Answer
$\dfrac{9C_0}{(3-x)^2}$
Work Step by Step
Given: $(x-3)y+2y=0$
Since, $y=\Sigma^\infty_{n=0}c_{n} x^{n}$
Here, we have
$y'=\Sigma^\infty_{n=1}n c_n x^{n-1}$
Then, the given equation $(x-3)y+2y=0$ becomes:
$(x-3) \Sigma^\infty_{n=1}nC_{n} x^{n-1}+2[ \Sigma^\infty_{n=0}C_{n}x^{n}]=0$
or, $-3\Sigma^\infty_{n=0}(n+1)c_{n+1}(n+1)x^{n}+\Sigma^\infty_{n=0}(n+2) c_{n}x^{n}=0$
or, $c_{n+1}=\dfrac{(n+2)}{3(n+1)}c_n$
For n=2 , we have $C_{2} =\dfrac{C_{0}}{3}$;
Thus, we get $C_{n} = \dfrac{n+1}{3^n}C_0$
We know that $ \Sigma^\infty_{n=0} \dfrac{x^n}{n!}=e^{x} $
$y= C_0\Sigma^\infty_{n=0} \dfrac{n+1}{3^n}x^n=C_0\Sigma^\infty_{n=0} (\dfrac{x}{3})^n+C_0\Sigma^\infty_{n=0} n(\dfrac{x}{3})^n$
or, $y=C_0\Sigma^\infty_{n=0} (\dfrac{x}{3})^n+C_0x \dfrac{d}{dx} [\Sigma^\infty_{n=0} n(\dfrac{x}{3})^n]$
or, $y=\dfrac{3(3-x)C_0}{(3-x)^2}+ \dfrac{3C_0x}{(3-x)^2}=\dfrac{9C_0}{(3-x)^2}$
