General solution of $C_{n+2}(x)=xC_n(x)+nC_{n-1}(x)$ Airy differential equation.
$y''(x)=xy(x)$
$y'''(x)=y(x)+x y'(x)$
$y'^v(x)=x^2y(x)+2 y'(x)$
$y^v(x)=4xy(x)+x^2 y'(x)$
$y^{(6)}(x)=(x^3+4)y(x)+6x y'(x)$
.
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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.
 A: Let $G(z)$ be the exponential generating function defining $G(z) = \sum_{n=0}^\infty C_n \frac{z^n}{n!}$. The recurrence equation $C_{n+2} = x C_n + n C_{n-1}$ translates into a differential equation:
$$
    G^{\prime\prime}(z)- (x+z) G(z)= C_2 - x C_0
$$
This equation admits a closed form solution:
$$
   G(z) = \kappa_1 Ai(x+z) + \kappa_2 Bi(x+z) + (x+z)^2(C_2 - x C_0) h(x+z)
$$
where 
$$
    h(u) = \frac{1}{2} \cdot {}_{0}F_{1}\left(\frac{2}{3}, \frac{u^3}{9}\right) \cdot {}_1 F_{2}\left(\frac{2}{3}; \frac{4}{3}, \frac{5}{3}; \frac{u^3}{9}\right) -
 {}_{0}F_{1}\left(\frac{4}{3}, \frac{u^3}{9}\right) \cdot {}_1 F_{2}\left(\frac{1}{3}; \frac{2}{3}, \frac{4}{3}; \frac{u^3}{9}\right)
$$
The initial condition $G(0) = C_0$ and $G^\prime(0) = C_1$ determines $\kappa_1$ and $\kappa_2$.

Added: After substitution of $C_0 = 1$, $C_1=0$, $C_2=x$ for $A_n(x)$ we get:
$$
   \sum_{n=0}^\infty \frac{z^n}{n!} A_n(x) = \pi \left( Bi^\prime(x) Ai(x+z) - Ai^\prime(x) Bi(x+z) \right)
$$
Similarly, for $B_n(x)$ with $C_0=0$, $C_1=1$, $C_2=0$ we obtain:
$$
   \sum_{n=0}^\infty \frac{z^n}{n!} B_n(x) =  \pi \left( Ai(x) Bi(x+z) - Bi(x) Ai(x+z) \right)
$$
The simplification is made possible by simple expression of ${}_0F_1$ in terms of Airy functions:
$$\begin{eqnarray}
    {}_{0}F_1\left(\frac{2}{3}; \frac{u^3}{9}\right) &=& -\frac{\Gamma\left(-\frac{1}{3}\right)}{ 3^{5/6}} \frac{1}{2} \left( \sqrt{3} Ai(u) + Bi(u) \right)  \\
    {}_{0}F_1\left(\frac{4}{3}; \frac{u^3}{9}\right) &=& -\frac{\Gamma\left(-\frac{2}{3}\right)}{ 3^{5/6}} \frac{1}{u} \left( \sqrt{3} Ai(u) - Bi(u) \right)
\end{eqnarray} $$

Of course, appearance of $y(x+z)$ should be anticipated, knowing the origin of the problem:
$$
    \left(\sum_{n=0}^\infty A_n(x) \frac{z^n}{n!} \right) y(x) + \left(\sum_{n=0}^\infty B_n(x) \frac{z^n}{n!} \right) y^\prime(x) = \sum_{n=0}^\infty \frac{z^n}{n!} y^{(n)}(x)  = y(x+z)
$$
A: I'm posting this as an answer, but it's more of a long-winded comment...
I started with the equation
$$C_{n+2}(x) = x C_n(x) + n C_{n-1}(x)$$
$$C_{n+3}(x) = x \left((C_{n+1}(x)) + n C_n(x)\right)$$
I realized that I already made a mistake in the previous step.  I will correct this!
Using Wilf's method for Ordinary Power Series Generating Functions(OPSGF) found in his book Generatingfunctionology, I use:
$$C_{n+3}(x) = \frac{C - (C_0 + x\cdot C_1 + x^2\cdot C_2 + x^3)}{x} \land$$
$$C_{n+1}(x) = \frac{C - C_0}{x}$$
$$C_n(x) = C$$
I realized here that I shoud have used a second variable, since we're taking a generating function of a series, and not of $C_n$s.
$$\frac{C - (C_0 + x\cdot C_1 + x^2\cdot C_2)}{x^3} = x \frac{C - C_0}{x} + x \frac{dC}{dx}$$
Solving for $\frac{dC}{dx}$, I then get
$$\left(x^3 \ne 1\right) \land \left(\frac{dC}{dx} = \frac{C(1 - x^3) + C_0(x^3 - 1) - xC_1 - x^2 C_2}{x^4} \right)$$
...as demonstrated here in Wolfram Aplha.  
The full equation for $C$, as demonstrated here (please be patient), is
$$C(x) = \frac{d_1 e^{-1/(3x^3)}}{x} + C_0 + $$
$$\frac{C_1 e^{-1/(3x^3)}(-1/(x^3))^{2/3} x \Gamma(1/3,-1/(3x^3))}{3^{2/3}} +$$
$$\frac{C_2 e^{-1/(3x^3)} Ei(1/(3x^3))}{3x}$$
where $\Gamma(a,b)$ is the incomplete gamma function and $Ei(a)$ is the exponential integral Ei.'
It looks like Sasha's solution is probably correct from here, and I'm not really sure that there's a good reason to keep this answer, which is full of mistakes, as it is.
