Airy's equation 2nd order ODE Suppose we have the equation 
$$ \frac{ d^2 y}{d z^2}= z\, y $$
 for function $y(z).$
We would like to find the 6 first coefficients of the equation given the boundary conditions 
$\dfrac{dy}{dz}=1$ at $ z=0,y=2,\,  y(z)=a_0+a_1z+a_2z^2+a_3z^3+...+a_5z^5.$
I've arrived at the recurrence relation
$$a_{k+3}=\frac{a_k}{(k+3)(k+2)} $$
Can anyone kindly help me with this? Would be really grateful.

You have a third order recurrence relation for the coefficients. Since the recursion is third order, it needs three initial conditions. $a_0$ and $a_1$ come directly from $y(0),y'(0)$. (If you had different boundary conditions, the problem would turn out differently.)
To find $a_2$, notice that putting the expansion into the equation gives you $2a_2=za_0$ at lowest order. Since Frobenius expansion is supposed to work in a neighborhood of $z=0$, you can in particular send $z \to 0$ in this equation to deduce what $a_2$ must be. Then you have all three initial conditions for the recurrence and you can iterate to solve it.
