2nd order differential equation with non-constant coefficients Consider the second order differential equation
$$y''-x^2y=0
$$
where $y$ itself is a function of $x$. I do not know how to solve this equation. I tried a series expansion and failed, and because the coefficients are not constant, I can not use the characteristic equation to solve it either. Hence, here I am, looking for any hints on how to solve this equation for $y$.
I know there are tons of questions already out there concerning second order differential equations looking like this one, and I looked through just about every one of them, however all the solutions provided seem to be very situational for the given DE, and I have yet to find a general method that I can use to solve the above. I though about reducing the order of the equation.
Thanks!
 A: This is a particular case of Weber differential equation
$$y''+\left( \nu+\frac 12-\frac {x^2}4\right)y=0$$ Have a look here.
The solution for your specific case is given by
$$y=c_1 D_{-\frac{1}{2}}\left(\sqrt{2} x\right)+c_2 D_{-\frac{1}{2}}\left(i \sqrt{2}
   x\right)$$ where appear the  parabolic cylinder function.
A: For a power series solution, let
$$y=\sum_{n=0}^{\infty}a_nx^n,\quad y'=\sum_{n=1}^{\infty}na_nx^{n-1}, \quad y''=\sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}$$
then substituting into $y''-x^2y=0$ forms
$$\sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}-x^2\sum_{n=0}^{\infty}a_nx^n=0$$
or
$$\sum_{n=2}^{\infty}n(n-1)a_{n}x^{n-2}-\sum_{n=0}^{\infty}a_nx^{n+2}=0$$
so that for the coefficients we find
$$x^0:\quad 2(1) a_2=0 \implies a_2=0$$
$$x^1:\quad 3(2) a_3=0 \implies a_3=0$$
$$x^2:\quad 4(3) a_4-a_0=0 \implies a_4=\frac{a_0}{12}$$
$$x^3:\quad 5(4) a_5-a_1=0 \implies a_5=\frac{a_1}{20}$$
$$x^4:\quad 6(5) a_6-a_2=0 \implies a_6=\frac{a_2}{30}=0$$
therefore our recurrence for $x^n$ is given by
$$(n+2)(n+1)a_{n+2}-a_{n-2}=0 \implies a_{n+2}=\frac{a_{n-2}}{(n+2)(n+1)},\quad n\ge 2$$
where
\begin{align}y&=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x_5+a_6x^6+\dots\\&=
a_0+a_1x+0x^2+0x^3+\frac{a_0}{12}x^4+\frac{a_1}{20}x^5+0x^6+\dots\\&=
a_0\left(1+\frac{x^4}{12}+\dots\right)+a_1\left(1+\frac{x^5}{20}+\dots\right)
\end{align}
You could then add more terms to simplify the representation of $a_0$ and $a_1$.
A: Note that this belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0205.pdf.
Let $y=e^{-\frac{x^2}{2}}u$ ,
Then $y'=e^{-\frac{x^2}{2}}u'-xe^{-\frac{x^2}{2}}u$
$y''=e^{-\frac{x^2}{2}}u''-xe^{-\frac{x^2}{2}}u'-xe^{-\frac{x^2}{2}}u'+(x^2-1)e^{-\frac{x^2}{2}}u=e^{-\frac{x^2}{2}}u''-2xe^{-\frac{x^2}{2}}u'+(x^2-1)e^{-\frac{x^2}{2}}u$
$\therefore e^{-\frac{x^2}{2}}u''-2xe^{-\frac{x^2}{2}}u'+(x^2-1)e^{-\frac{x^2}{2}}u-x^2e^{-\frac{x^2}{2}}u=0$
$e^{-\frac{x^2}{2}}u''-2xe^{-\frac{x^2}{2}}u'-e^{-\frac{x^2}{2}}u=0$
$u''-2xu'-u=0$
You can apply the procedure in Help on solving an apparently simple differential equation to get $u=c_1\int_0^\infty\dfrac{e^{-\frac{t^2}{4}+xt}}{\sqrt{t}}dt+c_2\int_0^\infty\dfrac{e^{-\frac{t^2}{4}-xt}}{\sqrt{t}}dt$
$\therefore y=c_1e^{-\frac{x^2}{2}}\int_0^\infty\dfrac{e^{-\frac{t^2}{4}+xt}}{\sqrt{t}}dt+c_2e^{-\frac{x^2}{2}}\int_0^\infty\dfrac{e^{-\frac{t^2}{4}-xt}}{\sqrt{t}}dt$
$y=c_1\int_0^\infty\dfrac{e^{-\frac{t^2}{4}+xt-\frac{x^2}{2}}}{\sqrt{t}}dt+c_2\int_0^\infty\dfrac{e^{-\frac{t^2}{4}-xt-\frac{x^2}{2}}}{\sqrt{t}}dt$
$y=C_1\int_0^\infty e^{-\frac{t^2}{4}+xt-\frac{x^2}{2}}~d(\sqrt{t})+C_2\int_0^\infty e^{-\frac{t^2}{4}-xt-\frac{x^2}{2}}~d(\sqrt{t})$
$y=C_1\int_0^\infty e^{-\frac{t^4}{4}+xt^2-\frac{x^2}{2}}~dt+C_2\int_0^\infty e^{-\frac{t^4}{4}-xt^2-\frac{x^2}{2}}~dt$
