# Thread: Power Series Problem

1. ## Power Series Problem

Find the recurrence relation for the coefficients of a power series of y about the given point Xo given the equation.

$(4+x^2)y^{ii} -xy^i+8y=0, Xo=0$

I am not sure what to do with the $(4+x^2)$

Thanks for any help.

2. Hello,

it's $y''$, not $y^{ii}$

Let $y=\sum_{n=0}^\infty a_n x^n$

Hence $y'=\sum_{n=1}^\infty n a_n x^{n-1}$

and $y''=\sum_{n=2}^\infty n(n-1)a_n x^{n-2}$

The equation is now :
$(4+x^2) \sum_{n=2}^\infty n(n-1)a_n x^{n-2}-x \sum_{n=1}^\infty n a_n x^{n-1}+8 \sum_{n=0}^\infty a_n x^n=0$

Distribute the factors :

$\sum_{n=2}^\infty 4n(n-1)a_n x^{n-2}+\sum_{n=2}^\infty n(n-1)a_n x^n-\sum_{n=1}^\infty na_n x^n+\sum_{n=0}^\infty 8a_n x^n=0$

Change the indices so that you have $x^n$ in all sums :

${\color{red}\sum_{n=0}^\infty 4(n+2)(n+1) a_{n+2} x^n}+\sum_{n=2}^\infty n(n-1)a_n x^n-\sum_{n=1}^\infty na_n x^n+\sum_{n=0}^\infty 8a_n x^n=0$

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Now, change the starting n's in the sums, so that they start at the same point (namely n=2)

$\sum_{n=0}^\infty 4(n+2)(n+1) a_{n+2} x^n$ $=4(0+2)(0+1)a_{0+2}+4(1+2)(1+1)a_{1+2} x+\sum_{{\color{red}n=2}}^\infty 4(n+2)(n+1) a_{n+2} x^n$ $=8a_2+24a_3 x+\sum_{n=2}^\infty 4(n+2)(n+1)a_{n+2} x^n$

Similarly, we have :
$\sum_{n=1}^\infty na_n x^n=a_1 x+\sum_{{\color{red}n=2}}^\infty na_n x^n$

$\sum_{n=0}^\infty 8a_n x^n=8a_0+8a_1x+\sum_{{\color{red}n=2}}^\infty 8a_n x^n$

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So we have the new equation :

$\left(8a_2+24a_3 x+\sum_{n=2}^\infty 4(n+2)(n+1)a_{n+2} x^n\right)+\left(\sum_{n=2}^\infty n(n-1)a_n x^n\right)$ $-\left(a_1 x+\sum_{n=2}^\infty na_n x^n\right)+\left(8a_0+8a_1x+\sum_{n=2}^\infty 8a_n x^n\right)=0$

Group the terms :

$\bigg(8a_2+8a_0+x(8a_1+24a_3-a_1)\bigg)$ $+\sum_{n=2}^\infty x^n \big(4(n+2)(n+1)a_{n+2}+n(n-1)a_n-n a_n+8a_n\big)=0$

$\bigg(8a_2+8a_0+x(8a_1+24a_3-a_1)\bigg)+\sum_{n=2}^\infty x^n \big(4(n+2)(n+1)a_{n+2}+a_n [n^2-2n+8]\big)=0$

Now you'll have to use the initial conditions to find a recursive relation for $a_n$

Hope that helps

3. Sorry I didn't know how to make the correct "prime" sign. But thanks for the incredible help