Answer
$y= \dfrac{9C_0}{(3-x)^2}$
Work Step by Step
tiWe have $y'=\Sigma^\infty_{n=1}n c_n x^{n-1}$
and $(x-3) \Sigma^\infty_{n=1}nC_{n} x^{n-1}+2 \times [ \Sigma^\infty_{n=0}C_{n}x^{n}]=0$
This gives: $-3\Sigma^\infty_{n=0}(n+1)c_{n+1}(n+1)x^{n}+\Sigma^\infty_{n=0}(n+2) c_{n}x^{n}=0$
$\implies c_{n+1}=\dfrac{(n+2)}{3(n+1)}c_n$
When n=2 , we have $C_{2} =\dfrac{C_{0}}{3}$;
$\implies C_{n} = \dfrac{n+1}{3^n}C_0$
Since, $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $
$y= C_0 \times \Sigma^\infty_{n=0} \dfrac{n+1}{3^n}x^n$
$\implies y=C_0\Sigma^\infty_{n=0} (\dfrac{x}{3})^n+C_0\Sigma^\infty_{n=0} n(\dfrac{x}{3})^n$
$\implies y=C_0\Sigma^\infty_{n=0} (\dfrac{x}{3})^n+C_0x \times \dfrac{d}{dx} [\Sigma^\infty_{n=0} n(\dfrac{x}{3})^n]$
$y=\dfrac{3(3-x)}{(3-x)^2}C_0+ \dfrac{3x}{(3-x)^2}C_0$
Hence,we get $y= \dfrac{9C_0}{(3-x)^2}$
