Find $y^{(n)}(0)$ for every $n$. Let $y(x)$ fulfill $y''-xy=0$. Furthermore: $y(0)=0,y'(0)=1$. Find $y^{(n)}(0)$ for every $n$. I tried different forms of recurrence relations but I couldn't do much with it without it becoming a complete mess. This is what I got:
$f^{(n+2)}=n\cdot y^{(n-1)}+x\cdot y^{(n)}$. 
I would really appreciate your simplifying. 
 A: Hint: write the solution $y$ as a power series $y\colon x\mapsto \sum_{n=0}^\infty a_n x^n$, and see what the differential equation and the initial conditions impose on the $a_n$'s.
Then, use the relation between $a_n$ and $y^{(n)}(0)$.
A: Why mess? Because you should calculate $y^{(n)}(0)$, not in other point, it could be not too messy:
$$0=y''-xy, \; \text{so} \; y''(0)=0$$
$$0=(0)'=(y''-xy)'=y^{(3)}-xy'-y \; \text{so} \; y^{(3)}(0)=y(0)=0$$
$$0=(y^{(3)}-xy'-y)'=y^{(4)}-y'-xy'-y^{'}=0  \; \text{so} \;y^{(4)}(0)=2y'(0)=2$$
Can you find a general pattern?
A: A hint:
Prove, using induction, that
$$y^{(n+3)}=(n+1) y^{(n)}+x y^{(n+1)}\qquad(n\geq0) \ ;$$
then put $x=0$.
A: Thank you all for your help. I post this answer as an attempt and application of your advises. Feel free to correct me. 
$y=a_0+a_1x+a_2x^2+...+anx^x+...$ 
Since $y(0)=0$, $y=0$ and we get $a_0=0$. Therefore $y=a_1x+a_2x^2+...+anx^x+...$.
$y'(0)=a_1+a_20+....=a_0=1$. Therefore $y=x+a_2x^2+...+anx^n+...$. 
$y''=2a_2+2\cdot 3\cdot  a_3+3\cdot4\cdot a_4 x^2+...+(n-1)\cdot n\cdot a_n x^{(n-2)}...=yx=x^2+a_2x^3+...+anx^{n+1}+...$. We get that $a_2=a_3=0,a_n=a_{n+3}\Rightarrow a_4={1\over 3\cdot 4},a_7={1\over 3\cdot 4\cdot6\cdot 7}...a_{3n+1}={1\over 3\cdot4\cdot6\cdot7....(3n-1)\cdot 3n}$ and that $a_2=a_3=a_5=a_6=...=a_{3n}=a_{3n+2}=0$. 
Therefore(for $n\ge 2$): $y=x+({1\over 3\cdot 4})\cdot x^ 4+({1\over 3\cdot 4\cdot 6\cdot 7})\cdot x^7+...+({1\over 3\cdot4\cdot6\cdot7....(3n-1)\cdot 3n})\cdot x^{3n+1}+...$
$y^{(1)}=1+({1\over 3})\cdot x^ 3...$ $\to 1$
$y^{(2)}= x^ 2...$ $\to 0$
$y^{(3)}= 2x...$ $\to 0$
$y^{(4)}= 2...$ $\to 2$
.
.
.
$({1\over 3\cdot 4\cdot 6\cdot 7})\cdot 7!...$ $\to 10$
And in 3 more cycles it is again not $0$ but $7!$. 
It is easy to see, hence, that:
$y^{(n)}(0) =
\begin{cases}
{n!\over 3\cdot4\cdot6\cdot7....(3k-1)\cdot 3k},  & \text{if $n=3k+1$ with $k\in\Bbb{N}$ } \\[2ex]
0, & \text{otherwise.}
\end{cases}$
