Solution of equation $y''+x^2y=0$

Question

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.

Answer 1

$\newcommand{\d}{\mathrm{d}}$ 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).