More power series: 2nd-order ODE

[Luminous Blob]

I felt pretty chuffed after getting the power series solution for a simple first-order ODE (with MathNerd's help) and thought I'd have a go at solving a second-order ODE using the same method. Then I realised I didn't understand it as well as I'd thought...

The differential I'm attempting to find the power series solution for is:

y'' - 2xy' + 4y = 0

My attempt at a solution:

y = \sum_{n=0}^\infty a_nx^n
y\'\; = \sum_{n=1}^\infty na_nx^{n-1}
y'' = \sum_{n=2}^\infty n(n-1)a_nx^{n-2}


substituting into the equation and shifting indices, etc. :

\sum_{n=2}^\infty n(n-1)a_nx^{n-2} -2x\sum_{n=1}^\infty na_nx^{n-1} + 4\sum_{n=0}^\infty a_nx^n = 0

\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n}-\sum_{n=1}^\infty 2na_nx^{n} + \sum_{n=0}^\infty 4a_nx^n = 0

4a_0 + 2a_2 + \sum_{n=1}^\infty (n+2)(n+1)a_{n+2}x^{n}-\sum_{n=1}^\infty 2na_nx^{n} + \sum_{n=1}^\infty 4a_nx^n = 0

then

4a_0 +2a_2 = 0

and

(n+2)(n+1)a_{n+2} - 2na_n + 4a_n = 0

the recursion relationship is:

a_{n+2} = 2a_n(n-2)/(n+2)(n+1)

Can anyone tell me:

i) if my recursion relationship is correct

and

ii) if so, how on Earth do you get from there to the answer of:

y = c_1(1 - 2x^2) + c_2(x-x^3/3 + ...)

[matt grime]

let c(1) =a(0), then you can find a(2) in terms of c(1), a(4) must then be zero by the relation, and hence so must all the other even terms.
now let c(2)=a(1), work out what a(3), a(5) etc are and see if you can prove a general rule about it, things ought to cancel out. alternatively you could look up somewhere how to solve recurrence relations ; they behave like differential equations but aren't taught anywhere near as well in school, if at all.

[Luminous Blob]

work out what a(3), a(5) etc are and see if you can prove a general rule about it

That's the part I'm having trouble with :)

[matt grime]

what do you think the general rule is, have you written down the first few terms and seen why the pattern develops the way it does? there are lots of cancellations, use the relation to write a(n+2) in terms of a(n-2), then a(n-4) then a(n-6).... etc