I felt pretty chuffed after getting the power series solution for a simple first-order ODE (with(adsbygoogle = window.adsbygoogle || []).push({}); MathNerd'shelp) 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:

[tex]y'' - 2xy' + 4y = 0[/tex]

My attempt at a solution:

[tex] y = \sum_{n=0}^\infty a_nx^n [/tex]

[tex]y\'\; = \sum_{n=1}^\infty na_nx^{n-1} [/tex]

[tex]y'' = \sum_{n=2}^\infty n(n-1)a_nx^{n-2}

[/tex]

substituting into the equation and shifting indices, etc. :

[tex]\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 [/tex]

[tex]\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 [/tex]

[tex]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 [/tex]

then

[tex]4a_0 +2a_2 = 0[/tex]

and

[tex](n+2)(n+1)a_{n+2} - 2na_n + 4a_n = 0[/tex]

the recursion relationship is:

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

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:

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

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# More power series: 2nd-order ODE

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