Let $y(x) = \sum_{n=0}^{\infty}a_nx^n$, I want to find a power series that satisfies $$ y''-xy = 0, y(0) = 1, y'(0) = 0 $$
My attempt:
$$ y(x) = \sum_{n=0}^{\infty}a_nx^n $$ $$ y'(x) = \sum_{n=1}^{\infty}na_nx^{n-1} $$ $$ y''(x) = \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2} $$ Reindexing the $n$ in $y''(x)$, we have $$ y''(x) = \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n} $$
Now I have $\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-\sum_{n=0}^{\infty}a_nx^{n+1} = 0$, but how can I proceed to find $a_n$? Or at least show that such a series exists?
Well, formally you can agroup the two sums in this way: $$ 2a_{2}+\sum_{n=1}^{\infty}\left((n+2)(n+1)a_{n+2}-a_{n-1}\right)x^n=0.$$ Differentiating succesively and putting $x=0$ we conclude that it's enough to compare de coefficients of each power of $x$ (remember the fact that the derivate of a power series converge in the same interval where the original serie does) This give us the following equations: $$a_{2}=0, 6a_{3}-a_{0}= 0, 12a_4-a_{1}=0, 20a_{5}-a_{2}=0,30a_{6}-a_{3}=0,...$$
Hence we necessarily have $$0=a_{2}=a_{5}=a_{8}=a_{11}=\cdots=a_{3n+2}=\cdots.$$ We also have $$a_{3}=\frac{a_0}{2\cdot 3}, a_{6}=\frac{a_3}{6\cdot 5}=\frac{a_0}{6\cdot 5\cdot 3\cdot 2}, a_9= \frac{a_6}{9\cdot 8}=\frac{a_0}{9\cdot 8\cdot 6\cdot 5\cdot 3\cdot 2},...$$ hence $$a_{3n}=\frac{a_0}{3n\cdot(3n-1)\cdot(3n-3)\cdot(3n-4)\cdots 3\cdot 2}$$
In the same way we have $$a_{4}=\frac{a_1}{4\cdot 3}, a_7=\frac{a_4}{7\cdot 6}=\frac{a_1}{7\cdot 6\cdot 4\cdot 3},...$$ so $a_{3n+1}=\frac{a_1}{(3n+1)\cdot 3n\cdots 4\cdot 3}.$ It remains to determine $a_0$ and $a_1.$ For this note that $a_0=y(0)=1$ and $0=y'(0)=a_{1}.$ Therefore, the power series solution should be $$y(x)= 1+ \frac{x^3}{3\cdot 2}+\frac{x^6}{6\cdot 5\cdot 3\cdot 2}+\cdots.$$
PS: note that this serie has infinite convergence radio and please excuse my english. By the way, I am not sure if it is neccesary but I think it should be checked if, with these $a_n$'s, the series $\sum (n+2)(n+1)a_{n+2}x^{n}$ and $\sum a_n x^{n+1}$ converge (to make the first line of my answer be valid)