## Calculus 8th Edition

$y= \dfrac{9C_0}{(3-x)^2}$
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 $y= \dfrac{9C_0}{(3-x)^2}$