1}{2}ia\right)}\right|^{\frac{%
1}{2}},\\
W'\left(a,0\right)&=-2^{-\frac{1}{4}}\left|\frac{\Gamma\left(\tfrac{3}{4}+%
\tfrac{1}{2}ia\right)}{\Gamma\left(\tfrac{1}{4}+\tfrac{1}{2}ia\right)}\right|^%
{\frac{1}{2}}
\end{align}$$
(DLMF 12.14.ii). However, in the special case $a=0$, $W$ is expressible in terms of fractional-order Bessel functions:
$$W\left(0,\pm x\right)=2^{-\frac{5}{4}}\sqrt{\pi x}\left(J_{-\frac{1}{4}}\left(%
\tfrac{1}{4}x^{2}\right)\mp J_{\frac{1}{4}}\left(\tfrac{1}{4}x^{2}\right)%
\right)$$
(DLMF 12.14.13).
