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

Answer 1

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

Answer 2

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
\[ a_{2n}=\frac{???a_0}{(n+2)! 2^n}\]

Answer 3

it should be 9 for n=2