Answer to Question #150268 in Differential Equations for Ernest

Question

y"+xy'+(x^2 +2)y=0

Answer 1

$y(x)=\displaystyle\sum_{n=0}^{\infin}a_nx^n$

$y'(x)=\displaystyle\sum_{n=1}^{\infin}na_nx^{n-1}$

$y''(x)=\displaystyle\sum_{n=2}^{\infin}n(n-1)a_nx^{n-2}$

$\displaystyle\sum_{n=2}^{\infin}n(n-1)a_nx^{n-2}+x\displaystyle\sum_{n=1}^{\infin}na_nx^{n-1}+(x^2+2)\displaystyle\sum_{n=0}^{\infin}a_nx^n=0$

$\displaystyle\sum_{n=2}^{\infin}n(n-1)a_nx^{n-2}+\displaystyle\sum_{n=1}^{\infin}na_nx^{n}+\displaystyle\sum_{n=0}^{\infin}a_nx^{n+2}+\displaystyle\sum_{n=0}^{\infin}2a_nx^n=0$

$\displaystyle\sum_{n=0}^{\infin}(n+2)(n+1)a_{n+2}x^{n}+\displaystyle\sum_{n=1}^{\infin}na_nx^{n}+\displaystyle\sum_{n=0}^{\infin}a_nx^{n+2}+\displaystyle\sum_{n=0}^{\infin}2a_nx^n=0$

$\displaystyle\sum_{n=0}^{\infin}(n+2)(n+1)a_{n+2}x^{n}+\displaystyle\sum_{n=1}^{\infin}na_nx^{n}+\displaystyle\sum_{n=2}^{\infin}a_{n-2}x^{n}+\displaystyle\sum_{n=0}^{\infin}2a_nx^n=0$

$2a_0+3a_1x+2a_2+6a_3x+\displaystyle\sum_{n=2}^{\infin}[(n+2)(n+1)a_{n+2}+na_n+a_{n-2}+2a_n]x^{n}=0$

$n=0\implies 2a_2+2a_0=0\implies a_0=-a_2$

$n=1\implies 2a_2+6a_3x+a_1x+2a_0+2a_1x=0\implies 6a_3+3a_1=0$

$a_1=-2a_3$

For $n=2,3,4,...$ :

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

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

$a_{n+2}=-\frac{a_{n-2}+(n+2)a_n}{(n+1)(n+2)}$

Answer:

$y(x)=a_0+a_1x-a_0x^2-\frac{a_1}{2}x^3-\frac{a_0+4a_2}{12}x^4+...$