$y''(x)=xy(x)$
$y'''(x)=y(x)+x y'(x)$
$y'^v(x)=x^2y(x)+2 y'(x)$
$y^v(x)=4xy(x)+x^2 y'(x)$
$y^{(6)}(x)=(x^3+4)y(x)+6x y'(x)$
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$y^{(n)}(x)=A_n(x)y(x)+B_n(x) y'(x)$
Where $A_n(x)$ and $B_n(x)$ are polynomials
($y^{(n)}(x)$ means $n$-th order derivative of $y(x)$ )
I have found that
$C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$
If $C_2(x)=x$ ;$C_3(x)=1$ ; $C_4(x)=x^2$ are initial condition
$C_n(x)=A_n(x)$
If $C_2(x)=0$;$C_3(x)=x$;$C_4(x)=2$ are initial condition
$C_n(x)=B_n(x)$
How can we find the general solution of $C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$ ?
Can we express $C_n(x)$ as Airy function?
Thanks for answers.
$EDIT:$ I tried generating function method as Sam recommended.
$g(z,x)=\sum_{n=3}^\infty z^n C_n(x)$ $g(z,x)=z^3C_3(x)+z^4C_4(x)+\sum_{n=5}^\infty z^n C_n(x)$
$g(z,x)=z^3C_3(x)+z^4C_4(x)+\sum_{n=3}^\infty z^{n+2} C_{n+2}(x)$
$g(z,x)=z^3C_3(x)+z^4C_4(x)+\sum_{n=3}^\infty z^{n+2}(xC_n(x)+nC_{n-1}(x))$
$g(z,x)=z^3C_3(x)+z^4C_4(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+z^4\sum_{n=3}^\infty nz^{n-2}C_{n-1}(x)$
$g(z,x)=z^3C_3(x)+z^4C_4(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+3z^5C_2(x)+z^4\sum_{n=4}^\infty nz^{n-2}C_{n-1}(x)$
$g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+z^4\sum_{n=3}^\infty (n+1)z^{n-1}C_{n}(x)$
$g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)+xz^2\sum_{n=3}^\infty z^{n}C_n(x)+z^4\sum_{n=3}^\infty nz^{n-1}C_{n}(x)+z^3\sum_{n=3}^\infty z^{n}C_{n}(x)$
We can get that
$\frac{\partial g(z,x)}{\partial z}=\sum_{n=3}^\infty n z^{n-1} C_n(x)$
And finally,
$g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)+xz^2g(z,x)+z^4\frac{\partial g(z,x)}{\partial z}+z^3g(z,x)$
$-z^4\frac{\partial g(z,x)}{\partial z}- (z^3+xz^2-1)g(z,x)=z^3C_3(x)+z^4C_4(x)+3z^5C_2(x)$
$\frac{\partial g(z,x)}{\partial z}+\frac{ z^3+xz^2-1}{z^4}g(z,x)=-\frac{1}{z}C_3(x)-C_4(x)-3zC_2(x)$
I will let you know after solving the first order linear differential equation.