Express the differential equation as power series 
My attempt: $y=\sum^{\infty}_{n=0}a_n x^n \rightarrow y'=\sum^{\infty}_{n=0}na_n x^{n-1}\rightarrow y''=\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}\\
(2+x^2)y"+x^2y'+3y=(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+x^2\sum^{\infty}_{n=0}na_n x^{n-1}+3\sum^{\infty}_{n=0}a_n x^n\\
2\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+\sum^{\infty}_{n=0}n(n-1)a_n x^{n}+\sum^{\infty}_{n=0}na_n x^{n+1}+3\sum^{\infty}_{n=0}a_n x^{n}$ 
from here can any help me
 A: You are asked to put the series on the form :
$$(2+x^2)y"+x^2y'+3y=\sum^{\infty}_{n=0}c_n x^n$$
ant to find the coefficients $c_n$.
$$(2+x^2)y"+x^2y'+3y=(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+x^2\sum^{\infty}_{n=0}na_n x^{n-1}+3\sum^{\infty}_{n=0}a_n x^n$$
So, you have to gather the terms of common power $n$.
$$---------------------$$
$$(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2} = 2\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+\sum^{\infty}_{n=0}n(n-1)a_{n} x^{n} $$
$$(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2} = 2\sum^{\infty}_{n=0}(n+2)(n+1)a_{n+2} x^{n}+\sum^{\infty}_{n=0}n(n-1)a_n x^{n} $$
$$(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2} = 4a_{2}+ 12a_{3}x+\sum^{\infty}_{n=2}\left( 2(n+2)(n+1)a_{n+2}+n(n-1)a_n \right)x^n $$
$$---------------------$$
$$x^2\sum^{\infty}_{n=0}na_n x^{n-1}=\sum^{\infty}_{n=0}na_n x^{n+1}=\sum^{\infty}_{n=2}(n-1)a_{n-1} x^{n} $$
$$---------------------$$
$$(2+x^2)y"+x^2y'+3y=4a_{2}+ 12a_{3}x+\sum^{\infty}_{n=2}\left( 2(n+2)(n+1)a_{n+2}+n(n-1)a_n \right)x^n +\sum^{\infty}_{n=2}(n-1)a_{n-1} x^{n} +3a_0+3a_1x+\sum^{\infty}_{n=2}3a_n x^n$$
$$---------------------$$
$$(2+x^2)y"+x^2y'+3y=\sum^{\infty}_{n=0}c_n x^n$$
$$c_0=4a_{2}+3a_0$$
$$c_1=12a_{3}+3a_1$$
$$c_n=2(n+2)(n+1)a_{n+2}+n(n-1)a_n +(n-1)a_{n-1}+3a_n\qquad n\geq 2$$
A: Note:
$$(2+x^2)y''+x^2y'+3y=$$
$$(2+x^2)(2a_2+3\cdot 2a_3x+4\cdot 3a_4x^2+5\cdot 4a_5x^3+\cdots)+$$
$$x^2(a_1+2a_2x+3a_3x^2+\cdots)+$$
$$3(a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots)=$$
$$4a_2+3a_0+(2(3\cdot 2)a_3+3a_1)x+$$
$$2(4\cdot 3)a_4+a_1+2a_2+3a_2)x^2+$$
$$(2(5\cdot 4)a_5+2a_2+3\cdot 2a_3+3a_3)x^3+$$
$$(2(6\cdot 5)a_6+3a_3+4\cdot 3a_4+3a_4)x^4+\cdots=$$
$$4a_2+3a_0+\sum_{n=1}^{\infty} [2(n+2)(n+1)a_{n+2}+(n-1)a_{n-1}+(3+(n-1)n)a_n]x^n.$$
