How to solve this differential equation via power series Studying Quantum Mechanics, I encountered the following differential equation:
how to solve $$y''-x^2y=0$$
Griffiths, the author of the book, just said that its solutions are
$$y=Ae^{x^2/2}+Be^{-x^2/2}$$
I can immediatelly see that this is true, but I do not know how to get to this solution, and I cannot find it solved anywhere else.
I tried to use the power series method:
$$y=\sum_{n=0}^{\infty}c_n x^n$$
So the equation becomes:
$$\sum_{n=0}^{\infty}c_{n+2}x^n(n+1)(n+2)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
Here I need to get both sums to the same power of n, so I removed the first two terms from the first sum, so that I can later change the index:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}c_{n+2}x^n(n+1)(n+2)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
now I change the index in the fist sum:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}c_{n+4}x^{n+2}(n+4)n(n+3)-\sum_{n=0}^{\infty}c_n x^{n+2}=0$$
Now I factor out the sum:
$$2c_2+6c_3x+\sum_{n=2}^{\infty}x^{n+2}(c_{n+4}(n+4)(n+3)-c_n)=0$$
Now all three terms must be zero at the same time, so $c_2=c_3=0$, and:
$$c_{n+4}(n+4)(n+3)-c_n=0$$
$$c_{n+4}=\frac{c_n}{(n+4)(n+3)}$$
The first terms of the series are:
$c_4=\frac{c_0}{4\times 3}$, $c_8=\frac{c_4}{8\times 7}=\frac{c_0}{8\times 7\times 4\times 3}$
However, I don't really know how to proceed from here. I tried putting the coefficients into the sum for $y$, but I can't really come close to the answer. Could you please guide me as how to advance from here?
 A: Check that
$y(x)=A e^{-x^2/2}$ is solution of $$\frac{y''(x)}{y(x)}=x^2-1$$
and $y(x)=B e^{x^2/2}$ is solution of $$\frac{y''(x)}{y(x)}=x^2+1$$
It is in this sense Griffith has claimed that $$\psi(x)=A e^{-x^2/2}+ B e^{x^2/2}~~~~(1)$$ is only APPROXIMATE solution of $$\psi''(x)-x^2\psi(x)=0~~~~~~~(2)$$
So please note that (1) is not the exact solution of (2) instead it is an APPROXIMATE solution of (2).
A: There is something wrong somewhere since, if
$$y=A e^{\frac{x^2}{2}}+B e^{-\frac{x^2}{2}}$$ then
$$y''-x^2 y=A e^{\frac{x^2}{2}}-B e^{-\frac{x^2}{2}}$$ which is obviously not $0$.
If fact, the solution of the given differential equation is much more complex
$$y=c_1 D_{-\frac{1}{2}}\left(\sqrt{2} x\right)+c_2 D_{-\frac{1}{2}}\left(i \sqrt{2}
   x\right)$$ where appear the parabolic cylinder functions.
It can also write
$$y=c_1 e^{-\frac{x^2}{2}} H_{-\frac{1}{2}}(x)+c_2 e^{\frac{x^2}{2}} H_{-\frac{1}{2}}(ix)$$ where appear Hermite polynomials.
So, the series solution is really the best way (at least, for the time being).
Edit
Coefficients $c_m$ are given by
$$c_m=\frac{2^{-m} \Gamma \left(\frac{3}{4}\right) \left(k_1+k_2 (-1)^m+\left(k_1-k_2\right)
   \sin \left(\frac{\pi  m}{2}\right)+\left(k_1+k_2\right) \cos \left(\frac{\pi 
   m}{2}\right)\right)}{\Gamma \left(\frac{m}{4}+1\right) \Gamma
   \left(\frac{m+3}{4}\right)}$$ which make
$$y=\frac{2 \pi  \sqrt{x}}{\Gamma \left(\frac{1}{4}\right)}\left(\left(k_1-k_2\right) I_{\frac{1}{4}}\left(\frac{x^2}{2}\right)+\left(k_1+k_2\right)
   I_{-\frac{1}{4}}\left(\frac{x^2}{2}\right)\right)$$
A: HINT
Perhaps easier to write
$$
\begin{split}
c_n
 &= \frac{c_{n-4}}{n(n-1)}\\
 &= \frac{c_{n-8}}{n(n-1)(n-4)(n-5)} \\
 &= \frac{c_{n-12}}{n(n-1)(n-4)(n-5)(n-8)(n-9)}
\end{split}
$$
So you can try to write a closed form for $c_n$ and you have 4 series driven by $c_0,c_1,c_2,c_3$ but $c_2=c_3=0$ makes life much simpler. You can finish it that way.

A different approach is to note that if differentiating twice adds two powers of $x$ in the front, perhaps differentiating once would add one, getting $y'=xy$ which is separable and implies
$$
\frac{dy}{y} = xdx \iff \ln y = \frac{x^2}{2}+C
$$
which is equivalent to one family of your solutions.
A: Since $c_2=0$ and $c_{n+4}=\frac {c_n}{(n+3)(n+4)}$, it follows that $c_6, c_{10},c_{14} ...=0$ and similarly $c_7, c_{11},c_{15} ...=0$. You can separate the remaining terms into $c_0x^0+c_4x^4+c_8x^8...$ and $c_1x^1+c_5x^5+c_9x^9...$. By relabeling the coefficients, we can get $y=
\sum_{n=0}^{\infty}a_nx^{4n}+\sum_{n=0}^{\infty}b_nx^{4n+1}$. Each of these sums has only one degree of freedom; the sequence $a_n$ is defined recursively from $a_0$ and similarly for $b_n$. This shows that $y = c_0f_0(x)+c_1f_1(x)$ for some $f_0,f_1$ that can be calculated numerically, if not in closed form.
