Answer
$\dfrac{9C_o}{(3-x)^2}$
Work Step by Step
The given equation is $(x-3)y+2y=0$
$(x-3) \Sigma^\infty_{n=1}nC_{n} x^{n-1}+2[ \Sigma^\infty_{n=0}C_{n}x^{n}]=0$
Re-write as: $-3\Sigma^\infty_{n=0}(n+1)c_{n+1}(n+1)x^{n}+\Sigma^\infty_{n=0}(n+2) c_{n}x^{n}=0$
and $c_{n+1}=\dfrac{(n+2)}{3(n+1)}c_n$
n=2 : we have $C_{2} =\dfrac{C_{0}}{3}$;
So, we have $C_{n} = \dfrac{n+1C_0}{3^n}$
$y= C_0\Sigma^\infty_{n=0} \dfrac{n+1}{3^n}x^n$
This implies that $=C_0\Sigma^\infty_{n=0} (\dfrac{x}{3})^n+C_0\Sigma^\infty_{n=0} n(\dfrac{x}{3})^n$
$=C_0\Sigma^\infty_{n=0} (\dfrac{x}{3})^n+C_0x \dfrac{d}{dx} [\Sigma^\infty_{n=0} n(\dfrac{x}{3})^n]$
$=\dfrac{3(3-x)}{(3-x)^2}C_0+ \dfrac{3x}{(3-x)^2}C_0=\dfrac{9C_o}{(3-x)^2}$
