Series Solution to Differential equation
(x^2+2)y''-xy+y=0 y(0)=-6 y'(0)=4
How would you use the initial conditions ? I found the recurrence formula (i will post below) and I am trying to find the coefficients

Question

Series Solution to Differential equation 
(x^2+2)y''-xy+y=0  y(0)=-6 y'(0)=4
How would you use the initial conditions ? I found the recurrence formula (i will post  below)  and I am trying to find the coefficients

anonymous · Answer

The recurrence formula i found: $$a_{n+2}= \frac{-(n-1)^2a_n}{2(n+2)(n+1)}$$

anonymous · Answer

I get the following terms ( all odd numbers are =0)
$$n=0\hspace{.5cm} a_2=\frac{-a_0}{2*2*1}$$
$$n=2 \hspace{0.5cm} a_4=\frac{a_0}{(2*3*4)*4}$$
$$n=4 \hspace{0.5cm} a_6=\frac{-9a_0}{(2*3*4*5*6)*8}$$

I  think the coefficient look something like this but I can't find the relation 
$$ a_{2n}=\frac{???a_0}{(n+2)! 2^n}$$

anonymous · Answer

it should be 9 for n=2